Find \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}}\) ;

Find \(\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}}\) .

\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x}}{{1 + 2x}}\)     M1A1A1

\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}} = \frac{1}{1} = 1\)     A1

[4 marks] 

\(\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 2x + 2x + 4x\ln x}}{{ - \frac{\pi }{2}\cos \frac{{\pi x}}{2}}}\)     M1A1A1

\( = \mathop {\lim }\limits_{x \to 1} \frac{{4 + 4\ln x}}{{\frac{{{\pi ^2}}}{4}\sin \frac{{\pi x}}{2}}}\)     M1A1A1

\(\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}} = \frac{4}{{\frac{{{\pi ^2}}}{4}}} = \frac{{16}}{{{\pi ^2}}}\)     A1

[7 marks]

This question was accessible to the vast majority of candidates, who recognised that L’Hopital’s rule was required. A few of the weaker candidates did not realise that it needed to be applied twice in part (b). Many fully correct solutions were seen.

This question was accessible to the vast majority of candidates, who recognised that L’Hopital’s rule was required. A few of the weaker candidates did not realise that it needed to be applied twice in part (b). Many fully correct solutions were seen.

By finding the values of successive derivatives when x = 0 , find the Maclaurin series for y as far as the term in \({x^3}\) .

(i)     Differentiate the function \({{\text{e}}^x}(\sin x + \cos x)\) and hence show that

\[\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c} .\]

(ii)     Find an integrating factor for the differential equation and hence find the solution in the form \(y = f(x)\) .

we note that \(y(0) = 1\) and \(y'(0) = 2\)     A1

\(y'' = 2{{\text{e}}^x} + y'\tan x + y{\sec ^2}x\)     M1

\(y''(0) = 3\)     A1

\(y''' = 2{{\text{e}}^x} + y''\tan x + 2y'{\sec ^2}x + 2y{\sec ^2}x\tan x\)     M1

\(y'''(0) = 6\)     A1

the maclaurin series solution is therefore

\(y = 1 + 2x + \frac{{3{x^2}}}{2} + {x^3} +  \ldots \)     A1

[6 marks]

(i)     \(\frac{{\text{d}}}{{{\text{d}}x}}\left( {{{\text{e}}^x}(\sin x + \cos x)} \right) = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x - \sin x)\)     M1

\( = 2{{\text{e}}^x}\cos x\)     A1

it follows that

\(\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c} \)     AG

(ii)     the differential equation can be written as

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\tan x = 2{{\text{e}}^x}\)     M1

\({\text{IF}} = {{\text{e}}^{\int { - \tan x{\text{d}}x} }} = {{\text{e}}^{\ln \cos x}} = \cos x\)     M1A1

\(\cos x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\sin x = 2{{\text{e}}^x}\cos x\)     M1

integrating,

\(y\cos x = {{\text{e}}^x}(\sin x + \cos x) + C\)     A1

y = 1 when x = 0 gives C = 0     M1

therefore

\(y = {{\text{e}}^x}(1 + \tan x)\)     A1

[9 marks]

(i)     Show that \(\int_1^\infty  {\frac{1}{{x(x + p)}}{\text{d}}x,{\text{ }}p \ne 0} \) is convergent if p > −1 and find its value in terms of p.

(ii)     Hence show that the following series is convergent.

\[\frac{1}{{1 \times 0.5}} + \frac{1}{{2 \times 1.5}} + \frac{1}{{3 \times 2.5}} + ...\]

Determine, for each of the following series, whether it is convergent or divergent.

(i)     \(\sum\limits_{n = 1}^\infty  {\sin \left( {\frac{1}{{n(n + 3)}}} \right)} \)

(ii)     \(\sqrt {\frac{1}{2}}  + \sqrt {\frac{1}{6}}  + \sqrt {\frac{1}{{12}}}  + \sqrt {\frac{1}{{20}}}  + …\)

(i)     the integrand is non-singular on the domain if p > –1 with the latter assumed, consider

\(\int_1^R {\frac{1}{{x(x + p)}}} {\text{d}}x = \frac{1}{p}\int_1^R {\frac{1}{x} - \frac{1}{{x + p}}{\text{d}}x} \)     M1A1

\( = \frac{1}{p}\left[ {\ln \left( {\frac{x}{{x + p}}} \right)} \right]_1^R,{\text{ }}p \ne 0\)     A1

this evaluates to

\(\frac{1}{p}\left( {\ln \frac{R}{{R + p}} - \ln \frac{1}{{1 + p}}} \right),{\text{ }}p \ne 0\)     M1

\( \to \frac{1}{p}\ln (1 + p)\)     A1

because \(\frac{R}{{R + p}} \to 1{\text{ as }}R \to \infty \)     R1

hence the integral is convergent     AG

(ii)     the given series is \(\sum\limits_{n = 1}^\infty  {f(n),{\text{ }}f(n) = \frac{1}{{n(n - 0.5)}}} \)     M1

the integral test and p = –0.5 in (i) establishes the convergence of the series     R1

[8 marks]

(i)     as we have a series of positive terms we can apply the comparison test, limit form

comparing with \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \)     M1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{{n(n + 3)}}} \right)}}{{\frac{1}{{{n^2}}}}} = 1\)     M1A1

as \(\sin \theta \approx \theta {\text{ for small }}\theta \)     R1

and \(\frac{{{n^2}}}{{n(n + 3)}} \to 1\)     R1

(so as the limit (of 1) is finite and non-zero, both series exhibit the same behavior)

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges, so this series converges     R1

(ii)     the general term is

\(\sqrt {\frac{1}{{n(n + 1)}}} \)     A1

\(\sqrt {\frac{1}{{n(n + 1)}}}  > \sqrt {\frac{1}{{(n + 1)(n + 1)}}} \)     M1

\(\sqrt {\frac{1}{{(n + 1)(n + 1)}}}  = \frac{1}{{n + 1}}\)     A1

the harmonic series diverges     R1

so by the comparison test so does the given series     R1

[11 marks]

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.

(a)     Write down the value of the constant term in the Maclaurin series for \(f(x)\) .

(b)     Find the first three derivatives of \(f(x)\) and hence show that the Maclaurin series for \(f(x)\) up to and including the \({x^3}\) term is \(x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\).

(c)     Use this series to find an approximate value for ln 2 .

(d)     Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.

(e)     How good is this upper bound as an estimate for the actual error?

(a)     Constant term = 0     A1

[1 mark]

(b)     \(f'(x) = \frac{1}{{1 - x}}\)     A1

\(f''(x) = \frac{1}{{{{(1 - x)}^2}}}\)     A1

\(f'''(x) = \frac{2}{{{{(1 - x)}^3}}}\)     A1

\(f'(0) = 1;{\text{ }}f''(0) = 1;{\text{ }}f'''(0) = 2\)     A1

Note: Allow FT on their derivatives.

\(f(x) = 0 + \frac{{1 \times x}}{{1!}} + \frac{{1 \times {x^2}}}{{2!}} + \frac{{2 \times {x^3}}}{{3!}} + …\)     M1A1

\( = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\)     AG

[6 marks]

(c)     \(\frac{1}{{1 - x}} = 2 \Rightarrow x = \frac{1}{2}\)     (A1)

\(\ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}}\)     M1

\( = \frac{2}{3}{\text{ (0.667)}}\)     A1

[3 marks]

(d)     Lagrange error \({\text{ = }}\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}} \times {\left( {\frac{1}{2}} \right)^{n + 1}}\)     (M1)

\( = \frac{6}{{{{(1 - c)}^4}}} \times \frac{1}{{24}} \times {\left( {\frac{1}{2}} \right)^4}\)     A1

\( < \frac{6}{{{{\left( {1 - \frac{1}{2}} \right)}^4}}} \times \frac{1}{{24}} \times \frac{1}{{16}}\)     A2

giving an upper bound of 0.25.     A1

[5 marks]

(e)     Actual error \( = \ln 2 - \frac{2}{3} = 0.0265\)     A1

The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate.     R1

[2 marks]

Total [17 marks]

In (a), some candidates appeared not to understand the term ‘constant term’. In (b), many candidates found the differentiation beyond them with only a handful realising that the best way to proceed was to rewrite the function as \(f(x) = - \ln (1 - x)\). In (d), many candidates were unable to use the Lagrange formula for the upper bound so that (e) became inaccessible.

Show that \(f''(x) = - \frac{1}{{(1 + \sin x)}}\) .

(i)     Find the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .

(ii)     Explain briefly why your result shows that f is neither an even function nor an odd function.

Determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}}\).

\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\)     A1

\(f''(x) = \frac{{ - \sin x(1 + \sin x) - \cos x\cos x}}{{{{(1 + \sin x)}^2}}}\)     M1A1

\( = \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}\)     A1

\( = - \frac{1}{{1 + \sin x}}\)     AG

[4 marks]

(i)     \(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\)     A1

\({f^{(4)}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\)     M1A1

\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = - 1\)     M1

\(f'''(0) = 1,{\text{ }}{f^{(4)}}(0) = - 2\)     A1

\(f(x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots \)     A1

(ii)     the series contains even and odd powers of x     R1

[7 marks]

\(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} +  \ldots  - x}}{{{x^2}}}\)     M1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} + \frac{x}{6} +  \ldots }}{1}\)     (A1)

\( = - \frac{1}{2}\)     A1

Note: Use of l’Hopital’s Rule is also acceptable.

[3 marks]

Find the set of values of x for which the series is convergent.

(i)     Show, by comparison with an appropriate geometric series, that

\[{{\text{e}}^x} - 1 < \frac{{2x}}{{2 - x}},{\text{ for }}0 < x < 2{\text{.}}\]

(ii)     Hence show that \({\text{e}} < {\left( {\frac{{2n + 1}}{{2n - 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

(i)     Write down the first three terms of the Maclaurin series for \(1 - {{\text{e}}^{ - x}}\) and explain why you are able to state that

\[1 - {{\text{e}}^{ - x}} > x - \frac{{{x^2}}}{2},{\text{ for }}0 < x < 2.\]

(ii)     Deduce that \({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} - 2n + 1}}} \right)^n}\), for \(n \in {\mathbb{Z}^ + }\).

Letting n = 1000, use the results in parts (b) and (c) to calculate the value of e correct to as many decimal places as possible.

using a ratio test,

\(\left| {\frac{{{T_{n + 1}}}}{{{T_n}}}} \right| = \left| {\frac{{{x^{n + 1}}}}{{(n + 1)!}}} \right| \times \left| {\frac{{n!}}{{{x^n}}}} \right| = \frac{{\left| x \right|}}{{n + 1}}\)     M1A1

Note: Condone omission of modulus signs.

\( \to 0{\text{ as }}n \to \infty \) for all values of x     R1

the series is therefore convergent for \(x \in \mathbb{R}\)     A1

[4 marks]

(i)     \({{\text{e}}^x} - 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 3}} + …\)     M1

\( < x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{2 \times 2}} + ...\,\,\,\,\,({\text{for }}x > 0)\)     A1

\( = \frac{x}{{1 - \frac{x}{2}}}\,\,\,\,\,({\text{for }}x < 2)\)     A1

\( = \frac{{2x}}{{2 - x}}\,\,\,\,\,({\text{for }}0 < x < 2)\)     AG

(ii)     \({{\text{e}}^x} < 1 + \frac{{2x}}{{2 - x}} = \frac{{2 + x}}{{2 - x}}\)     A1

\({\text{e}} < {\left( {\frac{{2 + x}}{{2 - x}}} \right)^{\frac{1}{x}}}\)     A1

replacing x by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) )     M1

\({\text{e}} < {\left( {\frac{{2n + 1}}{{2n - 1}}} \right)^n}\)     AG

[6 marks]

(i)     \(1 - {{\text{e}}^{ - x}} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + …\)     A1

for \(0 < x < 2\), the series is alternating with decreasing terms so that the sum is greater than the sum of an even number of terms     R1

therefore

\(1 - {{\text{e}}^{ - x}} > x - \frac{{{x^2}}}{2}\)     AG

(ii)     \({{\text{e}}^{ - x}} < 1 - x + \frac{{{x^2}}}{2}\)

\({{\text{e}}^x} > \frac{1}{{\left( {1 - x + \frac{{{x^2}}}{2}} \right)}}\)     M1

\({\text{e}} > {\left( {\frac{2}{{2 - 2x + {x^2}}}} \right)^{\frac{1}{x}}}\)     A1

replacing x by \(\frac{1}{n}\) (and noting that the result is true for \(n > \frac{1}{2}\) and therefore \({\mathbb{Z}^ + }\) )

\({\text{e}} > {\left( {\frac{{2{n^2}}}{{2{n^2} - 2n + 1}}} \right)^n}\)     AG

[4 marks]

from (b) and (c), \({\text{e}} < 2.718282…\) and \({\text{e}} > 2.718281…\)     A1

we conclude that e = 2.71828 correct to 5 decimal places     A1

[2 marks]

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Solutions to (a) were generally good although some candidates failed to reach the correct conclusion from correct application of the ratio test. Solutions to (b) and (c), however, were generally disappointing with many candidates unable to make use of the signposting in the question. Candidates who were unable to solve (b) and (c) often picked up marks in (d).

Determine whether or not the following series converge.

(a)     \(\sum\limits_{n = 0}^\infty  {\left( {\sin \frac{{n\pi }}{2} - \sin \frac{{(n + 1)\pi }}{2}} \right)} \)

(b)     \(\sum\limits_{n = 1}^\infty  {\frac{{{{\text{e}}^n} - 1}}{{{\pi ^n}}}} \)

(c)     \(\sum\limits_{n = 2}^\infty  {\frac{{\sqrt {n + 1} }}{{n(n - 1)}}} \)

(a)     \(\sum\limits_{n = 0}^\infty  {\left( {\sin \frac{{n\pi }}{2} - \sin \frac{{(n + 1)\pi }}{2}} \right)} \)

\( = \left( {\sin 0 - \sin \frac{\pi }{2}} \right) + \left( {\sin \frac{\pi }{2} - \sin \pi } \right) + \left( {\sin \pi  - \sin \frac{{3\pi }}{2}} \right) + \left( {\sin \frac{{3\pi }}{2} - \sin 2\pi } \right) + \ldots \)     (M1)

the \({n^{{\text{th}}}}\) term is ±1 for all n, i.e. the \({n^{{\text{th}}}}\) term does not tend to 0     A1

hence the series does not converge     A1

[3 marks]

(b)     EITHER

using the ratio test     (M1)

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}}}}{{{\pi ^{n + 1}}}}} \right)\left( {\frac{{{\pi ^n}}}{{{{\text{e}}^n} - 1}}} \right)\)     M1A1

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}} - 1}}{{{{\text{e}}^n} - 1}}} \right)\left( {\frac{{{\pi ^n}}}{{{\pi ^{n + 1}}}}} \right) = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\)     M1A1

\(\frac{{\text{e}}}{\pi } < 1\), hence the series converges     R1A1

OR

\(\sum\limits_{n = 1}^\infty  {\frac{{{{\text{e}}^n} - 1}}{{{\pi ^n}}} = \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} - {{\left( {\frac{1}{\pi }} \right)}^n} = \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} - \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{1}}}{\pi }} \right)}^n}} } } } \)     M1A1

the series is the difference of two geometric series, with \(r = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\)     M1A1

and \(\frac{1}{\pi }\,\,\,\,\,( \approx 0.318)\)     A1

for both \(\left| r \right| < 1\), hence the series converges     R1A1

OR

\(\forall n,{\text{ }}0 < \frac{{{{\text{e}}^n} - 1}}{{{\pi ^n}}} < \frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\)     (M1)A1A1

the series \(\frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\) converges since it is a geometric series such that \(\left| r \right| < 1\)     A1R1

therefore, by the comparison test, \(\frac{{{{\text{e}}^n} - 1}}{{{\pi ^n}}}\) converges     R1A1

[7 marks]

(c)     by limit comparison test with \(\frac{{\sqrt n }}{{{n^2}}}\),     (M1)

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\frac{{\sqrt {n + 1} }}{{n(n - 1)}}}}{{\frac{{\sqrt n }}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\sqrt {n + 1} }}{{n(n - 1)}} \times \frac{{{n^2}}}{{\sqrt n }}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n - 1}}\sqrt {\frac{{n + 1}}{n}}  = 1\)     M1A1

hence both series converge or both diverge     R1

by the p-test \(\sum\limits_{n = 1}^\infty  {\frac{{\sqrt n }}{{{n^2}}} = {n^{\frac{{ - 3}}{2}}}} \) converges, hence both converge     R1A1

[6 marks]

Total [16 marks]

This was the least successfully answered question on the paper. Candidates often did not know which convergence test to use; hence very few full successful solutions were seen. The communication of the method used was often quite poor.

a) Many candidates failed to see that this is a telescoping series. If this was recognized then the question was fairly straightforward. Often candidates unsuccessfully attempted to apply the standard convergence tests.

b) Many candidates used the ratio test, but some had difficulty in simplifying the expression. Others recognized that the series is the difference of two geometric series, and although the algebraic work was done correctly, some failed to communicate the conclusion that since the absolute value of the ratios are less than 1, hence the series converges. Some candidates successfully used the comparison test.

c) Although the limit comparison test was attempted by most candidates, it often failed through an inappropriate selection of a series.

Find \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos {x^6}}}{{{x^{12}}}}} \right)\).

METHOD 1

\(f(0) = \frac{0}{0}\), hence using l’Hôpital’s Rule,     (M1)

\(g(x) = 1 - \cos ({x^6}),{\text{ }}h(x) = {x^{12}};{\text{ }}\frac{{g'(x)}}{{h'(x)}} = \frac{{6{x^5}\sin ({x^6})}}{{12{x^{11}}}} = \frac{{\sin ({x^6})}}{{2{x^6}}}\)     A1A1

EITHER

\(\frac{{g'(0)}}{{h'(0)}} = \frac{0}{0}\), using l’Hôpital’s Rule again,     (M1)

\(\frac{{g''(x)}}{{h''(x)}} = \frac{{6{x^5}\cos ({x^6})}}{{12{x^5}}} = \frac{{\cos ({x^6})}}{2}\)     A1A1

\(\frac{{g''(0)}}{{h''(0)}} = \frac{1}{2}\), hence the limit is \(\frac{1}{2}\)     A1

OR

So \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\)     A1

\( = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\)     A1

\( = \frac{1}{2}{\text{ since }}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}} = 1\)     A1 (R1)

METHOD 2

substituting \({{x^6}}\) for x in the expansion \(\cos x = 1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} \ldots \)     (M1)

\(\frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \frac{{1 - \left( {1 - \frac{{{x^{12}}}}{2} + \frac{{{x^{24}}}}{{24}}} \right) \ldots }}{{{x^{12}}}}\)     M1A1

\( = \frac{1}{2} - \frac{{{x^{12}}}}{{24}} + ...\)     A1A1

\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \frac{1}{2}\)     M1A1

Note: Accept solutions using Maclaurin expansions.

[7 marks]

Surprisingly, some weaker candidates were more successful in answering this question than stronger candidates. If candidates failed to simplify the expression after the first application of L’Hôpital’s rule, they generally were not successful in correctly differentiating the expression a \({2^{{\text{nd}}}}\) time, hence could not achieve the final three A marks.

(a)     Given that \(y = \ln \cos x\) , show that the first two non-zero terms of the Maclaurin series for y are \( - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}}\).

(b)     Use this series to find an approximation in terms of \(\pi {\text{ for }}\ln 2\) .

(a)     \(f(x) = \ln \cos x\)

\(f'(x) = \frac{{ - \sin x}}{{\cos x}} = - \tan x\)     M1A1

\(f''(x) = - {\sec ^2}x\)     M1

\(f'''(x) = - 2\sec x\sec x\tan x\)     A1

\({f^{iv}}(x) = - 2{\sec ^2}x({\sec ^2}x) - 2\tan x(2{\sec ^2}x\tan x)\)

\( = - 2{\sec ^4}x - 4{\sec ^2}x{\tan ^2}x\)     A1

\(f(x) = f(0) + xf'(0) + \frac{{{x^2}}}{{2!}}f''(0) + \frac{{{x^3}}}{{3!}}f'''(0) + \frac{{{x^4}}}{{4!}}{f^{iv}}(0) + …\)

\(f(0) = 0,\)     M1

\(f'(0) = 0,\)

\(f''(0) = - 1,\)

\(f'''(0) = 0,\)

\({f^{iv}}(0) = - 2,\)     A1

Notes: Award the A1 if all the substitutions are correct.

Allow FT from their derivatives.

\(\ln (\cos x) \approx  - \frac{{{x^2}}}{{2!}} - \frac{{2{x^4}}}{{4!}}\)     A1

\( = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}}\)     AG

[8 marks]

(b)     Some consideration of the manipulation of ln 2     (M1)

Attempt to find an angle     (M1)

EITHER

Taking \(x = \frac{\pi }{3}\)     A1

\(\ln \frac{1}{2} \approx - \frac{{{{\left( {\frac{\pi }{3}} \right)}^2}}}{{2!}} - \frac{{2{{\left( {\frac{\pi }{3}} \right)}^4}}}{{4!}}\)     A1

\( - \ln 2 \approx - \frac{{\frac{{{\pi ^2}}}{9}}}{{2!}} - \frac{{2\frac{{{\pi ^4}}}{{81}}}}{{4!}}\)     A1

\(\ln 2 \approx \frac{{{\pi ^2}}}{{18}} + \frac{{{\pi ^4}}}{{972}} = \frac{{{\pi ^2}}}{9}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{108}}} \right)\)     A1

OR

Taking \(x = \frac{\pi }{4}\)     A1

\(\ln \frac{1}{{\sqrt 2 }} \approx - \frac{{{{\left( {\frac{\pi }{4}} \right)}^2}}}{{2!}} - \frac{{2{{\left( {\frac{\pi }{4}} \right)}^4}}}{{4!}}\)     A1

\( - \frac{1}{2}\ln 2 \approx - \frac{{\frac{{{\pi ^2}}}{{16}}}}{{2!}} - \frac{{2\frac{{{\pi ^4}}}{{256}}}}{{4!}}\)     A1

\(\ln 2 \approx \frac{{{\pi ^2}}}{{16}} + \frac{{{\pi ^4}}}{{1536}} = \frac{{{\pi ^2}}}{8}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{192}}} \right)\)     A1

[6 marks]

Total [14 marks]

Some candidates had difficulty organizing the derivatives but most were successful in getting the series. Using the series to find the approximation for \(\ln 2\) in terms of \(\pi \) was another story and it was rare to see a good solution.

Use Euler’s method with step length 0.1 to find an approximate value of y when x = 0.4.

Write down, giving a reason, whether your approximate value for y is greater than or less than the actual value of y .

use of \(y \to y + h\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     (M1)

approximate value of y = 1.57     A1

Note: Accept values in the tables correct to 3 significant figures.

[7 marks]

the approximate value is less than the actual value because it is assumed that \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) remains constant throughout each interval whereas it is actually an increasing function     R1

[1 mark]

Most candidates were familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures. Few candidates were able to answer (b) correctly with most believing incorrectly that the step length was a relevant factor.

Most candidates were familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures. Few candidates were able to answer (b) correctly with most believing incorrectly that the step length was a relevant factor.

Show that \(f’(0) = p\).

Show that \({f^{(n + 2)}}(0) = ({n^2} - {p^2}){f^{(n)}}(0)\).

For \(p \in \mathbb{R}\backslash \{  \pm 1,{\text{ }} \pm 3\} \), show that the Maclaurin series for \(f(x)\), up to and including the \({x^5}\) term, is

\(px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} + \frac{{p(9 - {p^2})(1 - {p^2})}}{{5!}}{x^5}\).

Hence or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x}\).

If \(p\) is an odd integer, prove that the Maclaurin series for \(f(x)\) is a polynomial of degree \(p\).

\(f’(x) = \frac{{p\cos (p\arcsin x)}}{{\sqrt {1 - {x^2}} }}\)     (M1)A1

Note: Award M1 for attempting to use the chain rule.

\(f’(0) = p\)     AG

[2 marks]

EITHER

\({f^{(n + 2)}}(0) + ({p^2} - {n^2}){f^{(n)}}(0) = 0\)     A1

OR

for eg, \((1 - {x^2}){f^{(n + 2)}}(x) = (2n + 1)x{f^{(n + 1)}}(x) - ({p^2} - {n^2}){f^{(n)}}(x)\)     A1

Note: Award A1 for eg, \((1 - {x^2}){f^{(n + 2)}}(x) - (2n + 1)x{f^{(n + 1)}}(x) =  - ({p^2} - {n^2}){f^{(n)}}(x)\).

THEN

\({f^{(n + 2)}}(0) = ({n^2} - {p^2}){f^{(n)}}(0)\)     AG

[1 mark]

considering \(f\) and its derivatives at \(x = 0\)     (M1)

\(f(0) = 0\) and \(f’(0) = p\) from (a)     A1

\(f’’(0) = 0,{\text{ }}{f^{(4)}}(0) = 0\)     A1

\({f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0) = (1 - {p^2})p\),

\({f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0) = (9 - {p^2})(1 - {p^2})p\)     A1

Note:     Only award the last A1 if either \({f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0)\) and \({f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0)\) have been stated or the general Maclaurin series has been stated and used.

\(px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} + \frac{{p(9 - {p^2})(1 - {p^2})}}{{5!}}{x^5}\)     AG

[4 marks]

METHOD 1

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} +  \ldots }}{3}\)     M1

\( = p\)     A1

METHOD 2

by l’Hôpital’s rule \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{p\cos (p\arcsin x)}}{{\sqrt {1 - {x^2}} }}\)     M1

\( = p\)     A1

[2 marks]

the coefficients of all even powers of \(x\) are zero     A1

the coefficient of \({x^p}\) for (\(p\) odd) is non-zero (or equivalent eg,

the coefficients of all odd powers of \(x\) up to \(p\) are non-zero)     A1

\({f^{(p + 2)}}(0) = ({p^2} - {p^2}){f^{(p)}}(0) = 0\) and so the coefficient of \({x^{p + 2}}\) is zero     A1

the coefficients of all odd powers of \(x\) greater than \(p + 2\) are zero (or equivalent)     A1

so the Maclaurin series for \(f(x)\) is a polynomial of degree \(p\)     AG

[4 marks]

Given that \(n > {\text{ln}}\,n\) for \(n > 0\), use the comparison test to show that the series \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is divergent.

Find the interval of convergence for \(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( {3x} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \).

METHOD 1

\({\text{ln}}\left( {n + 2} \right) < n + 2\)     (A1)

\( \Rightarrow \frac{1}{{{\text{ln}}\left( {n + 2} \right)}} > \frac{1}{{n + 2}}\) (for \(n \geqslant 0\))     A1

Note: Award A0 for statements such as \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}}  > \sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}} \). However condone such a statement if the above A1 has already been awarded.

\(\sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}} \) (is a harmonic series which) diverges     R1

Note: The R1 is independent of the A1s.

Award R0 for statements such as "\(\frac{1}{{n + 2}}\) diverges".

so \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test     AG

METHOD 2

\(\frac{1}{{{\text{ln}}\,n}} > \frac{1}{n}\) (for \(n \geqslant 2\))     A1

Note: Award A0 for statements such as \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}}  > \sum\limits_{n = 2}^\infty  {\frac{1}{n}} \). However condone such a statement if the above A1 has already been awarded.

a correct statement linking \(n\) and \(n + 2\) eg,

\(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}}  = \sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}} \) or \(\sum\limits_{n = 0}^\infty  {\frac{1}{{n + 2}}}  = \sum\limits_{n = 2}^\infty  {\frac{1}{n}} \)     A1

Note: Award A0 for \(\sum\limits_{n = 0}^\infty  {\frac{1}{n}} \)

\(\sum\limits_{n = 2}^\infty  {\frac{1}{n}} \) (is a harmonic series which) diverges

(which implies that \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{\text{ln}}\,n}}} \) diverges by the comparison test)      R1

Note: The R1 is independent of the A1s.

Award R0 for statements such as \(\sum\limits_{n = 0}^\infty  {\frac{1}{n}} \) deiverges and "\({\frac{1}{n}}\) diverges".

Award A1A0R1 for arguments based on \(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \).

so \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) diverges by the comparison test      AG

[3 marks]

applying the ratio test \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{{\left( {3x} \right)}^{n + 1}}}}{{{\text{ln}}\left( {n + 3} \right)}} \times \frac{{{\text{ln}}\left( {n + 2} \right)}}{{{{\left( {3x} \right)}^n}}}} \right|\)     M1

\( = \left| {3x} \right|\) (as \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left| {\frac{{{\text{ln}}\left( {n + 2} \right)}}{{{\text{ln}}\left( {n + 3} \right)}}} \right| = 1\)     A1

Note: Condone the absence of limits and modulus signs.

Note: Award M1A0 for \(3{x^n}\). Subsequent marks can be awarded.

series converges for \( - \frac{1}{3} < x < \frac{1}{3}\)

considering \(x =  - \frac{1}{3}\) and \(x = \frac{1}{3}\)     M1

Note: Award M1 to candidates who consider one endpoint.

when \(x = \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) which is divergent (from (a))      A1

Note: Award this A1 if \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{\text{ln}}\left( {n + 2} \right)}}} \) is not stated but reference to part (a) is.

when \(x =  - \frac{1}{3}\), series is \(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \)     A1

\(\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{{\text{ln}}\left( {n + 2} \right)}}} \) converges (conditionally) by the alternating series test      R1

(strictly alternating, \(\left| {{u_n}} \right| > \left| {{u_{n + 1}}} \right|\) for \(n \geqslant 0\) and \(\mathop {{\text{lim}}}\limits_{n \to \infty } \left( {{u_n}} \right) = 0\))

so the interval of convergence of S is \( - \frac{1}{3} \leqslant x < \frac{1}{3}\)     A1

Note: The final A1 is dependent on previous A1s – ie, considering correct series when \(x =  - \frac{1}{3}\) and \(x = \frac{1}{3}\) and on the final R1.

Award as above to candidates who firstly consider \(x =  - \frac{1}{3}\) and then state conditional convergence implies divergence at \(x = \frac{1}{3}\).

[7 marks]

Find series for \({\sec ^2}x\), in terms of \({a_1}\), \({a_3}\) and \({a_5}\), up to and including the \({x^4}\) term

by differentiating the above series for \(\tan x\);

Find series for \({\sec ^2}x\), in terms of \({a_1}\), \({a_3}\) and \({a_5}\), up to and including the \({x^4}\) term

by using the relationship \({\sec ^2}x = 1 + {\tan ^2}x\).

Hence, by comparing your two series, determine the values of \({a_1}\), \({a_3}\) and \({a_5}\).

\(({\sec ^2}x = ){\text{ }}{a_1} + 3{a_3}{x^2} + 5{a_5}{x^4} + \ldots \)     A1

[1 mark]

\({\sec ^2}x = 1 + {({a_1}x + {a_3}{x^3} + {a_5}{x^5} + \ldots )^2}\)

\( = 1 + a_1^2{x^2} + 2{a_1}{a_3}{x^4} + \ldots \)     M1A1

Note:     Condone the presence of terms with powers greater than four.

[2 marks]

equating constant terms: \({a_1} = 1\)     A1

equating \({x^2}\) terms: \(3{a_3} = a_1^2 = 1 \Rightarrow {a_3} = \frac{1}{3}\)     A1

equating \({x^4}\) terms: \(5{a_5} = 2{a_1}{a_3} = \frac{2}{3} \Rightarrow {a_5} = \frac{2}{{15}}\)     A1

[3 marks]

Show that putting \(z = {y^2}\) transforms the differential equation into \(\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2xz = 2x\).

By solving this differential equation in \(z\), obtain an expression for \(y\) in terms of \(x\).

METHOD 1

\(z = {y^2} \Rightarrow y = {z^{1/2}}\)

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{2{z^{1/2}}}}\frac{{{\text{d}}z}}{{{\text{d}}x}}\)    M1A1

substituting, \(\frac{1}{{2{z^{1/2}}}}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{x}{{{z^{1/2}}}} - x{z^{1/2}}\)     M1A1

\(\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2xz = 2x\)    AG

METHOD 2

\(z = {y^2}\)

\(\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}\)    M1A1

\(\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2x - 2x{y^2}\)    M1A1

\(\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2xz = 2x\)    AG

[4 marks]

METHOD 1

integrating factor \( = {{\text{e}}^{\int {2x{\text{d}}x} }} = {{\text{e}}^{{x^2}}}\)     (M1)A1

\({{\text{e}}^{{x^2}}}\frac{{{\text{d}}z}}{{{\text{d}}x}} + 2x{{\text{e}}^{{x^2}}}z = 2x{{\text{e}}^{{x^2}}}\)    (M1)

\(z{{\text{e}}^{{x^2}}} = \int {2x{{\text{e}}^{{x^2}}}{\text{d}}x} \)    A1

\( = {{\text{e}}^{{x^2}}} + C\)    A1

substitute \(y = 2\) therefore \(z = 4\) when \(x = 0\)     (M1)

\(4 = 1 + C\)

\(C = 3\)    (A1)

the solution is \(z = 1 + 3{{\text{e}}^{ - {x^2}}}\)     (M1)

Note:     This line may be seen before determining the value of \(C\).

so that \(y = \sqrt {1 + 3{{\text{e}}^{ - {x^2}}}} \)     A1

METHOD 2

\(\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2x(1 - z)\)

\(\int {\frac{1}{{1 - z}}{\text{d}}z = \int {2x{\text{d}}x} } \)    M1

\( - \ln (1 - z) = {x^2} + C\)    A1A1

\(1 - z = {{\text{e}}^{ - {x^2} - c}}\) (or \(1 - z = B{{\text{e}}^{ - {x^2}}}\))     M1A1

solving for \(z\)     (M1)

\(z = 1 + A{{\text{e}}^{ - {x^2}}}\)

\(z = 4\) when \(x = 0\)     (M1)

so \(A = 3\)     (A1)

the solution is \(z = 1 + 3{{\text{e}}^{ - {x^2}}}\)

so \(y = \sqrt {1 + 3{{\text{e}}^{ - {x^2}}}} \)     A1

[9 marks]

Several misconceptions were identified that showed poor understanding of the chain rule. Although many candidates were successful in establishing the result the presentation of their work was far from what is expected in a show that question.

Part (b) was well attempted using both method 1 (integration factor) and 2 (separation of variables). The most common error was omission of the constant of integration or errors in finding its value. Candidates that used method 2 often had difficulties in integrating \(\frac{1}{{(1 - z)}}\) correctly and making \(z\) the subject often losing out on accuracy marks.

Using l’Hôpital’s rule, find \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\).

(i) Find \({a_1}\) and \({a_2}\) and hence write down an expression for \({a_n}\).

(ii) Show that \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\).

Using a suitable test, determine whether this series converges or diverges.

\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\arcsin \left( {\frac{1}{{\sqrt {(x + 1)} }}} \right)}}{{\frac{1}{{\sqrt x }}}}} \right)\) is of the form \(\frac{0}{0}\)

and so will equal the limit of \(\frac{{\frac{{\frac{{ - 1}}{2}{{(x + 1)}^{ - \frac{3}{2}}}}}{{\sqrt {1 - \left( {\frac{1}{{x + 1}}} \right)} }}}}{{\frac{{ - 1}}{2}{x^{ - \frac{3}{2}}}}}\)     M1M1A1A1

Note: M1 for attempting differentiation of the top and bottom, M1A1 for derivative of top (only award M1 if chain rule is used), A1 for derivative of bottom.

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\frac{x}{{(x + 1)}}} \right)}^{\frac{3}{2}}}}}{{\sqrt {\frac{x}{{x + 1}}} }} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{x + 1}}} \right)\)    M1

Note: Accept any intermediate tidying up of correct derivative for the method mark.

\( = 1\)    A1

[6 marks]

(i)     \({a_1} = \sqrt 2 ,{\text{ }}{a_2} = \sqrt 3 \)     A1

\({a_n} = \sqrt {n + 1} \)    A1

(ii)     \(\sin {\theta _n} = \frac{1}{{{a_n}}} = \frac{1}{{\sqrt {n + 1} }}\)     A1

Note: Allow \({\theta _n} = \arcsin \left( {\frac{1}{{{a_n}}}} \right)\) if \({a_n} = \sqrt {n + 1} \) in b(i).

so \({\theta _n} = \arcsin \frac{1}{{\sqrt {(n + 1)} }}\)     AG

[3 marks]

for \(\sum\limits_{n = 1}^\infty  {\arcsin \frac{1}{{\sqrt {(n + 1)} }}} \) apply the limit comparison test (since both series of positive terms)     M1

with \(\sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt n }}} \)     A1

from (a) \(\mathop {\lim }\limits_{n \to \infty } \frac{{\arcsin \frac{1}{{\sqrt {(n + 1)} }}}}{{\frac{1}{{\sqrt n }}}} = 1\), so the two series either both converge or both diverge     M1R1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt 2 }}} \) diverges (as is a \(p\)-series with \(p = \frac{1}{2}\))     A1

hence \(\sum\limits_{n = 1}^\infty  {{\theta _n}} \) diverges     A1

[6 marks]

Show that the sequence converges to a limit L , the value of which should be stated.

Find the least value of the integer N such that \(\left| {{u_n} - L} \right| < \varepsilon \) , for all n > N where

(i)     \(\varepsilon  = 0.1\);

(ii)     \(\varepsilon  = 0.00001\).

For each of the sequences \(\left\{ {\frac{{{u_n}}}{n}} \right\},{\text{ }}\left\{ {\frac{1}{{2{u_n} - 2}}} \right\}\) and \(\left\{ {{{( - 1)}^n}{u_n}} \right\}\) , determine whether or not it converges.

Prove that the series \(\sum\limits_{n = 1}^\infty  {({u_n} - L)} \) diverges.

\({u_n} = \frac{{3 + \frac{2}{n}}}{{2 - \frac{1}{n}}}\) or \(\frac{3}{2} + \frac{A}{{2n - 1}}\)     M1

using \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\)     (M1)

obtain \(\mathop {\lim }\limits_{n \to \infty } {u_n} = \frac{3}{2} = L\)     A1     N1

[3 marks]

\({u_n} - L = \frac{7}{{2(2n - 1)}}\)     (A1)

\(\left| {{u_n} - L} \right| < \varepsilon  \Rightarrow n > \frac{1}{2}\left( {1 + \frac{7}{{2\varepsilon }}} \right)\)     (M1)

(i)     \(\varepsilon  = 0.1 \Rightarrow N = 18\)     A1

(ii)     \(\varepsilon  = 0.00001 \Rightarrow N = 175000\)     A1

[4 marks]

\({u_n} \to L\) and \(\frac{1}{n} \to 0\)     M1

\( \Rightarrow \frac{{{u_n}}}{n} \to (L \times 0) = 0\) , hence converges     A1

\(2{u_n} - 2 \to 2L - 2 = 1 \Rightarrow \frac{1}{{2{u_n} - 2}} \to 1\) , hence converges     M1A1 

Note: To award A1 the value of the limit and a statement of convergence must be clearly seen for each sequence.

\({( - 1)^n}{u_n}\) does not converge     A1

The sequence alternates (or equivalent wording) between values close to \( \pm L\)     R1

[6 marks]

\({u_n} - L > \frac{7}{{4n}}\) (re: harmonic sequence)     M1

\( \Rightarrow \sum\limits_{n = 1}^\infty  {({u_n} - L)} \) diverges by the comparison theorem     R1 

Note: Accept alternative methods.

[2 marks]

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks. 

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The “show that” in part (a) of this problem was not adequately dealt with by a significant minority of candidates and simply stating the limit and not demonstrating its existence lost marks. Part (b), whilst being possible without significant knowledge of limits, seemed to intimidate some candidates due to its unfamiliarity and the notation. Part (c) was somewhat disappointing as many candidates attempted to apply rules on the convergence of series to solve a problem that was dealing with the limits of sequences. The same confusion was seen on part (d) where also some errors in algebra prevented candidates from achieving full marks.

The function \(f\) is defined by

\[f(x) = \left\{ {\begin{array}{*{20}{l}} {{x^2} - 2,}&{x < 1} \\ {ax + b,}&{x \geqslant 1} \end{array}} \right.\]

where \(a\) and \(b\) are real constants.

Given that both \(f\) and its derivative are continuous at \(x = 1\), find the value of \(a\) and the value of \(b\).

considering continuity \(\mathop {\lim }\limits_{x \to {1^ - }} ({x^2} - 2) =  - 1\)     (M1)

\(a + b =  - 1\)     (A1)

considering differentiability \(2x = a\) when \(x = 1\)     (M1)

\( \Rightarrow a = 2\)     A1

\(b =  - 3\)     A1

[5 marks]

State the mean value theorem for a function that is continuous on the closed interval \([a,{\text{ }}b]\) and differentiable on the open interval \(]a,{\text{ }}b[\).

(i)     Find \(g(0)\).

(ii)     Find \(g(h)\).

(iii)     Apply the mean value theorem to the function \(g(x)\) on the closed interval \([0,{\text{ }}h]\) to show that there exists \(c\) in the open interval \(]0,{\text{ }}h[\) such that \(g'(c) = 0\).

(iv)     Find \(g'(x)\).

(v)     Hence show that \( - (h - c)f''(c) + \frac{{2(h - c)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = 0\).

(vi)     Deduce that \(f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}{\text{ }}f''(c)\).

Hence show that, for \(h > 0\)

\(1 - \cos (h) \leqslant \frac{{{h^2}}}{2}\).

there exists \(c\) in the open interval \(]a,{\text{ }}b[\) such that     A1

\(\frac{{f(b) - f(a)}}{{b - a}} = f'(c)\)    A1

Note: Open interval is required for the A1.

[2 marks]

(i)     \(g(0) = f(h) - f(0) - hf'(0) - \frac{{{h^2}}}{{{h^2}}}\left( {{\text{ }}f(h) - f(0) - hf'(0)} \right)\)

\( = 0\)    A1

(ii)     \(g(h) = f(h) - f(h) - 0 - 0\)

\( = 0\)    A1

(iii)     (\(g(x)\) is a differentiable function since it is a combination of other differentiable functions \(f\), \({f'}\) and polynomials.)

there exists \(c\) in the open interval \(]0,{\text{ }}h[\) such that

\(\frac{{g(h) - g(0)}}{h} = g'(c)\)    A1

\(\frac{{g(h) - g(0)}}{h} = 0\)    A1

hence \(g'(c) = 0\)     AG

(iv)     \(g'(x) =  - f'(x) + f'(x) - (h - x)f''(x) + \frac{{2(h - x)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right)\)     A1A1

Note: A1 for the second and third terms and A1 for the other terms (all terms must be seen).

\( =  - (h - x)f''(x) + \frac{{2(h - x)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right)\)

(v)     putting \(x = c\) and equating to zero     M1

\( - (h - c)f''(c) + \frac{{2(h - c)}}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = g'(c) = 0\)    AG

(vi)     \( - f''(c) + \frac{2}{{{h^2}}}\left( {f(h) - f(0) - hf'(0)} \right) = 0\)     A1

since \(h - c \ne 0\)     R1

\(\frac{{{h^2}}}{2}f''(c) = f(h) - f(0) - hf'(0)\)

\(f(h) = f(0) + hf'(0) + \frac{{{h^2}}}{2}f''(c)\)    AG

[9 marks]

letting \(f(x) = \cos (x)\)     M1

\(f'(x) =  - \sin (x)\)    \(f''(x) =  - \cos (x)\)     A1

\(\cos (h) = 1 + 0 - \frac{{{h^2}}}{2}\cos (c)\)     A1

\(1 - \cos (h) = \frac{{{h^2}}}{2}\cos (c)\)    (A1)

since \(\cos (c) \leqslant 1\)     R1

\(1 - \cos (h) \leqslant \frac{{{h^2}}}{2}\)    AG

Note: Allow \(f(x) = a \pm b\cos x\).

[5 marks]

Given that \(f(x) = \ln x\), use the mean value theorem to show that, for \(0 < a < b\), \(\frac{{b - a}}{b} < \ln \frac{b}{a} < \frac{{b - a}}{a}\).

Hence show that \(\ln (1.2)\) lies between \(\frac{1}{m}\) and \(\frac{1}{n}\), where \(m\), \(n\) are consecutive positive integers to be determined.

\(f'(x) = \frac{1}{x}\)    (A1)

using the MVT \(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\) (where \(c\) lies between \(a\) and \(b\))     (M1)

\(f'(c) = \frac{{\ln b - \ln a}}{{b - a}}\)    A1

\(\ln \frac{b}{a} = \ln b - \ln a\)    (M1)

\(f'(c) = \frac{{\ln \frac{b}{a}}}{{b - a}}\)

since \(f'(x)\) is a decreasing function or \(a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a}\)     R1

\(f'(b) < f'(c) < f'(a)\)    (M1)

\(\frac{1}{b} < \frac{{\ln \frac{b}{a}}}{{b - a}} < \frac{1}{a}\)    A1

\(\frac{{b - a}}{b} < \ln \frac{b}{a} < \frac{{b - a}}{a}\)    AG

[7 marks]

putting \(b = 1.2,{\text{ }}a = 1\), or equivalent     M1

\(\frac{1}{6} < \ln 1.2 < \frac{1}{5}\)    A1

\((m = 6,{\text{ }}n = 5)\)

[2 marks]

Although many candidates achieved at least a few marks in this question, the answers revealed difficulties in setting up a proof. The Mean value theorem was poorly quoted and steps were often skipped. The conditions under which the Mean value theorem is valid were largely ignored, as were the reasoned steps towards the answer.

There were inequalities everywhere, without a great deal of meaning or showing progress. A number of candidates attempted to work backwards and presented the work in a way that made it difficult to follow their reasoning; in part (b) many candidates ignored the instruction ‘hence’ and just used GDC to find the required values; candidates that did notice the link to part a) answered this question well in general. A number of candidates guessed the answer and did not present an analytical derivation as required.

Find the radius of convergence of the series \(\sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^n}{x^n}}}{{(n + 1){3^n}}}} \).

Determine whether the series \(\sum\limits_{n = 0}^\infty  {\left( {\sqrt[3]{{{n^3} + 1}} - n} \right)} \) is convergent or divergent.

The ratio test gives

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}(n + 1){3^n}}}{{(n + 2){3^{n + 1}}{{( - 1)}^n}{x^n}}}} \right|\)     M1A1

\( = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)x}}{{3(n + 2)}}} \right|\)     A1

\( = \frac{{\left| x \right|}}{3}\)     A1

So the series converges for \( \frac{{\left| x \right|}}{3} < 1,\)     A1

the radius of convergence is 3     A1

Note: Do not penalise lack of modulus signs.

[6 marks]

\({u_n} = \sqrt[3]{{{n^3} + 1}} - n\)

\( = n\left( {\sqrt[3]{{1 + \frac{1}{{{n^3}}} - 1}}} \right)\)     M1A1

\( = n\left( {1 + \frac{1}{{3{n^3}}} - \frac{1}{{9{n^6}}} + \frac{5}{{8{\text{l}}{n^9}}} - ... - 1} \right)\)     A1

using \({v_n} = \frac{1}{{{n^2}}}\) as the auxilliary series,     M1

since \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{v_n}}} = \frac{1}{3}{\text{ and }}\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + …\) converges     M1A1

then \(\sum {{u_n}} \) converges     A1

Note: Award M1A1A1M0M0A0A0 to candidates attempting to use the integral test.

[7 marks]

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.

Solve the differential equation

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + \frac{{{y^2}}}{{{x^2}}}\) (where x > 0 )

given that y = 2 when x = 1 . Give your answer in the form \(y = f(x)\) .

put y = vx so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     M1A1

the equation becomes \(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v + {v^2}\)     (A1)

leading to \(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2}\)     A1

separating variables, \(\int {\frac{{{\text{d}}x}}{x} = \int {\frac{{{\text{d}}v}}{{{v^2}}}} } \)     M1A1

hence \(\ln x = - {v^{ - 1}} + C\)     A1A1

substituting for v, \(\ln x = \frac{{ - x}}{y} + C\)     M1

Note: Do not penalise absence of C at the above stages.

substituting the boundary conditions,

\(0 = - \frac{1}{2} + C\)     M1

\(C = \frac{1}{2}\)     A1

the solution is \(\ln x = \frac{{ - x}}{y} + \frac{1}{2}\)     (A1)

leading to \(y = \frac{{2x}}{{1 - 2\ln x}}\) (or equivalent form)     A1

Note: Candidates are not required to note that \(x \ne \sqrt {\text{e}} \) .

[13 marks]

Many candidates were able to make a reasonable attempt at this question with many perfect solutions seen.

Find the set of values of k for which the improper integral \(\int_2^\infty {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) converges.

Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( - 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.

consider the limit as \(R \to \infty \) of the (proper) integral

\(\int_2^R {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \)     (M1)

substitute \(u = \ln x,{\text{ d}}u = \frac{1}{x}{\text{d}}x\)     (M1)

obtain \(\int_{\ln 2}^{\ln R} {\frac{1}{{{u^k}}}{\text{d}}u = \left[ { - \frac{1}{{k - 1}}\frac{1}{{{u^{k - 1}}}}} \right]_{\ln 2}^{\ln R}} \)     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

or \([\ln u]_{\ln 2}^{\ln R}\) if k = 1     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

because \(\ln R{\text{ }}({\text{and }}\ln \ln R) \to \infty {\text{ as }}R \to \infty \)     (M1)

converges in the limit if k > 1     A1

[6 marks]

C: \({\text{terms}} \to 0{\text{ as }}r \to \infty \)     A1

\(\left| {{u_{r + 1}}} \right| < \left| {{u_r}} \right|\) for all r     A1

convergence by alternating series test     R1

AC: \({(x\ln x)^{ - 1}}\) is positive and decreasing on \([2,\,\infty )\)     A1

not absolutely convergent by integral test using part (a) for k = 1     R1

[5 marks]

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

Determine whether the series \(\sum\limits_{n = 1}^\infty  {\frac{{{n^{10}}}}{{{{10}^n}}}} \) is convergent or divergent.

Consider

\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{{{(n + 1)}^{10}}}}{{{{10}^{n + 1}}}} \times \frac{{{{10}^n}}}{{{n^{10}}}}\)     M1A1

\( = \frac{1}{{10}}{\left( {1 + \frac{1}{n}} \right)^{10}}\)     A1

\( \to \frac{1}{{10}}{\text{ as }}n \to \infty \)     A1

\(\frac{1}{{10}} < 1\)     R1

So by the Ratio Test the series is convergent.     R1

[6 marks]

Most candidates used the Ratio Test successfully to establish convergence. Candidates who attempted to use Cauchy’s (Root) Test were often less successful although this was a valid method.

(a)     Using the diagram, show that \(\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + ... < \int_3^\infty  {\frac{1}{{{x^3}}}{\text{d}}x < \frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + ...} \) .

(b)     Hence find upper and lower bounds for \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}} \).

(a)     The area under the curve is sandwiched between the sum of the areas of the lower rectangles and the upper rectangles.     M2

Therefore

\(1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + 1 \times \frac{1}{{{6^3}}} + ... < \int_3^\infty  {\frac{{{\text{d}}x}}{{{x^3}}} < 1 \times \frac{1}{{{3^3}}} + 1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + ...} \)     A1

which leads to the printed result.

[3 marks]

(b)     We note first that

\(\int_3^\infty  {\frac{{{\text{d}}x}}{{{x^3}}} = \left[ { - \frac{1}{{2{x^2}}}} \right]_3^\infty  = \frac{1}{{18}}} \)     M1A1

Consider first

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}}  = 1 + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + \left( {\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + ...} \right)\)     M1A1

\( < 1 + \frac{1}{8} + \frac{1}{{27}} + \frac{1}{{18}}\)     M1A1

\( = \frac{{263}}{{216}}{\text{ (1.22)}}\) (which is an upper bound)     A1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^3}}}}  = 1 + \frac{1}{{{2^3}}} + \left( {\frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + ...} \right)\)     M1A1

\( > 1 + \frac{1}{8} + \frac{1}{{18}}\)     M1A1

\( = \frac{{85}}{{72}}\left( {\frac{{255}}{{216}}} \right){\text{ (1.18)}}\) (which is a lower bound)     A1

[12 marks]

Total [15 marks]

Many candidates failed to give a convincing argument to establish the inequality. In (b), few candidates progressed beyond simply evaluating the integral.

Figure 1 shows part of the graph of \(y = \frac{1}{x}\) together with line segments parallel to the coordinate axes.

(i)     By considering the areas of appropriate rectangles, show that

\[\frac{{2a + 1}}{{a(a + 1)}} < \ln \left( {\frac{{a + 1}}{{a - 1}}} \right) < \frac{{2a - 1}}{{a(a - 1)}}.\]

(ii)     Hence find lower and upper bounds for \(\ln (1.2)\).

An improved upper bound can be found by considering Figure 2 which again shows part of the graph of \(y = \frac{1}{x}\).

(i)     By considering the areas of appropriate regions, show that

\[\ln \left( {\frac{a}{{a - 1}}} \right) < \frac{{2a - 1}}{{2a(a - 1)}}.\]

(ii)     Hence find an upper bound for \(\ln (1.2)\).

(i)     the area under the curve between a – 1 and  a + 1

\( = \int_{a - 1}^{a + 1} {\frac{{{\text{d}}x}}{x}} \)     M1

\( = [\ln x]_{a - 1}^{a + 1}\)     A1

\( = \ln \left( {\frac{{a + 1}}{{a - 1}}} \right)\)     A1

lower sum \( = \frac{1}{a} + \frac{1}{{a + 1}}\)     M1A1

\( = \frac{{2a + 1}}{{a(a + 1)}}\)     AG

upper sum \( = \frac{1}{{a - 1}} + \frac{1}{a}\)     A1

\( = \frac{{2a - 1}}{{a(a - 1)}}\)     AG

it follows that

\(\frac{{2a + 1}}{{a(a + 1)}} < \ln \left( {\frac{{a + 1}}{{a - 1}}} \right) < \frac{{2a - 1}}{{a(a - 1)}}\)

because the area of the region under the curve lies between the areas of the regions defined by the lower and upper sums     R1

(ii)     putting

\(\left( {\frac{{a + 1}}{{a - 1}} = 1.2} \right) \Rightarrow a = 11\)     A1

therefore, \({\text{UB}} = \frac{{21}}{{110}}( = 0.191),{\text{ LB}} = \frac{{23}}{{132}}( = 0.174)\)     A1

[9 marks]

(i)     the area under the curve between a – 1 and a

\( = \int_{a - 1}^a {\frac{{{\text{d}}x}}{x}} \)     A1

\( = [\ln x]_{a - 1}^a = \ln \left( {\frac{a}{{a - 1}}} \right)\)

attempt to find area of trapezium     M1

area of trapezoidal “upper sum” \( = \frac{1}{2}\left( {\frac{1}{{a - 1}} + \frac{1}{a}} \right)\) or equivalent     A1

\( = \frac{{2a - 1}}{{2a(a - 1)}}\)

it follows that \(\ln \left( {\frac{a}{{a - 1}}} \right) < \frac{{2a - 1}}{{2a(a - 1)}}\)     AG

(ii)     putting

\(\left( {\frac{a}{{a - 1}} = 1.2} \right) \Rightarrow a = 6\)     A1

therefore, \({\text{UB}} = \frac{{11}}{{60}}( = 0.183)\)     A1 

[5 marks]

Many candidates made progress with this problem. This was pleasing since whilst being relatively straightforward it was not a standard problem. There were still some candidates who did not use the definite integral correctly to find the area under the curve in part (a) and part (b). Also candidates should take care to show all the required working in a “show that” question, even when demonstrating familiar results. The ability to find upper and lower bounds was often well done in parts (a) (ii) and (b) (ii).

Many candidates made progress with this problem. This was pleasing since whilst being relatively straightforward it was not a standard problem. There were still some candidates who did not use the definite integral correctly to find the area under the curve in part (a) and part (b). Also candidates should take care to show all the required working in a “show that” question, even when demonstrating familiar results. The ability to find upper and lower bounds was often well done in parts (a) (ii) and (b) (ii).

Find the exact values of \(a\) and \(b\) if \(f\) is continuous and differentiable at \(x = 1\).

(i)     Use Rolle’s theorem, applied to \(f\), to prove that \(2{x^4} - 4{x^3} - 5{x^2} + 4x + 1 = 0\) has a root in the interval \(\left] { - 1,1} \right[\).

(ii)     Hence prove that \(2{x^4} - 4{x^3} - 5{x^2} + 4x + 1 = 0\) has at least two roots in the interval \(\left] { - 1,1} \right[\).

\(\mathop {{\text{lim}}}\limits_{x \to {1^ - }} {{\text{e}}^{ - {x^2}}}\left( { - {x^3} + 2{x^2} + x} \right) = \mathop {{\text{lim}}}\limits_{x \to {1^ + }}  (ax + b)\)   \(( = a + b)\)     M1

\(2{{\text{e}}^{ - 1}} = a + b\)     A1

differentiability: attempt to differentiate both expressions     M1

\(f'(x) =  - 2x{{\text{e}}^{ - {x^2}}}\left( { - {x^3} + 2{x^2} + x} \right) + {{\text{e}}^{ - {x^2}}}\left( { - 3{x^2} + 4x + 1} \right)\)   \((x < 1)\)     A1

(or \(f'(x) = {{\text{e}}^{ - {x^2}}}\left( {2{x^4} - 4{x^3} - 5{x^2} + 4x + 1} \right)\))

\(f'(x) = a\)   \((x > 1)\)     A1

substitute \(x = 1\) in both expressions and equate

\( - 2{{\text{e}}^{ - 1}} = a\)     A1

substitute value of \(a\) and find \(b = 4{{\text{e}}^{ - 1}}\)     M1A1

[8 marks]

(i)     \(f'(x) = {{\text{e}}^{ - {x^2}}}\left( {2{x^4} - 4{x^3} - 5{x^2} + 4x + 1} \right)\)   (for \(x \leqslant 1\))     M1

\(f(1) = f( - 1)\)     M1

Rolle’s theorem statement     (A1)

by Rolle’s Theorem, \(f'(x)\) has a zero in \(\left] { - 1,1} \right[\)     R1

hence quartic equation has a root in \(\left] { - 1,1} \right[\)     AG

(ii)     let \(g(x) = 2{x^4} - 4{x^3} - 5{x^2} + 4x + 1\).

\(g( - 1) = g(1) < 0\) and \(g(0) > 0\)     M1

as \(g\) is a polynomial function it is continuous in \(\left[ { - 1,0} \right]\) and \(\left[ {0,{\text{ 1}}} \right]\).     R1

(or \(g\) is a polynomial function continuous in any interval of real numbers)

then the graph of \(g\) must cross the x-axis at least once in \(\left] { - 1,0} \right[\)     R1

and at least once in \(\left] {0,1} \right[\).

[7 marks]

(a)     Using the Maclaurin series for \({(1 + x)^n}\), write down and simplify the Maclaurin series approximation for \({(1 - {x^2})^{ - \frac{1}{2}}}\) as far as the term in \({x^4}\)

(b)     Use your result to show that a series approximation for arccos x is

\[\arccos x \approx \frac{\pi }{2} - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5}.\]

(c)     Evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} - \arccos ({x^2}) - {x^2}}}{{{x^6}}}\).

(d)     Use the series approximation for \(\arccos x\) to find an approximate value for

\[\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,\]

giving your answer to 5 decimal places. Does your answer give the actual value of the integral to 5 decimal places?

(a)     using or obtaining \({(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{2}{x^2} + \ldots \)     (M1)

\({(1 - {n^2})^{ - \frac{1}{2}}} = 1 + ( - {x^2}) \times \left( { - \frac{1}{2}} \right) + \frac{{{{( - {x^2})}^2}}}{2} \times \left( { - \frac{1}{2}} \right) \times \left( { - \frac{3}{2}} \right) + \ldots \)     (A1)

\( = 1 + \frac{1}{2}{x^2} + \frac{3}{8}{x^4} + \ldots \)     A1

[3 marks]

(b)     integrating, and changing sign

\(\arccos x = - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5} + C + \ldots \)     M1A1

put x = 0,

\(\frac{\pi }{2} = C\)     M1

\(\left( {\arccos x \approx \frac{\pi }{2} - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5}} \right)\)     AG

[3 marks]

(c)     EITHER

using \(\arccos {x^2} \approx \frac{\pi }{2} - {x^2} - \frac{1}{6}{x^6} - \frac{3}{{40}}{x^{10}}\)     M1A1

\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} - \arccos {x^2} - {x^2}}}{{{x^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^6}}}{6} + {\text{higher powers}}}}{{{x^6}}}\)     M1A1

\( = \frac{1}{6}\)     A1

OR

using l’Hôpital’s Rule     M1

\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^4}} }} \times 2x - 2x}}{{6{x^5}}}\)     M1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^4}} }} - 1}}{{3{x^4}}}\)     A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{ - \frac{1}{2} \times \frac{1}{{{{(1 - {x^4})}^{3/2}}}} \times - 4{x^3}}}{{12{x^3}}}\)     M1

\( = \frac{1}{6}\)     A1

[5 marks]

(d)     \(\int_0^{0.2} {\arccos \sqrt x {\text{d}}x \approx \int_0^{0.2} {\left( {\frac{\pi }{2} - {x^{\frac{1}{2}}} - \frac{1}{6}{x^{\frac{3}{2}}} - \frac{3}{{40}}{x^{\frac{5}{2}}}} \right){\text{d}}x} } \)     M1

\( = \left[ {\frac{\pi }{2}x - \frac{2}{3}{x^{\frac{3}{2}}} - \frac{1}{{15}}{x^{\frac{5}{2}}} - \frac{3}{{140}}{x^{\frac{7}{2}}}} \right]_0^{0.2}\)     (A1)

\( = \frac{\pi }{2} \times 0.2 - \frac{2}{3} \times {0.2^{\frac{3}{2}}} - \frac{1}{{15}} \times {0.2^{\frac{5}{2}}} - \frac{3}{{140}} \times {0.2^{\frac{7}{2}}}\)     (A1)

= 0.25326 (to 5 decimal places)     A1

Note: Accept integration of the series approximation using a GDC.

using a GDC, the actual value is 0.25325     A1

so the approximation is not correct to 5 decimal places     R1

[6 marks]

Total [17 marks]

Many candidates ignored the instruction in the question to use the series for \({(1 + x)^n}\) to deduce the series for \({(1 - {x^2})^{ - 1/2}}\) and attempted instead to obtain it by successive differentiation. It was decided at the standardisation meeting to award full credit for this method although in the event the algebra proved to be too difficult for many. Many candidates used l’Hopital’s Rule in (c) – this was much more difficult algebraically than using the series and it usually ended unsuccessfully. Candidates should realise that if a question on evaluating an indeterminate limit follows the determination of a Maclaurin series then it is likely that the series will be helpful in evaluating the limit. Part (d) caused problems for many candidates with algebraic errors being common. Many candidates failed to realise that the best way to find the exact value of the integral was to use the calculator.

(a)     Using l’Hopital’s Rule, show that \(\mathop {\lim }\limits_{x \to \infty } x{{\text{e}}^{ - x}} = 0\) .

(b)     Determine \(\int_0^a {x{{\text{e}}^{ - x}}{\text{d}}x} \) .

(c)     Show that the integral \(\int_0^\infty  {x{{\text{e}}^{ - x}}{\text{d}}x} \) is convergent and find its value.

(a)     \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\text{e}}^x}}}\)     M1A1

= 0     AG

[2 marks]

(b)     Using integration by parts     M1

\(\int_0^a {x{{\text{e}}^{ - x}}{\text{d}}x}  = \left[ { - x{{\text{e}}^{ - x}}} \right]_0^a + \int_0^a {{{\text{e}}^{ - x}}{\text{d}}x} \)     A1A1

\( = - a{{\text{e}}^{ - a}} - \left[ {{e^{ - x}}} \right]_0^a\)     A1

\( = 1 - a{{\text{e}}^{ - a}} - {{\text{e}}^{ - a}}\)     A1

[5 marks]

(c)     Since \({{\text{e}}^{ - a}}\) and \(a{{\text{e}}^{ - a}}\) are both convergent (to zero), the integral is convergent.     R1

Its value is 1.     A1

[2 marks]

Total [9 marks]

Most candidates made a reasonable attempt at (a). In (b), however, it was disappointing to note that some candidates were unable to use integration by parts to perform the integration. In (c), while many candidates obtained the correct value of the integral, proof of its convergence was often unconvincing.

By successive differentiation find the first four non-zero terms in the Maclaurin series for \(f(x) = (x + 1)\ln (1 + x) - x\).

Deduce that, for \(n \geqslant 2\), the coefficient of \({x^n}\) in this series is \({( - 1)^n}\frac{1}{{n(n - 1)}}\).

By applying the ratio test, find the radius of convergence for this Maclaurin series.

\(f(x) = (x + 1)\ln (1 + x) - x\)     \(f(0) = 0\)    A1

\(f'(x) = \ln (1 + x) + \frac{{x + 1}}{{1 + x}} - 1{\text{ }}\left( { = \ln (1 + x)} \right)\)     \(f'(0) = 0\)    M1A1A1

\(f''(x) = {(1 + x)^{ - 1}}\)     \(f''(0) = 1\)    A1A1

\(f'''(x) =  - {(1 + x)^{ - 2}}\)     \(f'''(0) =  - 1\)    A1

\({f^{(4)}}(x) = 2{(1 + x)^{ - 3}}\)     \({f^{(4)}}(0) = 2\)    A1

\({f^{(5)}}(x) =  - 3 \times 2{(1 + x)^{ - 4}}\)     \({f^{(5)}}(0) =  - 3 \times 2\)    A1

\(f(x) = \frac{{{x^2}}}{{2!}} - \frac{{1{x^3}}}{{3!}} + \frac{{2{x^4}}}{{4!}} - \frac{{6{x^5}}}{{5!}} \ldots \)    M1A1

\(f(x) = \frac{{{x^2}}}{{1 \times 2}} - \frac{{{x^3}}}{{2 \times 3}} + \frac{{{x^4}}}{{3 \times 4}} - \frac{{{x^5}}}{{4 \times 5}} \ldots \)

\(f(x) = \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} + \frac{{{x^4}}}{{12}} - \frac{{{x^5}}}{{20}} \ldots \)

Note: Allow follow through from the first error in a derivative (provided future derivatives also include the chain rule), no follow through after a second error in a derivative.

[11 marks]

\({f^{(n)}}(0) = {( - 1)^n}(n - 2)!\) So coefficient of \({x^n} = {( - 1)^n}\frac{{(n - 2)!}}{{n!}}\)     A1

coefficient of \({x^n}\) is \({( - 1)^n}\frac{1}{{n(n - 1)}}\)     AG

[1 mark]

applying the ratio test to the series of absolute terms

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{{\left| x \right|}^{n + 1}}}}{{(n + 1)n}}}}{{\frac{{{{\left| x \right|}^n}}}{{n(n - 1)}}}}\)    M1A1

\( = \mathop {\lim }\limits_{n \to \infty } \left| x \right|\frac{{(n - 1)}}{{(n + 1)}}\)    A1

\( = \left| x \right|\)    A1

so for convergence \(\left| x \right| < 1\), giving radius of convergence as 1     (M1)A1

[6 marks]

Find \(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} - {x^2}} \right)}}{{\cot \pi x}}} \right)\).

using l’Hôpital’s Rule     (M1)

\(\mathop {\lim }\limits_{x \to \frac{1}{2}} \left( {\frac{{\left( {\frac{1}{4} - {x^2}} \right)}}{{\cot \pi x}}} \right) = \mathop {\lim }\limits_{x \to \frac{1}{2}} \left[ {\frac{{ - 2}}{{ - \pi {\text{cose}}{{\text{c}}^2}\pi x}}} \right]\)     A1A1

\( = \frac{{ - 1}}{{ - \pi {\text{cose}}{{\text{c}}^2}\frac{\pi }{2}}} = \frac{1}{\pi }\)     (M1)A1

[5 marks]

This question was accessible to the vast majority of candidates, who recognised that L’Hôpital’s rule was required. However, some candidates omitted the factor \(\pi \) in the differentiation of \({\cot \pi x}\). Some candidates replaced \({\cot \pi x}\) by \(\cos \pi x{\text{/}}\sin \pi x\), which is a valid method but the extra algebra involved often led to an incorrect answer. Many fully correct solutions were seen.

Prove that f is continuous but not differentiable at the point (0, 0) .

Determine the value of \(\int_{ - a}^a {f(x){\text{d}}x} \) where \(a > 0\) .

we note that \(f(0) = 0,{\text{ }}f(x) = 3x\) for \(x > 0\) and \(f(x) = x{\text{ for }}x < 0\)

\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} x = 0\)     M1A1

\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 3x = 0\)     A1

since \(f(0) = 0\) , the function is continuous when x = 0     AG

\(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{h}{h} = 1\)     M1A1

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3h}}{h} = 3\)     A1

these limits are unequal     R1

so f is not differentiable when x = 0     AG

[7 marks]

\(\int_{ - a}^a {f(x){\text{d}}x = \int_{ - a}^0 {x{\text{d}}x + \int_0^a {3x{\text{d}}x} } } \)     M1

\( = \left[ {\frac{{{x^2}}}{2}} \right]_{ - a}^0 + \left[ {\frac{{3{x^2}}}{2}} \right]_0^a\)     A1

\( = {a^2}\)     A1

[3 marks]

By drawing a diagram and considering the area of a suitable region under the curve, show that for \(r > 0\),

\[\frac{1}{{r + 1}} < \ln \left( {\frac{{r + 1}}{r}} \right) < \frac{1}{r}.\]

Hence, given that \(n\) is a positive integer greater than one, show that

\(\sum\limits_{r = 1}^n {\frac{1}{r} > \ln (1 + n)} \);

Hence, given that \(n\) is a positive integer greater than one, show that

\(\sum\limits_{r = 1}^n {\frac{1}{r} < 1 + \ln n} \).

Hence, given that \(n\) is a positive integer greater than one, show that

\({U_n} > 0\);

Hence, given that \(n\) is a positive integer greater than one, show that

\({U_{n + 1}} < {U_n}\).

Explain why these two results prove that \(\{ {U_n}\} \) is a convergent sequence.

A1

Note:     Curve, both rectangles and correct \(x\)values required.

area of rectangles \(\frac{1}{r}\) and \(\frac{1}{{1 + r}}\)     A1

Note:     Correct values on the \(y\)-axis are sufficient evidence for this mark if not otherwise indicated.

in the above diagram, the area below the curve between \(x = r\) and \(x = r + 1\) is between the areas of the larger and smaller rectangle

or \(\frac{1}{{r + 1}} < \int\limits_r^{r + 1} {\frac{{{\text{d}}x}}{x} < \frac{1}{r}} \)     (R1)

integrating, \(\int_r^{r + 1} {\frac{{{\text{d}}x}}{x} = [\ln x]_r^{r + 1}\,\,\,\left( { = \ln (r + 1) - \ln (r)} \right)} \)     A1

\(\frac{1}{{r + 1}} < \ln \left( {\frac{{r + 1}}{r}} \right) < \frac{1}{r}\)     AG

[4 marks]

summing the right-hand part of the above inequality from \(r = 1\) to \(r = n\),

\(\sum\limits_{r = 1}^n {\frac{1}{r}} > \sum\limits_{r = 1}^n {\ln \left( {\frac{{r + 1}}{r}} \right)} \)     M1

\( = \ln \left( {\frac{2}{1}} \right) + \ln \left( {\frac{3}{2}} \right) + \ldots + \ln \left( {\frac{n}{{n - 1}}} \right) + \ln \left( {\frac{{n + 1}}{n}} \right)\)     (A1)

EITHER

\( = \ln \left( {\frac{2}{1} \times \frac{3}{2} \times \ldots \times \frac{n}{{n - 1}} \times \frac{{n + 1}}{n}} \right)\)     A1

OR

\(\ln 2 - \ln 1 + \ln 3 - \ln 2 + \ldots + \ln (n + 1) - \ln (n)\)     A1

\( = \ln (n + 1)\)     AG

[3 marks]

\(\sum\limits_{r = 1}^n {\frac{1}{r} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} < 1 + \ln \left( {\frac{2}{1}} \right) + \ln \left( {\frac{3}{2}} \right) + \ldots + \ln \left( {\frac{n}{{n - 1}}} \right)} \)     M1A1A1

\(\left( {1 + \sum\limits_{r = 1}^{n - 1} {\frac{1}{{r + 1}} < 1 + \sum\limits_{r = 1}^{n - 1} {\ln \left( {\frac{{r + 1}}{r}} \right)} } } \right)\)

Note:     M1 is for using the correct inequality from (a), A1 for both sides beginning with 1, A1 for completely correct expression.

Note:     The 1 might be added after the sums have been calculated.

\( = 1 + \ln n\)     AG

[3 marks]

from (b)(i) \({U_n} > \ln (1 + n) - \ln n > 0\)     A1

[1 mark]

\({U_{n + 1}} - {U_n} = \sum\limits_{r = 1}^{n + 1} {\frac{1}{r} - \ln (n + 1) - \sum\limits_{r = 1}^n {\frac{1}{r} + \ln n} } \)     M1

\( = \frac{1}{{n + 1}} - \ln \left( {\frac{{n + 1}}{n}} \right)\)     A1

\( < 0\) (using the result proved in (a))     A1

\({U_{n + 1}} < {U_n}\)     AG

[3 marks]

it follows from the two results that \(\{ {U_n}\} \) cannot be divergent either in the sense of tending to \( - \infty \) or oscillating therefore it must be convergent     R1

Note:     Accept the use of the result that a bounded (monotonically) decreasing sequence is convergent (allow “positive, decreasing sequence”).

[1 mark]

Find the value of \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right)\).

By using the series expansions for \({{\text{e}}^{{x^2}}}\) and cos x evaluate \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - {{\text{e}}^{{x^2}}}}}{{1 - \cos x}}} \right).\).

Using l’Hopital’s rule,

\(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{\frac{1}{x}}}{{2\pi \cos 2\pi x}}} \right)\)     M1A1

\( = \frac{1}{{2\pi }}\)     A1

[3 marks]

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - {{\text{e}}^{{x^2}}}}}{{1 - \cos x}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \left( {1 + {x^2} + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...} \right)}}{{1 - \left( {1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - ...} \right)}}} \right)\)     M1A1A1

Note: Award M1 for evidence of using the two series.

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - {x^2} - \frac{{{x^4}}}{{2!}} - \frac{{{x^6}}}{{3!}} - ...} \right)}}{{\left( {\frac{{{x^2}}}{{2!}} - \frac{{{x^4}}}{{4!}} + ...} \right)}}} \right)\)     A1

EITHER

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 1 - \frac{{{x^2}}}{{2!}} - \frac{{{x^4}}}{{3!}} - ...} \right)}}{{\left( {\frac{1}{{2!}} - \frac{{{x^2}}}{{4!}} + ...} \right)}}} \right)\)     M1A1

\( = \frac{{ - 1}}{{\frac{1}{2}}} = - 2\)     A1

OR

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 2x - \frac{{4{x^3}}}{{2!}} - \frac{{6{x^5}}}{{3!}} - ...} \right)}}{{\left( {\frac{{2x}}{{2!}} - \frac{{4{x^3}}}{{4!}} + ...} \right)}}} \right)\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 2 - \frac{{4{x^2}}}{{2!}} - \frac{{6{x^4}}}{{3!}} - ...} \right)}}{{\left( {1 - \frac{{4{x^2}}}{{4!}} + ...} \right)}}} \right)\)

\( = \frac{{ - 2}}{1} = - 2\)     A1

[7 marks]

Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed. 

Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.

(a)     (i)     Use Euler’s method to get an approximate value of y when x = 1.3 , taking steps of 0.1. Show intermediate steps to four decimal places in a table.

(ii)     How can a more accurate answer be obtained using Euler’s method?

(b)     Solve the differential equation giving your answer in the form y = f(x) .

(a)

(i)     \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} - y)\)

Note: Award A2 for complete table.

   Award A1 for a reasonable attempt.

\(f(1.3) = 2.14\,\,\,\,\,{\text{(accept 2.141)}}\)     A1

(ii)     Decrease the step size     A1

[5 marks]

(b)     \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} - y)\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = 2x(1 + {x^2})\)     M1

Integrating factor is \({{\text{e}}^{\int {2x{\text{d}}x} }} = {{\text{e}}^{{x^2}}}\)     M1A1

So, \({{\text{e}}^{{x^2}}}y = \int {(2x} {{\text{e}}^{{x^2}}} + 2x{{\text{e}}^{{x^2}}}{x^2}){\text{d}}x\)     A1

\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} - \int {2x{{\text{e}}^{{x^2}}}{\text{d}}x} \)     M1A1

\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} - {{\text{e}}^{{x^2}}} + k\)

\( = {x^2}{{\text{e}}^{{x^2}}} + k\)     A1

\(y = {x^2} + k{{\text{e}}^{ - {x^2}}}\)

\(x = 1,{\text{ }}y = 2 \to 2 = 1 + k{{\text{e}}^{ - 1}}\)     M1

\(k = {\text{e}}\)

\(y = {x^2} + {{\text{e}}^{1 - {x^2}}}\)     A1

[9 marks]

Total [14 marks]

Some incomplete tables spoiled what were often otherwise good solutions. Although the intermediate steps were asked to four decimal places the answer was not and the usual degree of IB accuracy was expected.

Some candidates surprisingly could not solve what was a fairly easy differential equation in part (b).

The variables x and y are related by \(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\tan x = \cos x\) .

(a)     Find the Maclaurin series for y up to and including the term in \({x^2}\) given that

\(y = - \frac{\pi }{2}\) when x = 0 .

(b)     Solve the differential equation given that y = 0 when \(x = \pi \) . Give the solution in the form \(y = f(x)\) .

(a)     from \(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\tan x + \cos x\) , \(f'(0) = 1\)     A1

now \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = y{\sec ^2}x + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x - \sin x\)     M1A1A1A1

Note: Award A1 for each term on RHS.

\( \Rightarrow f''(0) = - \frac{\pi }{2}\)     A1

\( \Rightarrow y = - \frac{\pi }{2} + x - \frac{{\pi {x^2}}}{4}\)     A1

[7 marks]

(b)     recognition of integrating factor     (M1)

integrating factor is \({{\text{e}}^{\int { - \tan x{\text{d}}x} }}\)

\( = {{\text{e}}^{\ln \cos x}}\)     (A1)

\( = \cos x\)     (A1)

\( \Rightarrow y\cos x = \int {{{\cos }^2}x{\text{d}}x} \)     M1

\( \Rightarrow y\cos x = \frac{1}{2}\int {(1 + \cos 2x){\text{d}}x} \)     A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} + k\)     A1

when \(x = \pi ,{\text{ }}y = 0 \Rightarrow k = - \frac{\pi }{2}\)     M1A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} - \frac{\pi }{2}\)     (A1)

\( \Rightarrow y = \sec x\left( {\frac{x}{2} + \frac{{\sin 2x}}{4} - \frac{\pi }{2}} \right)\)     A1

[10 marks]

Total [17 marks]

Part (a) of the question was set up in an unusual way, which caused a problem for a number of candidates as they tried to do part (b) first and then find the Maclaurin series by a standard method. Few were successful as they were usually weaker candidates and made errors in finding the solution \(y = f(x)\) . The majority of candidates knew how to start part (b) and recognised the need to use an integrating factor, but a number failed because they missed out the negative sign on the integrating factor, did not realise that \({{\text{e}}^{\ln \cos x}} = \cos x\) or were unable to integrate \({{{\cos }^2}x}\) . Having said this, a number of candidates succeeded in gaining full marks on this question.

Solve the differential equation given that \(y =  - 1\) when \(x = 1\). Give your answer in the form \(y = f\left( x \right)\).

Show that the \(x\)-coordinate(s) of the points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) satisfy the equation \({x^{p - 1}} = \frac{1}{p}\).

Deduce the set of values for \(p\) such that there are two points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\). Give a reason for your answer.

METHOD 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} = {x^{p - 1}} + \frac{1}{x}\)    (M1)

integrating factor \( = {{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }}\)     M1

\({\text{ = }}{{\text{e}}^{ - {\text{ln}}\,x}}\)     (A1)

= \(\frac{1}{x}\)     A1

\(\frac{1}{x}\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{y}{{{x^2}}} = {x^{p - 2}} + \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{x}} \right) = {x^{p - 2}} + \frac{1}{{{x^2}}}\)

\(\frac{y}{x} = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C\)    A1

Note: Condone the absence of C.

\(y = \frac{1}{{p - 1}}{x^p} + Cx - 1\)

substituting \(x = 1\), \(y =  - 1 \Rightarrow C =  - \frac{1}{{p - 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1\)      A1

[8 marks]

METHOD 2

put \(y = vx\) so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)    M1(A1)

substituting,       M1 

\(x\left( {v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}} \right) - vx = {x^p} + 1\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 1}} + \frac{1}{x}\)      M1

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 2}} + \frac{1}{{{x^2}}}\)

\(v = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C\)     A1

Note: Condone the absence of C.

\(y = \frac{1}{{p - 1}}{x^p} + Cx - 1\)

substituting \(x = 1\), \(y =  - 1 \Rightarrow C =  - \frac{1}{{p - 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1\)      A1

[8 marks]

METHOD 1

find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) and solve \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{p - 1}}\left( {p{x^{p - 1}} - 1} \right)\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow p{x^{p - 1}} - 1 = 0\)     A1

\(p{x^{p - 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p - 1}} = \frac{1}{p}\)     AG

METHOD 2

substitute \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) and their \(y\) into the differential equation and solve for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow  - \left( {\frac{{{x^p} - x}}{{p - 1}}} \right) + 1 = {x^p} + 1\)     M1

\({x^p} - x = {x^p} - p{x^p}\)     A1

\(p{x^{p - 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p - 1}} = \frac{1}{p}\)     AG

[2 marks]

there are two solutions for \(x\) when \(p\) is odd (and \(p > 1\)     A1

if \(p - 1\) is even there are two solutions (to \({x^{p - 1}} = \frac{1}{p}\))

and if \(p - 1\) is odd there is only one solution (to \({x^{p - 1}} = \frac{1}{p}\))   R1

Note: Only award the R1 if both cases are considered.

[4 marks]

Use the limit comparison test to prove that \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n(n + 1)}}} \) converges.

Using the Maclaurin series for \(\ln (1 + x)\) , show that the Maclaurin series for \(\left( {1 + x} \right)\ln \left( {1 + x} \right)\) is \(x + \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}} \).

apply the limit comparison test with \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \)     M1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{n(n + 1)}}}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{n(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1\)     M1A1

(since the limit is finite and \( \ne 0\) ) both series do the same     R1

we know that \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges and hence \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n(n + 1)}}} \) also converges     R1AG

[5 marks]

\((1 + x)\ln (1 + x) = (1 + x)\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4}...} \right)\)     A1

\( = \left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4}...} \right) + \left( {{x^2} - \frac{{x3}}{2} + \frac{{{x^4}}}{3} - \frac{{{x^5}}}{4}...} \right)\)

EITHER

\( = x + \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}{x^{n + 1}}}}{{n + 1}} + \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{n}} } \)     A1

\( = x + \sum\limits_{n = 1}^\infty  {{{( - 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{{ - 1}}{{n + 1}} + \frac{1}{n}} \right)} \)     M1

OR

\(x + \left( {1 - \frac{1}{2}} \right){x^2} - \left( {\frac{1}{2} - \frac{1}{3}} \right){x^3} + \left( {\frac{1}{3} - \frac{1}{4}} \right){x^4} - …\)     A1

\( = x + \sum\limits_{n = 1}^\infty  {{{( - 1)}^{n + 1}}{x^{n + 1}}\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} \)     M1

\( = x + \sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{{n(n + 1)}}} \)     AG

[3 marks]

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges.

Candidates and teachers need to be aware that the Limit comparison test is distinct from the comparison test. Quite a number of candidates lost most of the marks for this part by doing the wrong test.

Some candidates failed to state that because the result was finite and not equal to zero then the two series converge or diverge together. Others forgot to state, with a reason, that \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges.

In part (b) finding the partial fractions was well done. The second part involving the use of telescoping series was less well done, and students were clearly not as familiar with this technique as with some others.

Part (c) was the least well done of all the questions. It was expected that students would use explicitly the result from the first part of 4(b) or show it once again in order to give a complete answer to this question, rather than just assuming that a pattern spotted in the first few terms would continue.

Candidates need to be informed that unless specifically told otherwise they may use without proof any of the Maclaurin expansions given in the Information Booklet. There were many candidates who lost time and gained no marks by trying to derive the expansion for \(\ln (1 + x)\).

Consider the infinite geometric series

\[1 - {x^2} + {x^4} - {x^6} +  \ldots \;\;\;\left| x \right| < 1.\]

Show that the sum of the series is \(\frac{1}{{1 + {x^2}}}\).

Hence show that an expansion of \(\arctan x\) is \(\arctan x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} +  \ldots \)

\(f\) is a continuous function defined on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) with \(f'(x) > 0\) on \(]a,{\text{ }}b[\).

Use the mean value theorem to prove that for any \(x,{\text{ }}y \in [a,{\text{ }}b]\), if \(y > x\) then \(f(y) > f(x)\).

(i)     Given \(g(x) = x - \arctan x\), prove that \(g'(x) > 0\), for \(x > 0\).

(ii)     Use the result from part (c) to prove that \(\arctan x < x\), for \(x > 0\).

Use the result from part (c) to prove that \(\arctan x > x - \frac{{{x^3}}}{3}\), for \(x > 0\).

Hence show that \(\frac{{16}}{{3\sqrt 3 }} < \pi  < \frac{6}{{\sqrt 3 }}\).

\(r =  - {x^2},\;\;\;S = \frac{1}{{1 + {x^2}}}\)     A1AG

[1 mark]

\(\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} +  \ldots \)

EITHER

\(\int {\frac{1}{{1 + {x^2}}}{\text{d}}x}  = \int {1 - {x^2} + {x^4} - {x^6} +  \ldots } {\text{d}}x\)     M1

\(\arctan x = c + x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} +  \ldots \)     A1

Note:     Do not penalize the absence of \(c\) at this stage.

when \(x = 0\) we have \(\arctan 0 = c\) hence \(c = 0\)     M1A1

OR

\(\int_0^x {\frac{1}{{1 + {t^2}}}{\text{d}}t = } \int_0^x {1 - {t^2} + {t^4}}  - {t^6} +  \ldots {\text{d}}t\)     M1A1A1

Note: Allow \(x\) as the variable as well as the limit.

M1 for knowing to integrate, A1 for each of the limits.

\([\arctan t]_0^x = \left[ {t - \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} - \frac{{{t^7}}}{7} +  \ldots } \right]_0^x\)     A1

hence \(\arctan x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} +  \ldots \)     AG

[4 marks]

applying the \(MVT\) to the function \(f\) on the interval \([x,{\text{ }}y]\)     M1

\(\frac{{f(y) - f(x)}}{{y - x}} = f'(c)\;\;\;({\text{for some }}c \in ]x,{\text{ }}y[)\)     A1

\(\frac{{f(y) - f(x)}}{{y - x}} > 0\;\;\;({\text{as }}f'(c) > 0)\)     R1

\(f(y) - f(x) > 0{\text{ as }}y > x\)     R1

\( \Rightarrow f(y) > f(x)\)     AG

Note:     If they use \(x\) rather than \(c\) they should be awarded M1A0R0, but could get the next R1.

[4 marks]

(i)     \(g(x) = x - \arctan x \Rightarrow g'(x) = 1 - \frac{1}{{1 + {x^2}}}\)     A1

this is greater than zero because \(\frac{1}{{1 + {x^2}}} < 1\)     R1

so \(g'(x) > 0\)     AG

(ii)     (\(g\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(g'(x) > 0\) on \(]0,{\text{ }}b[\) for all \(b \in \mathbb{R}\))

(If \(x \in [0,{\text{ }}b]\) then) from part (c) \(g(x) > g(0)\)     M1

\(x - \arctan x > 0 \Rightarrow \arctan x < x\)     M1

(as \(b\) can take any positive value it is true for all \(x > 0\))     AG

[4 marks]

let \(h(x) = \arctan x - \left( {x - \frac{{{x^3}}}{3}} \right)\)     M1

(\(h\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(h'(x) > 0\) on \(]0,{\text{ }}b[\))

\(h'(x) = \frac{1}{{1 + {x^2}}} - (1 - {x^2})\)     A1

\( = \frac{{1 - (1 - {x^2})(1 + {x^2})}}{{1 + {x^2}}} = \frac{{{x^4}}}{{1 + {x^2}}}\)     M1A1

\(h'(x) > 0\) hence \(({\text{for }}x \in [0,{\text{ }}b]){\text{ }}h(x) > h(0)( = 0)\)     R1

\( \Rightarrow \arctan x > x - \frac{{{x^3}}}{3}\)     AG

Note: Allow correct working with \(h(x) = x - \frac{{{x^3}}}{3} - \arctan x\).

[5 marks]

use of \(x - \frac{{{x^3}}}{3} < \arctan x < x\)     M1

choice of \(x = \frac{1}{{\sqrt 3 }}\)     A1

\(\frac{1}{{\sqrt 3 }} - \frac{1}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\)     M1

\(\frac{8}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\)     A1

Note: Award final A1 for a correct inequality with a single fraction on each side that leads to the final answer.

\(\frac{{16}}{{3\sqrt 3 }} < \pi  < \frac{6}{{\sqrt 3 }}\)     AG

[4 marks]

Total [22 marks]

Most candidates picked up this mark for realizing the common ratio was \( - {x^2}\).

Quite a few candidates did not recognize the importance of ‘hence’ in this question, losing a lot of time by trying to work out the terms from first principles.

Of those who integrated the formula from part (a) only a handful remembered to include the ‘\( + c\)’ term, and to verify that this must be equal to zero.

Most candidates were able to achieve some marks on this question. The most commonly lost mark was through not stating that the inequality was unchanged when multiplying by \(y - x{\text{ as }}y > x\).

The first part of this question proved to be very straightforward for the majority of candidates.

In (ii) very few realized that they had to replace the lower variable in the formula from part (c) by zero.

Candidates found this part difficult, failing to spot which function was required.

Many candidates, even those who did not successfully complete (d) (ii) or (e), realized that these parts gave them the necessary inequality.

Show that \(1 + {x^2}\) is an integrating factor for this differential equation.

Hence solve this differential equation. Give the answer in the form \(y = f(x)\).

METHOD 1

attempting to find an integrating factor     (M1)

\(\int {\frac{{2x}}{{1 + {x^2}}}{\text{d}}x = \ln (1 + {x^2})} \)    (M1)A1

IF is \({{\text{e}}^{\ln (1 + {x^2})}}\)     (M1)A1

\( = 1 + {x^2}\)    AG

METHOD 2

multiply by the integrating factor

\((1 + {x^2})\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = {x^2}(1 + {x^2})\)    M1A1

left hand side is equal to the derivative of \((1 + {x^2})y\)

A3

[5 marks]

\((1 + {x^2})\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = (1 + {x^2}){x^2}\)    (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left[ {(1 + {x^2})y} \right] = {x^2} + {x^4}\)

\((1 + {x^2})y = \left( {\int {{x^2} + {x^4}{\text{d}}x = } } \right){\text{ }}\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5}( + c)\)    A1A1

\(y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c} \right)\)

\(x = 0,{\text{ }}y = 2 \Rightarrow c = 2\)    M1A1

\(y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + 2} \right)\)    A1

[6 marks]

Illustrate graphically the inequality, \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \).

Use the inequality in part (a) to find a lower and upper bound for \(\pi \).

Show that \(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}} = \frac{{1 + {{( - 1)}^{n - 1}}{x^{2n}}}}{{1 + {x^2}}}} \).

Hence show that \(\pi  = 4\left( {\sum\limits_{r = 0}^{n - 1} {\frac{{{{( - 1)}^r}}}{{2r + 1}} - {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\).

     A1A1A1

A1 for upper rectangles, A1 for lower rectangles, A1 for curve in between with \(0 \le x \le 1\)

hence \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \)     AG

[3 marks]

attempting to integrate from \(0\) to \(1\)     (M1)

\(\int_0^1 {f(x){\text{d}}x = [\arctan x]_0^1} \)

\( = \frac{\pi }{4}\)     A1

attempt to evaluate either summation     (M1)

\(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \frac{\pi }{4} < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)

hence \(\frac{4}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \pi  < \frac{4}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)

so \(2.93 < \pi  < 3.33\)     A1A1

Note:     Accept any answers that round to \(2.9\) and \(3.3\).

[5 marks]

EITHER

recognise \(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}}} \) as a geometric series with \(r =  - {x^2}\)     M1

sum of \(n\) terms is \(\frac{{1 - {{( - {x^2})}^n}}}{{1 -  - {x^2}}} = \frac{{1 + {{( - 1)}^{n - 1}}{x^{2n}}}}{{1 + {x^2}}}\)     M1AG

OR

\(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}(1 + {x^2}){x^{2r}} = (1 + {x^2}){x^0} - (1 + {x^2}){x^2} + (1 + {x^2}){x^4} +  \ldots } \)

\( + {( - 1)^{n - 1}}(1 + {x^2}){x^{2n - 2}}\)     M1

cancelling out middle terms     M1

\( = 1 + {( - 1)^{n - 1}}{x^{2n}}\)     AG

[2 marks]

\(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}} = \frac{1}{{1 + {x^2}}} + {{( - 1)}^{n - 1}}\frac{{{x^{2n}}}}{{1 + {x^2}}}} \)

integrating from \(0\) to \(1\)     M1

\(\left[ {\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} } \right]_0^1 = \int_0^1 {f(x){\text{d}}x + {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } \)     A1A1

\(\int_0^1 {f(x){\text{d}}x = \frac{\pi }{4}} \)     A1

so \(\pi  = 4\left( {\sum\limits_{r = 0}^{n - 1} {\frac{{{{( - 1)}^r}}}{{2r + 1}} - {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\)     AG

[4 marks]

Total [14 marks]

Use L’Hôpital’s Rule to find \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - 1 - x\cos x}}{{{{\sin }^2}x}}\) .

apply l’Hôpital’s Rule to a \(0/0\) type limit

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - 1 - x\cos x}}{{{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - \cos x + x\sin x}}{{2\sin x\cos x}}\)     M1A1

noting this is also a \(0/0\) type limit, apply l’Hôpital’s Rule again     (M1)

obtain \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} + \sin x + x\cos x + \sin x}}{{2\cos 2x}}\)     A1

substitution of x = 0     (M1)

= 0.5     A1

[6 marks]

The vast majority of candidates were familiar with L’Hôpitals rule and were also able to apply the technique twice as required by the problem. The errors that occurred were mostly due to difficulty in applying the differentiation rules correctly or errors in algebra. A small minority of candidates tried to use the quotient rule but it seemed that most candidates had a good understanding of L’Hôpital’s rule and its application to finding a limit.

Write down \(f'(x)\).

By differentiating \(f({x^2})\), obtain an expression for the derivative of \(\int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

Hence obtain an expression for the derivative of \(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

\(\ln (2 + \sin x)\)    A1

Note:     Do not accept \(\ln (2 + \sin t)\).

[1 mark]

attempt to use chain rule     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {f({x^2})} \right) = 2xf'({x^2})\)    (A1)

\( = 2x\ln \left( {2 + \sin ({x^2})} \right)\)    A1

[3 marks]

\(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t = \int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t - \int_0^x {\ln (2 + \sin t){\text{d}}t} } } \)    (M1)(A1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} } \right) = 2x\ln \left( {2 + \sin ({x^2})} \right) - \ln (2 + \sin x)\)    A1

[3 marks]

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

By finding a suitable number of derivatives of \(f\), determine the Maclaurin series for \(f(x)\) as far as the term in \({x^3}\).

Hence, or otherwise, determine the exact value of \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}}\).

(i)     Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in this approximation.

(ii)     Deduce from the Lagrange error term whether the approximation will be greater than or less than the actual value of \(f(0.5)\).

attempt to use product rule     (M1)

\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\)    A1

\(f''(x) = 2{{\text{e}}^x}\cos x\)    A1

\(f''(x) = 2{{\text{e}}^x}\cos x - 2{{\text{e}}^x}\sin x\)    A1

\(f(0) = 0,{\text{ }}f'(0) = 1\)

\(f''(0) = 2,{\text{ }}f'''(0) = 2\)    (M1)

\({{\text{e}}^x}\sin x = x + {x^2} + \frac{{{x^3}}}{3} +  \ldots \)    (M1)A1

[7 marks]

METHOD 1

\(\frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}} = \frac{{x + {x^2} + \frac{{{x^3}}}{3} +  \ldots  - x - {x^2}}}{{{x^3}}}\)    M1A1

\( \to \frac{1}{3}\) as \(x \to 0\)     A1

METHOD 2

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x - x - {x^2}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - 1 - 2x}}{{3{x^2}}}\)    A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x}\cos x - 2}}{{6x}}\)    A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x}\cos x - 2{{\text{e}}^x}\sin x}}{6} = \frac{1}{3}\)    A1

[3 marks]

(i)     attempt to find \({{\text{4}}^{{\text{th}}}}\) derivative from the \({{\text{3}}^{{\text{rd}}}}\) derivative obtained in (a)     M1

\(f''''(x) =  - 4{{\text{e}}^x}\sin x\)    A1

Lagrange error term \( = \frac{{{f^{(n + 1)}}(c){x^{n + 1}}}}{{(n + 1)!}}\) (where c lies between 0 and \(x\))

\( =  - \frac{{4{{\text{e}}^c}\sin c \times {{0.5}^4}}}{{4!}}\)    (M1)

the maximum absolute value of this expression occurs when \(c = 0.5\)     (A1)

Note:     This A1 is independent of previous M marks.

therefore

upper bound \( = \frac{{4{{\text{e}}^{0.5}}\sin 0.5 \times {{0.5}^4}}}{{4!}}\)     (M1)

\( = 0.00823\)    A1

(ii)     the approximation is greater than the actual value because the Lagrange error term is negative     R1

[7 marks]

This part of the question was well answered by most candidates. In a few cases candidates failed to follow instructions and attempted to use known series; in a few cases mistakes in the determination of the derivatives prevented other candidates from achieving full marks; part (b) was also well answered using both the Maclaurin expansion or L’Hôpital rule; again in most cases that candidates failed to achieve full marks were due to mistakes in the determination of derivatives.

Part (a) of the question was well answered by most candidates. In a few cases candidates failed to follow instructions and attempted to use known series; in a few cases mistakes in the determination of the derivatives prevented other candidates from achieving full marks; part (b) was also well answered using both the Maclaurin expansion or L’Hôpital rule; again in most cases that candidates failed to achieve full marks were due to mistakes in the determination of derivatives.

Part (c) was poorly answered with few candidates showing familiarity with this part of the option. Most candidates quoted the formula and managed to find the \({4^{{\text{th}}}}\) derivative of \(f\) but then could not use it to obtain the required answer; in other cases candidates did obtain an answer but showed little understanding of its meaning when answering (c)(ii).

Explain why the series is alternating.

(i)     Use the substitution \(T = t - \pi \) in the expression for \({u_{n + 1}}\) to show that \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\).

(ii)     Show that the series is convergent.

Show that \(S < 1.65\).

as \(t\) moves through the intervals \([0,{\text{ }}\pi ],{\text{ }}[\pi ,{\text{ }}2\pi ],{\text{ }}[2\pi ,{\text{ }}3\pi ],{\text{ }}[3\pi ,{\text{ }}4\pi ]\), etc, the sign of \(\sin t\), (and therefore the sign of the integral) alternates \( + ,{\text{ }} - ,{\text{ }} + ,{\text{ }} - \), etc, so that the series is alternating     R1

Note:     Award R1 only if it includes a clear reason that justifies that the sign of the integrand alternates between − and + and this pattern is valid for all the terms.

The change of signs can be justified by a labelled graph of \(y = \sin (x)\) or \(y = \frac{{\sin x}}{x}\) that shows the intervals \([0,{\text{ }}\pi ],{\text{ }}[\pi ,{\text{ }}2\pi ],{\text{ }}[2\pi ,{\text{ }}3\pi ],{\text{ }} \ldots \)

[1 mark]

(i)     \({u_{n + 1}} = \int_{(n + 1)\pi }^{(n + 2)\pi } {\frac{{\sin t}}{t}{\text{d}}t} \)

(M1)

put \(T = t--\pi \) and \({\text{d}}T = {\text{d}}t\)     (M1)

the limits change to \(n\pi ,{\text{ }}(n + 1)\pi \)

\(\left| {{u_{n + 1}}} \right| = \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin (T + \pi )} \right|}}{{T + \pi }}{\text{d}}T} \) (or equivalent)     A1

\(\left| {\sin (T + \pi )} \right| = \left| {\sin (T)} \right|\) or \(\sin (T + \pi ) =  - \sin (T)\)     (M1)

\( = \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin T} \right|}}{{T + \pi }}{\text{d}}T} \)

\( < \int_{n\pi }^{(n + 1)\pi } {\frac{{\left| {\sin T} \right|}}{T}{\text{d}}T = \left| {{u_n}} \right|} \)    A1AG

(ii)     \(\left| {{u_n}} \right| = \int_{n\pi }^{(n + 1)\pi } {\frac{{\sin t}}{t}{\text{d}}t} \)

\( < \int_{n\pi }^{(n + 1)\pi } {\frac{1}{t}{\text{d}}t} \)    M1

\( = [\ln t]_{n\pi }^{(n + 1)\pi }\)    A1

\( = \ln \left( {\frac{{n + 1}}{n}} \right)\)    A1

\( \to \ln 1 = 0\) as \(n \to \infty \)

from part (i) \(\left| {{u_n}} \right|\) is a decreasing sequence and since \(\mathop {\lim }\limits_{n \to \infty } \left| {{u_n}} \right| = 0\),     R1

the series is convergent     AG

[9 marks]

attempt to calculate the partial sums \(\sum\limits_{i = 0}^{n - 1} {{u_i} = \int_0^{n\pi } {\frac{{\sin t}}{t}{\text{d}}t} } \)     (M1)

the first partial sums are

two consecutive partial sums for \(n \geqslant 4\)     A1A1

(eg \({S_4} = 1.49\) and \({S_5} = 1.63\) or \({S_{100}} = 1.567 \ldots \) and \({S_{101}} = 1.573 \ldots \))

Note:     These answers must be given to a minimum of 3 significant figures.

the sum to infinity lies between any two consecutive partial sums,

eg between 1.49 and 1.63     R1

so that \(S < 1.65\)     AG

Note:     Award A1A1R1 to candidates who calculate at least two partial sums for only odd values of \(n\) and state that the upper bound is less than these values.

[4 marks]

Very few candidates presented a valid reason to justify the alternating nature of the series. In most cases candidates just reformulated the wording of the question by saying that it changed signs and completely ignored the interval over which the expression had to be integrated to obtain each term.

(i) Most candidates achieved 1 or 2 marks for attempting the given substitution; in most cases candidates failed to find the correct limits of integration for the new variable and then relate the expressions of the consecutive terms of the series. In part (ii) very few correct attempts were seen; in some cases candidates did recognize the conditions for the alternating series to be convergent but very few got close to establish that the limit of the general term was zero.

A few good attempts to use partial sums were seen although once again candidates showed difficulties in identifying what was needed to show the given answer. In most cases candidates just verified with GDC that in fact for high values of n the series was indeed less than the upper bound given but could not provide a valid argument that justified the given statement.

Show that the tangent to the curve \(y = f(x)\) at the point \((1,{\text{ }}0)\) is normal to the curve \(y = g(x)\) at the point \((1,{\text{ }}0)\).

Find \(g(x)\).

Use Euler’s method with steps of \(0.2\) to estimate \(f(2)\) to \(5\) decimal places.

Explain why \(y = f(x)\) cannot cross the isocline \(x - {y^2} = 0\), for \(x > 1\).

(i)     Sketch the isoclines \(x - {y^2} =  - 2,{\text{ }}0,{\text{ }}1\).

(ii)     On the same set of axes, sketch the graph of \(f\).

gradient of \(f\) at \((1,{\text{ }}0)\) is \(1 - {0^2} = 1\) and the gradient of \(g\) at \((1,{\text{ }}0)\) is \(0 - {1^2} =  - 1\)     A1

so gradient of normal is \(1\)     A1

= Gradient of the tangent of \(f\) at \((1,{\text{ }}0)\)     AG

[2 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y =  - {x^2}\)

integrating factor is \({{\text{e}}^{\int { - 1{\text{d}}x} }} = {{\text{e}}^{ - x}}\)     M1

\(y{{\text{e}}^{ - x}} = \int { - {x^2}{{\text{e}}^{ - x}}{\text{d}}x} \)     A1

\( = {x^2}{{\text{e}}^{ - x}} - \int {2x{{\text{e}}^{ - x}}{\text{d}}x} \)     M1

\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} - \int {2{{\text{e}}^{ - x}}{\text{d}}x} \)

\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} + 2{{\text{e}}^{ - x}} + c\)     A1

Note:     Condone missing \( + c\) at this stage.

\( \Rightarrow g(x) = {x^2} + 2x + 2 + c{{\text{e}}^x}\)

\(g(1) = 0 \Rightarrow c =  - \frac{5}{{\text{e}}}\)     M1

\( \Rightarrow g(x) = {x^2} + 2x + 2 - 5{{\text{e}}^{x - 1}}\)     A1

[6 marks]

use of \({y_{n + 1}} = {y_n} + hf'({x_n},{\text{ }}{y_n})\)     (M1)

\({x_0} = 1,{\text{ }}{y_0} = 0\)

\({x_1} = 1.2,{\text{ }}{y_1} = 0.2\)     A1

\({x_2} = 1.4,{\text{ }}{y_2} = 0.432\)     (M1)(A1)

\({x_3} = 1.6,{\text{ }}{y_3} = 0.67467 \ldots \)

\({x_4} = 1.8,{\text{ }}{y_4} = 0.90363 \ldots \)

\({x_5} = 2,{\text{ }}{y_5} = 1.1003255 \ldots \)

answer \( = 1.10033\)     A1     N3

Note:     Award A0 or N1 if \(1.10\) given as answer.

[5 marks]

at the point \((1,{\text{ }}0)\), the gradient of \(f\) is positive so the graph of \(f\) passes into the first quadrant for \(x > 1\)

in the first quadrant below the curve \(x - {y^2} = 0\) the gradient of \(f\) is positive     R1

the curve \(x - {y^2} = 0\) has positive gradient in the first quadrant     R1

if \(f\) were to reach \(x - {y^2} = 0\) it would have gradient of zero, and therefore would not cross     R1

[3 marks]

(i) and (ii)

     A4

Note:     Award A1 for 3 correct isoclines.

Award A1 for \(f\) not reaching \(x - {y^2} = 0\).

Award A1 for turning point of \(f\) on \(x - {y^2} = 0\).

Award A1 for negative gradient to the left of the turning point.

Note:     Award A1 for correct shape and position if curve drawn without any isoclines.

[4 marks]

Total [20 marks]

Use the integral test to determine whether the infinite series \(\sum\limits_{n = 2}^\infty {\frac{1}{{n\sqrt {\ln n} }}} \) is convergent or divergent.

consider \(I = \int\limits_2^N {\frac{{{\text{d}}x}}{{x\sqrt {\ln x} }}} \)     M1A1

Note:     Do not award A1 if \(n\) is used as the variable or if lower limit equal to 1, but some subsequent A marks can still be awarded. Allow \(\infty \) as upper limit.

let \(y = \ln x\)     (M1)

\({\text{d}}y = \frac{{{\text{d}}x}}{x},\)     (A1)

\(\left[ {2,{\text{ }}N} \right] \Rightarrow \left[ {\ln 2,{\text{ }}\ln N} \right]\)

\(I = \int\limits_{\ln \,2}^{\ln \,N} {\frac{{{\text{d}}y}}{{\sqrt y }}} \)    (A1)

Note:     Condone absence of limits, or wrong limits.

\( = \left[ {2\sqrt y } \right]_{\ln 2}^{\ln N}\)     A1

Note:     A1 is for the correct integral, irrespective of the limits used. Accept correct use of integration by parts.

\( = 2\sqrt {\ln N} - 2\sqrt {\ln 2} \)     (M1)

Note:     M1 is for substituting their limits into their integral and subtracting.

\( \to \infty {\text{ as }}N \to \infty \)     A1

Notes:     Allow “\( = \infty \)”, “limit does not exist”, “diverges” or equivalent.

Do not award if wrong limits substituted into the integral but allow \(N\) or \(\infty \) as an upper limit in place of \(\ln N\).

(by the integral test) the series is divergent (because the integral is divergent)     A1

Notes:     Do not award this mark if \(\infty \) used as upper limit throughout.

[9 marks]

Use the limit comparison test to show that the series \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}} \) is convergent.

Find the interval of convergence for \(S\).

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{n^2} + 2}}}}{{\frac{1}{{{n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 2}} = \left( {\mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{2}{{{n^2} + 2}}} \right)} \right)\)     M1

\( = 1\)     A1

since \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges (a \(p\)-series with \(p = 2\))     R1

by limit comparison test, \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}} \) also converges     AG

Notes:     The R1 is independent of the A1.

[3 marks]

applying the ratio test \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{(x - 3)}^{n + 1}}}}{{{{(n + 1)}^2} + 2}} \times \frac{{{n^2} + 2}}{{{{(x - 3)}^n}}}} \right|\)     M1A1

\( = \left| {x - 3} \right|{\text{ }}\left( {{\text{as }}\mathop {\lim }\limits_{n \to \infty } \frac{{({n^2} + 2)}}{{{{(n + 1)}^2} + 2}} = 1} \right)\)     A1

converges if \(\left| {x - 3} \right| < 1\) (converges for \(2 < x < 4\))     M1

considering endpoints \(x = 2\) and \(x = 4\)     M1

when \(x = 4\), series is \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}} \), convergent from (a)     A1

when \(x = 2\), series is \(\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{{n^2} + 2}}} \)     A1

EITHER

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2} + 2}}} \) is convergent therefore \(\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{{n^2} + 2}}} \) is (absolutely) convergent     R1

OR

\(\frac{1}{{{n^2} + 2}}\) is a decreasing sequence and \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + 2}} = 0\) so series converges by the alternating series test     R1

THEN

interval of convergence is \(2 \leqslant x \leqslant 4\)     A1

Note:     The final A1 is dependent on previous A1s – ie, considering correct series when \(x = 2\) and \(x = 4\) and on the final R1.

[9 marks]

For \(a = 0\) and \(b = 5\pi \), use the mean value theorem to find all possible values of \(c\) for the function \(g\).

Sketch the graph of \(y = g(x)\) on the interval \([0,{\text{ }}5\pi ]\) and hence illustrate the mean value theorem for the function \(g\).

\(\frac{{g(5\pi ) - g(0)}}{{5\pi  - 0}} =  - 0.6809 \ldots {\text{ }}\left( { = \cos \sqrt {5\pi } } \right)\) (gradient of chord)     (A1)

\(g’(x) = \cos \left( {\sqrt x } \right) - \frac{{\sqrt x \sin \left( {\sqrt x } \right)}}{2}\) (or equivalent)     (M1)(A1)

Note:     Award M1 to candidates who attempt to use the product and chain rules.

attempting to solve \(\cos \left( {\sqrt c } \right) - \frac{{\sqrt c \sin \left( {\sqrt c } \right)}}{2} =  - 0.6809 \ldots \) for \(c\)     (M1)

Notes:     Award M1 to candidates who attempt to solve their \(g’(c) = \) gradient of chord.

Do not award M1 to candidates who just attempt to rearrange their equation.

\(c = 2.26,{\text{ }}11.1\)     A1A1

Note:     Condone candidates working in terms of \(x\).

[6 marks]

correct graph: 2 turning points close to the endpoints, endpoints indicated and correct endpoint behaviour     A1

Notes:     Endpoint coordinates are not required. Candidates do not need to indicate axes scales.

correct chord     A1

tangents drawn at their values of \(c\) which are approximately parallel to the chord     A1A1

Notes:     Award A1A0A1A0 to candidates who draw a correct graph, do not draw a chord but draw 2 tangents at their values of \(c\). Condone the absence of their \(c - \) values stated on their sketch. However do not award marks for tangents if no \(c - \) values were found in (a).

[4 marks]

Find the radius of convergence.

Find the interval of convergence.

using the ratio test, \(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n{x^{n + 1}}}}{{{{(n + 1)}^2}{2^{n + 1}}}} \times \frac{{{n^2}{2^n}}}{{(n - 1){x^n}}}\)     M1

\( = \frac{{{n^3}}}{{{{(n + 1)}^2}(n - 1)}} \times \frac{x}{2}\)     A1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{x}{2}\)     A1

the radius of convergence R satisfies

\(\frac{R}{2} = 1\) so R = 2     A1

[4 marks]

considering x = 2 for which the series is

\(\sum\limits_{n = 1}^\infty  {\frac{{(n - 1)}}{{{n^2}}}} \)

using the limit comparison test with the harmonic series     M1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{n}} \), which diverges

consider

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n - 1}}{n} = 1\)     A1

the series is therefore divergent for x = 2     A1

when x = –2 , the series is

\(\sum\limits_{n = 1}^\infty  {\frac{{(n - 1)}}{{{n^2}}} \times {{( - 1)}^n}} \)

this is an alternating series in which the \({n^{{\text{th}}}}\) term tends to 0 as \(n \to \infty \)     A1

consider \(f(x) = \frac{{x - 1}}{{{x^2}}}\)     M1

\(f'(x) = \frac{{2 - x}}{{{x^3}}}\)     A1

this is negative for \(x > 2\) so the sequence \(\{ |{u_n}|\} \) is eventually decreasing     R1

the series therefore converges when x = –2 by the alternating series test     R1

the interval of convergence is therefore [–2, 2[     A1

[9 marks]

Use Euler’s method, with a step length of 0.1, to find an approximate value of y when x = 0.5.

(i)     Show that \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{2{y^3} - {y^2}}}{{{{(1 + x)}^2}}}\).

(ii)     Hence find the Maclaurin series for y, up to and including the term in \({x^2}\) .

(i)     Solve the differential equation.

(ii)     Find the value of a for which \(y \to \infty \) as \(x \to a\).

attempt the first step of

\({y_{n + 1}} = {y_n} + (0.1)f({x_n},\,{y_n})\) with \({y_0} = 1,{\text{ }}{x_0} = 0\)     (M1)

\({y_1} = 1.1\)     A1

\({y_2} = 1.1 + (0.1)\frac{{{{1.1}^2}}}{{1.1}} = 1.21\)     (M1)A1

\({y_3} = 1.332(0)\)     (A1)

\({y_4} = 1.4685\)     (A1)

\({y_5} = 1.62\)     A1

[7 marks]

(i)     recognition of both quotient rule and implicit differentiation     M1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{(1 + x)2y\frac{{{\text{d}}y}}{{{\text{d}}x}} - {y^2} \times 1}}{{{{(1 + x)}^2}}}\)     A1A1 

Note: Award A1 for first term in numerator, A1 for everything else correct.

\( = \frac{{(1 + x)2y\frac{{{y^2}}}{{1 + x}} - {y^2} \times 1}}{{{{(1 + x)}^2}}}\)     M1A1

\( = \frac{{2{y^3} - {y^2}}}{{{{(1 + x)}^2}}}\)     AG

(ii)     attempt to use \(y = y(0) + x\frac{{{\text{d}}y}}{{{\text{d}}x}}(0) + \frac{{{x^2}}}{{2!}}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}(0) + ...\)     (M1)

\( = 1 + x + \frac{{{x^2}}}{2}\)     A1A1 

Note: Award A1 for correct evaluation of \(y(0),{\text{ }}\frac{{dy}}{{dx}}(0),{\text{ }}\frac{{{d^2}y}}{{d{x^2}}}(0)\), A1 for correct series.

[8 marks]

(i)     separating the variables \(\int {\frac{1}{{{y^2}}}{\text{d}}y = \int {\frac{1}{{1 + x}}{\text{d}}x} } \)     M1

obtain \( - \frac{1}{y} = \ln (1 + x) + (c)\)     A1

impose initial condition \( - 1 = \ln 1 + c\)     M1

obtain \(y = \frac{1}{{1 - \ln (1 + x)}}\)     A1

(ii)     \(y \to \infty \) if \(\ln (1 + x) \to 1\) , so a = e – 1     (M1)A1 

Note: To award A1 must see either \(x \to e - 1\) or a = e – 1 . Do not accept x = e – 1.

[6 marks]

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

(a)     Show that the solution of the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = \cos x{\cos ^2}y{\text{,}}\]

given that \(y = \frac{\pi }{4}{\text{ when }}x = \pi {\text{, is }}y = \arctan (1 + \sin x){\text{.}}\)

(b)     Determine the value of the constant a for which the following limit exists

\[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\arctan (1 + \sin x) - a}}{{{{\left( {x - \frac{\pi }{2}} \right)}^2}}}\]

and evaluate that limit.

(a)     this separable equation has general solution

\(\int {{{\sec }^2}y{\text{d}}y = \int {\cos x{\text{d}}x} } \)     (M1)(A1)

\(\tan y = \sin x + c\)     A1

the condition gives

\(\tan \frac{\pi }{4} = \sin \pi  + c \Rightarrow c = 1\)     M1

the solution is \(\tan y = 1 + \sin x\)     A1

\(y = \arctan (1 + \sin x)\)     AG

[5 marks]

(b)     the limit cannot exist unless \(a = \arctan \left( {1 + \sin \frac{\pi }{2}} \right) = \arctan 2\)     R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is

\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\left( {\arctan (1 + \sin x)} \right)}^\prime }}}{{2\left( {x - \frac{\pi }{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y'}}{{2\left( {x - \frac{\pi }{2}} \right)}}\)     M1A1

where y is the solution of the differential equation

the numerator has zero limit (from the factor \(\cos x\) in the differential equation)     R1

so required limit is

\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y''}}{2}\)     M1A1

finally,

\(y'' = - \sin x{\cos ^2}y - 2\cos x\cos y\sin y \times y'(x)\)     M1A1

since \(\cos y\left( {\frac{\pi }{2}} \right) = \frac{1}{{\sqrt 5 }}\)     A1

\(y'' = - \frac{1}{5}{\text{ at }}x = \frac{\pi }{2}\)     A1

the required limit is \( - \frac{1}{{10}}\)     A1

[12 marks]

Total [17 marks]

Many candidates successfully obtained the displayed solution of the differential equation in part(a). Few complete solutions to part(b) were seen which used the result in part(a). The problem can, however, be solved by direct differentiation although this is algebraically more complicated. Some successful solutions using this method were seen.

Find the radius of convergence of the infinite series

\[\frac{1}{2}x + \frac{{1 \times 3}}{{2 \times 5}}{x^2} + \frac{{1 \times 3 \times 5}}{{2 \times 5 \times 8}}{x^3} + \frac{{1 \times 3 \times 5 \times 7}}{{2 \times 5 \times 8 \times 11}}{x^4} + \ldots {\text{ .}}\]

Determine whether the series \(\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{n} + n\pi } \right)} \) is convergent or divergent.

the nth term is

\({u_n} = \frac{{1 \times 3 \times 5 \ldots (2n - 1)}}{{2 \times 5 \times 8 \ldots (3n - 1)}}{x^n}\)     M1A1

(using the ratio test to test for absolute convergence)

\(\frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{(2n + 1)}}{{(3n + 2)}}\left| x \right|\)     M1A1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{2\left| x \right|}}{3}\)     A1

let R denote the radius of convergence

then \(\frac{{2R}}{3} = 1\) so \(r = \frac{3}{2}\)     M1A1

Note: Do not penalise the absence of absolute value signs.

[7 marks]

using the compound angle formula or a graphical method the series can be written in the form     (M1)

\(\sum\limits_{n = 1}^\infty {{u_n}} \) where \({u_n} = {( - 1)^n}\sin \left( {\frac{1}{n}} \right)\)     A2

since \(\frac{1}{n} < \frac{\pi }{2}\) i.e. an angle in the first quadrant,     R1

it is an alternating series     R1

\({u_n} \to 0{\text{ as }}n \to \infty \)     R1

and \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\)     R1

it follows that the series is convergent     R1

[8 marks]

Solutions to this question were generally disappointing. In (a), many candidates were unable even to find an expression for the nth term so that they could not apply the ratio test.

Solutions to this question were generally disappointing. In (b), few candidates were able to rewrite the nth term in the form \(\sum {{{( - 1)}^n}\sin \left( {\frac{1}{n}} \right)} \) so that most candidates failed to realise that the series was alternating.

Show that \(f'(x) = g(x)\) and \(g'(x) = f(x)\).

Find the first three non-zero terms in the Maclaurin expansion of \(f(x)\).

Hence find the value of \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}}\).

Find the value of the improper integral \(\int_0^\infty  {\frac{{g(x)}}{{{{\left[ {f(x)} \right]}^2}}}{\text{d}}x} \).

any correct step before the given answer     A1AG

eg, \(f'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } + {{\left( {{{\text{e}}^{ - x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2} = g(x)\)

any correct step before the given answer     A1AG

eg, \(g'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } - {{\left( {{{\text{e}}^{ - x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2} = f(x)\)

[2 marks]

METHOD 1

statement and attempted use of the general Maclaurin expansion formula     (M1)

\(f(0) = 1;{\text{ }}g(0) = 0\) (or equivalent in terms of derivative values)   A1A1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

METHOD 2

\({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

\({{\text{e}}^{ - x}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

adding and dividing by 2     M1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

Notes: Accept 1, \(\frac{{{x^2}}}{2}\) and \(\frac{{{x^4}}}{{24}}\) or 1, \(\frac{{{x^2}}}{{2!}}\) and \(\frac{{{x^4}}}{{4!}}\).

     Award A1 if two correct terms are seen.

[5 marks]

METHOD 1

attempted use of the Maclaurin expansion from (b)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - \left( {1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} +  \ldots } \right)}}{{{x^2}}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \left( { - \frac{1}{2} - \frac{{{x^2}}}{{24}} -  \ldots } \right)\)     A1

\( =  - \frac{1}{2}\)     A1

METHOD 2

attempted use of L’Hôpital and result from (a)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - g(x)}}{{2x}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - f(x)}}{2}\)     A1

\( =  - \frac{1}{2}\)     A1

[3 marks]

METHOD 1

use of the substitution \(u = f(x)\) and \(\left( {{\text{d}}u = g(x){\text{d}}x} \right)\)     (M1)(A1)

attempt to integrate \(\int_1^\infty  {\frac{{{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { - \frac{1}{u}} \right]_1^\infty \) or \(\left[ { - \frac{1}{{f(x)}}} \right]_0^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

METHOD 2

\(\int_0^\infty  {\frac{{\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}}}{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}} \right)}^2}}}{\text{d}}x = \int_0^\infty  {\frac{{2\left( {{{\text{e}}^x} - {{\text{e}}^{ - x}}} \right)}}{{{{\left( {{{\text{e}}^x} + {{\text{e}}^{ - x}}} \right)}^2}}}{\text{d}}x} } \)     (M1)

use of the substitution \(u = {{\text{e}}^x} + {{\text{e}}^{ - x}}\) and \(\left( {{\text{d}}u = {{\text{e}}^x} - {{\text{e}}^{ - x}}{\text{d}}x} \right)\)     (M1)

attempt to integrate \(\int_2^\infty  {\frac{{2{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { - \frac{2}{u}} \right]_2^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

[6 marks]

Solve this differential equation by separating the variables, giving your answer in the form y = f (x) .

Solve the same differential equation by using the standard homogeneous substitution y = vx .

Solve the same differential equation by the use of an integrating factor.

If y = 20 when x = 2 , find y when x = 5 .

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} \Rightarrow \int {\frac{1}{y}{\text{d}}y = \int {\frac{1}{x}{\text{d}}x} } \)     M1

\( \Rightarrow \ln y = \ln x + c\)     A1

\( \Rightarrow \ln y = \ln x + \ln k = \ln kx\)

\( \Rightarrow y = kx\)     A1

[3 marks]

\(y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (A1)

so \(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v\)     M1

\( \Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}v}}{{{\text{d}}x}} = 0\,\,\,\,\,({\text{as }}x \ne 0)\)     R1

\( \Rightarrow v = k\)

\( \Rightarrow \frac{y}{x} = k\,\,\,\,\,( \Rightarrow y = kx)\)     A1

[4 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{ - 1}}{x}} \right)y = 0\)     (M1)

\({\text{IF}} = {{\text{e}}^{\int {\frac{{ - 1}}{x}{\text{d}}x} }} = {{\text{e}}^{ - \ln x}} = \frac{1}{x}\)     M1A1

\({x^{ - 1}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - {x^{ - 2}}y = 0\)

\( \Rightarrow \frac{{{\text{d}}[{x^{ - 1}}y]}}{{{\text{d}}x}} = 0\)     (M1)

\( \Rightarrow {x^{ - 1}}y = k\,\,\,\,\,( \Rightarrow y = kx)\)     A1

[5 marks]

\(20 = 2k \Rightarrow k = 10{\text{  so  }}y(5) = 10 \times 5 = 50\)     A1

[1 mark]

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from \(\ln y = \ln x + c\) to \(y = x + c\)

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from \(\ln y = \ln x + c\) to \(y = x + c\)

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from \(\ln y = \ln x + c\) to \(y = x + c\)

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from \(\ln y = \ln x + c\) to \(y = x + c\)

The function \(f\) is defined by

\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\left| {x - 2} \right| + 1}&{x < 2} \\
{a{x^2} + bx}&{x \geqslant 2}
\end{array}} \right.\]

where \(a\) and \(b\) are real constants

Given that both \(f\) and its derivative are continuous at \(x = 2\), find the value of \(a\) and the value of \(b\).

considering continuity at \(x = 2\)

\(\mathop {{\text{lim}}}\limits_{x \to {2^ - }} f\left( x \right) = 1\) and \(\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f\left( x \right) = 4a + 2b\)    (M1)

\(4a + 2b = 1\)     A1

considering differentiability at \(x = 2\)

\(f'\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{ - 1}&{x < 2} \\
{2ax + b}&{x \geqslant 2}
\end{array}} \right.\)    (M1)

\(\mathop {{\text{lim}}}\limits_{x \to {2^ - }} f'\left( x \right) =  - 1\) and \(\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f'\left( x \right) = 4a + b\)     (M1)

Note: The above M1 is for attempting to find the left and right limit of their derived piecewise function at \(x = 2\).

\(4a + b =  - 1\)     A1

\(a =  - \frac{3}{4}\) and \(b = 2\)     A1

[6 marks]

The mean value theorem states that if \(f\) is a continuous function on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) then \(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\) for some \(c \in ]a,{\text{ }}b[\).

(i)     Find the two possible values of \(c\) for the function defined by \(f(x) = {x^3} + 3{x^2} - 2\) on the interval \([ - 3,{\text{ }}1]\).

(ii)     Illustrate this result graphically.

(i)     The function \(f\) is continuous on \([a,{\text{ }}b]\), differentiable on \(]a,{\text{ }}b[\) and \(f'(x) = 0\) for all \(x \in ]a,{\text{ }}b[\). Show that \(f(x)\) is constant on \([a,{\text{ }}b]\).

(ii)     Hence, prove that for \(x \in [0,{\text{ }}1],{\text{ }}2\arccos x + \arccos (1 - 2{x^2}) = \pi \).

(i)     \(f'(x) = 3{x^2} + 6x\)     A1

gradient of chord \( = 1\)     A1

\(3{c^2} + 6c = 1\)

\(c = \frac{{ - 3 \pm 2\sqrt 3 }}{3}{\text{ }}( =  - 2.15,{\text{ }}0.155)\)     A1A1

Note:     Accept any answers that round to the correct 2sf answers \(( - 2.2,{\text{ }}0.15)\).

(ii)     

award A1 for correct shape and clear indication of correct domain, A1 for chord (from \(x =  - 3\) to \(x = 1\)) and A1 for two tangents drawn at their values of \(c\)     A1A1A1

[7 marks]

(i)     METHOD 1

(if a theorem is true for the interval \([a,{\text{ }}b]\), it is also true for any interval \([{x_1},{\text{ }}{x_2}]\) which belongs to \([a,{\text{ }}b]\))

suppose \({x_1},{\text{ }}{x_2} \in [a,{\text{ }}b]\)     M1

by the \(MVT\), there exists \(c\) such that \(f'(c) = \frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = 0\)     M1A1

hence \(f({x_1}) = f({x_2})\)     R1

as \({x_1},{\text{ }}{x_2}\) are arbitrarily chosen, \(f(x)\) is constant on \([a,{\text{ }}b]\)

Note:     If the above is expressed in terms of \(a\) and \(b\) award M0M1A0R0.

METHOD 2

(if a theorem is true for the interval \([a,{\text{ }}b]\), it is also true for any interval \([{x_1},{\text{ }}{x_2}]\) which belongs to \([a,{\text{ }}b]\))

suppose \(x \in [a,{\text{ }}b]\)     M1

by the \(MVT\), there exists \(c\) such that \(f'(c) = \frac{{f(x) - f(a)}}{{x - a}} = 0\)     M1A1

hence \(f(x) = f(a) = \) constant     R1

(ii)     attempt to differentiate \((x) = 2\arccos x + \arccos (1 - 2{x^2})\)     M1

\( - 2\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{{ - 4x}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}\)     A1A1

\( =  - 2\frac{1}{{\sqrt {1 - {x^2}} }} + \frac{{4x}}{{\sqrt {4{x^2} - 4{x^4}} }} = 0\)     A1

Note:     Only award A1 for \(0\) if a correct attempt to simplify the denominator is also seen.

\(f(x) = f(0) = 2 \times \frac{\pi }{2} + 0 = \pi \)     A1AG

Note:     This A1 is not dependent on previous marks.

Note:     Allow any value of \(x \in [0,{\text{ }}1]\).

[9 marks]

Total [16 marks]

(i)     This was well done by most candidates.

(ii)     This was generally poorly done, with many candidates failing to draw the curve correctly as they did not appreciate the importance of the given domain. Another common error was to draw the graph of the derivative rather than the function.

(i)     This was very poorly done. A lot of the arguments seemed to be stating what was being required to be proved, eg ‘because the derivative is equal to 0 the line is flat’. Most candidates did not realise the importance of testing a point inside the interval, so the most common solutions seen involved the Mean Value Theorem applied to the end points. In addition there was some confusion between the Mean Value Theorem and Rolle’s Theorem.

(ii)     It was pleasing that so many candidates spotted the link with the previous part of the question. The most common error after this point was to differentiate incorrectly. Candidates should be aware this is a ‘prove’ question, and so it was not sufficient simply to state, for example, \(f(0) = \pi \).

Use l’Hôpital’s rule to determine the value of

\[\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{x\ln (1 + x)}}.\]

attempt to use l’Hôpital’s rule,     M1

\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin x\cos x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}\)\(\,\,\,\)or\(\,\,\,\)\(\frac{{\sin 2x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}\)     A1A1

Note:     Award A1 for numerator A1 for denominator.

this gives 0/0 so use the rule again     (M1)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\cos }^2}x - 2{{\sin }^2}x}}{{\frac{1}{{1 + x}} + \frac{{1 + x - x}}{{{{(1 + x)}^2}}}}}\)\(\,\,\,\)or\(\,\,\,\)\(\frac{{2\cos 2x}}{{\frac{{2 + x}}{{{{(1 + x)}^2}}}}}\)     A1A1

Note:     Award A1 for numerator A1 for denominator.

\( = 1\)     A1

Note:     This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.

[7 marks]

Solve the differential equation

\[(x - 1)\frac{{{\text{d}}y}}{{{\text{d}}x}} + xy = (x - 1){{\text{e}}^{ - x}}\]

given that y = 1 when x = 0. Give your answer in the form \(y = f(x)\).

writing the differential equation in standard form gives

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{x - 1}}y = {{\text{e}}^{ - x}}\)     M1

\(\int {\frac{x}{{x - 1}}{\text{d}}x = \int {\left( {1 + \frac{1}{{x - 1}}} \right){\text{d}}x = x + \ln (x - 1)} } \)     M1A1

hence integrating factor is \({{\text{e}}^{x + \ln (x - 1)}} = (x - 1){{\text{e}}^x}\)     M1A1

hence, \((x - 1){{\text{e}}^x}\frac{{{\text{d}}y}}{{{\text{d}}x}} + x{{\text{e}}^x}y = x - 1\)     (A1)

\( \Rightarrow \frac{{{\text{d}}\left[ {(x - 1){{\text{e}}^x}y} \right]}}{{{\text{d}}x}} = x - 1\)     (A1)

\( \Rightarrow (x - 1){{\text{e}}^x}y = \int {(x - 1){\text{d}}x} \)     A1

\( \Rightarrow (x - 1){{\text{e}}^x}y = \frac{{{x^2}}}{2} - x + c\)     A1

substituting (0, 1), c = –1     (M1)A1

\( \Rightarrow (x - 1){{\text{e}}^x}y = \frac{{{x^2} - 2x - 2}}{2}\)     (A1)

hence, \(y = \frac{{{x^2} - 2x - 2}}{{2(x - 1){{\text{e}}^x}}}\) (or equivalent)     A1

[13 marks]

Apart from some candidates who thought the differential equation was homogenous, the others were usually able to make a good start, and found it quite straightforward. Some made errors after identifying the correct integrating factor, and so lost accuracy marks.

Prove by induction that \(n! > {3^n}\), for \(n \ge 7,{\text{ }}n \in \mathbb{Z}\).

Hence use the comparison test to prove that the series \(\sum\limits_{r = 1}^\infty  {\frac{{{2^r}}}{{r!}}} \) converges.

if \(n = 7\) then \(7! > {3^7}\)     A1

so true for \(n = 7\)

assume true for \(n = k\)     M1

so \(k! > {3^k}\)

consider \(n = k + 1\)

\((k + 1)! = (k + 1)k!\)     M1

\( > (k + 1){3^k}\)

\( > 3.3k\;\;\;({\text{as }}k > 6)\)     A1

\( = {3^{k + 1}}\)

hence if true for \(n = k\) then also true for \(n = k + 1\). As true for \(n = 7\), so true for all \(n \ge 7\).     R1

Note:     Do not award the R1 if the two M marks have not been awarded.

[5 marks]

consider the series \(\sum\limits_{r = 7}^\infty  {{a_r}} \), where \({a_r} = \frac{{{2^r}}}{{r!}}\)     R1

Note:     Award the R1 for starting at \(r = 7\)

compare to the series \(\sum\limits_{r=7}^\infty  {{b_r}} \) where \({b_r} = \frac{{{2^r}}}{{{3^r}}}\)     M1

\(\sum\limits_{r = 7}^\infty  {{b_r}} \) is an infinite Geometric Series with \(r = \frac{2}{3}\) and hence converges     A1

Note:     Award the A1 even if series starts at \(r = 1\).

as \(r! > {3^r}\) so \((0 < ){a_r} < {b_r}\) for all \(r \ge 7\)     M1R1

as \(\sum\limits_{r = 7}^\infty  {{b_r}} \) converges and \({a_r} < {b_r}\) so \(\sum\limits_{r = 7}^\infty  {{a_r}} \) must converge

Note:     Award the A1 even if series starts at \(r = 1\).

as \(\sum\limits_{r = 1}^6 {{a_r}} \) is finite, so \(\sum\limits_{r = 1}^\infty  {{a_r}} \) must converge     R1

Note:     If the limit comparison test is used award marks to a maximum of R1M1A1M0A0R1.

[6 marks]

Total [11 marks]

Show that \(\sqrt {{x^2} + 1} \) is an integrating factor for this differential equation.

Solve the differential equation giving your answer in the form \(y = f(x)\).

METHOD 1

integrating factor \( = {{\text{e}}^{\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} }}\)     (M1)

\(\int {\frac{x}{{{x^2} + 1}}{\text{d}}x = \frac{1}{2}\ln ({x^2} + 1)} \)     (M1)

Note:     Award M1 for use of \(u = {x^2} + 1\) for example or \(\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \ln f(x)} \).

integrating factor \( = {{\text{e}}^{\frac{1}{2}\ln ({x^2} + 1)}}\)     A1

\( = {{\text{e}}^{\ln \left( {\sqrt {{x^2} + 1} } \right)}}\)     A1

Note:     Award A1 for \({{\text{e}}^{\ln \sqrt u }}\) where \(u = {x^2} + 1\).

\( = \sqrt {{x^2} + 1} \)     AG

METHOD 2

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {y\sqrt {{x^2} + 1} } \right) = \frac{{{\text{d}}y}}{{{\text{d}}x}}\sqrt {{x^2} + 1}  + \frac{x}{{\sqrt {{x^2} + 1} }}y\)     M1A1

\(\sqrt {{x^2} + 1} \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y} \right)\)     M1A1

Note:     Award M1 for attempting to express in the form \(\sqrt {{x^2} + 1}  \times {\text{(LHS of de)}}\).

so \(\sqrt {{x^2} + 1} \) is an integrating factor for this differential equation     AG

[4 marks]

\(\sqrt {{x^2} + 1} \frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{\sqrt {{x^2} + 1} }}y = x\sqrt {{x^2} + 1} \) (or equivalent)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {y\sqrt {{x^2} + 1} } \right) = x\sqrt {{x^2} + 1} \)

\(y\sqrt {{x^2} + 1}  = \int {x\sqrt {{x^2} + 1} {\text{d}}x{\text{ }}\left( {y = \frac{1}{{\sqrt {{x^2} + 1} }}\int {x\sqrt {{x^2} + 1} {\text{d}}x} } \right)} \)     A1

\( = \frac{1}{3}{({x^2} + 1)^{\frac{3}{2}}} + C\)     (M1)A1

Note:     Award M1 for using an appropriate substitution.

Note:     Condone the absence of \(C\).

substituting \(x = 0,{\text{ }}y = 1 \Rightarrow C = \frac{2}{3}\)     M1

Note:     Award M1 for attempting to find their value of \(C\).

\(y = \frac{1}{3}({x^2} + 1) + \frac{2}{{3\sqrt {{x^2} + 1} }}{\text{ }}\left( {y = \frac{{{{({x^2} + 1)}^{\frac{3}{2}}} + 2}}{{3\sqrt {{x^2} + 1} }}} \right)\)     A1

[6 marks]

Show that \(n! \geqslant {2^{n - 1}}\), for \(n \geqslant 1\).

Hence use the comparison test to determine whether the series \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n!}}} \) converges or diverges.

for \(n \geqslant 1,{\text{ }}n! = n(n - 1)(n - 2) \ldots 3 \times 2 \times 1 \geqslant 2 \times 2 \times 2 \ldots 2 \times 2 \times 1 = {2^{n - 1}}\)     M1A1

\( \Rightarrow n! \geqslant {2^{n - 1}}{\text{ for }}n \geqslant 1\)     AG

[2 marks]

\(n! \geqslant {2^{n - 1}} \Rightarrow \frac{1}{{n!}} \leqslant \frac{1}{{{2^{n - 1}}}}{\text{ for }}n \geqslant 1\)     A1

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{2^{n - 1}}}}} \) is a positive converging geometric series     R1

hence \(\sum\limits_{n = 1}^\infty  {\frac{1}{{n!}}} \) converges by the comparison test     R1

[3 marks]

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).

Show that the series \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}\ln n}}} \) converges.

(i)     Show that \(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln (n + 1)\).

(ii)     Using this result, show that an application of the ratio test fails to determine whether or not \(\sum\limits_{n = 2}^\infty  {\frac{1}{{n\ln n}}} \) converges.

(i)     State why the integral test can be used to determine the convergence or divergence of \(\sum\limits_{n = 2}^\infty  {\frac{1}{{n\ln n}}} \).

(ii)     Hence determine the convergence or divergence of \(\sum\limits_{n = 2}^\infty  {\frac{1}{{n\ln n}}} \).

METHOD 1

\((0 < )\frac{1}{{{n^2}\ln (n)}} < \frac{1}{{{n^2}}},{\text{ }}({\text{for }}n \ge 3)\)     A1

\(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}}}} \) converges     A1

by the comparison test ( \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}}}} \) converges implies) \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}\ln (n)}}} \) converges     R1

Note:     Mention of using the comparison test may have come earlier.

Only award R1 if previous 2 A1s have been awarded.

METHOD 2

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\frac{1}{{{n^2}\ln n}}}}{{\frac{1}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\ln n}} = 0\)     A1

\(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}}}} \) converges     A1

by the limit comparison test (if the limit is \(0\) and the series represented by the denominator converges, then so does the series represented by the numerator, hence) \(\sum\limits_{n = 2}^\infty  {\frac{1}{{{n^2}\ln (n)}}} \) converges     R1

Note:     Mention of using the limit comparison test may come earlier.

Do not award the R1 if incorrect justifications are given, for example the series “converge or diverge together”.

Only award R1 if previous 2 A1s have been awarded.

[3 marks]

(i)     EITHER

\(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln \left( {n\left( {1 + \frac{1}{n}} \right)} \right)\)     A1

OR

\(\ln (n) + \ln \left( {1 + \frac{1}{n}} \right) = \ln (n) + \ln \left( {\frac{{n + 1}}{n}} \right)\)

\( = \ln (n) + \ln (n + 1) - \ln (n)\)     A1

THEN

\( = \ln (n + 1)\)     AG

(ii)     attempt to use the ratio test \(\frac{n}{{(n + 1)}}\frac{{\ln (n)}}{{\ln (n + 1)}}\)     M1

\(\frac{n}{{n + 1}} \to 1{\text{ as }}n \to \infty \)     (A1)

\(\frac{{\ln (n)}}{{\ln (n + 1)}} = \frac{{\ln (n)}}{{\ln (n) + \ln \left( {1 + \frac{1}{n}} \right)}}\)     M1

\( \to 1\;\;\;({\text{as }}n \to \infty )\)     (A1)

\(\frac{n}{{(n + 1)}}\frac{{\ln (n)}}{{\ln (n + 1)}} \to 1\;\;\;({\text{as }}n \to \infty )\;\;\;\)hence ratio test is inconclusive     R1

Note:     A link with the limit equalling \(1\) and the result being inconclusive needs to be given for R1.

[6 marks]

(i)     consider \(f(x) = \frac{1}{{x\ln x}}\;\;\;({\text{for }}x > 1)\)     A1

\(f(x)\) is continuous and positive     A1

and is (monotonically) decreasing     A1

Note:     If a candidate uses \(n\) rather than \(x\), award as follows

\(\frac{1}{{n\ln n}}\) is positive and decreasing     A1A1

\(\frac{1}{{n\ln n}}\) is continuous for \(n \in \mathbb{R},{\text{ }}n > 1\) A1 (only award this mark if the domain has been explicitly changed).

(ii)     consider \(\int_2^R {\frac{1}{{x\ln x}}{\text{d}}x} \)     M1

\( = \left[ {\ln (\ln x)} \right]_2^R\)     (M1)A1

\( \to \infty {\text{ as }}R \to \infty \)     R1

hence series diverges     A1

Note:     Condone the use of \(\infty \) in place of \(R\).

Note:     If the lower limit is not equal to \(2\), but the expression is integrated correctly award M0M1A1R0A0.

[8 marks]

Total [17 marks]

In this part the required test was not given in the question. This led to some students attempting inappropriate methods. When using the comparison or limit comparision test many candidates wrote the incorrect statement \(\frac{1}{{{n^2}}}\) converges, (p-series) rather than the correct one with \(\sum {} \). This perhaps indicates a lack of understanding of the concepts involved.

There were many good, well argued answers to this part. Most candidates recognised the importance of the result in part (i) to find the limit in part (ii). Generally a standard result such as \(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{n + 1}}} \right) = 1\) can simply be quoted, but other limits such as \(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\ln n}}{{\ln (n + 1)}}} \right) = 1\) need to be carefully justified.

(i)     Candidates need to be aware of the necessary conditions for all the series tests.

(ii)     The integration was well done by the candidates. Most also made the correct link between the integral being undefined and the series diverging. In this question it was not necessary to initially take a finite upper limit and the use of \(\infty \) was acceptable. This was due to the command term being ‘determine’. In q4b a finite upper limit was required, as the command term was ‘show’. To ensure full marks are always awarded candidates should err on the side of caution and always use limit notation when working out indefinite integrals.

Find the values of \({a_0},{\text{ }}{a_1},{\text{ }}{a_2}\) and \({a_3}\).

Hence, or otherwise, find the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}}\).

let \(f(x) = \sqrt x ,{\text{ }}f(1) = 1\)     (A1)

\(f'(x) = \frac{1}{2}{x^{ - \frac{1}{2}}},{\text{ }}f'(1) = \frac{1}{2}\)     (A1)

\(f''(x) = - \frac{1}{4}{x^{ - \frac{3}{2}}},{\text{ }}f''(1) = - \frac{1}{4}\)     (A1)

\(f'''(x) = \frac{3}{8}{x^{ - \frac{5}{2}}},{\text{ }}f'''(1) = \frac{3}{8}\)     (A1)

\({a_1} = \frac{1}{2} \cdot \frac{1}{{1!}},{\text{ }}{a_2} = - \frac{1}{4} \cdot \frac{1}{{2!}},{\text{ }}{a_3} = \frac{3}{8} \cdot \frac{1}{{3!}}\)     (M1)

\({a_0} = 1,{\text{ }}{a_1} = \frac{1}{2},{\text{ }}{a_2} = - \frac{1}{8},{\text{ }}{a_3} = \frac{1}{{16}}\)     A1

Note: Accept \(y = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}{(x - 1)^2} + \frac{1}{{16}}{(x - 1)^3} +  \ldots \)

[6 marks]

METHOD 1

\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}(x - 1) - \frac{1}{8}{{(x - 1)}^2} +  \ldots }}{{x - 1}}\)     M1

\( = \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{2} - \frac{1}{8}(x - 1) +  \ldots } \right)\)     A1

\( = \frac{1}{2}\)     A1

METHOD 2

using l’Hôpital’s rule,     M1

\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}{x^{ - \frac{1}{2}}}}}{1}\)     A1

\( = \frac{1}{2}\)     A1

METHOD 3

\(\frac{{\sqrt x  - 1}}{{x + 1}} = \frac{1}{{\sqrt x  + 1}}\)     M1A1

\(\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{2}\)     A1

[3 marks]

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.

Use an integrating factor to show that the general solution for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} - \frac{x}{t} =  - \frac{2}{t},{\text{ }}t > 0\) is \(x = 2 + ct\), where \(c\) is a constant.

The weight in kilograms of a dog, \(t\) weeks after being bought from a pet shop, can be modelled by the following function:

\[w(t) = \left\{ {\begin{array}{*{20}{c}} {2 + ct}&{0 \le t \le 5} \\ {16 - \frac{{35}}{t}}&{t > 5} \end{array}.} \right.\]

Given that \(w(t)\) is continuous, find the value of \(c\).

Write down

(i)     the weight of the dog when bought from the pet shop;

(ii)     an upper bound for the weight of the dog.

Prove from first principles that \(w(t)\) is differentiable at \(t = 5\).

integrating factor \({e^{\int { - \frac{1}{2}{\text{d}}t} }} = {e^{ - \ln t}}\left( { = \frac{1}{t}} \right)\)     M1A1

\(\frac{x}{t} = \int { - \frac{2}{{{t^2}}}{\text{d}}t = \frac{2}{t} + c} \)     A1A1

Note:     Award A1 for \(\frac{x}{t}\) and A1 for \({\frac{2}{t} + c}\).

\(x = 2 + ct\)     AG

[4 marks]

given continuity at \(x = 5\)

\(5c + 2 = 16 - \frac{{35}}{5} \Rightarrow c = \frac{7}{5}\)     M1A1

[2 marks]

(i)     \(2\)     A1

(ii)     any value \( \ge 16\)     A1

Note:     Accept values less than \(16\) if fully justified by reference to the maximum age for a dog.

[2 marks]

\(\mathop {\lim }\limits_{h \to 0 - } \left( {\frac{{\frac{7}{5}(5 + h) + 2 - \frac{7}{5}(5) - 2}}{h}} \right) = \frac{7}{5}\)     M1A1

\(\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{16 - \frac{{35}}{{5 + h}} - 16 + \frac{{35}}{5}}}{h}} \right)\;\;\;\left( { = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ - 35}}{{5 + h}} + 7}}{h}} \right)} \right)\)     M1

\( = \)\(\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ - 35 + 35 + 7h}}{{(5 + h)}}}}{h}} \right) = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{7}{{5 + h}}} \right) = \frac{7}{5}\)     M1A1

both limits equal so differentiable at \(t = 5\)     R1AG

Note:     The limits \(t \to 5\) could also be used.

For each value of \(\frac{7}{5}\) obtained by standard differentiation award     A1.

To gain the other 4 marks a rigorous explanation must be given on how you can get from the left and right hand derivatives to the derivative.

Note:     If the candidate works with \(t\) and then substitutes \(t = 5\) at the end award as follows

First M1 for using formula with \(t\) in the linear case, A1 for \(\frac{7}{5}\)

Award next 2 method marks even if \(t = 5\) not substituted, A1 for \(\frac{7}{5}\)

[6 marks]

Total [14 marks]

This was generally well done. Some candidates did not realize \({e^{ - \ln t}}\) could be simplified to \(\frac{1}{t}\).

This part was well done by the majority of candidates.

This part was well done by the majority of candidates.

Some candidates ignored the instruction to prove from first principles and instead used standard differentiation. Some candidates also only found a derivative from one side.

Consider the power series \(\sum\limits_{k = 1}^\infty {k{{\left( {\frac{x}{2}} \right)}^k}} \).

(i)     Find the radius of convergence.

(ii)     Find the interval of convergence.

Consider the infinite series \(\sum\limits_{k = 1}^\infty  {{{( - 1)}^{k + 1}} \times \frac{k}{{2{k^2} + 1}}} \).

(i)     Show that the series is convergent.

(ii)     Show that the sum to infinity of the series is less than 0.25.

(i)     consider \(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{\left| {\frac{{(n - 1){x^{n + 1}}}}{{{2^{n + 1}}}}} \right|}}{{\left| {\frac{{n{x^n}}}{{{2^n}}}} \right|}}\)     M1

\( = \frac{{(n + 1)\left| x \right|}}{{2n}}\)     A1

\( \to \frac{{\left| x \right|}}{2}{\text{ as }}n \to \infty \)     A1

the radius of convergence satisfies

\(\frac{R}{2} = 1\), i.e. R = 2     A1

(ii)     the series converges for \( - 2 < x < 2\), we need to consider \(x = \pm 2\)     (R1)

when x = 2, the series is \(1 + 2 + 3 + \ldots \)     A1

this is divergent for any one of several reasons e.g. finding an expression for or a comparison test with the harmonic series or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) etc.     R1

when x = – 2, the series is \( - 1 + 2 - 3 + 4 \ldots \)     A1

this is divergent for any one of several reasons

e.g. partial sums are

\( - 1,{\text{ }}1,{\text{ }} - 2,{\text{ }}2,{\text{ }} - 3,{\text{ }}3 \ldots \) or noting that \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\) etc.     R1

the interval of convergence is \( - 2 < x < 2\)     A1

[10 marks]

(i)     this alternating series is convergent because the moduli of successive terms are monotonic decreasing     R1

and the \({n^{{\text{th}}}}\) term tends to zero as \(n \to \infty \)     R1

(ii)     consider the partial sums

0.333, 0.111, 0.269, 0.148, 0.246     M1A1

since the sum to infinity lies between any pair of successive partial sums, it follows that the sum to infinity lies between 0.148 and 0.246 so that it is less than 0.25     R1

Note: Accept a solution which looks only at 0.333, 0.269, 0.246 and states that these are successive upper bounds.

[5 marks]

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

Most candidates found the radius of convergence correctly but examining the situation when \(x = \pm 2\) often ended in loss of marks through inadequate explanations. In (b)(i) many candidates were able to justify the convergence of the given series. In (b)(ii), however, many candidates seemed unaware of the fact the sum to infinity lies between any pair of successive partial sums.

(a)     Show that the solution of the homogeneous differential equation

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + 1,{\text{ }}x > 0,\)

given that \(y = 0{\text{ when }}x = {\text{e, is }}y = x(\ln x - 1)\).

(b)     (i)     Determine the first three derivatives of the function \(f(x) = x(\ln x - 1)\).

  (ii)     Hence find the first three non-zero terms of the Taylor series for f(x) about x = 1.

(a)     EITHER

use the substitution y = vx

\(\frac{{{\text{d}}y}}{{{\text{d}}x}}x + v = v + 1\)     M1A1

\(\int {{\text{d}}v = \int {\frac{{{\text{d}}x}}{x}} } \)

by integration

\(v = \frac{y}{x} = \ln x + c\)     A1

OR

the equation can be rearranged as first order linear

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{x}y = 1\)     M1

the integrating factor I is

\({{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }} = {{\text{e}}^{ - \ln x}} = \frac{1}{x}\)     A1

multiplying by I gives

\(\frac{{{\text{d}}}}{{{\text{d}}x}}\left( {\frac{1}{x}y} \right) = \frac{1}{x}\)

\(\frac{1}{x}y = \ln x + c\)     A1

THEN

the condition gives c = –1

so the solution is \(y = x(\ln x - 1)\)     AG

[5 marks]

(b)     (i)     \(f'(x) = \ln x - 1 + 1 = \ln x\)     A1

\(f''(x) = \frac{1}{x}\)     A1

\(f'''(x) = - \frac{1}{{{x^2}}}\)     A1

(ii)     the Taylor series about x = 1 starts

\(f(x) \approx f(1) + f'(1)(x - 1) + f''(1)\frac{{{{(x - 1)}^2}}}{{2!}} + f'''(1)\frac{{{{(x - 1)}^3}}}{{3!}}\)     (M1)

\( =  - 1 + \frac{{{{(x - 1)}^2}}}{{2!}} - \frac{{{{(x - 1)}^3}}}{{3!}}\)     A1A1A1

[7 marks]

Total: [12 marks]

Part(a) was well done by many candidates. In part(b)(i), however, it was disappointing to see so many candidates unable to differentiate \(x(\ln x - 1)\) correctly. Again, too many candidates were able to quote the general form of a Taylor series expansion, but not how to apply it to the given function.

Show that \(f'\left( 0 \right) = 0\).

By differentiating the above equation twice, show that

\[\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0\]

where \({f^{\left( 3 \right)}}\left( x \right)\) and \({f^{\left( 4 \right)}}\left( x \right)\) denote the 3rd and 4th derivative of \(f\left( x \right)\) respectively.

Hence show that the Maclaurin series for \(f\left( x \right)\) up to and including the term in \({x^4}\) is \({x^2} + \frac{1}{3}{x^4}\).

Use this series approximation for \(f\left( x \right)\) with \(x = \frac{1}{2}\) to find an approximate value for \({\pi ^2}\).

\(f'\left( x \right) = \frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}\)     M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for \(f'\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)\).

\(f'\left( 0 \right) = 0\)     AG

[2 marks]

differentiating gives \(\left( {1 - {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) - 2xf''\left( x \right) - f'\left( x \right) - xf''\left( x \right)\left( { = 0} \right)\)      M1A1

differentiating again gives \(\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 2x{f^{\left( 3 \right)}}\left( x \right) - 3f''\left( x \right) - 3x{f^{\left( 3 \right)}}\left( x \right) - f''\left( x \right)\left( { = 0} \right)\)     M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

\(\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0\)      AG

[4 marks]

attempting to find one of \(f''\left( 0 \right)\), \({f^{\left( 3 \right)}}\left( 0 \right)\) or \({f^{\left( 4 \right)}}\left( 0 \right)\) by substituting \(x = 0\) into relevant differential equation(s)       (M1)

Note: Condone \(f''\left( 0 \right)\) found by calculating \(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}} \right)\) at \(x = 0\).

\(\left( {f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0} \right)\)

\(f''\left( 0 \right) = 2\) and \({f^{\left( 4 \right)}}\left( 0 \right) - 4f''\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8\)      A1

\({f^{\left( 3 \right)}}\left( 0 \right) = 0\) and so \(\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}\)     A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to \({f^{\left( 3 \right)}}\left( 0 \right) = 0\) stated or \({f^{\left( 3 \right)}}\left( 0 \right) = 0\) seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if \({f^{\left( 4 \right)}}\left( 0 \right) = 8\) is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for \(f\left( x \right)\) up to and including the term in \({x^4}\) is \({x^2} + \frac{1}{3}{x^4}\)     AG

[3 marks]

substituting \(x = \frac{1}{2}\) into \({x^2} + \frac{1}{3}{x^4}\)      M1

the series approximation gives a value of \(\frac{{13}}{{48}}\)

so \({\pi ^2} \simeq \frac{{13}}{{48}} \times 36\)

\( \simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)\)     A1

Note: Accept 9.76.

[2 marks]

A function \(f\) is defined in the interval \(\left] { - k,{\text{ }}k} \right[\), where \(k > 0\). The gradient function \({f'}\) exists at each point of the domain of \(f\).

The following diagram shows the graph of \(y = f(x)\), its asymptotes and its vertical symmetry axis.

(a)     Sketch the graph of \(y = f'(x)\).

Let \(p(x) = a + bx + c{x^2} + d{x^3} +  \ldots \) be the Maclaurin expansion of \(f(x)\).

(b)     (i)     Justify that \(a > 0\).

(ii)     Write down a condition for the largest set of possible values for each of the parameters \(b\), \(c\) and \(d\).

(c)     State, with a reason, an upper bound for the radius of convergence.

(a)

A1 for shape, A1 for passing through origin     A1A1

Note: Asymptotes not required.

[2 marks]

(b)     \(p(x) = \underbrace {f(0)}_a + \underbrace {f'(0)}_bx + \underbrace {\frac{{f''(0)}}{{2!}}}_c{x^2} + \underbrace {\frac{{{f^{(3)}}(0)}}{{3!}}}_d{x^3} +  \ldots \)

(i)     because the y-intercept of \(f\) is positive     R1

(ii)     \(b = 0\)     A1

\(c \geqslant 0\)     A1A1

Note: A1 for \( > \) and A1 for \( = \).

\(d = 0\)     A1

[5 marks]

(c)     as the graph has vertical asymptotes \(x =  \pm k,{\text{ }}k > 0\),     R1

the radius of convergence has an upper bound of \(k\)     A1

Note: Accept \(r < k\).

[2 marks]

Overall candidates made good attempts to parts (a) and most candidates realized that the graph contained the origin; however many candidates had difficulty rendering the correct shape of the graph of \(f'\). Part b(i) was also well answered although some candidates where not very clear and digressed a lot. Part (ii) was less successful with most candidates scoring just part of the marks. A small number of candidates answered part (c) correctly with a valid reason.

(a)     Find an integrating factor for this differential equation.

(b)     Solve the differential equation given that \(y = 1\) when \(x = 1\) , giving your answer in the forms \(y = f(x)\) .

(a)     Rewrite the equation in the form

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{2}{x}y = \frac{{{x^2}}}{{{x^2} + 1}}\)     M1A1

Integrating factor \( = {{\text{e}}^{\int { - \frac{2}{x}{\text{d}}x} }}\)     M1

\( = {{\text{e}}^{ - 2\ln x}}\)     A1

\( = \frac{1}{{{x^2}}}\)     A1

Note: Accept \(\frac{1}{{{x^3}}}\) as applied to the original equation.

[5 marks]

(b)     Multiplying the equation,

\(\frac{1}{{{x^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{2}{{{x^3}}}y = \frac{1}{{{x^2} + 1}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{{{x^2}}}} \right) = \frac{1}{{{x^2} + 1}}\)     (M1)(A1)

\(\frac{y}{{{x^2}}} = \int {\frac{{{\text{d}}x}}{{{x^2} + 1}}} \)     M1

\( = \arctan x + C\)     A1

Substitute \(x = 1,{\text{ }}y = 1\) .     M1

\(1 = \frac{\pi }{4} + C \Rightarrow C = 1 - \frac{\pi }{4}\)     A1

\(y = {x^2}\left( {\arctan x + 1 - \frac{\pi }{4}} \right)\)     A1

[8 marks]

Total [13 marks]

The response to this question was often disappointing. Many candidates were unable to find the integrating factor successfully. 

The real and imaginary parts of a complex number \(x + {\text{i}}y\) are related by the differential equation \((x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x - y) = 0\).

By solving the differential equation, given that \(y = \sqrt 3 \) when x =1, show that the relationship between the modulus r and the argument \(\theta \) of the complex number is \(r = 2{{\text{e}}^{\frac{\pi }{3} - \theta }}\).

\((x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x - y) = 0\)

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y - x}}{{x + y}}\)

let \(y = vx\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     A1

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx - x}}{{x + vx}}\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{v - 1}}{{v + 1}} - v = \frac{{v - 1 - {v^2} - v}}{{v + 1}} = \frac{{ - 1 - {v^2}}}{{1 + v}}\)     A1

\(\int {\frac{{v + 1}}{{1 + {v^2}}}{\text{d}}v = - \int {\frac{1}{x}{\text{d}}x} } \)     M1

\(\int {\frac{v}{{1 + {v^2}}}{\text{d}}v + \int {\frac{1}{{1 + {v^2}}}{\text{d}}v = - \int {\frac{1}{x}{\text{d}}x} } } \)     M1

\( \Rightarrow \frac{1}{2}\ln \left| {1 + {v^2}} \right| + \arctan v = - \ln \left| x \right| + k\)     A1A1

Notes: Award A1 for \(\frac{1}{2}\ln \left| {1 + {v^2}} \right|\), A1 for the other two terms.

Do not penalize missing k or missing modulus signs at this stage.

\( \Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = - \ln \left| x \right| + k\)     M1

\( \Rightarrow \frac{1}{2}\ln 4 + \arctan \sqrt 3 = - \ln 1 + k\)     (M1)

\( \Rightarrow k = \ln 2 + \frac{\pi }{3}\)     A1

\( \Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = - \ln \left| x \right| + \ln 2 + \frac{\pi }{3}\)

attempt to combine logarithms     M1

\( \Rightarrow \frac{1}{2}\ln \left| {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right| + \frac{1}{2}\ln \left| {{x^2}} \right| = \ln 2 + \frac{\pi }{3} - \arctan \frac{y}{x}\)

\( \Rightarrow \frac{1}{2}\ln \left| {{y^2} + {x^2}} \right| = \ln 2 + \frac{\pi }{3} - \arctan \frac{y}{x}\)     (A1)

\( \Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2 + \frac{\pi }{3}\arctan \frac{y}{x}}}\)     (A1)

\( \Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2}} \times {{\text{e}}^{\frac{\pi }{3} - \arctan \frac{y}{x}}}\)     A1

\( \Rightarrow r = 2{{\text{e}}^{\frac{\pi }{3} - \theta }}\)     AG

[15 marks]

Most candidates realised that this was a homogeneous differential equation and that the substitution \(y = vx\) was the way forward. Many of these candidates reached as far as separating the variables correctly but integrating \(\frac{{v + 1}}{{{v^2} + 1}}\) proved to be too difficult for many candidates – most failed to realise that the expression had to be split into two separate integrals. Some candidates successfully evaluated the arbitrary constant but the combination of logs and the subsequent algebra necessary to obtain the final result proved to be beyond the majority of candidates.

Consider the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {\frac{y}{x}} \right),{\text{ }}x > 0.\]

Use the substitution \(y = vx\) to show that the general solution of this differential equation is

\[\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant.}}\]

Hence, or otherwise, solve the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{x^2} + 3xy + {y^2}}}{{{x^2}}},{\text{ }}x > 0,\]

given that \(y = 1\) when \(x = 1\). Give your answer in the form \(y = g(x)\).

\(y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     M1

the differential equation becomes

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = f(v)\)     A1

\(\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \int {\frac{{{\text{d}}v}}{x}} } \)     A1

integrating, Constant \(\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant}}\)     AG

[3 marks]

EITHER

\(f(v) = 1 + 3v + {v^2}\)     (A1)

\(\left( {\int {\frac{{{\text{d}}v}}{{f(v) - v}} = } } \right)\,\,\,\int {\frac{{{\text{d}}v}}{{1 + 3v + {v^2} - v}} = \ln x + C} \)     M1A1

\(\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}} = (\ln x + C)} \)     A1

Note:     A1 is for correct factorization.

\( - \frac{1}{{1 + v}}\,\,\,( = \ln x + C)\)     A1

OR

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 + 3v + {v^2}\)     A1

\(\int {\frac{{{\text{d}}v}}{{1 + 2v + {v^2}}} = \int {\frac{1}{x}{\text{d}}x} } \)     M1

\(\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}}\,\,\,\left( { = \int {\frac{1}{x}{\text{d}}x} } \right)} \)     (A1)

Note:     A1 is for correct factorization.

\( - \frac{1}{{1 + v}} = \ln x( + C)\)     A1A1

THEN

substitute \(y = 1\) or \(v = 1\) when \(x = 1\)     (M1)

therefore \(C = - \frac{1}{2}\)     A1

Note:     This A1 can be awarded anywhere in their solution.

substituting for \(v\),

\( - \frac{1}{{\left( {1 + \frac{y}{x}} \right)}} = \ln x - \frac{1}{2}\)     M1

Note:     Award for correct substitution of \(\frac{y}{x}\) into their expression.

\(1 + \frac{y}{x} = \frac{1}{{\frac{1}{2} - \ln x}}\)     (A1)

Note:     Award for any rearrangement of a correct expression that has \(y\) in the numerator.

\(y = x\left( {\frac{1}{{\left( {\frac{1}{2} - \ln x} \right)}} - 1} \right)\,\,\,{\text{(or equivalent)}}\)     A1

\(\left( { = x\left( {\frac{{1 + 2\ln x}}{{1 - 2\ln x}}} \right)} \right)\)

[10 marks]

Solve the differential equation

\({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} + 3xy + 2{x^2}\)

given that y = −1 when x =1. Give your answer in the form \(y = f(x)\) .

put y = vx so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     M1

substituting,     M1

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} + 3v{x^2} + 2{x^2}}}{{{x^2}}}{\text{ }}( = {v^2} + 3v + 2)\)     (A1)