Consider the function \(f:x \to \sqrt {\frac{\pi }{4} - \arccos x} \).

(a)     Find the largest possible domain of f.

(b)     Determine an expression for the inverse function, \({f^{ - 1}}\), and write down its domain.

(a)     \(\frac{\pi }{4} - \arccos x \geqslant 0\)

\(\arccos x \leqslant \frac{\pi }{4}\)     (M1)

\(x \geqslant \frac{{\sqrt 2 }}{2}\,\,\,\,\,\left( {{\text{accept }}x \geqslant \frac{1}{{\sqrt 2 }}} \right)\)     (A1)

since \( - 1 \leqslant x \leqslant 1\)     (M1)

\( \Rightarrow \frac{{\sqrt 2 }}{2} \leqslant x \leqslant 1\,\,\,\,\,\left( {{\text{accept }}\frac{1}{{\sqrt 2 }} \leqslant x \leqslant 1} \right)\)     A1

Note: Penalize the use of \( < \) instead of \( \leqslant \) only once.

(b)     \(y = \sqrt {\frac{\pi }{4} - \arccos x}  \Rightarrow x = \cos \left( {\frac{\pi }{4} - {y^2}} \right)\)     M1A1

\({f^{ - 1}}:x \to \cos \left( {\frac{\pi }{4} - {x^2}} \right)\)     A1

\(0 \leqslant x \leqslant \sqrt {\frac{\pi }{4}} \)     A1

[8 marks]

Very few correct solutions were seen to (a). Many candidates realised that \(\arccos x \leqslant \frac{\pi }{4}\) but then concluded incorrectly, not realising that cos is a decreasing function, that \(x \leqslant \cos \left( {\frac{\pi }{4}} \right)\). In (b) candidates often gave an incorrect domain.

Find an expression for \({f^{ - 1}}(x)\).

Given that \(f(x)\) can be written in the form \(f(x) = A + \frac{B}{{2x - 1}}\), find the values of the constants \(A\) and \(B\).

Hence, write down \(\int {\frac{{3x - 2}}{{2x - 1}}} {\text{d}}x\).

\(f:x \to y = \frac{{3x - 2}}{{2x - 1}}\;\;\;{f^{ - 1}}:y \to x\)

\(y = \frac{{3x - 2}}{{2x - 1}} \Rightarrow 3x - 2 = 2xy - y\)     M1

\( \Rightarrow 3x - 2xy =  - y + 2\)     M1

\(x(3 - 2y) = 2 - y\)

\(x = \frac{{2 - y}}{{3 - 2y}}\)     A1

\(\left( {{f^{ - 1}}(y) = \frac{{2 - y}}{{3 - 2y}}} \right)\)

\({f^{ - 1}}(x) = \frac{{2 - x}}{{3 - 2x}}\;\;\;\left( {x \ne \frac{3}{2}} \right)\)     A1

Note:     \(x\) and \(y\) might be interchanged earlier.

Note:     First M1 is for interchange of variables second M1 for manipulation

Note:     Final answer must be a function of \(x\)

[4 marks]

\(\frac{{3x - 2}}{{2x - 1}} = A + \frac{B}{{2x - 1}} \Rightarrow 3x - 2 = A(2x - 1) + B\)

equating coefficients \(3 = 2A\) and \( - 2 =  - A + B\)     (M1)

\(A = \frac{3}{2}\) and \(B =  - \frac{1}{2}\)     A1

Note:     Could also be done by division or substitution of values.

[2 marks]

\(\int {f(x){\text{d}}x = \frac{3}{2}x - \frac{1}{4}\ln \left| {2x - 1} \right| + c} \)     A1

Note:     accept equivalent e.g. \(\ln \left| {4x - 2} \right|\)

[1 mark]

Total [7 marks]

Well done. Only a few candidates confused inverse with derivative or reciprocal.

Not enough had the method of polynomial division.

Reasonable if they had an answer to (b) (follow through was given) usual mistakes with not allowing for the derivative of the bracket.

Write \(\ln ({x^2} - 1) - 2\ln (x + 1) + \ln ({x^2} + x)\) as a single logarithm, in its simplest form.

\(\ln ({x^2} - 1) - \ln {(x + 1)^2} + \ln x(x + 1)\)     (A1)

\( = \ln \frac{{x({x^2} - 1)(x + 1)}}{{{{(x + 1)}^2}}}\)     (M1)A1

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

\( = \ln x(x - 1)\,\,\,\,\,\left( { = \ln ({x^2} - x)} \right)\)     A1

[5 marks]

There were fewer correct solutions to this question than might be expected. A significant number of students managed to combine the terms to form one logarithm, but rather than factorising, then expanded the brackets, which left them unable to gain an answer in its simplest form.

Find the set of values of y for which this equation has real roots.

Hence determine the range of the function \(f:x \to \frac{{x + 1}}{{{x^2} + x + 1}}\).

Explain why f has no inverse.

for the equation to have real roots

\({(y - 1)^2} - 4y(y - 1) \geqslant 0\)     M1

\( \Rightarrow 3{y^2} - 2y - 1 \leqslant 0\)

(by sign diagram, or algebraic method)     M1

\( - \frac{1}{3} \leqslant y \leqslant 1\)     A1A1

Note: Award first A1 for \( - \frac{1}{3}\) and 1, and second A1 for inequalities. These are independent marks.

[4 marks]

\(f:x \to \frac{{x + 1}}{{{x^2} + x + 1}} \Rightarrow x + 1 = y{x^2} + yx + y\)     (M1)

\( \Rightarrow 0 = y{x^2} + (y - 1)x + (y - 1)\)     A1

hence, from (a) range is \( - \frac{1}{3} \leqslant y \leqslant 1\)     A1

[3 marks]

a value for y would lead to 2 values for x from (a)     R1

Note: Do not award R1 if (b) has not been tackled.

[1 mark]

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).

Using the information shown in the diagram, find the values of a , b and c .

If g(x) = 3f(x − 2) ,

(i)     state the coordinates of the points where the graph of g intercepts the x-axis.

(ii)     Find the y-intercept of the graph of g .

METHOD 1

f(x) = (x + 1)(x − 1)(x − 2)     M1

\( = {x^3} - 2{x^2} - x + 2\)     A1A1A1

a = −2 , b = −1 and c = 2

METHOD 2

from the graph or using f(0) = 2

c = 2     A1

setting up linear equations using f(1) = 0 and f(–1) = 0 (or f(2) = 0)     M1

obtain a = −2 , b = −1     A1A1

[4 marks]

(i)     (1, 0) , (3, 0) and (4, 0)     A1

(ii)     g(0) occurs at 3f(−2)     (M1)

= −36     A1

[3 marks]

This question was well answered in general. Part b(ii) was often the most problematic, usually because of candidates going to the trouble of finding an explicit and sometimes incorrect expression for f(x − 2).

This question was well answered in general. Part b(ii) was often the most problematic, usually because of candidates going to the trouble of finding an explicit and sometimes incorrect expression for f(x − 2).

Sketch the graph of \(y = f(x)\), indicating clearly any asymptotes and points of intersection with the \(x\) and \(y\) axes.

Find an expression for \({f^{ - 1}}(x)\).

Find all values of \(x\) for which \(f(x) = {f^{ - 1}}(x)\).

Solve the inequality \(\left| {f(x)} \right| < \frac{3}{2}\).

Solve the inequality \(f\left( {\left| x \right|} \right) < \frac{3}{2}\).

Note: In the diagram, points marked \(A\) and \(B\) refer to part (d) and do not need to be seen in part (a).

shape of curve     A1

Note:     This mark can only be awarded if there appear to be both horizontal and vertical asymptotes.

intersection at \((0,{\text{ }}0)\)     A1

horizontal asymptote at \(y = 3\)     A1

vertical asymptote at \(x = 2\)     A1

[4 marks]

\(y = \frac{{3x}}{{x - 2}}\)

\(xy - 2y = 3x\)     M1A1

\(xy - 3x = 2y\)

\(x = \frac{{2y}}{{y - 3}}\)

\(\left( {{f^{ - 1}}(x)} \right) = \frac{{2x}}{{x - 3}}\)     M1A1

Note:     Final M1 is for interchanging of \(x\) and \(y\), which may be seen at any stage.

[4 marks]

METHOD 1

attempt to solve \(\frac{{2x}}{{x - 3}} = \frac{{3x}}{{x - 2}}\)     (M1)

\(2x(x - 2) = 3x(x - 3)\)

\(x\left[ {2(x - 2) - 3(x - 3)} \right] = 0\)

\(x(5 - x) = 0\)

\(x = 0\;\;\;\)or\(\;\;\;x = 5\)     A1A1

METHOD 2

\(x = \frac{{3x}}{{x - 2}}\;\;\;\)or\(\;\;\;x = \frac{{2x}}{{x - 3}}\)     (M1)

\(x = 0\;\;\;\)or\(\;\;\;x = 5\)     A1A1

[3 marks]

METHOD 1

at \({\text{A}}:\frac{{3x}}{{x - 2}} = \frac{3}{2}\) AND at \({\text{B}}:\frac{{3x}}{{x - 2}} =  - \frac{3}{2}\)     M1

\(6x = 3x - 6\)

\(x =  - 2\)     A1

\(6x = 6 - 3x\)

\(x = \frac{2}{3}\)     A1

solution is \( - 2 < x < \frac{2}{3}\)     A1

METHOD 2

\({\left( {\frac{{3x}}{{x - 2}}} \right)^2} < {\left( {\frac{3}{2}} \right)^2}\)     M1

\(9{x^2} < \frac{9}{4}{(x - 2)^2}\)

\(3{x^2} + 4x - 4 < 0\)

\((3x - 2)(x + 2) < 0\)

\(x =  - 2\)     (A1)

\(x = \frac{2}{3}\)     (A1)

solution is \( - 2 < x < \frac{2}{3}\)     A1

[4 marks]

\( - 2 < x < 2\)     A1A1

Note:     A1 for correct end points, A1 for correct inequalities.

Note:     If working is shown, then A marks may only be awarded following correct working.

[2 marks]

Total [17 marks]

A function is defined by \(h(x) = 2{{\text{e}}^x} - \frac{1}{{{{\text{e}}^x}}},{\text{ }}x \in \mathbb{R}\) . Find an expression for \({h^{ - 1}}(x)\) .

\(x = 2{{\text{e}}^y} - \frac{1}{{{{\text{e}}^y}}}\)     M1

Note: The M1 is for switching the variables and may be awarded at any stage in the process and is awarded independently. Further marks do not rely on this mark being gained.

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

\(2{{\text{e}}^{2y}} - x{{\text{e}}^y} - 1 = 0\)     A1

\({{\text{e}}^y} = \frac{{x \pm \sqrt {{x^2} + 8} }}{4}\)     M1A1

\(y = \ln \left( {\frac{{x \pm \sqrt {{x^2} + 8} }}{4}} \right)\)

therefore \({h^{ - 1}}(x) = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\)     A1

since ln is undefined for the second solution     R1

Note: Accept \(y = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\).

Note: The R1 may be gained by an appropriate comment earlier.

[6 marks]

A significant number of candidates did not recognise the need for the quadratic formula in order to find the inverse. Even when they did most candidates who got this far did not recognise the need to limit the solution to the positive only. This question was done well by a very limited number of candidates.

The polynomial \(P(x) = {x^3} + a{x^2} + bx + 2\) is divisible by (x +1) and by (x − 2) .

Find the value of a and of b, where \(a,{\text{ }}b \in \mathbb{R}\) .

METHOD 1

As (x +1) is a factor of P(x), then P(−1) = 0     (M1)

\( \Rightarrow a - b + 1 = 0\) (or equivalent)     A1

As (x − 2) is a factor of P(x), then P(2) = 0     (M1)

\( \Rightarrow 4a + 2b + 10 = 0\) (or equivalent)     A1

Attempting to solve for a and b     M1

a = −2 and b = −1     A1     N1

[6 marks]

METHOD 2

By inspection third factor must be x −1.     (M1)A1

\((x + 1)(x - 2)(x - 1) = {x^3} - 2{x^2} - x + 2\)     (M1)A1

Equating coefficients a = −2, b = −1     (M1)A1     N1

[6 marks]

METHOD 3

Considering \(\frac{{P(x)}}{{{x^2} - x - 2}}\) or equivalent     (M1)

\(\frac{{P(x)}}{{{x^2} - x - 2}} = (x + a + 1) + \frac{{(a + b + 3)x + 2(a + 2)}}{{{x^2} - x - 2}}\)     A1A1

Recognising that \((a + b + 3)x + 2(a + 2) = 0\)     (M1)

Attempting to solve for a and b     M1

a = −2 and b = −1     A1     N1

[6 marks]

Most candidates successfully answered this question. The majority used the factor theorem, but a few employed polynomial division or a method based on inspection to determine the third linear factor.

Showing any \(x\) and \(y\) intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

\(y = f(x)\);

Showing any \(x\) and \(y\) intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

\(y = \frac{1}{{f(x)}}\);

Showing any \(x\) and \(y\) intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

\(y = \left| {\frac{1}{{f(x)}}} \right|\).

Find \(\int {f(x)\cos x{\text{d}}x} \).

By finding \(g'(x)\) explain why \(g\) is an increasing function.

A1 for correct shape

A1 for correct \(x\) and \(y\) intercepts and minimum point

[2 marks]

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

A1 for reflecting negative branch from (ii) in the \(x\)-axis

A1 for correctly labelled minimum point

[2 marks]

EITHER

attempt at integration by parts     (M1)

\(\int {({x^2} - {a^2})\cos x{\text{d}}x = ({x^2} - {a^2})\sin x - \int {2x\sin x{\text{d}}x} } \)     A1A1

\( = ({x^2} - {a^2})\sin x - 2\left[ { - x\cos x + \int {\cos x{\text{d}}x} } \right]\)     A1

\( = ({x^2} - {a^2})\sin x + 2x\cos - 2\sin x + c\)     A1

OR

\(\int {({x^2} - {a^2})\cos x{\text{d}}x = \int {{x^2}\cos x{\text{d}}x - \int {{a^2}\cos x{\text{d}}x} } } \)

attempt at integration by parts     (M1)

\(\int {{x^2}\cos x{\text{d}}x = {x^2}\sin x - \int {2x\sin x{\text{d}}x} } \)     A1A1

\( = {x^2}\sin x - 2\left[ { - x\cos x + \int {\cos x{\text{d}}x} } \right]\)     A1

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

\( - \int {{a^2}\cos x{\text{d}}x = - {a^2}\sin x} \)

\(\int {({x^2} - {a^2})\cos x{\text{d}}x = ({x^2} - {a^2})\sin x + 2x\cos x - 2\sin x + c} \)     A1

[5 marks]

\(g(x) = x{({x^2} - {a^2})^{\frac{1}{2}}}\)

\(g'(x) = {({x^2} - {a^2})^{\frac{1}{2}}} + \frac{1}{2}x{({x^2} - {a^2})^{ - \frac{1}{2}}}(2x)\)     M1A1A1

Note:     Method mark is for differentiating the product. Award A1 for each correct term.

\(g'(x) = {({x^2} - {a^2})^{\frac{1}{2}}} + {x^2}{({x^2} - {a^2})^{ - \frac{1}{2}}}\)

both parts of the expression are positive hence \(g'(x)\) is positive     R1

and therefore \(g\) is an increasing function (for \(\left| x \right| > a\))     AG

[4 marks]

Show that \(g \circ f(x) = 3\sin \left( {2x + \frac{\pi }{5}} \right) + 4\).

Find the range of \(g \circ f\).

Given that \(g \circ f\left( {\frac{{3\pi }}{{20}}} \right) = 7\), find the next value of \(x\), greater than \({\frac{{3\pi }}{{20}}}\), for which \(g \circ f(x) = 7\).

The graph of \(y = g \circ f(x)\) can be obtained by applying four transformations to the graph of \(y = \sin x\). State what the four transformations represent geometrically and give the order in which they are applied.

\(g \circ f(x) = g\left( {f(x)} \right)\)     M1

\( = g\left( {2x + \frac{\pi }{5}} \right)\)

\( = 3\sin \left( {2x + \frac{\pi }{5}} \right) + 4\)     AG

[1 mark]

since \( - 1 \le sin \theta  \le  + 1\), range is \([1,{\text{ }}7]\)     (R1)A1

[2 marks]

\(3\sin \left( {2x + \frac{\pi }{5}} \right) + 4 = 7 \Rightarrow 2x + \frac{\pi }{5} = \frac{\pi }{2} + 2n\pi  \Rightarrow x = \frac{{3\pi }}{{20}} + n\pi \)     (M1)

so next biggest value is \(\frac{{23\pi }}{{20}}\)     A1

Note:     Allow use of period.

[2 marks]

Note:     Transformations can be in any order but see notes below.

stretch scale factor \(3\) parallel to \(y\) axis (vertically)     A1

vertical translation of \(4\) up     A1

Note:     Vertical translation is \(\frac{4}{3}\) up if it occurs before stretch parallel to \(y\) axis.

stretch scale factor \(\frac{1}{2}\) parallel to \(x\) axis (horizontally)     A1

horizontal translation of \(\frac{\pi }{{10}}\) to the left     A1

Note:     Horizontal translation is \(\frac{\pi }{{5}}\) to the left if it occurs before stretch parallel to \(x\) axis.

Note:     Award A1 for magnitude and direction in each case.

Accept any correct terminology provided that the meaning is clear eg shift for translation.

[4 marks]

Total [9 marks]

Well done.

Generally well done, some used more complicated methods rather than considering the range of sine.

Fine if they realised the period was \(\pi \), not if they thought it was \(2\pi \).

Typically 3 marks were gained. It was the shift in the axis \(\chi \) of \(\frac{\pi }{{10}}\) that caused the problem.

Find an expression for \(g \circ f(x)\), stating its domain.

Hence show that \(g \circ f(x) = \frac{{\sin x + \cos x}}{{\sin x - \cos x}}\).

Let \(y = g \circ f(x)\), find an exact value for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) at the point on the graph of \(y = g \circ f(x)\) where \(x = \frac{\pi }{6}\), expressing your answer in the form \(a + b\sqrt 3 ,{\text{ }}a,{\text{ }}b \in \mathbb{Z}\).

Show that the area bounded by the graph of \(y = g \circ f(x)\), the \(x\)-axis and the lines \(x = 0\) and \(x = \frac{\pi }{6}\) is \(\ln \left( {1 + \sqrt 3 } \right)\).

\(g \circ f(x) = \frac{{\tan x + 1}}{{\tan x - 1}}\)     A1

\(x \ne \frac{\pi }{4},{\text{ }}0 \le x < \frac{\pi }{2}\)     A1

[2 marks]

\(\frac{{\tan x + 1}}{{\tan x - 1}} = \frac{{\frac{{\sin x}}{{\cos x}} + 1}}{{\frac{{\sin x}}{{\cos x}} - 1}}\)     M1A1

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

[2 marks]

METHOD 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}\)     M1(A1)

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

\( = \frac{{ - 2}}{{1 - \sin 2x}}\)

Substitute \(\frac{\pi }{6}\) into any formula for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1

\(\frac{{ - 2}}{{1 - \sin \frac{\pi }{3}}}\)

\( = \frac{{ - 2}}{{1 - \frac{{\sqrt 3 }}{2}}}\)     A1

\( = \frac{{ - 4}}{{2 - \sqrt 3 }}\)

\( = \frac{{ - 4}}{{2 - \sqrt 3 }}\left( {\frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}} \right)\)     M1

\( = \frac{{ - 8 - 4\sqrt 3 }}{1} =  - 8 - 4\sqrt 3 \)     A1

METHOD 2

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

\( = \frac{{ - 2{{\sec }^2}x}}{{{{(\tan x - 1)}^2}}}\)     A1

\( = \frac{{ - 2{{\sec }^2}\frac{\pi }{6}}}{{{{\left( {\tan \frac{\pi }{6} - 1} \right)}^2}}} = \frac{{ - 2\left( {\frac{4}{3}} \right)}}{{{{\left( {\frac{1}{{\sqrt 3 }} - 1} \right)}^2}}} = \frac{{ - 8}}{{{{\left( {1 - \sqrt 3 } \right)}^2}}}\)     M1

Note: Award M1 for substitution \(\frac{\pi }{6}\).

\(\frac{{ - 8}}{{{{\left( {1 - \sqrt 3 } \right)}^2}}} = \frac{{ - 8}}{{\left( {4 - 2\sqrt 3 } \right)}}\frac{{\left( {4 + 2\sqrt 3 } \right)}}{{\left( {4 + 2\sqrt 3 } \right)}} =  - 8 - 4\sqrt 3 \)     M1A1

[6 marks]

Area \(\left| {\int_0^{\frac{\pi }{6}} {\frac{{\sin x + \cos x}}{{\sin x - \cos x}}{\text{d}}x} } \right|\)     M1

\( = \left| {\left[ {\ln \left| {\sin x - \cos x} \right|} \right]_0^{\frac{\pi }{6}}} \right|\)     A1

Note:     Condone absence of limits and absence of modulus signs at this stage.

\( = \left| {\ln \left| {\sin \frac{\pi }{6} - \cos \frac{\pi }{6}} \right| - \ln \left| {\sin 0 - \cos 0} \right|} \right|\)     M1

\( = \left| {\ln \left| {\frac{1}{2} - \frac{{\sqrt 3 }}{2}} \right| - 0} \right|\)

\( = \left| {\ln \left( {\frac{{\sqrt 3  - 1}}{2}} \right)} \right|\)     A1

\( =  - \ln \left( {\frac{{\sqrt 3  - 1}}{2}} \right) = \ln \left( {\frac{2}{{\sqrt 3  - 1}}} \right)\)     A1

\( = \ln \left( {\frac{2}{{\sqrt 3  - 1}} \times \frac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}}} \right)\)     M1

\( = \ln \left( {\sqrt 3  + 1} \right)\)     AG

[6 marks]

Total [16 marks]

Express \({x^2} + 3x + 2\) in the form \({(x + h)^2} + k\).

Factorize \({x^2} + 3x + 2\).

Sketch the graph of \(f(x)\), indicating on it the equations of the asymptotes, the coordinates of the \(y\)-intercept and the local maximum.

Show that \(\frac{1}{{x + 1}} - \frac{1}{{x + 2}} = \frac{1}{{{x^2} + 3x + 2}}\).

Hence find the value of \(p\) if \(\int_0^1 {f(x){\text{d}}x = \ln (p)} \).

Sketch the graph of \(y = f\left( {\left| x \right|} \right)\).

Determine the area of the region enclosed between the graph of \(y = f\left( {\left| x \right|} \right)\), the \(x\)-axis and the lines with equations \(x = - 1\) and \(x = 1\).

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

[1 mark]

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

[1 mark]

A1 for the shape

A1 for the equation \(y = 0\)

A1 for asymptotes \(x = - 2\) and \(x = - 1\)

A1 for coordinates \(\left( { - \frac{3}{2},{\text{ }} - 4} \right)\)

A1 \(y\)-intercept \(\left( {0,{\text{ }}\frac{1}{2}} \right)\)

[5 marks]

\(\frac{1}{{x + 1}} - \frac{1}{{x + 2}} = \frac{{(x + 2) - (x + 1)}}{{(x + 1)(x + 2)}}\)     M1

\( = \frac{1}{{{x^2} + 3x + 2}}\)     AG

[1 mark]

\(\int\limits_0^1 {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}{\text{d}}x} \)

\( = \left[ {\ln (x + 1) - \ln (x + 2)} \right]_0^1\)     A1

\( = \ln 2 - \ln 3 - \ln 1 + \ln 2\)     M1

\( = \ln \left( {\frac{4}{3}} \right)\)     M1A1

\(\therefore p = \frac{4}{3}\)

[4 marks]

symmetry about the \(y\)-axis     M1

correct shape     A1

Note:     Allow FT from part (b).

[2 marks]

\(2\int_0^1 {f(x){\text{d}}x} \)     (M1)(A1)

\( = 2\ln \left( {\frac{4}{3}} \right)\)     A1

Note:     Do not award FT from part (e).

[3 marks]

For the polynomial equation \(p(x) = 0\), state

(i)     the sum of the roots;

(ii)     the product of the roots.

A new polynomial is defined by \(q(x) = p(x + 4)\).

Find the sum of the roots of the equation \(q(x) = 0\).

(i)     \(\left( { - \frac{{{a_{n - 1}}}}{{{a_n}}} = } \right) - \frac{1}{2}\)     A1

(ii)     \(\left( {{{( - 1)}^n}\frac{{{a_0}}}{{{a_n}}} = } \right) - \frac{{36}}{2} = ( - 18)\)     A1A1

Note:     First A1 is for the negative sign.

[3 marks]

METHOD 1

if \(\lambda \) satisfies \(p(\lambda ) = 0\) then \(q(\lambda  - 4) = 0\)

so the roots of \(q(x)\) are each \(4\) less than the roots of \(p(x)\)     (R1)

so sum of roots is \( - \frac{1}{2} - 4 \times 5 =  - 20.5\)     A1

METHOD 2

\(p(x + 4) = 2{x^5} + 2 \times 5 \times 4{x^4} \ldots  + {x^4} \ldots  = 2{x^5} + 41{x^4} \ldots \)     (M1)

so sum of roots is \( - \frac{{41}}{2} =  - 20.5\)     A1

[2 marks]

Tofal [5 marks]

Both parts fine if they used the formula, some tried to use the quadratic equivalent formula. Surprisingly some even found all the roots.

Some notation problems for weaker candidates. Good candidates used either of the methods shown in the Markscheme.

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).

Prove by induction that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n - 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\) for \(n \in {\mathbb{Z}^ + }\).

Find the coordinates of any local maximum and minimum points on the graph of \(y(x)\).

Justify whether any such point is a maximum or a minimum.

Find the coordinates of any points of inflexion on the graph of \(y(x)\). Justify whether any such point is a point of inflexion.

Hence sketch the graph of \(y(x)\), indicating clearly the points found in parts (c) and (d) and any intercepts with the axes.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = ({{\text{e}}^{3x}} + 3x{{\text{e}}^{3x}})\)     M1A1

[2 marks]

let \(P(n)\) be the statement \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n - 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\)

prove for \(n = 1\)     M1

\(LHS\) of \(P(1)\) is \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) which is \(1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}}\) and \(RHS\) is \({3^0}{{\text{e}}^{3x}} + x{3^1}{{\text{e}}^{3x}}\)     R1

as \({\text{LHS}} = {\text{RHS, }}P(1)\) is true

assume \(P(k)\) is true and attempt to prove \(P(k + 1)\) is true     M1

assuming \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = k{3^{k - 1}}{{\text{e}}^{3x}} + x{3^k}{{\text{e}}^{3x}}\)

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

\( = k{3^{k - 1}} \times 3{{\text{e}}^{3x}} + 1 \times {3^k}{{\text{e}}^{3x}} + x{3^k} \times 3{{\text{e}}^{3x}}\)     A1

\( = (k + 1){3^k}{{\text{e}}^{3x}} + x{3^{k + 1}}{{\text{e}}^{3x}}\;\;\;\)(as required)     A1

Note:     Can award the A marks independent of the M marks

since \(P(1)\) is true and \(P(k)\) is true \( \Rightarrow P(k + 1)\) is true

then (by \(PMI\)), \(P(n)\) is true \((\forall n \in {\mathbb{Z}^ + })\)     R1

Note: To gain last R1 at least four of the above marks must have been gained.

[7 marks]

\({{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = 0 \Rightarrow 1 + 3x = 0 \Rightarrow x =  - \frac{1}{3}\)     M1A1

point is \(\left( { - \frac{1}{3},{\text{ }} - \frac{1}{{3{\text{e}}}}} \right)\)     A1

EITHER

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)

when \(x =  - \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} > 0\) therefore the point is a minimum     M1A1

OR

nature table shows point is a minimum     M1A1

[5 marks]

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)     A1

\(2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}} = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x =  - \frac{2}{3}\)     M1A1

point is \(\left( { - \frac{2}{3},{\text{ }} - \frac{2}{{3{{\text{e}}^2}}}} \right)\)     A1

since the curvature does change (concave down to concave up) it is a point of inflection     R1

Note:     Allow \({3^{{\text{rd}}}}\) derivative is not zero at \( - \frac{2}{3}\)

[5 marks]

(general shape including asymptote and through origin)     A1

showing minimum and point of inflection     A1

Note:     Only indication of position of answers to (c) and (d) required, not coordinates.

[2 marks]

Total [21 marks]

Well done.

The logic of an induction proof was not known well enough. Many candidates used what they had to prove rather than differentiating what they had assumed. They did not have enough experience in doing Induction proofs.

Good, some forgot to test for min/max, some forgot to give the \(y\) value.

Again quite good, some forgot to check for change in curvature and some forgot the \(y\) value.

Some accurate sketches, some had all the information from earlier parts but could not apply it. The asymptote was often missed.

The quadratic equation \({x^2} - 2kx + (k - 1) = 0\) has roots \(\alpha \) and \(\beta \) such that \({\alpha ^2} + {\beta ^2} = 4\). Without solving the equation, find the possible values of the real number \(k\).

\(\alpha  + \beta  = 2k\)    A1

\(\alpha \beta  = k - 1\)    A1

\({(\alpha  + \beta )^2} = 4{k^2} \Rightarrow {\alpha ^2} + {\beta ^2} + 2\underbrace {\alpha \beta }_{k - 1} = 4{k^2}\)    (M1)

\({\alpha ^2} + {\beta ^2} = 4{k^2} - 2k + 2\)

\({\alpha ^2} + {\beta ^2} = 4 \Rightarrow 4{k^2} - 2k - 2 = 0\)    A1

attempt to solve quadratic     (M1)

\(k = 1,{\text{ }} - \frac{1}{2}\)    A1

[6 marks]

the degree of the polynomial;

the value of \({a_0}\).

the sum of the roots of the polynomial \( = \frac{{63}}{{16}}\)     (A1)

\(2\left( {\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}}} \right) = \frac{{63}}{{16}}\)     M1A1

Note:     The formula for the sum of a geometric sequence must be equated to a value for the M1 to be awarded.

\(1 - {\left( {\frac{1}{2}} \right)^n} = \frac{{63}}{{64}} \Rightarrow {\left( {\frac{1}{2}} \right)^n} = \frac{1}{{64}}\)

\(n = 6\)     A1

[4 marks]

\(\frac{{{a_0}}}{{{a_n}}} = 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}},{\text{ (}}{{\text{a}}_n} = 16)\)     M1

\({a_0} = 16 \times 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}}\)

\({a_0} = {2^{ - 5}}\;\;\;\left( { = \frac{1}{{32}}} \right)\)     A1

[2 marks]

Total [6 marks]

Consider the function f , where \(f(x) = \arcsin (\ln x)\).

(a)     Find the domain of f .

(b)     Find \({f^{ - 1}}(x)\).

(a)     \( - 1 \leqslant \ln x \leqslant 1\)     (M1)

\( \Rightarrow \frac{1}{{\text{e}}} \leqslant x \leqslant {\text{e}}\)     A1A1

(b)     \(y = \arcsin (\ln x) \Rightarrow \ln x = \sin y\)     (M1)

\(\ln y = \sin x \Rightarrow y = {{\text{e}}^{\sin x}}\)     (M1)

\( \Rightarrow {f^{ - 1}}(x) = {{\text{e}}^{\sin x}}\)     A1

[6 marks]

Very few candidates attempted part (a), and of those that did, few were successful. Part (b) was answered fairly well by most candidates.

Show that \(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \sqrt {n + 1}  - \sqrt n \) where \(n \ge 0,{\text{ }}n \in \mathbb{Z}\).

Hence show that \(\sqrt 2  - 1 < \frac{1}{{\sqrt 2 }}\).

Prove, by mathematical induction, that \(\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n } \) for \(n \ge 2,{\text{ }}n \in \mathbb{Z}\).

\(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \frac{1}{{\sqrt n  + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt {n + 1}  - \sqrt n }}\)     M1

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

\( = \sqrt {n + 1}  - \sqrt n \)     AG

[2 marks]

\(\sqrt 2  - 1 = \frac{1}{{\sqrt 2  + \sqrt 1 }}\)     A2

\( < \frac{1}{{\sqrt 2 }}\)     AG

[2 marks]

consider the case \(n = 2\): required to prove that \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \)     M1

from part (b) \(\frac{1}{{\sqrt 2 }} > \sqrt 2  - 1\)

hence \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) is true for \(n = 2\)     A1

now assume true for \(n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k } \)     M1

\(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k \)

attempt to prove true for \(n = k + 1:\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

from assumption, we have that \(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k  + \frac{1}{{\sqrt {k + 1} }}\)     M1

so attempt to show that \(\sqrt k  + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

EITHER

\(\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}  - \sqrt k \)     A1

\(\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k  + \sqrt {k + 1} }}\), (from part a), which is true     A1

OR

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

\( > \frac{{\sqrt k \sqrt k  + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1} \)     A1

THEN

so true for \(n = 2\) and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \ge 2\)     R1

Note:     Award R1 only if all previous M marks have been awarded.

[9 marks]

Total [13 marks]

Let \(g(x) = {\log _5}\left| {2{{\log }_3}x} \right|\) . Find the product of the zeros of g .

\(g(x) = 0\)

\({\log _5}\left| {2{{\log }_3}x} \right| = 0\)     (M1)

\(\left| {2{{\log }_3}x} \right| = 1\)     A1

\({\log _3}x =  \pm \frac{1}{2}\)     (A1)

\(x = {3^{ \pm \frac{1}{2}}}\)     A1

so the product of the zeros of g is \({3^{\frac{1}{2}}} \times {3^{ - \frac{1}{2}}} = 1\)     A1     N0

[5 marks]

There were many candidates showing difficulties in manipulating logarithms and the absolute value to solve the equation.

Given that \(f\) is an even function, show that \(b = 0\).

Given that \(g\) is an odd function, find the value of \(r\).

The function \(h\) is both odd and even, with domain \(\mathbb{R}\).

Find \(h(x)\).

EITHER

\(f( - x) = f(x)\)     M1

\( \Rightarrow a{x^2} - bx + c = a{x^2} + bx + c \Rightarrow 2bx = 0,{\text{ }}(\forall x \in \mathbb{R})\)     A1

OR

\(y\)-axis is eqn of symmetry     M1

so \(\frac{{ - b}}{{2a}} = 0\)     A1

THEN

\( \Rightarrow b = 0\)     AG

[2 marks]

\(g( - x) =  - g(x) \Rightarrow p\sin ( - x) - qx + r =  - p\sin x - qx - r\)

\( \Rightarrow  - p\sin x - qx + r =  - p\sin x - qx - r\)     M1

Note:     M1 is for knowing properties of sin.

\( \Rightarrow 2r = 0 \Rightarrow r = 0\)     A1

Note:     In (a) and (b) allow substitution of a particular value of \(x\)

[2 marks]

\(h( - x) = h(x) =  - h(x) \Rightarrow 2h(x) = 0 \Rightarrow h(x) = 0,{\text{ }}(\forall x)\)     M1A1

Note:     Accept geometrical explanations.

[2 marks]

Total [6 marks]

Sometimes backwards working but many correct approaches.

Some candidates did not know what odd and even functions were. Correct solutions from those who applied the definition.

Some realised: just apply the definitions. Some did very strange things involving \(f\) and \(g\).

Given that \(q(x)\) has a factor \((x - 4)\), find the value of \(k\).

Hence or otherwise, factorize \(q(x)\) as a product of linear factors.

\(q(4) = 0\)     (M1)

\(192 - 176 + 4k + 8 = 0{\text{ }}(24 + 4k = 0)\)     A1

\(k =  - 6\)     A1

[3 marks]

\(3{x^3} - 11{x^2} - 6x + 8 = (x - 4)(3{x^2} + px - 2)\)

equate coefficients of \({x^2}\):     (M1)

\( - 12 + p =  - 11\)

\(p = 1\)

\((x - 4)(3{x^2} + x - 2)\)     (A1)

\((x - 4)(3x - 2)(x + 1)\)     A1

Note:     Allow part (b) marks if any of this work is seen in part (a).

Note:     Allow equivalent methods (eg, synthetic division) for the M marks in each part.

[3 marks]

Write down the sum and the product of the roots of \(P(z) = 0\).

Show that \((z - 1)\) is a factor of \(P(z)\).

Find the value of \(b\) and the value of \(c\).

Hence find the complex roots of \(P(z) = 0\).

Show that the graph of \(y = q(x)\) is concave up for \(x > 1\).

Sketch the graph of \(y = q(x)\) showing clearly any intercepts with the axes.

\({\text{sum}} = 0\)     A1

\({\text{product}} = 6\)     A1

[2 marks]

\(P(1) = 1 - 10 + 15 - 6 = 0\)     M1A1

\( \Rightarrow (z - 1)\) is a factor of \(P(z)\)     AG

Note:     Accept use of division to show remainder is zero.

[2 marks]

METHOD 1

\({(z - 1)^3}\left( {{z^2} + bz + c} \right) = {z^5} - 10{z^2} + 15z - 6\)     (M1)

by inspection \(c = 6\)     A1

\(({z^3} - 3{z^2} + 3z - 1)({z^2} + bz + 6) = {z^5} - 10{z^2} + 15z - 6\)     (M1)(A1)

\(b = 3\)     A1

METHOD 2

\(\alpha \), \(\beta \) are two roots of the quadratic

\(b = - (\alpha + \beta ),{\text{ }}c = \alpha \beta \)     (A1)

from part (a) \(1 + 1 + 1 + \alpha + \beta = 0\)     (M1)

\( \Rightarrow b = 3\)     A1

\(1 \times 1 \times 1 \times \alpha \beta = 6\)     (M1)

\( \Rightarrow c = 6\)     A1

Note:     Award FT if \(b = - 7\) following through from their sum \( = 10\).

METHOD 3

\(({z^5} - 10{z^2} + 15z - 6) \div (z - 1) = {z^4} + {z^3} + {z^2} - 9z + 6\)     (M1)A1

Note:     This may have been seen in part (b).

\({z^4} + {z^3} + {z^2} - 9z + 6 \div (z - 1) = {z^3} + 2{z^2} + 3z - 6\)     (M1)

\({z^3} + 2{z^2} + 3z - 6 \div (z - 1) = {z^2} + 3z + 6\)     A1A1

[5 marks]

\({z^2} + 3z + 6 = 0\)     M1

\(z = \frac{{ - 3 \pm \sqrt {9 - 4 \bullet 6} }}{2}\)     M1

\( = \frac{{ - 3 \pm \sqrt { - 15} }}{2}\)

\(z = - \frac{3}{2} \pm \frac{{{\text{i}}\sqrt {15} }}{2}\)     A1

(or \(z = 1\))

Notes:     Award the second M1 for an attempt to use the quadratic formula or to complete the square.

Do not award FT from (c).

[3 marks]

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 20{x^3} - 20\)     M1A1

for \(x > 1,{\text{ }}20{x^3} - 20 > 0 \Rightarrow \) concave up     R1AG

[3 marks]

\(x\)-intercept at \((1,{\text{ }}0)\)     A1

\(y\)-intercept at \((0,{\text{ }} - 6)\)     A1

stationary point of inflexion at \((1,{\text{ }}0)\) with correct curvature either side     A1

[3 marks]

The function \(f\)  is defined as \(f(x) = \frac{{3x + 2}}{{x + 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne  - 1\).

Sketch the graph of \(y = f(x)\), clearly indicating and stating the equations of any asymptotes and the coordinates of any axes intercepts.

     A1A1A1A1A1

Note:     Award A1 for correct shape, A1 for \(x =  - 1\) clearly stated and asymptote shown, A1 for \(y = 3\) clearly stated and asymptote shown, A1 for \(\left( { - \frac{2}{3},{\text{ }}0} \right)\) and A1 for \((0,{\text{ }}2)\).

[5 marks]

Another standard question. On this occasion, specific coordinates were asked for, so some otherwise good candidates missed out on a couple of marks which they would have gained through greater care.

(i)     \(p =  - (\alpha  + \beta  + \gamma )\);

(ii)     \(q = \alpha \beta  + \beta \gamma  + \gamma \alpha \);

(iii)     \(c =  - \alpha \beta \gamma \).

It is now given that \(p =  - 6\) and \(q = 18\) for parts (b) and (c) below.

(i)     In the case that the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form an arithmetic sequence, show that one of the roots is \(2\).

(ii)     Hence determine the value of \(c\).

In another case the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form a geometric sequence. Determine the value of \(c\).

(i)-(iii) given the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \), we have

\({x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )\)     M1

\( = \left( {{x^2} - (\alpha  + \beta )x + \alpha \beta } \right)(x - \gamma )\)     A1

\( = {x^3} - (\alpha  + \beta  + \gamma ){x^2} + (\alpha \beta  + \beta \gamma  + \gamma \alpha )x - \alpha \beta \gamma \)     A1

comparing coefficients:

\(p =  - (\alpha  + \beta  + \gamma )\)     AG

\(q = (\alpha \beta  + \beta \gamma  + \gamma \alpha )\)     AG

\(c =  - \alpha \beta \gamma \)     AG

[3 marks]

METHOD 1

(i)     Given \( - \alpha  - \beta  - \gamma  =  - 6\)

And \(\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18\)

Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \)

So \(\beta  - \alpha  = \gamma  - \beta \)     M1

or \(2\beta  = \alpha  + \gamma \)

Attempt to solve simultaneous equations:     M1

\(\beta  + 2\beta  = 6\)     A1

\(\beta  = 2\)     AG

(ii)     \(\alpha  + \gamma  = 4\)

\(2\alpha  + 2\gamma  + \alpha \gamma  = 18\)

\( \Rightarrow {\gamma ^2} - 4\gamma  + 10 = 0\)

\( \Rightarrow \gamma  = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}\)     (A1)

Therefore \(c =  - \alpha \beta \gamma  =  - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 =  - 20\)     A1

METHOD 2

(i)     let the three roots be \(\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d\)     M1

adding roots     M1

to give \(3\alpha  = 6\)     A1

\(\alpha  = 2\)     AG

(ii)     \(\alpha \) is a root, so \({2^3} - 6 \times {2^2} + 18 \times 2 + c = 0\)     M1

\(8 - 24 + 36 + c = 0\)

\(c =  - 20\)     A1

METHOD 3

(i)     let the three roots be \(\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d\)     M1

adding roots     M1

to give \(3\alpha  = 6\)     A1

\(\alpha  = 2\)     AG

(ii)     \(q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)\)     M1

\({d^2} =  - 6 \Rightarrow d = \sqrt 6 {\text{i}}\)

\( \Rightarrow c =  - 20\)     A1

[5 marks]

METHOD 1

Given \( - \alpha  - \beta  - \gamma  =  - 6\)

And \(\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18\)

Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \).

So \(\frac{\beta }{\alpha } = \frac{\gamma }{\beta }\)     M1

or \({\beta ^2} = \alpha \gamma \)

Attempt to solve simultaneous equations:     M1

\(\alpha \beta  + \gamma \beta  + {\beta ^2} = 18\)

\(\beta (\alpha  + \beta  + \gamma ) = 18\)

\(6\beta  = 18\)

\(\beta  = 3\)     A1

\(\alpha  + \gamma  = 3,{\text{ }}\alpha  = \frac{9}{\gamma }\)

\( \Rightarrow {\gamma ^2} - 3\gamma  + 9 = 0\)

\( \Rightarrow \gamma  = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}\)     (A1)(A1)

Therefore \(c =  - \alpha \beta \gamma  =  - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 =  - 27\)     A1

METHOD 2

let the three roots be \(a,{\text{ }}ar,{\text{ }}a{r^2}\)     M1

attempt at substitution of \(a,{\text{ }}ar,{\text{ }}a{r^2}\) and \(p\) and \(q\) into equations from (a)     M1

\(6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)\)     A1

\(18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)\)     A1

therefore \(3 = ar\)     A1

therefore \(c =  - {a^3}{r^3} =  - {3^3} =  - 27\)     A1

[6 marks]

Total [14 marks]

Sketch the graph of \(y = h(x)\).

Find an expression for the composite function \(h \circ g(x)\) and state its domain.

Given that \(f(x) = h(x) + h \circ g(x)\),

(i)     find \(f'(x)\) in simplified form;

(ii)     show that \(f(x) = \frac{\pi }{2}\) for \(x > 0\).

Nigel states that \(f\) is an odd function and Tom argues that \(f\) is an even function.

(i)     State who is correct and justify your answer.

(ii)     Hence find the value of \(f(x)\) for \(x < 0\).

    A1A1

Note:     A1 for correct shape, A1 for asymptotic behaviour at \(y =  \pm \frac{\pi }{2}\).

[2 marks]

\(h \circ g(x) = \arctan \left( {\frac{1}{x}} \right)\)     A1

domain of \(h \circ g\) is equal to the domain of \(g:x \in  \circ ,{\text{ }}x \ne 0\)     A1

[2 marks]

(i)     \(f(x) = \arctan (x) + \arctan \left( {\frac{1}{x}} \right)\)

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

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

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

\( = 0\)     A1

(ii)     METHOD 1

f is a constant     R1

when \(x > 0\)

\(f(1) = \frac{\pi }{4} + \frac{\pi }{4}\)     M1A1

\( = \frac{\pi }{2}\)     AG

METHOD 2

from diagram

\(\theta  = \arctan \frac{1}{x}\)     A1

\(\alpha  = \arctan x\)     A1

\(\theta  + \alpha  = \frac{\pi }{2}\)     R1

hence \(f(x) = \frac{\pi }{2}\)     AG

METHOD 3

\(\tan \left( {f(x)} \right) = \tan \left( {\arctan (x) + \arctan \left( {\frac{1}{x}} \right)} \right)\)     M1

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

denominator = 0, so \(f(x) = \frac{\pi }{2}{\text{ (for }}x > 0)\)     R1

[7 marks]

(i)     Nigel is correct.     A1

METHOD 1

\(\arctan (x)\) is an odd function and \(\frac{1}{x}\) is an odd function

composition of two odd functions is an odd function and sum of two odd functions is an odd function     R1

METHOD 2

\(f( - x) = \arctan ( - x) + \arctan \left( { - \frac{1}{x}} \right) =  - \arctan (x) - \arctan \left( {\frac{1}{x}} \right) =  - f(x)\)

therefore f is an odd function.     R1

(ii)     \(f(x) =  - \frac{\pi }{2}\)     A1

[3 marks]

Write down the range of \(f\).

Find an expression for \({f^{ - 1}}(x)\).

Write down the domain and range of \({f^{ - 1}}\).

\( - 11 \leqslant f(x) \leqslant 21\)     A1A1

Note:     A1 for correct end points, A1 for correct inequalities.

[2 marks]

\({f^{ - 1}}(x) = \sqrt[3]{{\frac{{x - 5}}{2}}}\)     (M1)A1

[2 marks]

\( - 11 \leqslant x \leqslant 21,{\text{ }} - 2 \leqslant {f^{ - 1}}(x) \leqslant 2\)     A1A1

[2 marks]

Show that \(f\) is an odd function.

Find \(f'(x)\).

Hence find the \(x\)-coordinates of any local maximum or minimum points.

Find the range of \(f\).

Sketch the graph of \(y = f(x)\) indicating clearly the coordinates of the \(x\)-intercepts and any local maximum or minimum points.

Find the area of the region enclosed by the graph of \(y = f(x)\) and the \(x\)-axis for \(x \ge 0\).

Show that \(\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > \left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right|} \).

\(f( - x) = ( - x)\sqrt {1 - {{( - x)}^2}} \)     M1

\( =  - x\sqrt {1 - {x^2}} \)

\( =  - f(x)\)     R1

hence \(f\) is odd     AG

[2 marks]

\(f'(x) = x \bullet \frac{1}{2}{(1 - {x^2})^{ - \frac{1}{2}}} \bullet  - 2x + {(1 - {x^2})^{\frac{1}{2}}}\)     M1A1A1

[3 marks]

\(f'(x) = \sqrt {1 - {x^2}}  - \frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}\;\;\;\left( { = \frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)\)     A1

Note:     This may be seen in part (b).

Note:     Do not allow FT from part (b).

\(f'(x) = 0 \Rightarrow 1 - 2{x^2} = 0\)     M1

\(x =  \pm \frac{1}{{\sqrt 2 }}\)     A1

[3 marks]

\(y\)-coordinates of the Max Min Points are \(y =  \pm \frac{1}{2}\)     M1A1

so range of \(f(x)\) is \(\left[ { - \frac{1}{2},{\text{ }}\frac{1}{2}} \right]\)     A1

Note:     Allow FT from (c) if values of \(x\), within the domain, are used.

[3 marks]

Shape: The graph of an odd function, on the given domain, s-shaped,

where the max(min) is the right(left) of \(0.5{\text{ }}( - 0.5)\)     A1

\(x\)-intercepts     A1

turning points     A1

[3 marks]

\({\text{area}} = \int_0^1 {x\sqrt {1 - {x^2}} {\text{d}}x} \)     (M1)

attempt at “backwards chain rule” or substitution     M1

\( =  - \frac{1}{2}\int_0^1 {( - 2x)\sqrt {1 - {x^2}} {\text{d}}x} \)

Note:     Condone absence of limits for first two marks.

\( = \left[ {\frac{2}{3}{{(1 - {x^2})}^{\frac{3}{2}}} \bullet  - \frac{1}{2}} \right]_0^1\)     A1

\( = \left[ { - \frac{1}{3}{{(1 - {x^2})}^{\frac{3}{2}}}} \right]_0^1\)

\( = 0 - \left( { - \frac{1}{3}} \right) = \frac{1}{3}\)     A1

[4 marks]

\(\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > 0} \)     R1

\(\left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right| = 0\)     R1

so \(\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > \left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right|} \)     AG

[2 marks]

Total [20 marks]

Given that f and its derivative, \(f'\) , are continuous for all values in the domain of f , find the values of a and b .

Show that f is a one-to-one function.

Obtain expressions for the inverse function \({f^{ - 1}}\) and state their domains.

f continuous \( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f(x)\)     M1

\(4a + 2b = 8\)     A1

\(f'(x) = \left\{ {\begin{array}{*{20}{c}}
  {2,}&{x < 2} \\
  {2ax + b,}&{2 < x < 3}
\end{array}} \right.\)     A1

\(f'{\text{ continuous}} \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f'(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f'(x)\)

\(4a + b = 2\)     A1

solve simultaneously     M1

to obtain a = –1 and b = 6     A1

[6 marks]

for \(x \leqslant 2,{\text{ }}f'(x) = 2 > 0\)     A1

for \(2 < x < 3,{\text{ }}f'(x) = - 2x + 6 > 0\)     A1

since \(f'(x) > 0\) for all values in the domain of f , f is increasing     R1

therefore one-to-one     AG

[3 marks]

\(x = 2y - 1 \Rightarrow y = \frac{{x + 1}}{2}\)     M1

\(x = - {y^2} + 6y - 5 \Rightarrow {y^2} - 6y + x + 5 = 0\)     M1

\(y = 3 \pm \sqrt {4 - x} \)

therefore

\({f^{ - 1}}(x) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{x + 1}}{2},}&{x \leqslant 3} \\
  {3 - \sqrt {4 - x} ,}&{3 < x < 4}
\end{array}} \right.\)     A1A1A1

Note: Award A1 for the first line and A1A1 for the second line.

[5 marks]

Determine whether or not \(f\)is continuous.

The graph of the function \(g\) is obtained by applying the following transformations to the graph of \(f\):

a reflection in the \(y\)–axis followed by a translation by the vector \(\left( \begin{array}{l}2\\0\end{array} \right)\).

Find \(g(x)\).

\(1 - 2(2) = -  3\) and \(\frac{3}{4}{(2 - 2)^2} - 3 =  - 3\)     A1

both answers are the same, hence f is continuous (at \(x = 2\))     R1

Note:     R1 may be awarded for justification using a graph or referring to limits. Do not award A0R1.

[2 marks]

reflection in the y-axis

\(f( - x) = \left\{ \begin{array}{r}1 + 2x,\\{\textstyle{3 \over 4}}{(x + 2)^2} - 3,\end{array} \right.\begin{array}{*{20}{c}}{x \ge  - 2}\\{x <  - 2}\end{array}\)     (M1)

Note:     Award M1 for evidence of reflecting a graph in y-axis.

translation \(\left( \begin{array}{l}2\\0\end{array} \right)\)

\(g(x) = \left\{ \begin{array}{r}2x - 3,\\{\textstyle{3 \over 4}}{x^2} - 3,\end{array} \right.\begin{array}{*{20}{c}}{x \ge 0}\\{x < 0}\end{array}\)     (M1)A1A1

Note:     Award (M1) for attempting to substitute \((x - 2)\) for x, or translating a graph along positive x-axis.

     Award A1 for the correct domains (this mark can be awarded independent of the M1).

     Award A1 for the correct expressions.

[4 marks]

Sketch the graph of \(y = \frac{{1 - 3x}}{{x - 2}}\), showing clearly any asymptotes and stating the coordinates of any points of intersection with the axes.

Hence or otherwise, solve the inequality \(\left| {\frac{{1 - 3x}}{{x - 2}}} \right| < 2\).

correct vertical asymptote     A1

shape including correct horizontal asymptote     A1

\(\left( {\frac{1}{3},{\text{ }}0} \right)\)     A1

\(\left( {0,{\text{ }} - \frac{1}{2}} \right)\)     A1

Note:     Accept \(x = \frac{1}{3}\) and \(y =  - \frac{1}{2}\) marked on the axes.

[4 marks]

METHOD 1

\(\frac{{1 - 3x}}{{x - 2}} = 2\)     (M1)

\( \Rightarrow x = 1\)    A1

\( - \left( {\frac{{1 - 3x}}{{x - 2}}} \right) = 2\)     (M1)

Note:     Award this M1 for the line above or a correct sketch identifying a second critical value.

\( \Rightarrow x =  - 3\)     A1

solution is \( - 3 < x < 1\)     A1

METHOD 2

\(\left| {1 - 3x} \right| < 2\left| {x - 2} \right|,{\text{ }}x \ne 2\)

\(1 - 6x + 9{x^2} < 4({x^2} - 4x + 4)\)     (M1)A1

\(1 - 6x + 9{x^2} < 4{x^2} - 16x + 16\)

\(5{x^2} + 10x - 15 < 0\)

\({x^2} + 2x - 3 < 0\)     A1

\((x + 3)(x - 1) < 0\)     (M1)

solution is \( - 3 < x < 1\)     A1

METHOD 3

\( - 2 < \frac{{1 - 3x}}{{x - 2}} < 2\)

consider \(\frac{{1 - 3x}}{{x - 2}} < 2\)     (M1)

Note:     Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at \(x = 1\)     A1

consider \( - 2 < \frac{{1 - 3x}}{{x - 2}}\)     (M1)

Note:     Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at \(x =  - 3\)     A1

solution is \( - 3 < x < 1\)     A1

[5 marks]

(i)     Find an expression for \(f'(x)\).

(ii)     Hence determine the coordinates of the point A, where \(f'(x) = 0\).

Find an expression for \(f''(x)\) and hence show the point A is a maximum.

Find the coordinates of B, the point of inflexion.

The graph of the function \(g\) is obtained from the graph of \(f\) by stretching it in the x-direction by a scale factor 2.

          (i)     Write down an expression for \(g(x)\).

          (ii)     State the coordinates of the maximum C of \(g\).

          (iii)     Determine the x-coordinates of D and E, the two points where \(f(x) = g(x)\).

Sketch the graphs of \(y = f(x)\) and \(y = g(x)\) on the same axes, showing clearly the points A, B, C, D and E.

Find an exact value for the area of the region bounded by the curve \(y = g(x)\), the x-axis and the line \(x = 1\).

(i)     \(f'(x) = {{\text{e}}^{ - x}} - x{{\text{e}}^{ - x}}\)     M1A1

(ii)     \(f'(x) = 0 \Rightarrow x = 1\)

coordinates \(\left( {1,{\text{ }}{{\text{e}}^{ - 1}}} \right)\)     A1

[3 marks]

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

substituting \(x = 1\) into \(f''(x)\)     M1

\(f''(1){\text{ }}\left( { =  - {{\text{e}}^{ - 1}}} \right) < 0\) hence maximum     R1AG

[3 marks]

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

coordinates \(\left( {2,{\text{ 2}}{{\text{e}}^{ - 2}}} \right)\)     A1

[2 marks]

(i)     \(g(x) = \frac{x}{2}{{\text{e}}^{ - \frac{x}{2}}}\)     A1

(ii)     coordinates of maximum \(\left( {2,{\text{ }}{{\text{e}}^{ - 1}}} \right)\)     A1

(iii)     equating \(f(x) = g(x)\) and attempting to solve \(x{{\text{e}}^{ - x}} = \frac{x}{2}{{\text{e}}^{ - \frac{x}{2}}}\)

                    \( \Rightarrow x\left( {2{{\text{e}}^{\frac{x}{2}}} - {{\text{e}}^x}} \right) = 0\)     (A1)

                    \( \Rightarrow x = 0\)     A1

                    or \(2{{\text{e}}^{\frac{x}{2}}} = {{\text{e}}^x}\)

                    \( \Rightarrow {{\text{e}}^{\frac{x}{2}}} = 2\)

                    \( \Rightarrow x = 2\ln 2\)   \((\ln 4)\)     A1

Note:     Award first (A1) only if factorisation seen or if two correct

solutions are seen.

    A4

Note:     Award A1 for shape of \(f\), including domain extending beyond \(x = 2\).

     Ignore any graph shown for \(x < 0\).

     Award A1 for A and B correctly identified.

     Award A1 for shape of \(g\), including domain extending beyond \(x = 2\).

     Ignore any graph shown for \(x < 0\). Allow follow through from \(f\).

     Award A1 for C, D and E correctly identified (D and E are interchangeable).

[4 marks]

\(A = \int_0^1 {\frac{x}{2}{{\text{e}}^{ - \frac{x}{2}}}{\text{d}}x} \)     M1

\( = \left[ { - x{{\text{e}}^{ - \frac{x}{2}}}} \right]_0^1 - \int_0^1 { - {{\text{e}}^{ - \frac{x}{2}}}{\text{d}}x} \)     A1

Note:     Condone absence of limits or incorrect limits.

\( =  - {{\text{e}}^{ - \frac{1}{2}}} - \left[ {2{{\text{e}}^{ - \frac{x}{2}}}} \right]_0^1\)

\( =  - {{\text{e}}^{ - \frac{1}{2}}} - \left( {2{{\text{e}}^{ - \frac{1}{2}}} - 2} \right) = 2 - 3{{\text{e}}^{ - \frac{1}{2}}}\)     A1

[3 marks]

Part a) proved to be an easy start for the vast majority of candidates.

Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so.

Part c) again proved to be an easily earned 2 marks.

Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so.

Part c) again proved to be an easily earned 2 marks.

Many candidates lost their way in part d). A variety of possibilities for \(g(x)\) were suggested, commonly \(2x{{\text{e}}^{ - 2x}}\), \(\frac{{x{{\text{e}}^{ - 1}}}}{2}\) or similar variations. Despite section ii) being worth only one mark, (and ‘state’ being present in the question), many laborious attempts at further differentiation were seen. Part diii was usually answered well by those who gave the correct function for \(g(x)\).

Part e) was also answered well by those who had earned full marks up to that point.

While the integration by parts technique was clearly understood, it was somewhat surprising how many careless slips were seen in this part of the question. Only a minority gained full marks for part f).

state the value of \(a\) and the value of \(c\);

find the value of \(b\).

\(a = 1\)    A1

\(c = 3\)    A1

[2 marks]

use the coordinates of \((1,{\text{ }}0)\) on the graph     M1

\(f(1) = 0 \Rightarrow 1 + \frac{b}{{1 - 3}} = 0 \Rightarrow b = 2\)    A1

[2 marks]

The equation \(5{x^3} + 48{x^2} + 100x + 2 = a\) has roots \({r_1}\), \({r_2}\) and \({r_3}\).

Given that \({r_1} + {r_2} + {r_3} + {r_1}{r_2}{r_3} = 0\), find the value of a.

\({r_1} + {r_2} + {r_3} = \frac{{ - 48}}{5}\)     (M1)(A1)

\({r_1}{r_2}{r_3} = \frac{{a - 2}}{5}\)     (M1)(A1)

\(\frac{{ - 48}}{5} + \frac{{a - 2}}{5} = 0\)     M1

\(a = 50\)     A1

Note:     Award M1A0M1A0M1A1 if answer of 50 is found using \(\frac{{48}}{5}\) and \(\frac{{2 - a}}{5}\).

[6 marks]

Write down the numerical value of the sum and of the product of the roots of this equation.

The roots of this equation are three consecutive terms of an arithmetic sequence.

Taking the roots to be \(\alpha {\text{ , }}\alpha  \pm \beta \) , solve the equation.

\({\text{sum}} = \frac{{45}}{9},{\text{ product}} = \frac{{40}}{9}\)     A1

[1 mark]

it follows that \(3\alpha = \frac{{45}}{9}\) and \(\alpha ({\alpha ^2} - {\beta ^2}) = \frac{{40}}{9}\)     A1A1

solving, \(\alpha = \frac{5}{3}\)     A1

\(\frac{5}{3}\left( {\frac{{25}}{9} - {\beta ^2}} \right) = \frac{{40}}{9}\)     M1

\(\beta = ( \pm )\frac{1}{3}\)     A1

the other two roots are 2, \(\frac{4}{3}\)     A1

[6 marks]

Let f(x) = x⁴ + px³ + qx + 5 where p, q are constants.

The remainder when f(x) is divided by (x + 1) is 7, and the remainder when f(x) is divided by (x − 2) is 1. Find the value of p and the value of q.

attempt to substitute x = −1 or x = 2 or to divide polynomials      (M1)

1 − p − q + 5 = 7, 16 + 8p + 2q + 5 = 1 or equivalent      A1A1

attempt to solve their two equations M1

p = −3, q = 2      A1

[5 marks]

The cubic polynomial \(3{x^3} + p{x^2} + qx - 2\) has a factor \((x + 2)\) and leaves a remainder 4 when divided by \((x + 1)\). Find the value of p and the value of q.

\(f( - 2) = 0{\text{ }}( \Rightarrow  - 24 + 4p - 2q - 2 = 0)\)     M1

\(f( - 1) = 4{\text{ }}( \Rightarrow  - 3 + p - q - 2 = 4)\)     M1

Note:     In each case award the M marks if correct substitution attempted and right-hand side correct.

attempt to solve simultaneously \((2p - q = 13,{\text{ }}p - q = 9)\)     M1

\(p = 4\)     A1

\(q =  - 5\)     A1

[5 marks]

Many candidates scored full marks on what was thought to be an easy first question. However, a number of candidates wrote down two correct equations but proceeded to make algebraic errors and thus found incorrect values for p and q. A small number also attempted to answer this question using long division, but fully correct answers using this technique were rarely seen.

Without solving the equation, find the value of

(i)     \(\alpha  + \beta \);

(ii)     \(\alpha \beta \).

Another quadratic equation \({x^2} + px + q = 0,{\text{ }}p,{\text{ }}q \in \mathbb{Z}\) has roots \(\frac{2}{\alpha }\) and \(\frac{2}{\beta }\).

Find the value of \(p\) and the value of \(q\).

using the formulae for the sum and product of roots:

(i)     \(\alpha  + \beta  = 4\)     A1

(ii)     \(\alpha \beta  = \frac{1}{2}\)     A1

Note:     Award A0A0 if the above results are obtained by solving the original equation (except for the purpose of checking).

[2 marks]

METHOD 1

required quadratic is of the form \({x^2} - \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)x + \left( {\frac{2}{\alpha }} \right)\left( {\frac{2}{\beta }} \right)\)     (M1)

\(q = \frac{4}{{\alpha \beta }}\)

\(q = 8\)     A1

\(p =  - \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)\)

\( =  - \frac{{2(\alpha  + \beta )}}{{\alpha \beta }}\)     M1

\( =  - \frac{{2 \times 4}}{{\frac{1}{2}}}\)

\(p =  - 16\)     A1

Note:     Accept the use of exact roots

METHOD 2

replacing \(x\) with \(\frac{2}{x}\)     M1

\(2{\left( {\frac{2}{x}} \right)^2} - 8\left( {\frac{2}{x}} \right) + 1 = 0\)

\(\frac{8}{{{x^2}}} - \frac{{16}}{x} + 1 = 0\)     (A1)

\({x^2} - 16x + 8 = 0\)

\(p =  - 16\) and \(q = 8\)     A1A1

Note:     Award A1A0 for \({x^2} - 16x + 8 = 0\) ie, if \(p =  - 16\) and \(q = 8\) are not explicitly stated.

[4 marks]

Total [6 marks]

Most candidates obtained full marks.

Many candidates obtained full marks, but some responses were inefficiently expressed. A very small minority attempted to use the exact roots, usually unsuccessfully.

State the range of f and of g .

Find an expression for the composite function \(f \circ g(x)\) in the form \(\frac{{a{x^2} + bx + c}}{{3750}}\), where \(a,{\text{ }}b{\text{ and }}c \in \mathbb{Z}\) .

(i)     Find an expression for the inverse function \({f^{ - 1}}(x)\) .

(ii)     State the domain and range of \({f^{ - 1}}\) .

The domains of f and g are now restricted to {0, 1, 2, 3, 4} .

By considering the values of f and g on this new domain, determine which of f and g could be used to find a probability distribution for a discrete random variable X , stating your reasons clearly.

Using this probability distribution, calculate the mean of X .

\(f(x) \geqslant \frac{1}{{25}}\)     A1 

\(g(x) \in \mathbb{R},{\text{ }}g(x) \geqslant 0\)     A1

[2 marks]

\(f \circ g(x) = \frac{{2{{\left( {\frac{{3x - 4}}{{10}}} \right)}^2} + 3}}{{75}}\)     M1A1

\( = \frac{{\frac{{2(9{x^2} - 24x + 16)}}{{100}} + 3}}{{75}}\)     (A1)

\( = \frac{{9{x^2} - 24x + 166}}{{3750}}\)     A1

[4 marks]

(i)     METHOD 1

\(y = \frac{{2{x^2} + 3}}{{75}}\)

\({x^2} = \frac{{75y - 3}}{2}\)     M1

\(x = \sqrt {\frac{{75y - 3}}{2}} \)     (A1)

\( \Rightarrow {f^{ - 1}}(x) = \sqrt {\frac{{75x - 3}}{2}} \)     A1  

Note: Accept ± in line 3 for the (A1) but not in line 4 for the A1. 

Award the A1 only if written in the form \({f^{ - 1}}(x) = \) .

METHOD 2

\(y = \frac{{2{x^2} + 3}}{{75}}\)

\(x = \frac{{2{y^2} + 3}}{{75}}\)     M1

\(y = \sqrt {\frac{{75x - 3}}{2}} \)     (A1)

\( \Rightarrow {f^{ - 1}}(x) = \sqrt {\frac{{75x - 3}}{2}} \)     A1  

Note: Accept ± in line 3 for the (A1) but not in line 4 for the A1. 

Award the A1 only if written in the form \({f^{ - 1}}(x) = \) .

(ii)     domain: \(x \geqslant \frac{1}{{25}}\) ; range: \({f^{ - 1}}(x) \geqslant 0\)     A1

[4 marks]

probabilities from \(f(x)\) :

     A2

Note: Award A1 for one error, A0 otherwise.

probabilities from \(g(x)\) :

     A2

Note: Award A1 for one error, A0 otherwise. 

only in the case of \(f(x)\) does \(\sum {P(X = x) = 1} \) , hence only \(f(x)\) can be used as a probability mass function     A2

[6 marks]

\(E(x) = \sum {x \cdot {\text{P}}(X = x)} \)     M1

\( = \frac{5}{{75}} + \frac{{22}}{{75}} + \frac{{63}}{{75}} + \frac{{140}}{{75}} = \frac{{230}}{{75}}\left( { = \frac{{46}}{{15}}} \right)\)     A1

[2 marks]

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

Sketch the graphs on the same set of axes.

Given that the graphs enclose a region of area 18 square units, find the value of b.

graphs sketched correctly (condone missing b)     A1A1

[2 marks]

\(\frac{{{b^2}}}{2} = 18\)     (M1)A1

\(b = 6\)     A1

[3 marks]

(i)     Find \(\left( {g \circ f} \right)\left( x \right)\) and write down the domain of the function.

(ii)     Find \(\left( {f \circ g} \right)\left( x \right)\) and write down the domain of the function.

Find the coordinates of the point where the graph of \(y = f(x)\) and the graph of \(y = \left( {{g^{ - 1}} \circ f \circ g} \right)(x)\) intersect.

(i)     \(\left( {g \circ f} \right)\left( x \right) = \frac{1}{{2x + 3}}\), \(x \ne - \frac{3}{2}\) (or equivalent)     A1

(ii)     \(\left( {f \circ g} \right)\left( x \right) = \frac{2}{x} + 3\), \(x \ne 0\) (or equivalent)     A1

[2 marks]

EITHER

\(f(x) = \left( {{g^{ - 1}} \circ f \circ g} \right)(x) \Rightarrow \left( {f \circ g} \right)\left( x \right)\)     (M1)

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

OR

\(\left( {{g^{ - 1}} \circ f \circ g} \right)(x) = \frac{1}{{\frac{2}{x} + 3}}\)     A1

\(2x + 3 = \frac{1}{{\frac{2}{x} + 3}}\)     M1

THEN

\(6{x^2} + 12x + 6 = 0\) (or equivalent)     A1

\(x = - 1\), \(y = 1\) (coordinates are (−1, 1) )     A1

[4 marks]

Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.

Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.

Determine whether \({f_n}\) is an odd or even function, justifying your answer.

By using mathematical induction, prove that

\({f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}\) where \(m \in \mathbb{Z}\).

Hence or otherwise, find an expression for the derivative of \({f_n}(x)\) with respect to \(x\).

Show that, for \(n > 1\), the equation of the tangent to the curve \(y = {f_n}(x)\) at \(x = \frac{\pi }{4}\) is \(4x - 2y - \pi  = 0\).

even function     A1

since \(\cos kx = \cos ( - kx)\) and \({f_n}(x)\) is a product of even functions     R1

OR

even function     A1

since \((\cos 2x)(\cos 4x) \ldots  = \left( {\cos ( - 2x)} \right)\left( {\cos ( - 4x)} \right) \ldots \)     R1

Note:     Do not award A0R1.

[2 marks]

consider the case \(n = 1\)

\(\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x\)     M1

hence true for \(n = 1\)     R1

assume true for \(n = k\), ie, \((\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\)     M1

Note:     Do not award M1 for “let \(n = k\)” or “assume \(n = k\)” or equivalent.

consider \(n = k + 1\):

\({f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)\)     (M1)

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

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

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

so \(n = 1\) true and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

attempt to use \(f’ = \frac{{vu' - uv'}}{{{v^2}}}\) (or correct product rule)     M1

\({f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) - (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}\)     A1A1

Note:     Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) - \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}\)     (M1)(A1)

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}\)     (A1)

\( = 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n - 1}}\pi )\)     A1

\({f’_n}\left( {\frac{\pi }{4}} \right) = 2\)     A1

\({f_n}\left( {\frac{\pi }{4}} \right) = 0\)     A1

Note:     This A mark is independent from the previous marks.

\(y = 2\left( {x - \frac{\pi }{4}} \right)\)     M1A1

\(4x - 2y - \pi  = 0\)     AG

[8 marks]

A function f is defined by \(f(x) = \frac{{2x - 3}}{{x - 1}},{\text{ }}x \ne 1\).

(a)     Find an expression for \({f^{ - 1}}(x)\).

(b)     Solve the equation \(\left| {{f^{ - 1}}(x)} \right| = 1 + {f^{ - 1}}(x)\).

(a)     Note: Interchange of variables may take place at any stage.

for the inverse, solve for x in

\(y = \frac{{2x - 3}}{{x - 1}}\)

\(y(x - 1) = 2x - 3\)     M1

\(yx - 2x = y - 3\)

\(x(y - 2) = y - 3\)     (A1)

\(x = \frac{{y - 3}}{{y - 2}}\)

\( \Rightarrow {f^{ - 1}}(x) = \frac{{x - 3}}{{x - 2}}\,\,\,\,\,(x \ne 2)\)     A1

Note: Do not award final A1 unless written in the form \({f^{ - 1}}(x) =  \ldots \)

(b)     \( \pm {f^{ - 1}}(x) = 1 + {f^{ - 1}}(x)\) leads to

\(2\frac{{x - 3}}{{x - 2}} = - 1\)     (M1)

\(x = \frac{8}{3}\)     A1

[6 marks]

Many candidates gained the correct answer to part (a), although a significant minority left the answer in the form \(y = \ldots {\text{ or }}x =  \ldots \) rather than \({f^{ - 1}}(x) =  \ldots \). Only the better candidates were able to make significant progress in part (b).

Find the value of p and the value of q .

The graph of f(x) is translated 3 units in the positive direction parallel to the x-axis. Determine the equation of the new graph.

METHOD 1

\(f'(x) = q - 2x = 0\)     M1

\(f'(3) = q - 6 = 0\)

q = 6     A1

f(3) = p + 18 − 9 = 5     M1

p = −4     A1 

METHOD 2

\(f(x) = - {(x - 3)^2} + 5\)     M1A1

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

q = 6, p = −4     A1A1

[4 marks]

\(g(x) = - 4 + 6(x - 3) - {(x - 3)^2}{\text{ }}( = - 31 + 12x - {x^2})\)     M1A1

Note: Accept any alternative form which is correct.

Award M1A0 for a substitution of (x + 3) .

[2 marks]

In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values.

In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values.

Find an expression for \((f \circ f)(x)\) .

Let \({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}}\), where \(0 \leqslant x \leqslant 1\).

Use mathematical induction to show that for any \(n \in {\mathbb{Z}^ + }\)

\[\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\] .

Show that \({F_{ - n}}(x)\) is an expression for the inverse of \({F_n}\) .

(i)     State \({F_n}(0){\text{ and }}{F_n}(1)\) .

(ii)     Show that \({F_n}(x) < x\) , given 0 < x < 1, \(n \in {\mathbb{Z}^ + }\) .

(iii)     For \(n \in {\mathbb{Z}^ + }\) , let \({A_n}\) be the area of the region enclosed by the graph of \(F_n^{ - 1}\) , the x-axis and the line x = 1. Find the area \({B_n}\) of the region enclosed by \({F_n}\) and \(F_n^{ - 1}\) in terms of \({A_n}\) .

\((f \circ f)(x) = f\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{2 - \frac{x}{{2 - x}}}}\)     M1A1

\((f \circ f)(x) = \frac{x}{{4 - 3x}}\)     A1

[3 marks]

\(P(n):\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\)

\(P(1):{\text{ }}f(x) = {F_1}(x)\)

\(LHS = f(x) = \frac{x}{{2 - x}}{\text{ and }}RHS = {F_1}(x) = \frac{x}{{{2^1} - ({2^1} - 1)x}} = \frac{x}{{2 - x}}\)     A1A1

\(\therefore P(1){\text{ true}}\)

assume that P(k) is true, i.e., \(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k times}}}(x) = {F_k}(x)\)     M1

consider \(P(k + 1)\)

EITHER

\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = f\left( {{F_k}(x)} \right)\)     (M1)

\( = f\left( {\frac{x}{{{2^k} - ({2^k} - 1)x}}} \right) = \frac{{\frac{x}{{{2^k} - ({2^k} - 1)x}}}}{{2 - \frac{x}{{{2^k} - ({2^k} - 1)x}}}}\)     A1

\( = \frac{x}{{2\left( {{2^k} - ({2^k} - 1)x} \right) - x}} = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 2)x - x}}\)     A1

OR

\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = {F_k}(f(x))\)     (M1)

\( = {F_k}\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{{2^k} - ({2^k} - 1)\frac{x}{{2 - x}}}}\)     A1

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

THEN

\( = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 1)x}} = {F_{k + 1}}(x)\)     A1

P(k) true implies P(k + 1) true, P(1) true so P(n) true for all \(n \in {\mathbb{Z}^ + }\)     R1

[8 marks]

METHOD 1

\(x = \frac{y}{{{2^n} - ({2^n} - 1)y}} \Rightarrow {2^n}x - ({2^n} - 1)xy = y\)     M1A1

\( \Rightarrow {2^n}x = \left( {({2^n} - 1)x + 1} \right)y \Rightarrow y = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\)     A1

\(F_n^{ - 1}(x) = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\)     A1

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

\(F_n^{ - 1}(x) = \frac{x}{{(1 - {2^{ - n}})x + {2^{ - n}}}}\)     A1

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

METHOD 2

attempt \({F_{ - n}}\left( {{F_n}(x)} \right)\)     M1

\( = {F_{ - n}}\left( {\frac{x}{{{2^n} - ({2^n} - 1)x}}} \right) = \frac{{\frac{x}{{{2^n} - ({2^n} - 1)x}}}}{{{2^{ - n}} - ({2^{ - n}} - 1)\frac{x}{{{2^n} - ({2^n} - 1)x}}}}\)     A1A1

\( = \frac{x}{{{2^{ - n}}({2^n} - ({2^n} - 1)x) - ({2^{ - n}} - 1)x}}\)     A1A1

Note: Award A1 marks for numerators and denominators.

\( = \frac{x}{1} = x\)     A1AG

METHOD 3

attempt \({F_n}\left( {{F_{ - n}}(x)} \right)\)     M1

\( = {F_n}\left( {\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}} \right) = \frac{{\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}{{{2^n} - ({2^n} - 1)\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}\)     A1A1

\( = \frac{x}{{{2^n}({2^{ - n}} - ({2^{ - n}} - 1)x) - ({2^n} - 1)x}}\)     A1A1

Note: Award A1 marks for numerators and denominators.

\( = \frac{x}{1} = x\)     A1AG

[6 marks]

(i)     \({F_n}(0) = 0,{\text{ }}{F_n}(1) = 1\)     A1

(ii)     METHOD 1

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

\( > 0{\text{ if }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\)     A1

so \({2^n} - ({2^n} - 1)x > 1{\text{ and }}{F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < \frac{x}{1}( < x)\)     R1

\({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < x{\text{ for }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\)     AG

METHOD 2

\(\frac{x}{{{2^n} - ({2^n} - 1)x}} < x \Leftrightarrow {2^n} - ({2^n} - 1)x > 1\)     (M1)

\( \Leftrightarrow ({2^n} - 1)x < {2^n} - 1\)     A1

\( \Leftrightarrow x < \frac{{{2^n} - 1}}{{{2^n} - 1}} = 1\) true in the interval \(\left] {0,{\text{ }}1} \right[\)     R1

(iii)     \({B_n} = 2\left( {{A_n} - \frac{1}{2}} \right){\text{ }}( = 2{A_n} - 1)\)     (M1)A1

[6 marks]

Part b) was often answered well, and candidates were well prepared in this session for this type of question. Candidates still need to take care when showing explicitly that P(1) is true, and some are still writing ‘Let n = k’ which gains no marks. The inductive step was often well argued, and given in clear detail, though the final inductive reasoning step was incorrect, or appeared rushed, even from the better candidates. ‘True for n =1, n = k and n = k + 1’ is still disappointingly seen, as were some even more unconvincing variations.

Part c) was again very well answered by the majority. A few weaker candidates attempted to find an inverse for the individual case n = 1 , but gained no credit for this.

Part d) was not at all well understood, with virtually no candidates able to tie together the hints given by connecting the different parts of the question. Rash, and often thoughtless attempts were made at each part, though by this stage some seemed to be struggling through lack of time. The inequality part of the question tended to be ‘fudged’, with arguments seen by examiners being largely unconvincing and lacking clarity. A tiny number of candidates provided the correct answer to the final part, though a surprising number persisted with what should have been recognised as fruitless working – usually in the form of long-winded integration attempts.

State the set of values of \(a\) for which the function \(x \mapsto {\log _a}x\) exists, for all \(x \in {\mathbb{R}^ + }\).

Given that \({\log _x}y = 4{\log _y}x\), find all the possible expressions of \(y\) as a function of \(x\).

\(a > 0\)     A1

\(a \ne 0\)     A1

[2 marks]

METHOD 1

\({\log _x}y = \frac{{\ln y}}{{\ln x}}\) and \({\log _y}x = \frac{{\ln x}}{{\ln y}}\)     M1A1

Note:     Use of any base is permissible here, not just “e”.

\({\left( {\frac{{\ln y}}{{\ln x}}} \right)^2} = 4\)     A1

\(\ln y =  \pm 2\ln x\)     A1

\(y = {x^2}\;\;\;\)or\(\;\;\;\frac{1}{{{x^2}}}\)     A1A1

METHOD 2

\({\log _y}x = \frac{{{{\log }_x}x}}{{{{\log }_x}y}} = \frac{1}{{{{\log }_x}y}}\)     M1A1

\({({\log _x}y)^2} = 4\)     A1

\({\log _x}y =  \pm 2\)     A1

\(y = {x^2}\;\;\;\)or\(\;\;\;y = \frac{1}{{{x^2}}}\)     A1A1

Note: The final two A marks are independent of the one coming before.

[6 marks]

Total [8 marks]

When the function \(q(x) = {x^3} + k{x^2} - 7x + 3\) is divided by (x + 1) the remainder is seven times the remainder that is found when the function is divided by (x + 2) .

Find the value of k .

\(q( - 1) = k + 9\)     M1A1

\(q( - 2) = 4k + 9\)     A1

\(k + 9 = 7(4k + 9)\)     M1

\(k = - 2\)     A1

Notes: The first M1 is for one substitution and the consequent equations.

Accept expressions for \(q( - 1)\) and \(q( - 2)\) that are not simplified.

[5 marks] 

Most candidates were able to access this question although the number who used either synthetic division or long division was surprising as this often lead to difficulty and errors. The most common error was in applying the factor of 7 to the wrong side of the equation. It was also disappointing the number of students who made simple algebraic errors late in the question.

Solve \({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}\).

\({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)\)

EITHER

\({\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}\)     M1

\( = \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}\)     A1

OR

\(\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)\)     M1A1

THEN

\({\text{ln}}\,x = 2\,{\text{ln}}\,2\) or \( - {\text{ln}}\,2\)     A1

\( \Rightarrow x = 4\) or \(x = \frac{1}{2}\)       (M1)A1   

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is \(\frac{1}{2} < x < 4\)     A1

[6 marks]

The roots of a quadratic equation \(2{x^2} + 4x - 1 = 0\) are \(\alpha \) and \(\beta \).

Without solving the equation,

(a)     find the value of \({\alpha ^2} + {\beta ^2}\);

(b)     find a quadratic equation with roots \({\alpha ^2}\) and \({\beta ^2}\).

(a) using the formulae for the sum and product of roots:

\(\alpha  + \beta  =  - 2\)     A1

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

\({\alpha ^2} + {\beta ^2} = {(\alpha  + \beta )^2} - 2\alpha \beta \)     M1

\( = {( - 2)^2} - 2\left( { - \frac{1}{2}} \right)\)

\( = 5\)     A1

Note:     Award M0 for attempt to solve quadratic equation.

[4 marks]

(b)     \((x - {\alpha ^2})(x - {\beta ^2}) = {x^2} - ({\alpha ^2} + {\beta ^2})x + {\alpha ^2}{\beta ^2}\)     M1

\({x^2} - 5x + {\left( { - \frac{1}{2}} \right)^2} = 0\)     A1

\({x^2} - 5x + \frac{1}{4} = 0\)

Note:     Final answer must be an equation. Accept alternative correct forms.

[2 marks]

Total [6 marks]

When the polynomial \(3{x^3} + ax + b\) is divided by \((x - 2)\), the remainder is 2, and when divided by \((x + 1)\), it is 5. Find the value of a and the value of b.

\(P(2) = 24 + 2a + b = 2,{\text{ }}P( - 1) =  - 3 - a + b = 5\)     M1A1A1

\((2a + b =  - 22,{\text{ }} - a + b = 8)\)

Note:     Award M1 for substitution of 2 or −1 and equating to remainder, A1 for each correct equation.

attempt to solve simultaneously     M1

\(a =  - 10,{\text{ }}b =  - 2\)     A1

[5 marks]

Sketch on the same axes the curve \(y = \left| {\frac{7}{{x - 4}}} \right|\) and the line \(y = x + 2\), clearly indicating any axes intercepts and any asymptotes.

Find the exact solutions to the equation \(x + 2 = \left| {\frac{7}{{x - 4}}} \right|\).

A1 for vertical asymptote and for the \(y\)-intercept \(\frac{7}{4}\)

A1 for general shape of \(y = \left| {\frac{7}{{x - 4}}} \right|\) including the \(x\)-axis as asymptote

A1 for straight line with \(y\)-intercept 2 and \(x\)-intercept of \( - 2\)     A1A1A1

[3 marks]

METHOD 1

for \(x > 4\)

\((x + 2)(x - 4) = 7\)     (M1)

\({x^2} - 2x - 8 = 7 \Rightarrow {x^2} - 2x - 15 = 0\)

\((x - 5)(x + 3) = 0\)

\(({\text{as }}x > 4{\text{ then}}){\text{ }}x = 5\)     A1

Note:     Award A0 if \(x =  - 3\) is also given as a solution.

for \(x < 4\)

\((x + 2)(x - 4) =  - 7\)     M1

\( \Rightarrow {x^2} - 2x - 1 = 0\)

\(x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2 \)     (M1)A1

Note:     Second M1 is dependent on first M1.

[5 marks]

METHOD 2

\({(x + 2)^2} = \frac{{49}}{{{{(x - 4)}^2}}}\)     M1

\({x^4} - 4{x^3} - 12{x^2} + 32x + 15 = 0\)     A1

\((x + 3)(x - 5)({x^2} - 2x - 1) = 0\)

\(x = 5\)     A1

Note:     Award A0 if \(x =  - 3\) is also given as a solution.

\(x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2 \)     (M1)A1

[5 marks]

Though generally well done, some candidates lost marks unnecessarily by not heeding the instruction to clearly indicate the axes intercepts and asymptotes.

Though this was generally well done, quite a few of the candidates failed to use the graph drawn in part (a) to discount one of the solutions obtained in part (b).

Sketch the graph of \(y = {f^{ - 1}}(x)\) on the same axes.

State the range of \({f^{ - 1}}\).

Given that \(f(x) = \ln (ax + b),{\text{ }}x > 1\), find the value of \(a\) and the value of \(b\).

(a)     

shape with y-axis intercept (0, 4)     A1

Note:     Accept curve with an asymptote at \(x = 1\) suggested.

correct asymptote \(y = 1\)     A1

[2 marks]

range is \({f^{ - 1}}(x) > 1{\text{ (or }}\left] {1,{\text{ }}\infty } \right[)\)     A1

Note:     Also accept \(\left] {1,{\text{ 10}}} \right]\) or \(\left] {1,{\text{ 10}}} \right[\).

Note:     Do not allow follow through from incorrect asymptote in (a).

[1 mark]

\((4,{\text{ }}0) \Rightarrow \ln (4a + b) = 0\)     M1

\( \Rightarrow 4a + b = 1\)     A1

asymptote at \(x = 1 \Rightarrow a + b = 0\)     M1

\( \Rightarrow a = \frac{1}{3},{\text{ }}b =  - \frac{1}{3}\)     A1

[4 marks]

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

Find the set of values of x for which \(\left| {x - 1} \right| > \left| {2x - 1} \right|\).

EITHER

\(\left| {x - 1} \right| > \left| {2x - 1} \right| \Rightarrow {(x - 1)^2} > {(2x - 1)^2}\)     M1

\({x^2} - 2x + 1 > 4{x^2} - 4x + 1\)

\(3{x^2} - 2x < 0\)     A1

\(0 < x < \frac{2}{3}\)     A1A1     N2

Note: Award A1A0 for incorrect inequality signs.

OR

\(\left| {x - 1} \right| > \left| {2x - 1} \right|\)

\(x - 1 = 2x - 1\)     \(x - 1 = 1 - 2x\)     M1A1

\( - x = 0\)     \(3x = 2\)

\(x = 0\)     \(x = \frac{2}{3}\)

Note: Award M1 for any attempt to find a critical value. If graphical methods are used, award M1 for correct graphs, A1 for correct values of x.

\(0 < x < \frac{2}{3}\)     A1A1     N2

Note: Award A1A0 for incorrect inequality signs.

[4 marks]

This question turned out to be more difficult than expected. Candidates who squared both sides or drew a graph generally gave better solutions than those who relied on performing algebraic operations on terms involving modulus signs.

Let \(f(x) = \frac{4}{{x + 2}},{\text{ }}x \ne - 2{\text{ and }}g(x) = x - 1\).

If \(h = g \circ f\) , find

(a)     h(x) ;

(b)     \({h^{ - 1}}(x)\) , where \({h^{ - 1}}\) is the inverse of h.

(a)     \(h(x) = g\left( {\frac{4}{{x + 2}}} \right)\)     (M1)

\( = \frac{4}{{x + 2}} - 1\,\,\,\,\,\left( { = \frac{{2 - x}}{{2 + x}}} \right)\)     A1

(b)     METHOD 1

\(x = \frac{4}{{y + 2}} - 1\,\,\,\,\,\)(interchanging x and y)     M1

Attempting to solve for y     M1

\((y + 2)(x + 1) = 4\,\,\,\,\,\left( {y + 2 = \frac{4}{{x + 1}}} \right)\)     (A1)

\({h^{ - 1}}(x) = \frac{4}{{x + 1}} - 2\,\,\,\,\,(x \ne - 1)\)     A1     N1

METHOD 2

\(x = \frac{{2 - y}}{{2 + y}}\,\,\,\,\,\)(interchanging x and y)     M1

Attempting to solve for y     M1

\(xy + y = 2 - 2x\,\,\,\,\,\left( {y(x + 1) = 2(1 - x)} \right)\)     (A1)

\({h^{ - 1}}(x) = \frac{{2(1 - x)}}{{x + 1}}\,\,\,\,\,(x \ne - 1)\)     A1     N1

Note: In either METHOD 1 or METHOD 2 rearranging first and interchanging afterwards is equally acceptable.

[6 marks]

This question was generally well done, with very few candidates calculating \(f \circ g\) rather than \(g \circ f\).

Show that \(f(x) > 1\) for all x > 0.

Solve the equation \(f(x) = 4\).

EITHER

\(f(x) - 1 = \frac{{1 + {3^{ - x}}}}{{{3^x} - {3^{ - x}}}}\)     M1A1

> 0 as both numerator and denominator are positive     R1

OR

\({3^x} + 1 > {3^x} > {3^x} - {3^{ - x}}\)     M1A1

Note: Accept a convincing valid argument the numerator is greater than the denominator.

numerator and denominator are positive     R1

hence \(f(x) > 1\)     AG

[3 marks]

one line equation to solve, for example, \(4({3^x} - {3^{ - x}}) = {3^x} + 1\), or equivalent     A1

\(\left( {3{y^2} - y - 4 = 0} \right)\)

attempt to solve a three-term equation     M1

obtain \(y = \frac{4}{3}\)     A1

\(x = {\log _3}\left( {\frac{4}{3}} \right)\) or equivalent     A1

Note: Award A0 if an extra solution for x is given.

[4 marks]

(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for \(x > 0\). Candidates were more successful with part (b) than with part (a).

(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for \(x > 0\). Candidates were more successful with part (b) than with part (a).

(i)     Express each of the complex numbers \({z_1} = \sqrt 3  + {\text{i, }}{z_2} = - \sqrt 3  + {\text{i}}\) and \({z_3} = - 2{\text{i}}\) in modulus-argument form.

(ii)     Hence show that the points in the complex plane representing \({z_1}\), \({z_2}\) and \({z_3}\) form the vertices of an equilateral triangle.

(iii)     Show that \({\text{z}}_1^{3n} + z_2^{3n} = 2z_3^{3n}\) where \(n \in \mathbb{N}\).

(i)     State the solutions of the equation \({z^7} = 1\) for \(z \in \mathbb{C}\), giving them in modulus-argument form.

(ii)     If w is the solution to \({z^7} = 1\) with least positive argument, determine the argument of 1 + w. Express your answer in terms of \(\pi \).

(iii)     Show that \({z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) is a factor of the polynomial \({z^7} - 1\). State the two other quadratic factors with real coefficients.

(i)     \({z_1} = 2{\text{cis}}\left( {\frac{\pi }{6}} \right),{\text{ }}{z_2} = 2{\text{cis}}\left( {\frac{{5\pi }}{6}} \right),{\text{ }}{z_3} = 2{\text{cis}}\left( { - \frac{\pi }{2}} \right){\text{ or }}2{\text{cis}}\left( {\frac{{3\pi }}{2}} \right)\)     A1A1A1

Note: Accept modulus and argument given separately, or the use of exponential (Euler) form.

Note: Accept arguments given in rational degrees, except where exponential form is used.

(ii)     the points lie on a circle of radius 2 centre the origin     A1

differences are all \(\frac{{2\pi }}{3}(\bmod 2\pi )\)     A1

\( \Rightarrow \) points equally spaced \( \Rightarrow \) triangle is equilateral     R1AG

Note: Accept an approach based on a clearly marked diagram.

(iii)     \({\text{z}}_1^{3n} + z_2^{3n} = {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right) + {2^{3n}}{\text{cis}}\left( {\frac{{5n\pi }}{2}} \right)\)     M1

\( = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\)     A1

\(2z_3^{3n} = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{9n\pi }}{2}} \right) = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\)     A1AG

[9 marks]

(i)     attempt to obtain seven solutions in modulus argument form     M1

\(z = {\text{cis}}\left( {\frac{{2k\pi }}{7}} \right),{\text{ }}k = 0,{\text{ }}1 \ldots 6\)     A1

(ii)     w has argument \(\frac{{2\pi }}{7}\) and 1 + w has argument \(\phi \),

then \(\tan (\phi ) = \frac{{\sin \left( {\frac{{2\pi }}{7}} \right)}}{{1 + \cos \left( {\frac{{2\pi }}{7}} \right)}}\)     M1

\( = \frac{{2\sin \left( {\frac{\pi }{7}} \right)\cos \left( {\frac{\pi }{7}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{7}} \right)}}\)     A1

\( = \tan \left( {\frac{\pi }{7}} \right) \Rightarrow \phi  = \frac{\pi }{7}\)     A1

Note: Accept alternative approaches.

(iii)     since roots occur in conjugate pairs,     (R1)

\({z^7} - 1\) has a quadratic factor \(\left( {z - {\text{cis}}\left( {\frac{{2\pi }}{7}} \right)} \right) \times \left( {z - {\text{cis}}\left( { - \frac{{2\pi }}{7}} \right)} \right)\)     A1

\( = {z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\)     AG

other quadratic factors are \({z^2} - 2z\cos \left( {\frac{{4\pi }}{7}} \right) + 1\)     A1

and \({z^2} - 2z\cos \left( {\frac{{6\pi }}{7}} \right) + 1\)     A1

[9 marks]

(i) A disappointingly large number of candidates were unable to give the correct arguments for the three complex numbers. Such errors undermined their efforts to tackle parts (ii) and (iii).

Many candidates were successful in part (i), but failed to capitalise on that – in particular, few used the fact that roots of \({z^7} - 1 = 0\) come in complex conjugate pairs.

Write down the exact values of \(\left| {{z_1}} \right|\) and \(\arg ({z_2})\).

Find the minimum value of \(\left| {{z_1} + \alpha{z_2}} \right|\), where \(\alpha \in \mathbb{R}\).

\(\left| {{z_1}} \right| = \sqrt {10} ;{\text{ }}\arg ({z_2}) = - \frac{{3\pi }}{4}{\text{ }}\left( {{\text{accept }}\frac{{5\pi }}{4}} \right)\)     A1A1

[2 marks]

\(\left| {{z_1} + \alpha{z_2}} \right| = \sqrt {{{(1 - \alpha )}^2} + {{(3 - \alpha )}^2}} \) or the squared modulus     (M1)(A1)

attempt to minimise \(2{\alpha ^2} - 8\alpha  + 10\) or their quadratic or its half or its square root     M1

obtain \(\alpha  = 2\) at minimum     (A1)

state \(\sqrt 2 \) as final answer     A1

[5 marks]

Disappointingly, few candidates obtained the correct argument for the second complex number, mechanically using arctan(1) but not thinking about the position of the number in the complex plane.

Most candidates obtained the correct quadratic or its square root, but few knew how to set about minimising it.

The same remainder is found when \(2{x^3} + k{x^2} + 6x + 32\) and \({x^4} - 6{x^2} - {k^2}x + 9\) are divided by \(x + 1\) . Find the possible values of k .

let \(f(x) = 2{x^3} + k{x^2} + 6x + 32\)

let \(g(x) = {x^4} - 6{x^2} - {k^2}x + 9\)

\(f( - 1) =  - 2 + k - 6 + 32( = 24 + k)\)     A1

\(g( - 1) = 1 - 6 + {k^2} + 9( = 4 + {k^2})\)     A1

\( \Rightarrow 24 + k = 4 + {k^2}\)     M1

\( \Rightarrow {k^2} - k - 20 = 0\)

\( \Rightarrow (k - 5)(k + 4) = 0\)     (M1)

\( \Rightarrow k = 5,\, - 4\)     A1A1

[6 marks]

Candidates who used the remainder theorem usually went on to find the two possible values of k. Some candidates, however, attempted to find the remainders using long division. While this is a valid method, the algebra involved proved to be too difficult for most of these candidates. 

When \(3{x^5} - ax + b\) is divided by x −1 and x +1 the remainders are equal. Given that a , \(b \in \mathbb{R}\) , find

(a)     the value of a ;

(b)     the set of values of b .

(a)     \(f(1) = 3 - a + b\)     (A1)

\(f( - 1) = - 3 + a + b\)     (A1)

\(3 - a + b = - 3 + a + b\)     M1

\(2a = 6\)

\(a = 3\)     A1     N4

(b)     b is any real number     A1

[5 marks]

Many candidates answered part (a) successfully. For part (b), some candidates did not consider that the entire set of real numbers was asked for.

(a)     Express the quadratic \(3{x^2} - 6x + 5\) in the form \(a{(x + b)^2} + c\), where a, b, c \( \in \mathbb{Z}\).

(b)     Describe a sequence of transformations that transforms the graph of \(y = {x^2}\) to the graph of \(y = 3{x^2} - 6x + 5\).

(a)     attempt at completing the square     (M1)

\(3{x^2} - 6x + 5 = 3({x^2} - 2x) + 5 = 3{(x - 1)^2} - 1 + 5\)     (A1)

\( = 3{(x - 1)^2} + 2\)     A1

\((a = 3,{\text{ }}b = - 1,{\text{ }}c = 2)\)

(b)     definition of suitable basic transformations:

\({{\text{T}}_1} = \) stretch in y direction scale factor 3     A1

\({{\text{T}}_2} = \) translation \(\left( {\begin{array}{*{20}{c}}
  1 \\
  0
\end{array}} \right)\)     A1

\({{\text{T}}_3} = \) translation \(\left( {\begin{array}{*{20}{c}}
  0 \\
  2
\end{array}} \right)\)     A1

[6 marks]

There were fewer correct solutions to this question than might be expected with a significant minority of candidates unable to complete the square successfully and a number of candidates unable to describe the transformations. A minority of candidates knew the correct terminology for the transformations and this potentially highlights the need for teachers to teach students appropriate terminology.

Find an expression for \(g(x)\).

State the equations of the asymptotes of the graph of \(g\).

\(g(x) = \frac{1}{{x + 3}} + 1\)     A1A1

Note:     Award A1 for \(x + 3\) in the denominator and A1 for the “\( + 1\)”.

[2 marks]

\(x =  - 3\)     A1

\(y = 1\)     A1

[2 marks]

Total [4 marks]

This question was generally well done. A few candidates made a sign error for the horizontal translation. A few candidates expressed the required equations for the asymptotes as ‘inequalities’, which received no marks.

This question was generally well done. A few candidates made a sign error for the horizontal translation. A few candidates expressed the required equations for the asymptotes as ‘inequalities’, which received no marks.

Factorize \({z^3} + 1\) into a linear and quadratic factor.

Let \(\gamma = \frac{{1 + {\text{i}}\sqrt 3 }}{2}\).

(i)     Show that \(\gamma \) is one of the cube roots of −1.

(ii)     Show that \({\gamma ^2} = \gamma - 1\).

(iii)     Hence find the value of \({(1 - \gamma )^6}\).

using the factor theorem z +1 is a factor     (M1)

\({z^3} + 1 = (z + 1)({z^2} - z + 1)\)     A1

[2 marks]

(i)     METHOD 1

\({z^3} = - 1 \Rightarrow {z^3} + 1 = (z + 1)({z^2} - z + 1) = 0\)     (M1)

solving \({z^2} - z + 1 = 0\)     M1

\(z = \frac{{1 \pm \sqrt {1 - 4} }}{2} = \frac{{1 \pm {\text{i}}\sqrt 3 }}{2}\)     A1

therefore one cube root of −1 is \(\gamma \)     AG

METHOD 2

\({\gamma ^2} = \left( {{{\frac{{1 + i\sqrt 3 }}{2}}^2}} \right) = \frac{{ - 1 + i\sqrt 3 }}{2}\)     M1A1

\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2} \times \frac{{1 + i\sqrt 3 }}{2} = \frac{{ - 1 - 3}}{4}\)     A1

= −1     AG

METHOD 3

\(\gamma  = \frac{{1 + i\sqrt 3 }}{2} = {e^{i\frac{\pi }{3}}}\)     M1A1

\({\gamma ^3} = {e^{i\pi }} = - 1\)     A1

(ii)     METHOD 1

as \(\gamma \) is a root of \({z^2} - z + 1 = 0\) then \({\gamma ^2} - \gamma + 1 = 0\)     M1R1

\(\therefore {\gamma ^2} = \gamma - 1\)     AG

Note: Award M1 for the use of \({z^2} - z + 1 = 0\) in any way.

Award R1 for a correct reasoned approach.

METHOD 2

\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2}\)     M1

\(\gamma - 1 = \frac{{1 + i\sqrt 3 }}{2} - 1 = \frac{{ - 1 + i\sqrt 3 }}{2}\)     A1

(iii)     METHOD 1

\({(1 - \gamma )^6} = {( - {\gamma ^2})^6}\)     (M1)

\( = {(\gamma )^{12}}\)     A1

\( = {({\gamma ^3})^4}\)     (M1)

\( = {( - 1)^4}\)

\( = 1\)     A1

METHOD 2

\({(1 - \gamma )^6}\)

\( = 1 - 6\gamma + 15{\gamma ^2} - 20{\gamma ^3} + 15{\gamma ^4} - 6{\gamma ^5} + {\gamma ^6}\)     M1A1

Note: Award M1 for attempt at binomial expansion.

use of any previous result e.g. \( = 1 - 6\gamma + 15{\gamma ^2} + 20 - 15\gamma  + 6{\gamma ^2} + 1\)     M1

= 1     A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.

[9 marks]

In part a) the factorisation was, on the whole, well done.

Part (b) was done well by most although using a substitution method rather than the result above. This used much m retime than was necessary but was successful. A number of candidates did not use the previous results in part (iii) and so seemed to not understand the use of the ‘hence’.

Given that the curve passes through the point \((a,{\text{ }}0)\), state the value of \(a\).

Use the substitution \(u = \ln x\) to find the area of the region \(R\).

(i)     Find the value of \({I_0}\).

(ii)     Prove that \({I_n} = \frac{1}{{\text{e}}} + n{I_{n - 1}},{\text{ }}n \in {\mathbb{Z}^ + }\).

(iii)     Hence find the value of \({I_1}\).

Find the volume of the solid formed when the region \(R\) is rotated through \(2\pi \) about the \(x\)-axis.

\(a = 1\)    A1

[1 mark]

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}\)    (A1)

\(\int {\frac{{{{(\ln x)}^2}}}{x}{\text{d}}x = \int {{u^2}{\text{d}}u} } \)    M1A1

area \( = \left[ {\frac{1}{3}{u^3}} \right]_0^1\) or \(\left[ {\frac{1}{3}{{(\ln x)}^3}} \right]_1^{\text{e}}\)     A1

\( = \frac{1}{3}\)    A1

[5 marks]

(i)     \({I_0} = \left[ { - \frac{1}{x}} \right]_1^{\text{e}}\)     (A1)

\( = 1 - \frac{1}{{\text{e}}}\)    A1

(ii)     use of integration by parts     M1

\({I_n} = \left[ { - \frac{1}{x}{{(\ln x)}^n}} \right]_1^{\text{e}} + \int_1^{\text{e}} {\frac{{n{{(\ln x)}^{n - 1}}}}{{{x^2}}}{\text{d}}x} \)    A1A1

\( =  - \frac{1}{{\text{e}}} + n{I_{n - 1}}\)    AG

Note:     If the substitution \(u = \ln x\) is used A1A1 can be awarded for \({I_n} = [ - {{\text{e}}^{ - u}}{u^n}]_0^1 + \int_0^1 n {{\text{e}}^{ - u}}{u^{n - 1}}{\text{d}}u\).

(iii)     \({I_1} =  - \frac{1}{{\text{e}}} + 1 \times {I_0}\)     (M1)

\( = 1 - \frac{2}{{\text{e}}}\)    A1

[7 marks]

(d)     volume \( = \pi \int_1^{\text{e}} {\frac{{{{(\ln x)}^4}}}{{{x^2}}}{\text{d}}x{\text{ }}( = \pi {I_4})} \)     (A1)

EITHER

\({I_4} =  - \frac{1}{{\text{e}}} + 4{I_3}\)    M1A1

\( =  - \frac{1}{{\text{e}}} + 4\left( { - \frac{1}{{\text{e}}} + 3{I_2}} \right)\)    M1

\( =  - \frac{5}{{\text{e}}} + 12{I_2} =  - \frac{5}{{\text{e}}} + 12\left( { - \frac{1}{{\text{e}}} + 2{I_1}} \right)\)

OR

using parts \(\int_1^{\text{e}} {\frac{{{{(\ln x)}^4}}}{{{x^2}}}{\text{d}}x =  - \frac{1}{{\text{e}}} + 4\int_1^{\text{e}} {\frac{{{{(\ln x)}^3}}}{{{x^2}}}{\text{d}}x} } \)     M1A1

\( =  - \frac{1}{{\text{e}}} + 4\left( { - \frac{1}{{\text{e}}} + 3\int_1^{\text{e}} {\frac{{{{(\ln x)}^2}}}{{{x^2}}}{\text{d}}x} } \right)\)    M1

THEN

\( =  - \frac{{17}}{{\text{e}}} + 24\left( {1 - \frac{2}{{\text{e}}}} \right) = 24 - \frac{{65}}{{\text{e}}}\)    A1

volume \( = \pi \left( {24 - \frac{{65}}{{\text{e}}}} \right)\)

[5 marks]

(a) and (b) were well done. Most candidates could integrate by substitution, though many did not change the limits during the substitution and, though they changed back to \(x\) at the end of their solution, under a different markscheme they might have lost marks for this in the intermediate stages.

(a) and (b) were well done. Most candidates could integrate by substitution, though many did not change the limits during the substitution and, though they changed back to \(x\) at the end of their solution, under a different markscheme they might have lost marks for this in the intermediate stages.

(c)(i) This part was well done by the candidates.

(c)(ii) This proved to be the part that was done by fewest candidates. Those who spotted that they should use integration by parts obtained the answer fairly easily.

(c)(iii) Many candidates displayed good exam technique in this question and obtained full marks without being able to do part (ii).

The same good exam technique was on show here as many students who failed to prove the expression in (c)(ii) were able to use it to obtain full marks in this question. A few candidates failed to remember correctly the formula for a volume of revolution.

Given that \(A{x^3} + B{x^2} + x + 6\) is exactly divisible by \((x +1)(x − 2)\), find the value of A and the value of B .

using the factor theorem or long division     (M1)
\( - A + B - 1 + 6 = 0 \Rightarrow A - B = 5\)     (A1)
\(8A + 4B + 2 + 6 = 0 \Rightarrow 2A + B = - 2\)     (A1)
\(3A = 3 \Rightarrow A = 1\)     (A1)
\(B = - 4\)     (A1)     (N3)

Note: Award M1A0A0A1A1 for using \((x - 3)\) as the third factor, without justification that the leading coefficient is 1.

[5 marks]

Most candidates attempted this question and it was the best done question on the paper with many fully correct answers. It was good to see a range of approaches used (mainly factor theorem or long division). A number of candidates assumed \((x - 3)\) was the missing factor without justification.

Determine whether f is even, odd or neither even nor odd.

Show that \(f''(0) = 0\) .

John states that, because \(f''(0) = 0\) , the graph of f has a point of inflexion at the point (0, 2) . Explain briefly whether John’s statement is correct or not.

\(f( - x) = 2\cos ( - x) + ( - x)\sin ( - x)\)     M1

\( = 2\cos x + x\sin x\,\,\,\,\,\left( { = f(x)} \right)\)     A1

therefore f is even     A1

[3 marks]

\(f'(x) = - 2\sin x + \sin x + x\cos x\,\,\,\,\,( = - \sin x + x\cos x)\)     A1

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

so \(f''(0) = 0\)     AG

[2 marks]

John’s statement is incorrect because

either; there is a stationary point at (0, 2) and since f is an even function and therefore symmetrical about the y-axis it must be a maximum or a minimum

or; \(f''(x)\) is even and therefore has the same sign either side of (0, 2)     R2

[2 marks]

Sketch the graph of \(y = \left| {\cos \left( {\frac{x}{4}} \right)} \right|\) for \(0 \leqslant x \leqslant 8\pi \).

Solve \(\left| {\cos \left( {\frac{x}{4}} \right)} \right| = \frac{1}{2}\) for \(0 \leqslant x \leqslant 8\pi \).

    A1A1

Note:     Award A1 for correct shape and A1 for correct domain and range.

[2 marks]

\(\left| {\cos \left( {\frac{x}{4}} \right)} \right| = \frac{1}{2}\)

\(x = \frac{{4\pi }}{3}\)     A1

attempting to find any other solutions     M1

Note:     Award (M1) if at least one of the other solutions is correct (in radians or degrees) or clear use of symmetry is seen.

\(x = 8\pi  - \frac{{4\pi }}{3} = \frac{{20 \pi }}{3}\)

\(x = 4\pi  - \frac{{4\pi }}{3} = \frac{{8\pi }}{3}\)

\(x = 4\pi  + \frac{{4\pi }}{3} = \frac{{16\pi }}{3}\)     A1

Note:     Award A1 for all other three solutions correct and no extra solutions.

Note:     If working in degrees, then max A0M1A0.

[3 marks]

The functions f and g are defined as:

\[f(x) = {{\text{e}}^{{x^2}}},{\text{ }}x \geqslant 0\]

\[g(x) = \frac{1}{{x + 3}},{\text{ }}x \ne - 3.\]

(a)     Find \(h(x){\text{ where }}h(x) = g \circ f(x)\) .

(b)     State the domain of \({h^{ - 1}}(x)\) .

(c)     Find \({h^{ - 1}}(x)\) .

(a)     \(h(x) = g \circ f(x) = \frac{1}{{{{\text{e}}^{{x^2}}} + 3}},{\text{ }}(x \geqslant 0)\)     (M1)A1

(b)     \(0 < x \leqslant \frac{1}{4}\)     A1A1

Note: Award A1 for limits and A1 for correct inequality signs.

(c)     \(y = \frac{1}{{{{\text{e}}^{{x^2}}} + 3}}\)

\(y{{\text{e}}^{{x^2}}} + 3y = 1\)     M1

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

\({x^2} = \ln \frac{{1 - 3y}}{y}\)     M1

\(x = \pm \sqrt {\ln \frac{{1 - 3y}}{y}} \)

\( \Rightarrow {h^{ - 1}}(x) = \sqrt {\ln \frac{{1 - 3x}}{x}} {\text{ }}\left( { = \sqrt {\ln \left( {\frac{1}{x} - 3} \right)} } \right)\)     A1

[8 marks]

Part (a) was correctly done by the vast majority of candidates. In contrast, only the very best students gave the correct answer to part (b). Part (c) was correctly started by a majority of candidates, but many did not realise that they needed to use logarithms and were careless about the use of notation

When \(f(x) = {x^4} + 3{x^3} + p{x^2} - 2x + q\) is divided by (x − 2) the remainder is 15, and (x + 3) is a factor of f(x) .

Find the values of p and q .

\(f(2) = 16 + 24 + 4p - 4 + q = 15\)     M1

\( \Rightarrow 4p + q = - 21\)     A1

\(f( - 3) = 81 - 81 + 9p + 6 + q = 0\)     M1

\( \Rightarrow 9p + q = - 6\)     A1

\( \Rightarrow p = 3{\text{ and }}q = - 33\)     A1A1     N0

[6 marks]

Most candidates made a meaningful attempt at this question. Weaker candidates often made arithmetic errors and a few candidates tried using long division, which also often resulted in arithmetic errors. Overall there were many fully correct solutions.

Solve the following equations:

(a)     \({\log _2}(x - 2) = {\log _4}({x^2} - 6x + 12)\);

(b)     \({x^{\ln x}} = {{\text{e}}^{{{(\ln x)}^3}}}\).

(a)     \({\log _2}(x - 2) = {\log _4}({x^2} - 6x + 12)\)

EITHER

\({\log _2}(x - 2) = \frac{{{{\log }_2}({x^2} - 6x + 12)}}{{{{\log }_2}4}}\)     M1

\(2{\log _2}(x - 2) = {\log _2}({x^2} - 6x + 12)\)

OR

\(\frac{{{{\log }_4}(x - 2)}}{{{{\log }_4}2}} = {\log _4}({x^2} - 6x + 12)\)     M1

\(2{\log _4}(x - 2) = {\log _4}({x^2} - 6x + 12)\)

THEN

\({(x - 2)^2} = {x^2} - 6x + 12\)     A1

\({x^2} - 4x + 4 = {x^2} - 6x + 12\)

\(x = 4\)     A1     N1

[3 marks]

(b)     \({x^{\ln x}} = {{\text{e}}^{{{(\ln x)}^3}}}\)

taking ln of both sides or writing \(x = {{\text{e}}^{\ln x}}\)     M1

\({(\ln x)^2} = {(\ln x)^3}\)     A1

\({(\ln x)^2}(\ln x - 1) = 0\)     (A1)

\(x = 1,{\text{ }}x = {\text{e}}\)     A1A1     N2

Note:     Award second (A1) only if factorisation seen or if two correct

solutions are seen.

[5 marks]

Total [8 marks]

Part a) was answered well, and a very large proportion of candidates displayed familiarity and confidence with this type of change-of base equation.

In part b), good candidates were able to solve this proficiently. A number obtained only one solution, either through observation or mistakenly cancelling a \(\ln x\) term. An incorrect solution \(x = {{\text{e}}^3}\) was somewhat prevalent amongst the weaker candidates.

Find the inverse function \({f^{ - 1}}\), stating its domain.

Express \(g\left( x \right)\) in the form \(A + \frac{B}{{x - 2}}\) where A, B are constants.

Sketch the graph of \(y = g\left( x \right)\). State the equations of any asymptotes and the coordinates of any intercepts with the axes.

The function \(h\) is defined by \(h\left( x \right) = \sqrt x \), for \(x\) ≥ 0.

State the domain and range of \(h \circ g\).

attempt to make \(x\) the subject of \(y = \frac{{ax + b}}{{cx + d}}\)      M1

\(y\left( {cx + d} \right) = ax + b\)      A1

\(x = \frac{{dy - b}}{{a - cy}}\)     A1

\({f^{ - 1}}\left( x \right) = \frac{{dx - b}}{{a - cx}}\)     A1

Note: Do not allow \(y = \) in place of \({f^{ - 1}}\left( x \right)\).

\(x \ne \frac{a}{c},\,\,\,\left( {x \in \mathbb{R}} \right)\)     A1

Note: The final A mark is independent.

[5 marks]

\(g\left( x \right) = 2 + \frac{1}{{x - 2}}\)     A1A1

[2 marks]

hyperbola shape, with single curves in second and fourth quadrants and third quadrant blank, including vertical asymptote \(x = 2\)     A1

horizontal asymptote \(y = 2\)     A1

intercepts \(\left( {\frac{3}{2},\,0} \right),\,\left( {0,\,\frac{3}{2}} \right)\)     A1

[3 marks]

the domain of \(h \circ g\) is \(x \leqslant \frac{3}{2},\,\,x > 2\)     A1A1

the range of \(h \circ g\) is \(y \geqslant 0,\,\,y \ne \sqrt 2 \)     A1A1

[4 marks]

Express \(4{x^2} - 4x + 5\) in the form \(a{(x - h)^2} + k\) where a, h, \(k \in \mathbb{Q}\).

The graph of \(y = {x^2}\) is transformed onto the graph of \(y = 4{x^2} - 4x + 5\). Describe a sequence of transformations that does this, making the order of transformations clear.

Sketch the graph of \(y = f(x)\).

Find the range of f.

By using a suitable substitution show that \(\int {f(x){\text{d}}x = \frac{1}{4}\int {\frac{1}{{{u^2} + 1}}{\text{d}}u} } \).

Prove that \(\int_1^{3.5} {\frac{1}{{4{x^2} - 4x + 5}}{\text{d}}x = \frac{\pi }{{16}}} \).

\(4{(x - 0.5)^2} + 4\)     A1A1

Note: A1 for two correct parameters, A2 for all three correct.

[2 marks]

translation \(\left( {\begin{array}{*{20}{c}}
  {0.5} \\
  0
\end{array}} \right)\) (allow “0.5 to the right”)     A1

stretch parallel to y-axis, scale factor 4 (allow vertical stretch or similar)     A1

translation \(\left( {\begin{array}{*{20}{c}}
  0 \\
  4
\end{array}} \right)\) (allow “4 up”)     A1

Note: All transformations must state magnitude and direction.

Note: First two transformations can be in either order.

It could be a stretch followed by a single translation of \(\left( {\begin{array}{*{20}{c}}
  {0.5} \\
  4
\end{array}} \right)\). If the vertical translation is before the stretch it is \(\left( {\begin{array}{*{20}{c}}
  0 \\
  1
\end{array}} \right)\).

[3 marks]

general shape (including asymptote and single maximum in first quadrant),     A1

intercept \(\left( {0,\frac{1}{5}} \right)\) or maximum \(\left( {\frac{1}{2},\frac{1}{4}} \right)\) shown     A1

[2 marks]

\(0 < f(x) \leqslant \frac{1}{4}\)     A1A1

Note: A1 for \( \leqslant \frac{1}{4}\), A1 for \(0 < \).

[2 marks]

let \(u = x - \frac{1}{2}\)     A1

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1\,\,\,\,\,{\text{(or d}}u = {\text{d}}x)\)     A1

\(\int {\frac{1}{{4{x^2} - 4x + 5}}{\text{d}}x = \int {\frac{1}{{4{{\left( {x - \frac{1}{2}} \right)}^2} + 4}}{\text{d}}x} } \)     A1

\(\int {\frac{1}{{4{u^2} + 4}}{\text{d}}u = \frac{1}{4}\int {\frac{1}{{{u^2} + 1}}{\text{d}}u} } \)     AG

Note: If following through an incorrect answer to part (a), do not award final A1 mark.

[3 marks]

\(\int_1^{3.5} {\frac{1}{{4{x^2} - 4x + 5}}{\text{d}}x = \frac{1}{4}\int_{0.5}^3 {\frac{1}{{{u^2} + 1}}{\text{d}}u} } \)     A1

Note: A1 for correct change of limits. Award also if they do not change limits but go back to x values when substituting the limit (even if there is an error in the integral).

\(\frac{1}{4}\left[ {\arctan (u)} \right]_{0.5}^3\)     (M1)

\(\frac{1}{4}\left( {\arctan (3) - \arctan \left( {\frac{1}{2}} \right)} \right)\)     A1

let the integral = I

\(\tan 4I = \tan \left( {\arctan (3) - \arctan \left( {\frac{1}{2}} \right)} \right)\)     M1

\(\frac{{3 - 0.5}}{{1 + 3 \times 0.5}} = \frac{{2.5}}{{2.5}} = 1\)     (M1)A1

\(4I = \frac{\pi }{4} \Rightarrow I = \frac{\pi }{{16}}\)     A1AG

[7 marks]

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e).

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e) but the following points caused some difficulties.

(b) Exam technique would have helped those candidates who could not get part (a) correct as any solution of the form given in the question could have led to full marks in part (b). Several candidates obtained expressions which were not of this form in (a) and so were unable to receive any marks in (b) Many missed the fact that if a vertical translation is performed before the vertical stretch it has a different magnitude to if it is done afterwards. Though on this occasion the markscheme was fairly flexible in the words it allowed to be used by candidates to describe the transformations it would be less risky to use the correct expressions.

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e) but the following points caused some difficulties.

(c) Generally the sketches were poor. The general rule for all sketch questions should be that any asymptotes or intercepts should be clearly labelled. Sketches do not need to be done on graph paper, but a ruler should be used, particularly when asymptotes are involved.

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e).

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e) but the following points caused some difficulties.

(e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to use the compound angle formula.

This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) – (e) but the following points caused some difficulties.

(e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to use the compound angle formula.

Express \(f(x)\) in the form \(A + \frac{B}{{x + 2}}\), where \(A\) and \(B \in \mathbb{Z}\).

Hence show that \(f'(x) > 0\) on D.

State the range of f.

(i)     Find an expression for \({f^{ - 1}}(x)\).

(ii)     Sketch the graph of \(y = f(x)\), showing the points of intersection with both axes.

(iii)     On the same diagram, sketch the graph of \(y = f'(x)\).

(i)     On a different diagram, sketch the graph of \(y = f(|x|)\) where \(x \in D\).

(ii)     Find all solutions of the equation \(f(|x|) = - \frac{1}{4}\).

by division or otherwise

\(f(x) = 2 - \frac{5}{{x + 2}}\)     A1A1

[2 marks]

\(f'(x) = \frac{5}{{{{(x + 2)}^2}}}\)     A1

> 0 as \({(x + 2)^2} > 0\) (on D)     R1AG

Note: Do not penalise candidates who use the original form of the function to compute its derivative.

[2 marks]

\(S = \left[ { - 3,\frac{3}{2}} \right]\)     A2

Note: Award A1A0 for the correct endpoints and an open interval.

[2 marks]

(i)     EITHER

rearrange \(y = f(x)\) to make x the subject     M1

obtain one-line equation, e.g. \(2x - 1 = xy + 2y\)     A1

\(x = \frac{{2y + 1}}{{2 - y}}\)     A1

OR

interchange x and y     M1

obtain one-line equation, e.g. \(2y - 1 = xy + 2x\)     A1

\(y = \frac{{2x + 1}}{{2 - x}}\)     A1

THEN

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

Note: Accept \(\frac{5}{{2 - x}} - 2\)

(ii), (iii)

     A1A1A1A1

[8 marks]

Note: Award A1 for correct shape of \(y = f(x)\).

Award A1 for x intercept \(\frac{1}{2}\) seen. Award A1 for y intercept \( - \frac{1}{2}\) seen.

Award A1 for the graph of \(y = {f^{ - 1}}(x)\) being the reflection of \(y = f(x)\) in the line \(y = x\). Candidates are not required to indicate the full domain, but \(y = f(x)\) should not be shown approaching \(x = - 2\). Candidates, in answering (iii), can FT on their sketch in (ii).

(i)

     A1A1A1

Note: A1 for correct sketch \(x > 0\), A1 for symmetry, A1 for correct domain (from –1 to +8).

Note: Candidates can FT on their sketch in (d)(ii).

(ii)     attempt to solve \(f(x) = - \frac{1}{4}\)     (M1)

obtain \(x = \frac{2}{9}\)     A1

use of symmetry or valid algebraic approach     (M1)

obtain \(x = - \frac{2}{9}\)     A1

[7 marks]

Generally well done.

In their answers to Part (b), most candidates found the derivative, but many assumed it was obviously positive.

Part (d)(i) Generally well done, but some candidates failed to label their final expression as \({f^{ - 1}}(x)\). Part (d)(ii) Marks were lost by candidates who failed to mark the intercepts with values.

Marks were also lost in this part and in part (e)(i) for graphs that went beyond the explicitly stated domain.

(i)     Find \({f^{ - 1}}(x)\).

(ii)     State the domain of \({f^{ - 1}}\).

The function \(g\) is defined as \(g(x) = \ln x,{\text{ }}x \in {\mathbb{R}^ + }\).

The graph of \(y = g(x)\) and the graph of \(y = {f^{ - 1}}(x)\) intersect at the point \(P\).

Find the coordinates of \(P\).

The graph of \(y = g(x)\) intersects the \(x\)-axis at the point \(Q\).

Show that the equation of the tangent \(T\) to the graph of \(y = g(x)\) at the point \(Q\) is \(y = x - 1\).

A region \(R\) is bounded by the graphs of \(y = g(x)\), the tangent \(T\) and the line \(x = {\text{e}}\).

Find the area of the region \(R\).

A region \(R\) is bounded by the graphs of \(y = g(x)\), the tangent \(T\) and the line \(x = {\text{e}}\).

(i)     Show that \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\).

(ii)     By replacing \(x\) with \(\frac{1}{x}\) in part (e)(i), show that \(\frac{{x - 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }\).

(i)     \(x = {{\text{e}}^{3y + 1}}\)     M1

Note:     The M1 is for switching variables and can be awarded at any stage.

Further marks do not rely on this mark being awarded.

taking the natural logarithm of both sides and attempting to transpose     M1

\(\left( {{f^{ - 1}}(x)} \right) = \frac{1}{3}(\ln x - 1)\)     A1

(ii)     \(x \in {\mathbb{R}^ + }\) or equivalent, for example \(x > 0\).     A1

[4 marks]

\(\ln x = \frac{1}{3}(\ln x - 1) \Rightarrow \ln x - \frac{1}{3}\ln x =  - \frac{1}{3}\) (or equivalent)     M1A1

\(\ln x =  - \frac{1}{2}\) (or equivalent)     A1

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

coordinates of \(P\) are \(\left( {{{\text{e}}^{ - \frac{1}{2}}},{\text{ }} - \frac{1}{2}} \right)\)     A1

[5 marks]

coordinates of \(Q\) are (\(1,{\rm{ }}0\)) seen anywhere     A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{x}\)     M1

at \({\text{Q, }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1\)     A1

\(y = x - 1\)     AG

[3 marks]

let the required area be \(A\)

\(A = \int_1^e {x - 1{\text{d}}x - \int_1^e {\ln x{\text{d}}x} } \)     M1

Note:     The M1 is for a difference of integrals. Condone absence of limits here.

attempting to use integration by parts to find \(\int {\ln x{\text{d}}x} \)     (M1)

\( = \left[ {\frac{{{x^2}}}{2} - x} \right]_1^{\text{e}} - [x\ln x - x]_1^{\text{e}}\)     A1A1

Note:     Award A1 for \(\frac{{{x^2}}}{2} - x\) and A1 for \(x\ln x - x\).

Note:     The second M1 and second A1 are independent of the first M1 and the first A1.

\( = \frac{{{{\text{e}}^2}}}{2} - {\text{e}} - \frac{1}{2}\left( { = \frac{{{{\text{e}}^2} - 2{\text{e}} - 1}}{2}} \right)\)     A1

[5 marks]

(i)     METHOD 1

consider for example \(h(x) = x - 1 - \ln x\)

\(h(1) = 0\;\;\;{\text{and}}\;\;\;h'(x) = 1 - \frac{1}{x}\)     (A1)

as \(h'(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1\)     R1

as \(h'(x) \le 0\;\;\;{\text{for}}\;\;\;0 < x \le 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;0 < x \le 1\)     R1

so \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\)     AG

METHOD 2

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

\(g''(x) < 0\;\;\;\)(concave down) for\(\;\;\;x \in {\mathbb{R}^ + }\)     R1

the graph of \(y = g(x)\) is below its tangent \((y = x - 1\;\;\;{\text{at}}\;\;\;x = 1)\)     R1

so \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\)     AG

Note:     The reasoning may be supported by drawn graphical arguments.

METHOD 3

clear correct graphs of \(y = x - 1\;\;\;{\text{and}}\;\;\;\ln x\;\;\;{\text{for}}\;\;\;x > 0\)     A1A1

statement to the effect that the graph of \(\ln x\) is below the graph of its tangent at \(x = 1\)     R1AG

(ii)     replacing \(x\) by \(\frac{1}{x}\) to obtain \(\ln \left( {\frac{1}{x}} \right) \le \frac{1}{x} - 1\left( { = \frac{{1 - x}}{x}} \right)\)     M1

\( - \ln x \le \frac{1}{x} - 1\left( { = \frac{{1 - x}}{x}} \right)\)     (A1)

\(\ln x \ge 1 - \frac{1}{x}\left( { = \frac{{x - 1}}{x}} \right)\)     A1

so \(\frac{{x - 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }\)     AG

[6 marks]

Total [23 marks]

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

A productive question for many candidates, but some didn’t realise that a difference of areas/integrals was required.

(i)     Many candidates adopted a graphical approach, but sometimes with unconvincing reasoning.

(ii)     Poorly answered. Many candidates applied the suggested substitution only to one side of the inequality, and then had to fudge the answer.

Sketch the graph of the function. You are not required to find the coordinates of the maximum.

Find the value of k .

     A1

Note: Award A1 for intercepts of 0 and 2 and a concave down curve in the given domain .

Note: Award A0 if the cubic graph is extended outside the domain [0, 2] .

[1 mark]

\(\int_0^2 {kx(x + 1)(2 - x){\text{d}}x = 1} \)     (M1)

Note: The correct limits and =1 must be seen but may be seen later.

\(k\int_0^2 {( - {x^3} + {x^2} + 2x){\text{d}}x = 1} \)     A1

\(k\left[ { - \frac{1}{4}{x^4} + \frac{1}{3}{x^3} + {x^2}} \right]_0^2 = 1\)     M1

\(k\left( { - 4 + \frac{8}{3} + 4} \right) = 1\)     (A1)

\(k = \frac{3}{8}\)     A1

[5 marks]

Most candidates completed this question well. A number extended the graph beyond the given domain.

Most candidates completed this question well. A number extended the graph beyond the given domain.

The graph of \(y = f\left( x \right)\) has a local maximum at A. Find the coordinates of A.

Show that there is exactly one point of inflexion, B, on the graph of \(y = f\left( x \right)\).

The coordinates of B can be expressed in the form B\(\left( {{2^a},\,b \times {2^{ - 3a}}} \right)\) where a, b\( \in \mathbb{Q}\). Find the value of a and the value of b.

Sketch the graph of \(y = f\left( x \right)\) showing clearly the position of the points A and B.

attempt to differentiate      (M1)

\(f'\left( x \right) =  - 3{x^{ - 4}} - 3x\)     A1

Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example \(f'\left( x \right) = \frac{{ - 15{x^4} \times 2{x^3} - 6{x^2}\left( {2 - 3{x^5}} \right)}}{{{{\left( {2{x^3}} \right)}^2}}}\).

\( - \frac{3}{{{x^4}}} - 3x = 0\)     M1

\( \Rightarrow {x^5} =  - 1 \Rightarrow x =  - 1\)     A1

\({\text{A}}\left( { - 1,\, - \frac{5}{2}} \right)\)     A1

[5 marks]

\(f''\left( x \right) = 0\)     M1

\(f''\left( x \right) = 12{x^{ - 5}} - 3\left( { = 0} \right)\)     A1

Note: Award A1 for correct derivative seen even if not simplified.

\( \Rightarrow x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)\)     A1

hence (at most) one point of inflexion      R1

Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.

\(f''\left( x \right)\) changes sign at \(x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)\)      R1

so exactly one point of inflexion

[5 marks]

\(x = \sqrt[5]{4} = {2^{\frac{2}{5}}}\left( { \Rightarrow a = \frac{2}{5}} \right)\)      A1

\(f\left( {{2^{\frac{2}{5}}}} \right) = \frac{{2 - 3 \times {2^2}}}{{2 \times {2^{\frac{6}{5}}}}} =  - 5 \times {2^{ - \frac{6}{5}}}\left( { \Rightarrow b =  - 5} \right)\)     (M1)A1

Note: Award M1 for the substitution of their value for \(x\) into \(f\left( x \right)\).

[3 marks]

A1A1A1A1

A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.

Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.

[4 marks]

Shown below are the graphs of \(y = f(x)\) and \(y = g(x)\).

If \((f \circ g)(x) = 3\), find all possible values of x.

\(g(x) = 0{\text{ or 3}}\)     (M1)(A1)

x = –1 or 4 or 1 or 2     A1A1

Notes: Award A1A1 for all four correct values,

A1A0 for two or three correct values,

A0A0 for less than two correct values.

Award M1 and corresponding A marks for correct attempt to find expressions for f and g.

[4 marks]

A small number of candidates gave correct and well explained answers. Many candidates answered the question without showing any kind of work and in many cases it was clear that candidates were guessing and clearly did not know about composition of functions. A number of candidates attempted to find expressions for both functions but made little progress and wasted time.

(i)     Find an expression for \(f'(x)\) .

(ii)     Given that the equation \(f'(x) = 0\) has two roots, state their values.

Sketch the graph of f , showing clearly the coordinates of the maximum and minimum.