Show that the points (0, 0) and (\(\sqrt {2\pi } \) , \( - \sqrt {2\pi } \)) on the curve \({{\text{e}}^{\left( {x + y} \right)}} = \cos \left( {xy} \right)\) have a common tangent.

attempt at implicit differentiation     M1

\({{\text{e}}^{\left( {x + y} \right)}}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = - \sin \left( {xy} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right)\)     A1A1

let \(x = 0\), \(y = 0\)     M1

\({{\text{e}}^0}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = 0\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1\)     A1

let \(x = \sqrt {2\pi } \) , \(y = - \sqrt {2\pi } \)

\({{\text{e}}^0}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = - \sin \left( {- 2\pi } \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right) = 0\)

so \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1\)     A1

since both points lie on the line \(y = - x\) this is a common tangent     R1

Note: \(y = - x\) must be seen for the final R1. It is not sufficient to note that the gradients are equal.

[7 marks]

Implicit differentiation was attempted by many candidates, some of whom obtained the correct value for the gradient of the tangent. However, very few noticed the need to go further and prove that both points were on the same line.

Show that \(f'(x) = \frac{1}{2}{x^{ - \frac{1}{2}}}{(1 - x)^{ - \frac{3}{2}}}\) and deduce that f is an increasing function.

Show that the curve \(y = f(x)\) has one point of inflexion, and find its coordinates.

 Use the substitution \(x = {\sin ^2}\theta \) to show that \(\int {f(x){\text{d}}x}  = \arcsin \sqrt x  - \sqrt {x - {x^2}}  + c\) .

EITHER

derivative of \(\frac{x}{{1 - x}}\) is \(\frac{{(1 - x) - x( - 1)}}{{{{(1 - x)}^2}}}\)     M1A1

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

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

\(f'(x) > 0\) (for all \(0 < x < 1\)) so the function is increasing     R1

OR

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

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

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

\( = \frac{1}{2}{x^{ - \frac{1}{2}}}{(1 - x)^{ - \frac{3}{2}}}[1 - x + x]\)     M1

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

\(f'(x) > 0\) (for all \(0 < x < 1\)) so the function is increasing     R1

[5 marks]

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

\( \Rightarrow f''(x) = -\frac{1}{4}{x^{ - \frac{3}{2}}}{(1 - x)^{ - \frac{3}{2}}} + \frac{3}{4}{x^{ - \frac{1}{2}}}{(1 - x)^{ - \frac{5}{2}}}\)     M1A1

\( = -\frac{1}{4}{x^{ - \frac{3}{2}}}{(1 - x)^{ - \frac{5}{2}}}[1 - 4x]\)

\(f''(x) = 0 \Rightarrow x = \frac{1}{4}\)     M1A1

\(f''(x)\) changes sign at \(x = \frac{1}{4}\) hence there is a point of inflexion     R1

\(x = \frac{1}{4} \Rightarrow y = \frac{1}{{\sqrt 3 }}\)     A1

the coordinates are \(\left( {\frac{1}{4},\frac{1}{{\sqrt 3 }}} \right)\)

[6 marks] 

\(x = {\sin ^2}\theta  \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sin \theta \cos \theta \)     M1A1

\(\int {\sqrt {\frac{x}{{1 - x}}} {\text{d}}x = \int {\sqrt {\frac{{{{\sin }^2}\theta }}{{1 - {{\sin }^2}\theta }}} 2\sin \theta \cos \theta {\text{d}}\theta } } \)     M1A1

\( = \int {2{{\sin }^2}\theta {\text{d}}\theta } \)     A1

\( = \int {1 - \cos 2\theta } {\text{d}}\theta \)     M1A1

\( = \theta  - \frac{1}{2}\sin 2\theta  + c\)     A1

\(\theta  = \arcsin \sqrt x \)     A1

\(\frac{1}{2}\sin 2\theta  = \sin \theta \cos \theta  = \sqrt x \sqrt {1 - x}  = \sqrt {x - {x^2}} \)     M1A1

hence \(\int {\sqrt {\frac{x}{{1 - x}}} {\text{d}}x = \arcsin \sqrt x }  - \sqrt {x - {x^2}}  + c\)     AG

[11 marks] 

 Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.

Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.

Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.

The first set of axes below shows the graph of \(y = {\text{ }}f(x)\) for \( - 4 \leqslant x \leqslant 4\).

Let \(g(x) = \int_{ - 4}^x {f(t){\text{d}}t} \) for \( - 4 \leqslant x \leqslant 4\).

(a)     State the value of x at which \(g(x)\) is a minimum.

(b)     On the second set of axes, sketch the graph of \(y = g(x)\).

(a)     \(x = 1\)     A1

[1 mark]

(b)     A1 for point (–4, 0)

A1 for (0, − 4)

A1 for min at \(x = 1\) in approximately the correct place

A1 for (4, 0)

A1 for shape including continuity at \(x = 0\)

[5 marks]

Total [6 marks]

The function f is defined by \(f(x) = {{\text{e}}^{{x^2} - 2x - 1.5}}\).

(a)     Find \(f'(x)\).

(b)     You are given that \(y = \frac{{f(x)}}{{x - 1}}\) has a local minimum at x = a, a > 1. Find the

value of a.

(a)     \(\left( {u = {x^2} - 2x - 1.5;\frac{{{\text{d}}u}}{{{\text{d}}x}} = 2x - 2} \right)\)

\(\frac{{{\text{d}}f}}{{{\text{d}}x}} = \frac{{{\text{d}}f}}{{{\text{d}}u}}\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^u}(2x - 2)\)     (M1)

\( = 2(x - 1){{\text{e}}^{{x^2} - 2x - 1.5}}\)     A1

(b)     \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(x - 1) \times 2(x - 1){{\text{e}}^{{x^2} - 2x - 1.5}} - 1 \times {{\text{e}}^{{x^2} - 2x - 1.5}}}}{{{{(x - 1)}^2}}}\)     M1A1

\( = \frac{{2{x^2} - 4x + 1}}{{{{(x - 1)}^2}}}{{\text{e}}^{{x^2} - 2x - 1.5}}\)     (A1)

minimum occurs when \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     (M1)

\(x = 1 \pm \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}x = \frac{{4 \pm \sqrt 8 }}{4}} \right)\)     A1

\(a = 1 + \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}a = \frac{{4 + \sqrt 8 }}{4}} \right)\)     R1

[8 marks]

Part (a) was successfully answered by most candidates. Most candidates were able to make significant progress with part (b) but were then let down by being unable to simplify the expression or by not understanding the significance of being told that a > 1.

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.

Find \(f'\left( x \right)\).

Find \(g'\left( x \right)\).

Hence, or otherwise, find \(\int\limits_0^\pi  {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x} \).

attempt at product rule      M1

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

[2 marks]

\(g'\left( x \right) =  - {{\text{e}}^{ - x}}\,{\text{cos}}\,x - {{\text{e}}^{ - x}}\,{\text{sin}}\,x\)      A1

[1 mark]

METHOD 1

Attempt to add \(f'\left( x \right)\) and \(g'\left( x \right)\)      (M1)

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

\(\int\limits_0^\pi  {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x}  = \left[ { - \frac{{{{\text{e}}^{ - x}}}}{2}\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)} \right]_0^\pi \) (or equivalent)      A1

Note: Condone absence of limits.

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

METHOD 2

\(I = \int {{{\text{e}}^{ - x}}} \,{\text{sin}}\,x\,{\text{d}}x\)

\( =  - {{\text{e}}^{ - x}}\,{\text{cos}}\,x - \int {{{\text{e}}^{ - x}}} \,{\text{cos}}\,x\,{\text{d}}x\) OR \( =  - {{\text{e}}^{ - x}}\,{\text{sin}}\,x + \int {{{\text{e}}^{ - x}}} \,{\text{cos}}\,x\,{\text{d}}x\)     M1A1

\( =  - {{\text{e}}^{ - x}}\,{\text{sin}}\,x - {{\text{e}}^{ - x}}\,{\text{cos}}\,x - \int {{{\text{e}}^{ - x}}} \,{\text{sin}}\,x\,{\text{d}}x\)

\(I = \frac{1}{2}{{\text{e}}^{ - x}}\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)\)     A1

\(\int_0^\pi  {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x = \frac{1}{2}\left( {1 + {{\text{e}}^{ - \pi }}} \right)} \)    A1

[4 marks]

A drinking glass is modelled by rotating the graph of \(y = {{\text{e}}^x}\) about the y-axis, for \(1 \leqslant y \leqslant 5\) . Find the volume of the glass.

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

volume \( = \pi \int_1^5 {{{(\ln y)}^2}{\text{d}}y} \)     (M1)A1

using integration by parts     (M1)

\(\pi \int_1^5 {{{(\ln y)}^2}{\text{d}}y} = \pi \left[ {y{{(\ln y)}^2}} \right]_1^5 - 2\int_1^5 {\ln y{\text{d}}y} \)     A1A1

\( = \pi \left[ {y{{(\ln y)}^2} - 2y\ln y + 2y} \right]_1^5\)     A1A1

Note: Award A1 marks if \(\pi \) is present in at least one of the above lines.

\( \Rightarrow \pi \int_1^5 {{{(\ln y)}^2}{\text{d}}y} = \pi {\text{ }}5{(\ln 5)^2} - 10\ln 5 + 8\)     A1

[8 marks]

Only the best candidates were able to make significant progress with this question. Quite a few did not consider rotation about the y-axis. Others wrote the correct expression, but seemed daunted by needing to integrate by parts twice.

Show that there is no point where the tangent to the curve is horizontal.

Find the coordinates of the points where the tangent to the curve is vertical.

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

a horizontal tangent occurs if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) so \(y = 0\)     M1

we can see from the equation of the curve that this solution is not possible \((0 = 4)\) and so there is not a horizontal tangent     R1

[4 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y - x}}\) or equivalent with \(\frac{{{\text{d}}x}}{{{\text{d}}y}}\)

the tangent is vertical when \(2y = x\)     M1

substitute into the equation to give \(2{y^2} = {y^2} + 4\)     M1

\(y =  \pm 2\)     A1

coordinates are \((4,{\text{ }}2),{\text{ }}( - 4,{\text{ }} - 2)\)     A1

[4 marks]

Total [8 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]

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.

Using integration by parts find \(\int {x\sin x{\text{d}}x} \).

attempt to integrate one factor and differentiate the other, leading to a sum of two terms     M1

\(\int {x\sin x{\text{d}}x = x( - \cos x) + } \int {\cos x{\text{d}}x} \)     (A1)(A1)

\( =  - x\cos x + \sin x + c\)     A1

Note:     Only award final A1 if \( + {\text{ }}c\) is seen.

[4 marks]

By using the substitution \(u = {{\text{e}}^x} + 3\), find \(\int {\frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 6{{\text{e}}^x} + 13}}{\text{d}}x} \).

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^x}\)     (A1)

EITHER

integral is \(\int {\frac{{{{\text{e}}^x}}}{{{{({{\text{e}}^x} + 3)}^2} + {2^2}}}{\text{d}}x} \)     M1A1

\( = \frac{1}{{{u^2} + {2^2}}}{\text{d}}u\)     M1A1

Note:     Award M1 only if the integral has completely changed to one in \(u\).

Note:     \({\text{d}}u\) needed for final A1

OR

\({{\text{e}}^x} = u - 3\)

integral is \(\int {\frac{1}{{{{(u - 3)}^2} + 6(u - 3) + 13}}{\text{d}}u} \)     M1A1

Note: Award M1 only if the integral has completely changed to one in \(u\).

\( = \int {\frac{1}{{{u^2} + {2^2}}}{\text{d}}u} \)     M1A1

Note:     In both solutions the two method marks are independent.

THEN

\( = \frac{1}{2}\arctan \left( {\frac{u}{2}} \right)( + c)\)     (A1)

\( = \frac{1}{2}\arctan \left( {\frac{{{{\text{e}}^x} + 3}}{2}} \right)( + c)\)     A1

Total [7 marks]

Many good complete answers. Some did not realise it was arctan. Some had poor understanding of the method.

Determine the values of \(x\) for which \(f(x)\) is a decreasing function.

There is a point of inflexion, \(P\), on the curve \(y = f(x)\).

Find the coordinates of \(P\).

attempt to differentiate \(f(x) = {x^3} - 3{x^2} + 4\)     M1

\(f'(x) = 3{x^2} - 6x\)     A1

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

(Critical values occur at) \(x = 0,{\text{ }}x = 2\)     (A1)

so \(f\) decreasing on \(x \in ]0,{\text{ }}2[\;\;\;({\text{or }}0 < x < 2)\)     A1

[4 marks]

\(f''(x) = 6x - 6\)     (A1)

setting \(f''(x) = 0\)     M1

\( \Rightarrow x = 1\)

coordinate is \((1,{\text{ }}2)\)     A1

[3 marks]

Total [7 marks]

Show that \(\sin \left( {\theta  + \frac{\pi }{2}} \right) = \cos \theta \).

Consider \(f(x) = \sin (ax)\) where \(a\) is a constant. Prove by mathematical induction that \({f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)\) where \(n \in {\mathbb{Z}^ + }\) and \({f^{(n)}}(x)\) represents the \({{\text{n}}^{{\text{th}}}}\) derivative of \(f(x)\).

\(\sin \left( {\theta  + \frac{\pi }{2}} \right) = \sin \theta \cos \frac{\pi }{2} + \cos \theta \sin \frac{\pi }{2}\)     M1

\( = \cos \theta \)     AG

Note:     Accept a transformation/graphical based approach.

[1 mark]

consider \(n = 1,{\text{ }}f'(x) = a\cos (ax)\)     M1

since \(\sin \left( {ax + \frac{\pi }{2}} \right) = \cos ax\) then the proposition is true for \(n = 1\)     R1

assume that the proposition is true for \(n = k\) so \({f^{(k)}}(x) = {a^k}\sin \left( {ax + \frac{{k\pi }}{2}} \right)\)     M1

\({f^{(k + 1)}}(x) = \frac{{{\text{d}}\left( {{f^{(k)}}(x)} \right)}}{{{\text{d}}x}}\;\;\;\left( { = a\left( {{a^k}\cos \left( {ax + \frac{{k\pi }}{2}} \right)} \right)} \right)\)     M1

\( = {a^{k + 1}}\sin \left( {ax + \frac{{k\pi }}{2} + \frac{\pi }{2}} \right)\) (using part (a))     A1

\( = {a^{k + 1}}\sin \left( {ax + \frac{{(k + 1)\pi }}{2}} \right)\)     A1

given that the proposition is true for \(n = k\) then we have shown that the proposition is true for \(n = k + 1\). Since we have shown that the proposition is true for \(n = 1\) then the proposition is true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     Award final R1 only if all prior M and R marks have been awarded.

[7 marks]

Total [8 marks]

Find the value of the integral \(\int_0^{\sqrt 2 } {\sqrt {4 - {x^2}} {\text{d}}x} \) .

Find the value of the integral \(\int_0^{0.5} {\arcsin x {\text{d}}x} \) .

Using the substitution \(t = \tan \theta \) , find the value of the integral

\[\int_0^{\frac{\pi }{4}} {\frac{{{\text{d}}\theta }}{{3{{\cos }^2}\theta + {{\sin }^2}\theta }}} {\text{ }}.\]

let \(x = 2\sin \theta \)     M1

\({\text{d}}x = 2\cos \theta {\text{d}}\theta \)     A1

\(I = \int_0^{\frac{\pi }{4}} {2\cos \theta \times 2\cos \theta {\text{d}}\theta \,\,\,\,\,\left( { = 4\int_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right)} \)     A1A1

Note: Award A1 for limits and A1 for expression.

\( = 2\int_0^{\frac{\pi }{4}} {(1 + \cos 2\theta ){\text{d}}\theta } \)     A1

\( = 2\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_0^{\frac{\pi }{4}}\)     A1

\( = 1 + \frac{\pi }{2}\)     A1

[7 marks]

\(I = [x\arcsin x]_0^{0.5} - \int_0^{0.5} {x \times \frac{1}{{\sqrt {1 - {x^2}} }}{\text{d}}x} \)     M1A1A1

\( = [x\arcsin x]_0^{0.5} + \left[ {\sqrt {1 - {x^2}} } \right]_0^{0.5}\)     A1

\( = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1\)     A1

[5 marks]

\({\text{d}}t = {\sec ^2}\theta {\text{d}}\theta {\text{ , }}\left[ {0,\frac{\pi }{4}} \right] \to [0,{\text{ 1]}}\)     A1(A1)

\(I = \int_0^1 {\frac{{\frac{{{\text{d}}t}}{{(1 + {t^2})}}}}{{\frac{3}{{(1 + {t^2})}} + \frac{{{t^2}}}{{(1 + {t^2})}}}}} \)     M1(A1)

\( = \int_0^1 {\frac{{{\text{d}}t}}{{3 + {t^2}}}} \)     A1

\( = \frac{1}{{\sqrt 3 }}\left[ {\arctan \left( {\frac{x}{{\sqrt 3 }}} \right)} \right]_0^1\)     A1

\( = \frac{\pi }{{6\sqrt 3 }}\)     A1

[7 marks]

Find \(f'(x)\).

Hence find the \(x\)-coordinates of the points where the gradient of the graph of \(f\) is zero.

Find \(f''(x)\) expressing your answer in the form \(\frac{{p(x)}}{{{{({x^2} + 1)}^3}}}\), where \(p(x)\) is a polynomial of degree 3.

Find the \(x\)-coordinates of the other two points of inflexion.

Find the area of the shaded region. Express your answer in the form \(\frac{\pi }{a} - \ln \sqrt b \), where \(a\) and \(b\) are integers.

(a)     \(f'(x) = \frac{{\left( {{x^2} + 1} \right) - 2x(x + 1)}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( { = \frac{{ - {x^2} - 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)\)     M1A1

[2 marks]

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

\(x =  - 1 \pm \sqrt 2 \)     A1

[1 mark]

\(f''(x) = \frac{{( - 2x - 2){{\left( {{x^2} + 1} \right)}^2} - 2(2x)\left( {{x^2} + 1} \right)\left( { - {x^2} - 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}\)     A1A1

Note:     Award A1 for \(( - 2x - 2){\left( {{x^2} + 1} \right)^2}\) or equivalent.

Note:     Award A1 for \( - 2(2x)\left( {{x^2} + 1} \right)\left( { - {x^2} - 2x + 1} \right)\) or equivalent.

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

\( = \frac{{2{x^3} + 6{x^2} - 6x - 2}}{{{{\left( {{x^2} + 1} \right)}^3}}}\)     A1

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

[3 marks]

recognition that \((x - 1)\) is a factor     (R1)

\((x - 1)\left( {{x^2} + bx + c} \right) = \left( {{x^3} + 3{x^2} - 3x - 1} \right)\)     M1

\( \Rightarrow {x^2} + 4x + 1 = 0\)     A1

\(x =  - 2 \pm \sqrt 3 \)     A1

Note:     Allow long division / synthetic division.

[4 marks]

\(\int_{ - 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x} \)     M1

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

\( = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)\)     A1A1

\( = \left[ {\frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)} \right]_{ - 1}^0 = \frac{1}{2}\ln 1 + \arctan 0 - \frac{1}{2}\ln 2 - \arctan ( - 1)\)     M1

\( = \frac{\pi }{4} - \ln \sqrt 2 \)     A1

[6 marks]

Use De Moivre’s theorem to show that \({z^n} + {z^{ - n}} = 2\cos n\theta ,{\text{ }}n \in {\mathbb{Z}^ + }\).

Expand \({\left( {z + {z^{ - 1}}} \right)^4}\).

Hence show that \({\cos ^4}\theta  = p\cos 4\theta  + q\cos 2\theta  + r\), where \(p,{\text{ }}q\) and \(r\) are constants to be determined.

Show that \({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\).

Hence find the value of \(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta } \).

S is rotated through \(2\pi \) radians about the x-axis. Find the value of the volume generated.

(i)     Write down an expression for the constant term in the expansion of \({\left( {z + {z^{ - 1}}} \right)^{2k}}\), \(k \in {\mathbb{Z}^ + }\).

(ii)     Hence determine an expression for \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta } \) in terms of k.

\({z^n} + {z^{ - n}} = \cos n\theta  + i\sin n\theta  + \cos ( - n\theta ) + i\sin ( - n\theta )\)     M1

\( = \cos n\theta  + \cos n\theta  + i\sin n\theta  - i\sin n\theta \)     A1

\( = 2\cos n\theta \)     AG

[2 marks]

(b)     \({\left( {z + {z^{ - 1}}} \right)^4} = {z^4} + 4{z^3}\left( {\frac{1}{z}} \right) + 6{z^2}\left( {\frac{1}{{{z^2}}}} \right) + 4z\left( {\frac{1}{{{z^3}}}} \right) + \frac{1}{{{z^4}}}\)     A1

Note:     Accept \({\left( {z + {z^{ - 1}}} \right)^4} = 16{\cos ^4}\theta \).

[1 mark]

METHOD 1

\({\left( {z + {z^{ - 1}}} \right)^4} = \left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 4\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 6\)     M1

\({(2\cos \theta )^4} = 2\cos 4\theta  + 8\cos 2\theta  + 6\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

METHOD 2

\({\cos ^4}\theta  = {\left( {\frac{{\cos 2\theta  + 1}}{2}} \right)^2}\)     M1

\( = \frac{1}{4}({\cos ^2}2\theta  + 2\cos 2\theta  + 1)\)     A1

\( = \frac{1}{4}\left( {\frac{{\cos 4\theta  + 1}}{2} + 2\cos 2\theta  + 1} \right)\)     A1

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

[4 marks]

\({\left( {z + {z^{ - 1}}} \right)^6} = {z^6} + 6{z^5}\left( {\frac{1}{z}} \right) + 15{z^4}\left( {\frac{1}{{{z^2}}}} \right) + 20{z^3}\left( {\frac{1}{{{z^3}}}} \right) + 15{z^2}\left( {\frac{1}{{{z^4}}}} \right) + 6z\left( {\frac{1}{{{z^5}}}} \right) + \frac{1}{{{z^6}}}\)     M1

\({\left( {z + {z^{ - 1}}} \right)^6} = \left( {{z^6} + \frac{1}{{{z^6}}}} \right) + 6\left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 15\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 20\)

\({(2\cos \theta )^6} = 2\cos 6\theta  + 12\cos 4\theta  + 30\cos 2\theta  + 20\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\)     AG

Note:     Accept a purely trigonometric solution as for (c).

[3 marks]

\(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta  = \int_0^{\frac{\pi }{2}} {\left( {\frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}} \right){\text{d}}\theta } } \)

\( = \left[ {\frac{1}{{192}}\sin 6\theta  + \frac{3}{{64}}\sin 4\theta  + \frac{{15}}{{64}}\sin 2\theta  + \frac{5}{{16}}\theta } \right]_0^{\frac{\pi }{2}}\)     M1A1

\( = \frac{{5\pi }}{{32}}\)     A1

[3 marks]

\({\text{V}} = \pi \int_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^4}x{\text{d}}x} \)     M1

\( = \pi \int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x - \pi \int_0^{\frac{\pi }{2}} {{{\cos }^6}x{\text{d}}x} } \)     M1

\(\int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x}  = \frac{{3\pi }}{{16}}\)     A1

\({\text{V}} = \frac{{3{\pi ^2}}}{{16}} - \frac{{5{\pi ^2}}}{{32}} = \frac{{{\pi ^2}}}{{32}}\)     A1

Note:     Follow through from an incorrect r in (c) provided the final answer is positive.

(i)     constant term = \(\left( \begin{array}{c}2k\\k\end{array} \right)\) \( = \frac{{(2k)!}}{{k!k!}} = \frac{{(2k)!}}{{{{(k!)}^2}}}{\text{ (accept }}C_k^{2k})\)     A1

(ii)     \({2^{2k}}\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{{(k!)}^2}}}\frac{\pi }{2}} \)     A1

          \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{2^{2k + 1}}{{(k!)}^2}}}} \) \(\left( {{\rm{or}}\frac{{\left( \begin{array}{c}2k\\k\end{array} \right)\pi }}{{{2^{2k + 1}}}}} \right)\)     A1

[3 marks]

Part a) has appeared several times before, though with it again being a ‘show that’ question, some candidates still need to be more aware of the need to show every step in their working, including the result that \(\sin ( - n\theta ) =  - \sin (n\theta )\).

Part b) was usually answered correctly.

Part c) was again often answered correctly, though some candidates often less successfully utilised a trig-only approach rather than taking note of part b).

Part d) was a good source of marks for those who kept with the spirit of using complex numbers for this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions using this approach were extremely rare.

Part e) was well answered, though numerical slips were often common. A small number integrated \(\sin n\theta \) as \(n\cos n\theta \).

A large number of candidates did not realise the help that part e) inevitably provided for part f). Some correctly expressed the volume as \(\pi \int {{{\cos }^4}x{\text{d}}x - \pi \int {{{\cos }^6}x{\text{d}}x} } \) and thus gained the first 2 marks but were able to progress no further. Only a small number of able candidates were able to obtain the correct answer of \(\frac{{{\pi ^2}}}{{32}}\).

Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest scoring candidates gain all 3 marks.

The graphs of \(f(x) = - {x^2} + 2\) and \(g(x) = {x^3} - {x^2} - bx + 2,{\text{ }}b > 0\), intersect and create two closed regions. Show that these two regions have equal areas.

to find the points of intersection of the two curves

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

\({x^3} - bx = x({x^2} - b) = 0\)

\( \Rightarrow x = 0;{\text{ }}x = \pm \sqrt b \)     A1A1

\({A_1} = \int_{ - \sqrt b }^0 {\left[ {({x^3} - {x^2} - bx + 2) - ( - {x^2} + 2)} \right]} {\text{d}}x\left( { = \int_{ - \sqrt b }^0 {({x^3} - bx){\text{d}}x} } \right)\)     M1

\( = \left[ {\frac{{{x^4}}}{4} - \frac{{b{x^2}}}{2}} \right]_{ - \sqrt b }^0\)

\( = - \left( {\frac{{{{( - \sqrt b )}^4}}}{4} - \frac{{b{{( - \sqrt b )}^2}}}{2}} \right) = - \frac{{{b^2}}}{4} + \frac{{{b^2}}}{2} = \frac{{{b^2}}}{4}\)     A1

\({A_2} = \int_0^{\sqrt b } {\left[ {( - {x^2} + 2) - ({x^3} - {x^2} - bx + 2)} \right]{\text{d}}x} \)     M1

\( = \int_0^{\sqrt b } {( - {x^3} + bx){\text{d}}x} \)

\( = \left[ { - \frac{{{x^4}}}{4} + \frac{{b{x^2}}}{2}} \right]_0^{\sqrt b } = \frac{{{b^2}}}{4}\)     A1

therefore \({A_1} = {A_2} = \frac{{{b^2}}}{4}\)     AG

[7 marks]

Most candidates knew how to tackle this question. The most common error was in giving +b and –b as the x-coordinates of the point of intersection.

Consider the part of the curve \(4{x^2} + {y^2} = 4\) shown in the diagram below.

(a)     Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y .

(b)     Find the gradient of the tangent at the point \(\left( {\frac{2}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)\).

(c)     A bowl is formed by rotating this curve through \(2\pi \) radians about the x-axis.

Calculate the volume of this bowl.

(a)     \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1A1

Note: Award M1A0 for \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4\) .

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{4x}}{y}\)     A1

(b)     – 4     A1

(c)     \(V = \int {\pi {y^2}{\text{d}}x} \) or equivalent     M1

\(V = \pi \int_0^1 {(4 - 4{x^2}){\text{d}}x} \)     A1

\( = \pi \left[ {4x - \frac{4}{3}{x^3}} \right]_0^1\)     A1

\( = \frac{{8\pi }}{3}\)     A1

Note: If it is correct except for the omission of \(\pi \) , award 2 marks.

[8 marks]

The first part of this question was done well by many, the only concern being the number that did not simplify the result from \( - \frac{{8x}}{{2y}}\). There were many variations on the formula for the volume in part c), the most common error being a multiple of \(2\pi \) rather than \(\pi \). On the whole this question was done well by many.

Find the equation of the tangent to C at the point (2, e).

Find \(f(x)\).

Determine the largest possible domain of f.

Show that the equation \(f(x) = f'(x)\) has no solution.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\text{e}}}{{\ln {\text{e}}}}(2 + 2) = 4{\text{e}}\)     A1

at (2, e) the tangent line is \(y - {\text{e}} = 4{\text{e}}(x - 2)\)     M1

hence \(y = 4{\text{e}}x - 7{\text{e}}\)     A1

[3 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{\ln y}}(x + 2) \Rightarrow \frac{{\ln y}}{y}{\text{d}}y = (x + 2){\text{d}}x\)     M1

\(\int {\frac{{\ln y}}{y}{\text{d}}y = \int {(x + 2){\text{d}}x} } \)

using substitution \(u = \ln y;{\text{ d}}u = \frac{1}{y}{\text{d}}y\)     (M1)(A1)

\( \Rightarrow \int {\frac{{\ln y}}{y}{\text{d}}y = \int {u{\text{d}}u = \frac{1}{2}{u^2}} } \)     (A1)

\( \Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x + c\)     A1A1

at (2, e), \(\frac{{{{(\ln {\text{e}})}^2}}}{2} = 6 + c\)     M1

\( \Rightarrow c = - \frac{{11}}{2}\)     A1

\( \Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x - \frac{{11}}{2} \Rightarrow {(\ln y)^2} = {x^2} + 4x - 11\)

\(\ln y = \pm \sqrt {{x^2} + 4x - 11}  \Rightarrow y = {{\text{e}}^{ \pm \sqrt {{x^2} + 4x - 11} }}\)     M1A1

since y > 1, \(f(x) = {{\text{e}}^{\sqrt {{x^2} + 4x - 11} }}\)     R1

Note:M1 for attempt to make y the subject.

[11 marks]

EITHER

\({x^2} + 4x - 11 > 0\)     A1

using the quadratic formula     M1

critical values are \(\frac{{ - 4 \pm \sqrt {60} }}{2}{\text{ }}\left( { = - 2 \pm \sqrt {15} } \right)\)     A1

using a sign diagram or algebraic solution     M1

\(x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15} \)     A1A1

OR

\({x^2} + 4x - 11 > 0\)     A1

by methods of completing the square     M1

\({(x + 2)^2} > 15\)     A1

\( \Rightarrow x + 2 < - \sqrt {15} {\text{ or }}x + 2 > \sqrt {15} \)     (M1)

\(x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15} \)     A1A1

[6 marks]

\(f(x) = f'(x) \Rightarrow f(x) = \frac{{f(x)}}{{\ln f(x)}}(x + 2)\)     M1

\( \Rightarrow \ln \left( {f(x)} \right) = x + 2\,\,\,\,\,\left( { \Rightarrow x + 2 = \sqrt {{x^2} + 4x - 11} } \right)\)     A1

\( \Rightarrow {(x + 2)^2} = {x^2} + 4x - 11 \Rightarrow {x^2} + 4x + 4 = {x^2} + 4x - 11\)     A1

\( \Rightarrow 4 = - 11,{\text{ hence }}f(x) \ne f'(x)\)     R1AG

[4 marks]

Nearly always correctly answered.

Most candidates separated the variables and attempted the integrals. Very few candidates made use of the condition y > 1, so losing 2 marks.

Part (c) was often well answered, sometimes with follow through.

Only the best candidates were successful on part (d).

METHOD 1

\({\text{area}} = \mathop \smallint \limits_0^{\sqrt 3 } \arctan x{\text{d}}x\)     A1

attempting to integrate by parts     M1

\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } - \mathop \smallint \limits_0^{\sqrt 3 } x\frac{1}{{1 + {x^2}}}{\text{d}}x\)     A1A1

\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } - \left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right]_0^{\sqrt 3 }\)     A1

Note: Award A1 even if limits are absent.

\( = \frac{\pi }{{\sqrt 3 }} - \frac{1}{2}\ln 4\)     A1

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

METHOD 2

\({\text{area}} = \frac{{\pi \sqrt 3 }}{3} - \mathop \smallint \limits_0^{\frac{\pi }{3}} \tan y{\text{d}}y\)     M1A1A1

\(\ { = \frac{{\pi \sqrt 3 }}{3} + \left[ {\ln \left| {\cos y} \right|} \right]_0^{\frac{\pi }{3}}} \)     M1A1

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

[6 marks]

Many candidates were able to write down the correct expression for the required area, although in some cases with incorrect integration limits. However, very few managed to achieve any further marks due to a number of misconceptions, in particular \(\arctan x = \cot x = \frac{{\cos x}}{{\sin x}}\). Candidates who realised they should use integration by parts were in general very successful in answering this question. It was pleasing to see a few alternative correct approaches to this question.

Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of \(x\) and \(y\).

Find the equations of the tangents to this curve at the points where the curve intersects the line \(x = 1\).

attempt to differentiate implicitly     M1

\(3 - \left( {4y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2{y^2}} \right){{\text{e}}^{x - 1}} = 0\)    A1A1A1

Note: Award A1 for correctly differentiating each term.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 \bullet {{\text{e}}^{1 - x}} - 2{y^2}}}{{4y}}\)    A1

Note: This final answer may be expressed in a number of different ways.

[5 marks]

\(3 - 2{y^2} = 2 \Rightarrow {y^2} = \frac{1}{2} \Rightarrow y =  \pm \sqrt {\frac{1}{2}} \)    A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 - 2 \bullet \frac{1}{2}}}{{ \pm 4\sqrt {\frac{1}{2}} }} =  \pm \frac{{\sqrt 2 }}{2}\)    M1

at \(\left( {1,{\text{ }}\sqrt {\frac{1}{2}} } \right)\) the tangent is \(y - \sqrt {\frac{1}{2}}  = \frac{{\sqrt 2 }}{2}(x - 1)\) and     A1

at \(\left( {1,{\text{ }} - \sqrt {\frac{1}{2}} } \right)\) the tangent is \(y + \sqrt {\frac{1}{2}}  =  - \frac{{\sqrt 2 }}{2}(x - 1)\)     A1

Note: These equations simplify to \(y =  \pm \frac{{\sqrt 2 }}{2}x\).

Note: Award A0M1A1A0 if just the positive value of \(y\) is considered and just one tangent is found.

[4 marks]

A curve has equation \(\arctan {x^2} + \arctan {y^2} = \frac{\pi }{4}\).

(a)     Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.

(b)     Find the gradient of the curve at the point where \(x = \frac{1}{{\sqrt 2 }}\) and \(y < 0\).

(a)     METHOD 1

\(\frac{{2x}}{{1 + {x^4}}} + \frac{{2y}}{{1 + {y^4}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1A1A1

Note:     Award M1 for implicit differentiation, A1 for LHS and A1 for RHS.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{x\left( {1 + {y^4}} \right)}}{{y\left( {1 + {x^4}} \right)}}\)     A1

METHOD 2

\({y^2} = \tan \left( {\frac{\pi }{4} - \arctan {x^2}} \right)\)

\( = \frac{{\tan \frac{\pi }{4} - \tan \left( {\arctan {x^2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan {x^2}} \right)} \right)}}\)     (M1)

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

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

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

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

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

[4 marks]

(b)     \({y^2} = \tan \left( {\frac{\pi }{4} - \arctan \frac{1}{2}} \right)\)     (M1)

\( = \frac{{\tan \frac{\pi }{4} - \tan \left( {\arctan \frac{1}{2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan \frac{1}{2}} \right)} \right)}}\)     (M1)

Note:     The two M1s may be awarded for working in part (a).

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

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

substitution into \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)

\( = \frac{{4\sqrt 6 }}{9}\)     A1

Note: Accept \(\frac{{8\sqrt 3 }}{{9\sqrt 2 }}\) etc.

[5 marks]

Total [9 marks]

Let \({x^3}y = a\sin nx\) . Using implicit differentiation, show that

\[{x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\] .

\({x^3}y = a\sin nx\)

attempt to differentiate implicitly     M1

\( \Rightarrow 3{x^2}y + {x^3}\frac{{{\text{d}}y}}{{{\text{d}}x}} = an\cos nx\)     A2 

Note: Award A1 for two out of three correct, A0 otherwise.

\( \Rightarrow 6xy + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\)     A2 

Note: Award A1 for three or four out of five correct, A0 otherwise.

\( \Rightarrow 6xy + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\)

\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 6xy + {n^2}{x^3}y = 0\)     A1

\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\)     AG

[6 marks]

Candidates who are comfortable using implicit differentiation found this to be a fairly straightforward question and were able to answer it in just a few lines. Many candidates, however, were unable to differentiate \({x^3}y\) with respect to x and were therefore unable to proceed. Candidates whose first step was to write \(y = \frac{{a\sin nx}}{{{x^3}}}\) were given no credit since the question required the use of implicit differentiation.

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

Show that \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^x}\cos x\).

Show that the function \(f\) has a local maximum value when \(x = \frac{{3\pi }}{4}\).

Find the \(x\)-coordinate of the point of inflexion of the graph of \(f\).

Sketch the graph of \(f\), clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.

Find the area of the region enclosed by the graph of \(f\) and the \(x\)-axis.

The curvature at any point \((x,{\text{ }}y)\) on a graph is defined as \(\kappa  = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\).

Find the value of the curvature of the graph of \(f\) at the local maximum point.

Find the value \(\kappa \) for \(x = \frac{\pi }{2}\) and comment on its meaning with respect to the shape of the graph.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x{\text{ }}\left( { = {{\text{e}}^x}(\sin x + \cos x)} \right)\)    M1A1

[2 marks]

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x - \sin x)\)    M1A1

\( = 2{{\text{e}}^x}\cos x\)    AG

[2 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{\frac{{3\pi }}{4}}}\left( {\sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4}} \right) = 0\)    R1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} < 0\)    R1

hence maximum at \(x = \frac{{3\pi }}{4}\)     AG

[2 marks]

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0 \Rightarrow 2{{\text{e}}^x}\cos x = 0\)    M1

\( \Rightarrow x = \frac{\pi }{2}\)    A1

Note: Award M1A0 if extra zeros are seen.

[2 marks]

correct shape and correct domain     A1

max at \(x = \frac{{3\pi }}{4}\), point of inflexion at \(x = \frac{\pi }{2}\)     A1

zeros at \(x = 0\) and \(x = \pi \)     A1

Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.

[3 marks]

EITHER

\(\int_0^x {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi  - \int_0^\pi  {{{\text{e}}^x}\cos x{\text{d}}x} } \)    M1A1

\(\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi  - \left( {[{{\text{e}}^x}\cos x]_0^x + \int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x} } \right)} \)    A1

OR

\(\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = [ - {{\text{e}}^x}\cos x]_0^\pi  + \int_0^\pi  {{{\text{e}}^x}\cos x{\text{d}}x} } \)    M1A1

\(\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = [ - {{\text{e}}^x}\cos x]} _0^\pi  + \left( {[{{\text{e}}^x}\sin x]_0^\pi  - \int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x} } \right)\)    A1

THEN

\(\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}\left( {[{{\text{e}}^x}\sin x]_0^x - [{{\text{e}}^x}\cos x]_0^x} \right)} \)    M1A1

\(\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}({{\text{e}}^x} + 1)} \)    A1

[6 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)    (A1)

 \(\frac{{{d^2}y}}{{d{x^2}}} = 2{e^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} =  - \sqrt 2 {e^{\frac{{3\pi }}{4}}}\) (A1)

\(\kappa  = \frac{{\left| { - \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}} \right|}}{1} = \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}\)    A1

[3 marks]

\(\kappa  = 0\)    A1

the graph is approximated by a straight line     R1

[2 marks]

Show that \(t = \frac{2}{3}(2\cos \theta + \theta )\).

Find the value of \(\theta \) for which \(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = 0\).

What route should Jorg take to travel from A to B in the least amount of time?

Give reasons for your answer.

angle APB is a right angle

\( \Rightarrow \cos \theta = \frac{{{\text{AP}}}}{4} \Rightarrow {\text{AP}} = 4\cos \theta \)     A1

Note: Allow correct use of cosine rule.

\({\text{arc PB}} = 2 \times 2\theta = 4\theta \)     A1

\(t = \frac{{{\text{AP}}}}{3} + \frac{{{\text{PB}}}}{6}\)     M1

Note: Allow use of their AP and their PB for the M1.

\( \Rightarrow t = \frac{{4\cos \theta }}{3} + \frac{{4\theta }}{6} = \frac{{4\cos \theta }}{3} + \frac{{2\theta }}{3} = \frac{2}{3}(2\cos \theta + \theta )\)     AG

[3 marks]

\(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = \frac{2}{3}( - 2\sin \theta + 1)\)     A1

\(\frac{2}{3}( - 2\sin \theta + 1) = 0 \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi }{6}\) (or 30 degrees)     A1

[2 marks]

\(\frac{{{{\text{d}}^2}t}}{{{\text{d}}{\theta ^2}}} = - \frac{4}{3}\cos \theta < 0\,\,\,\,\left( {{\text{at }}\theta = \frac{\pi }{6}} \right)\)     M1

\( \Rightarrow t\) is maximized at \(\theta = \frac{\pi }{6}\)     R1

time needed to walk along arc AB is \(\frac{{2\pi }}{6}{\text{ (}} \approx {\text{1 hour)}}\)

time needed to row from A to B is \(\frac{4}{3}{\text{ (}} \approx {\text{1.33 hour)}}\)

hence, time is minimized in walking from A to B     R1

[3 marks]

The fairly easy trigonometry challenged a large number of candidates.

Part (b) was very well done.

Satisfactory answers were very rarely seen for (c). Very few candidates realised that a minimum can occur at the beginning or end of an interval.

Using the definition of a derivative as \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\) , show that the derivative of \(\frac{1}{{2x + 1}}{\text{ is }}\frac{{ - 2}}{{{{(2x + 1)}^2}}}\).

Prove by induction that the \({n^{{\text{th}}}}\) derivative of \({(2x + 1)^{ - 1}}\) is \({( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\).

let \(f(x) = \frac{1}{{2x + 1}}\) and using the result \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\)

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{1}{{2(x + h) + 1}} - \frac{1}{{2x + 1}}}}{h}} \right)\)     M1A1

\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{[2x + 1] - [2(x + h) + 1]}}{{h[2(x + h) + 1][2x + 1]}}} \right)\)     A1

\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - 2}}{{[2(x + h) + 1][2x + 1]}}} \right)\)     A1

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

[4 marks]

let \(y = \frac{1}{{2x + 1}}\)

we want to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = {( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\)

let \(n = 1 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = {( - 1)^1}\frac{{{2^1}1!}}{{{{(2x + 1)}^{1 + 1}}}}\)     M1

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2}}{{{{(2x + 1)}^2}}}\) which is the same result as part (a)

hence the result is true for \(n = 1\)     R1

assume the result is true for \(n = k{\text{ : }}\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = {( - 1)^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}\)     M1

\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}} \right]\)     M1

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}{2^k}k!{{(2x + 1)}^{ - k - 1}}} \right]\)     (A1)

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^k}{2^k}k!( - k - 1){(2x + 1)^{ - k - 2}} \times 2\)     A1

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}{2^{k + 1}}(k + 1)!{(2x + 1)^{ - k - 2}}\)     (A1)

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}\frac{{{2^{k + 1}}(k + 1)!}}{{{{(2x + 1)}^{k + 2}}}}\)     A1

hence if the result is true for \(n = k\) , it is true for \(n = k + 1\)

since the result is true for \(n = 1\) , the result is proved by mathematical induction     R1 

Note: Only award final R1 if all the M marks have been gained.

[9 marks]

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.

Calculate the exact value of \(\int_1^{\text{e}} {{x^2}\ln x{\text{d}}x} \) .

Recognition of integration by parts     M1

\(\int {{x^2}\ln x{\text{d}}x = \left[ {\frac{{{x^3}}}{3}\ln x} \right] - \int {\frac{{{x^3}}}{3} \times \frac{1}{x}{\text{d}}x} } \)     A1A1

\( = \left[ {\frac{{{x^3}}}{3}\ln x} \right] - \int {\frac{{{x^2}}}{3}{\text{d}}x} \)

\( = \left[ {\frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9}} \right]\)     A1

\( \Rightarrow \int_1^{\text{e}} {{x^2}\ln x{\text{d}}x}  = \left( {\frac{{{{\text{e}}^3}}}{3} - \frac{{{{\text{e}}^3}}}{9}} \right) - \left( {0 - \frac{1}{9}} \right)\,\,\,\,\,\left( { = \frac{{2{{\text{e}}^3} + 1}}{9}} \right)\)     A1

[5 marks]

Most candidates recognised that a method of integration by parts was appropriate for this question. However, although a good number of correct answers were seen, a number of candidates made algebraic errors in the process. A number of students were also unable to correctly substitute the limits.

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]

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]

Find the area of the window in terms of P and \(r\).

Find the width of the window in terms of P when the area is a maximum, justifying that this is a maximum.

Show that in this case the height of the rectangle is equal to the radius of the semicircle.

the width of the rectangle is \(2r\) and let the height of the rectangle be \(h\)

\(P = 2r + 2h + \pi r\)     (A1)

\(A = 2rh + \frac{{\pi {r^2}}}{2}\)     (A1)

\(h = \frac{{{\text{P}} - 2r - \pi r}}{2}\)

\(A = 2r\left( {\frac{{{\text{P}} - 2r - \pi r}}{2}} \right) + \frac{{\pi {r^2}}}{2}\,\,\,\left( { = \operatorname{P} r - 2{r^2} - \frac{{\pi {r^2}}}{2}} \right)\)     M1A1

[4 marks]

\(\frac{{{\text{d}}A}}{{{\text{d}}r}} = {\text{P}} - 4r - \pi r\)     A1

\(\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0\)     M1

\( \Rightarrow r = \frac{{\text{P}}}{{4 + \pi }}\)     (A1)

hence the width is \(\frac{{2{\text{P}}}}{{4 + \pi }}\)     A1

\(\frac{{{{\text{d}}^2}A}}{{{\text{d}}{r^2}}} = - 4 - \pi < 0\)     R1

hence maximum     AG

[5 marks]

EITHER

\(h = \frac{{{\text{P}} - 2r - \pi r}}{2}\)

\(h = \frac{{{\text{P}} - \frac{{2{\text{P}}}}{{4 + \pi }} - \frac{{{\text{P}}\pi }}{{4 + \pi }}}}{2}\)     M1

\(h = \frac{{4{\text{P}} + \pi {\text{P}} - 2{\text{P}} - \pi {\text{P}}}}{{2(4 + \pi )}}\)    A1

\(h = \frac{{\text{P}}}{{(4 + \pi )}} = r\)     AG

OR

\(h = \frac{{{\text{P}} - 2r - \pi r}}{2}\)

\(P = r(4 + \pi )\)     M1

\(h = \frac{{r(4 + \pi ) - 2r - \pi r}}{2}\)     A1

\(h = \frac{{4r + \pi r - 2r - \pi r}}{2} = r\)     AG

[2 marks]

A tranquilizer is injected into a muscle from which it enters the bloodstream.

The concentration \(C\) in \({\text{mg}}{{\text{l}}^{ - 1}}\), of tranquilizer in the bloodstream can be modelled by the function \(C(t) = \frac{{2t}}{{3 + {t^2}}},{\text{ }}t \ge 0\) where \(t\) is the number of minutes after the injection.

Find the maximum concentration of tranquilizer in the bloodstream.

use of the quotient rule or the product rule     M1

\(C'(t) = \frac{{(3 + {t^2}) \times 2 - 2t \times 2t}}{{{{\left( {3 + {t^2}} \right)}^2}}}\;\;\;\left( { = \frac{{6 - 2{t^2}}}{{{{\left( {3 + {t^2}} \right)}^2}}}} \right)\;\;\;{\text{or}}\;\;\;\frac{2}{{3 + {t^2}}} - \frac{{4{t^2}}}{{{{\left( {3 + {t^2}} \right)}^2}}}\)     A1A1

Note:     Award A1 for a correct numerator and A1 for a correct denominator in the quotient rule, and A1 for each correct term in the product rule.

attempting to solve \(C'(t) = 0\;\;\;{\text{for }}t\)     (M1)

\(t =  \pm \sqrt 3 \;\;\;{\text{(minutes)}}\)     A1

\(C\left( {\sqrt 3 } \right) = \frac{{\sqrt 3 }}{3}\;\;\;\left( {{\text{mg}}{{\text{l}}^{ - 1}}} \right)\;\;\;{\text{or equivalent.}}\)     A1

[6 marks]

This question was generally well done. A significant number of candidates did not calculate the maximum value of \(C\).

Show that \(\cot \alpha  = \tan \left( {\frac{\pi }{2} - \alpha } \right)\) for \(0 < \alpha  < \frac{\pi }{2}\).

Hence find \(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha  < \frac{\pi }{2}} \).

EITHER

use of a diagram and trig ratios

eg,

\(\tan \alpha  = \frac{O}{A} \Rightarrow \cot \alpha  = \frac{A}{O}\)

from diagram, \(\tan \left( {\frac{\pi }{2} - \alpha } \right) = \frac{A}{O}\)     R1

OR

use of \(\tan \left( {\frac{\pi }{2} - \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} - \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} - \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}\)     R1

THEN

\(\cot \alpha  = \tan \left( {\frac{\pi }{2} - \alpha } \right)\)     AG

[1 mark]

\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x}  = [\arctan x]_{\tan \alpha }^{\cot \alpha }\)     (A1)

Note:     Limits (or absence of such) may be ignored at this stage.

\( = \arctan (\cot \alpha ) - \arctan (\tan \alpha )\)     (M1)

\( = \frac{\pi }{2} - \alpha  - \alpha \)     (A1)

\( = \frac{\pi }{2} - 2\alpha \)     A1

[4 marks]

This was generally well done.

This was generally well done. Some weaker candidates tried to solve part (b) through use of a substitution, though the standard result \(\arctan x\) was well known. A small number used \(\arctan x + c\) and went on to obtain an incorrect final answer.

\(\int_{ - 2}^0 {\left( {f\left( x \right){\text{ + 2}}} \right){\text{d}}x} \).

\(\int_{ - 2}^0 {f\left( {x{\text{ + 2}}} \right){\text{d}}x} \).

\(\int_{ - 2}^0 {f\left( x \right){\text{d}}x = 10}  - 12 =  - 2\)     (M1)(A1)

\(\int_{ - 2}^0 {2{\text{d}}x = \left[ {2x} \right]} _{ - 2}^0 = 4\)     A1

\(\int_{ - 2}^0 {\left( {f\left( x \right){\text{ + 2}}} \right){\text{d}}x}  = 2\)     A1

[4 marks]

\(\int_{ - 2}^0 {f\left( {x{\text{ + 2}}} \right){\text{d}}x}  = \int_0^2 {f\left( x \right){\text{d}}x} \)    (M1)

= 12     A1

[2 marks]

By using the substitution \(t = \tan x\), find \(\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}}} \).

Express your answer in the form \(m\arctan (n\tan x) + c\), where \(m\), \(n\) are constants to be determined.

EITHER

\(x = \arctan t\)     (M1)

\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{1}{{1 + {t^2}}}\)     A1

OR

\(t = \tan x\)

\(\frac{{{\text{d}}t}}{{{\text{d}}x}} = {\sec ^2}x\)     (M1)

\( = 1 + {\tan ^2}x\)     A1

\( = 1 + {t^2}\)

THEN

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

Note:     This A1 is independent of the first two marks

\(\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}} = \int {\frac{{\frac{{{\text{d}}t}}{{1 + {t^2}}}}}{{1 + {{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}} } \)     M1A1

Note:     Award M1 for attempting to obtain integral in terms of \(t\) and \({\text{d}}t\)

\( = \int {\frac{{{\text{d}}t}}{{(1 + {t^2}) + {t^2}}} = \int {\frac{{{\text{d}}t}}{{1 + 2{t^2}}}} } \)     A1

\( = \frac{1}{2}\int {\frac{{{\text{d}}t}}{{\frac{1}{2} + {t^2}}} = \frac{1}{2} \times \frac{1}{{\frac{1}{{\sqrt 2 }}}}\arctan \left( {\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right)} \)     A1

\( = \frac{{\sqrt 2 }}{2}\arctan \left( {\sqrt 2 \tan x} \right)( + c)\)     A1

[8 marks]

Find the equation of the normal to the curve \(5x{y^2} - 2{x^2} = 18\) at the point (1, 2) .

\(5{y^2} + 10xy\frac{{{\text{d}}y}}{{{\text{d}}x}} - 4x = 0\)     A1A1A1

Note: Award A1A1 for correct differentiation of \(5x{y^2}\).

A1 for correct differentiation of \( - 2{x^2}\) and 18.

At the point (1, 2), \(20 + 20\frac{{{\text{d}}y}}{{{\text{d}}x}} - 4 = 0\)

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{4}{5}\)     (A1)

Gradient of normal \( = \frac{5}{4}\)     A1

Equation of normal \(y - 2 = \frac{5}{4}(x - 1)\)     M1

\(y = \frac{5}{4}x - \frac{5}{4} + \frac{8}{4}\)

\(y = \frac{5}{4}x + \frac{3}{4}\,\,\,\,\,(4y = 5x + 3)\)     A1

[7 marks]

It was pleasing to see that a significant number of candidates understood that implicit differentiation was required and that they were able to make a reasonable attempt at this. A small number of candidates tried to make the equation explicit. This method will work, but most candidates who attempted this made either arithmetic or algebraic errors, which stopped them from gaining the correct answer.

Show that \(f'(x) = \frac{{1 - \ln x}}{{{x^2}}}\).

Find the coordinates of B, at which the curve reaches its maximum value.

Find the coordinates of C, the point of inflexion on the curve.

The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.

Find the equation of the tangent to the graph of \(f\) at the point A.

The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.

Find the area enclosed by the curve \(y = f(x)\), the tangent at A, and the line \(x = {\text{e}}\).

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

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

[2 marks]

\(\frac{{1 - \ln x}}{{{x^2}}} = 0\) has solution \(x = {\text{e}}\)     M1A1

\(y = \frac{1}{{\text{e}}}\)     A1

hence maximum at the point \(\left( {{\text{e, }}\frac{1}{{\text{e}}}} \right)\)

[3 marks]

\(f''(x) = \frac{{{x^2}\left( { - \frac{1}{x}} \right) - 2x(1 - \ln x)}}{{{x^4}}}\)     M1A1

\( = \frac{{2\ln x - 3}}{{{x^3}}}\)

Note:     The M1A1 should be awarded if the correct working appears in part (b).

point of inflexion where \(f''(x) = 0\)     M1

so \(x = {{\text{e}}^{\frac{3}{2}}},{\text{ }}y = \frac{3}{2}{{\text{e}}^{\frac{{ - 3}}{2}}}\)     A1A1

C has coordinates \(\left( {{{\text{e}}^{\frac{3}{2}}},{\text{ }}\frac{3}{2}{{\text{e}}^{\frac{{ - 3}}{2}}}} \right)\)

[5 marks]

\(f(1) = 0\)     A1

\(f'(1) = 1\)     (A1)

\(y = x + c\)     (M1)

through (1, 0)

equation is \(y = x - 1\)     A1

[4 marks]

METHOD 1

area \( = \int_1^{\text{e}} {x - 1 - \frac{{\ln x}}{x}{\text{d}}x} \)     M1A1A1

Note:     Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.

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

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

\( = \left[ {\frac{1}{2}{x^2} - x - \frac{1}{2}{{(\ln x)}^2}} \right]_1^{\text{e}}\)

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

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

METHOD 2

area = area of triangle \( - \int_1^e {\frac{{\ln x}}{x}{\text{d}}x} \)     M1A1

Note:     A1 is for correct integral with limits and is dependent on the M1.

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

area of triangle \( = \frac{1}{2}(e - 1)(e - 1)\)     M1A1

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

[7 marks]

The region enclosed between the curves \(y = \sqrt x {{\text{e}}^x}\) and \(y = {\text{e}}\sqrt x \) is rotated through \(2\pi \) about the x-axis. Find the volume of the solid obtained.

\(\sqrt x {{\text{e}}^x} = {\text{e}}\sqrt x  \Rightarrow x = 0{\text{ or 1}}\)     (A1)

attempt to find \(\int {{y^2}{\text{d}}x} \)     M1

\({V_1} = \pi \int_0^1 {{{\text{e}}^2}x{\text{d}}x} \)

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

\( = \frac{{\pi {{\text{e}}^2}}}{2}\)     A1

\({V_2} = \pi \int_0^1 {x{{\text{e}}^{2x}}{\text{d}}x} \)

\( = \pi \left( {\left[ {\frac{1}{2}x{{\text{e}}^{2x}}} \right]_0^1 - \int_0^1 {\frac{1}{2}{{\text{e}}^{2x}}{\text{d}}x} } \right)\)     M1A1

Note: Award M1 for attempt to integrate by parts.

\( = \frac{{\pi {{\text{e}}^2}}}{2} - \pi \left[ {\frac{1}{4}{{\text{e}}^{2x}}} \right]_0^1\)

finding difference of volumes     M1

volume\( = {V_1} - {V_2}\)

\( = \pi \left[ {\frac{1}{4}{{\text{e}}^{2x}}} \right]_0^1\)

\( = \frac{1}{4}\pi ({{\text{e}}^2} - 1)\)     A1

[7 marks]

While only a minority of candidates achieved full marks in this question, many candidates made good attempts. Quite a few candidates obtained the limits correctly and many realized a square was needed in the integral, though a number of them subtracted then squared rather than squaring and then subtracting. There was evidence that quite a few knew about integration by parts. One common mistake was to have \(2\pi \), rather than \(\pi \) in the integral.

Show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\cos }^2}x - {x^2}\sin x}}{{{{(x + \cos x)}^2}}},{\text{ }}x \geqslant 0\).

Find the equation of the tangent to C at the point \(\left( {\frac{\pi }{2},0} \right)\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(x + \cos x)(\cos x - x\sin x) - x\cos x(1 - \sin x)}}{{{{(x + \cos x)}^2}}}\)     M1A1A1

Note: Award M1 for attempt at differentiation of a quotient and a product condoning sign errors in the quotient formula and the trig differentiations, A1 for correct derivative of “u”, A1 for correct derivative of “v”.

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

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

[4 marks]

the derivative has value –1     (A1)

the equation of the tangent line is \((y - 0) = ( - 1)\left( {x - \frac{\pi }{2}} \right)\left( {y = \frac{\pi }{2} - x} \right)\)     M1A1

[3 marks]

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining \( - \sin x\) rather than \(1 - \sin x\). A disappointing number of candidates failed to calculate the correct gradient at the specified point.

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining \( - \sin x\) rather than \(1 - \sin x\). A disappointing number of candidates failed to calculate the correct gradient at the specified point.

Find \(\int {(1 + {{\tan }^2}x){\text{d}}x} \).

Find \(\int {{{\sin }^2}x{\text{d}}x} \).

\(\int {(1 + {{\tan }^2}x){\text{d}}x}  = \int {{{\sec }^2}x{\text{d}}x = \tan x( + c)} \)     M1A1

[2 marks]

\(\int {{{\sin }^2}x{\text{d}}x}  = \int {\frac{{1 - \cos 2x}}{2}{\text{d}}x} \)     M1A1

\( = \frac{x}{2} - \frac{{\sin 2x}}{4}( + c)\)     A1

Note:     Allow integration by parts followed by trig identity.

Award M1 for parts, A1 for trig identity, A1 final answer.

[3 marks]

Total [5 marks]

Some correct answers but too many candidates had a poor approach and did not use the trig identity.

Same as (a).

By using the substitution \(u = 1 + \sqrt x \), find \(\int {\frac{{\sqrt x }}{{1 + \sqrt x }}{\text{d}}x} \).

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{{2\sqrt x }}\)     A1

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

Note:     Award the A1 for any correct relationship between \({\text{d}}x\) and \({\text{d}}u\).

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

Note:     Award the M1 for an attempt at substitution resulting in an integral only involving \(u\) .

\( = 2\int {u - 2 + \frac{1}{u}{\text{d}}u} \)     (A1)

\( = {u^2} - 4u + 2\ln u( + C)\)     A1

\( = x - 2\sqrt x  - 3 + 2\ln \left( {1 + \sqrt x } \right)( + C)\)     A1

Note:     Award the A1 for a correct expression in \(x\), but not necessarily fully expanded/simplified.

[6 marks]

Many candidates worked through this question successfully. A significant minority either made algebraic mistakes with the substitution or tried to work with an integral involving both \(x\) and \(u\).

Find \(\frac{{{\text{d}}y}}{{{\text{d}}\theta }}\)

Hence find the values of θ for which \(\frac{{{\text{d}}y}}{{{\text{d}}\theta }} = 2y\).

attempt at chain rule or product rule     (M1)

\(\frac{{{\text{d}}y}}{{{\text{d}}\theta }} = 2\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta \)     A1

[2 marks]

\(2\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta  = 2{\text{si}}{{\text{n}}^2}\theta \)

sin θ = 0     (A1)

θ = 0, \(\pi \)     A1

obtaining cos θ = sin θ     (M1)

tan θ = 1     (M1)

\(\theta  = \frac{\pi }{4}\)     A1

[5 marks]

Find \({t_1}\) and \({t_2}\).

Find the displacement of the particle when \(t = {t_1}\)

\(s = t + \cos 2t\)

\(\frac{{{\text{d}}s}}{{{\text{d}}t}} = 1 - 2\sin 2t\)     M1A1

\( = 0\)     M1

\( \Rightarrow \sin 2t = \frac{1}{2}\)

\({t_1} = \frac{\pi }{{12}}(s),{\text{ }}{t_2} = \frac{{5\pi }}{{12}}(s)\)     A1A1

Note:     Award A0A0 if answers are given in degrees.

[5 marks]

\(s = \frac{\pi }{{12}} + \cos \frac{\pi }{6}\,\,\,\left( {s = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2}(m)} \right)\)     A1A1

[2 marks]

Using the substitution \(x = \tan \theta \) show that \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x = } \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } \).

Hence find the value of \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x} \).

let \(x = \tan \theta \)

\( \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = {\sec ^2}\theta \)     (A1)

\(\int {\frac{1}{{{{({x^2} + 1)}^2}}}{\text{d}}x = \int {\frac{{{{\sec }^2}\theta }}{{{{({{\tan }^2}\theta + 1)}^2}}}{\text{d}}\theta } } \)     M1

Note:     The method mark is for an attempt to substitute for both \(x\) and \({\text{d}}x\).

\( = \int {\frac{1}{{{{\sec }^2}\theta }}{\text{d}}\theta } \) (or equivalent)     A1

when \(x = 0,{\text{ }}\theta = 0\) and when \(x = 1,{\text{ }}\theta = \frac{\pi }{4}\)     M1

\(\int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } \)    AG

[4 marks]

\(\left( {\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x}  = \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right) = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \cos 2\theta } \right){\text{d}}\theta } \)   M1

\( = \frac{1}{2}\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right]_0^{\frac{\pi }{4}}\)     A1

\( = \frac{\pi }{8} + \frac{1}{4}\)     A1

[3 marks]

Use the substitution \(u = \ln x\) to find the value of \(\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}} \).

METHOD 1

\(\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}}  = \left[ {\ln (\ln x)} \right]_{\text{e}}^{{{\text{e}}^2}}\)     (M1)A1

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

\( = \ln 2\)     A1

[4 marks]

METHOD 2

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

\( = \int_1^2 {\frac{{{\text{d}}u}}{u}} \)     A1

Note:     Condone absent or incorrect limits here.

\( = [\ln u]_1^2\) or equivalent in \(x( = \ln 2 - \ln 1)\)     (A1)

\( = \ln 2\)     A1

[4 marks]

Use the identity \(\cos 2\theta  = 2{\cos ^2}\theta  - 1\) to prove that \(\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} ,{\text{ }}0 \leqslant x \leqslant \pi \).

Find a similar expression for \(\sin \frac{1}{2}x,{\text{ }}0 \leqslant x \leqslant \pi \).

Hence find the value of \(\int_0^{\frac{\pi }{2}} {\left( {\sqrt {1 + \cos x}  + \sqrt {1 - \cos x} } \right){\text{d}}x} \).

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

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

positive as \(0 \leqslant x \leqslant \pi \)     R1

\(\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} \)     AG

[2 marks]

\(\cos 2\theta  = 1 - 2{\sin ^2}\theta \)     (M1)

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

[2 marks]

\(\sqrt 2 \int_0^{\frac{\pi }{2}} {\cos \frac{1}{2}x + \sin \frac{1}{2}x{\text{d}}x} \)     A1

\( = \sqrt 2 \left[ {2\sin \frac{1}{2}x - 2\cos \frac{1}{2}x} \right]_0^{\frac{\pi }{2}}\)     A1

\( = \sqrt 2 (0) - \sqrt 2 (0 - 2)\)     A1

\( = 2\sqrt 2 \)     A1

[4 marks]

Paint is poured into a tray where it forms a circular pool with a uniform thickness of 0.5 cm. If the paint is poured at a constant rate of \(4{\text{ c}}{{\text{m}}^3}{{\text{s}}^{ - 1}}\), find the rate of increase of the radius of the circle when the radius is 20 cm.

\(V = 0.5\pi {r^2}\)     (A1)

EITHER

\(\frac{{dV}}{{dr}} = \pi r\)     A1

\(\frac{{dV}}{{dt}} = 4\)     (A1)

applying chain rule     M1

for example \(\frac{{dr}}{{dt}} = \frac{{dV}}{{dt}} \times \frac{{dr}}{{dV}}\)

OR

\(\frac{{dV}}{{dt}} = \pi r\frac{{dr}}{{dt}}\)     M1A1

\(\frac{{dV}}{{dt}} = 4\)     (A1)

THEN

\(\frac{{dr}}{{dt}} = 4 \times \frac{1}{{\pi r}}\)     A1

when \(r = 20,{\text{ }}\frac{{dr}}{{dt}} = \frac{4}{{20\pi }}{\text{ or }}\frac{1}{{5\pi }}{\text{ (cm}}\,{{\text{s}}^{ - 1}})\)     A1

Note: Allow h instead of 0.5 up until the final A1.

[6 marks]

There was a large variety of methods used in this question, with most candidates choosing to implicitly differentiate the expression for volume in terms of r.

\(p'(3)\);

\(h'(2)\).

\(p'(3) = f'(3)g(3) + g'(3)f(3)\)     (M1)

Note:     Award M1 if the derivative is in terms of \(x\) or \(3\).

\( = 2 \times 4 + 3 \times 1\)

\( = 11\)     A1

[2 marks]

\(h'(x) = g'\left( {f(x)} \right)f'(x)\)     (M1)(A1)

\(h'(2) = g'(1)f'(2)\)     A1

\( = 4 \times 4\)

\( = 16\)     A1

[4 marks]

Total [6 marks]

This was a problem question for many candidates. Some quite strong candidates, on the evidence of their performance on other questions, did not realise that ‘composite functions’ and ‘functions of a function’ were the same thing, and therefore that the chain rule applied.

This was a problem question for many candidates. Some quite strong candidates, on the evidence of their performance on other questions, did not realise that ‘composite functions’ and ‘functions of a function’ were the same thing, and therefore that the chain rule applied.

The region bounded by the curve \(y = \frac{{\ln (x)}}{x}\) and the lines x = 1, x = e, y = 0 is rotated through \(2\pi \) radians about the x-axis.

Find the volume of the solid generated.

METHOD 1

\(V = \pi \int_1^{\text{e}} {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \)     M1

Integrating by parts:

\(u = {(\ln x)^2},{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{{2\ln x}}{x},{\text{ }}v = - \frac{1}{x}\)

\( \Rightarrow V = \pi \left( { - \frac{{{{(\ln x)}^2}}}{x} + 2\int {\frac{{\ln x}}{{{x^2}}}{\text{d}}x} } \right)\)     A1

\(u = \ln x,{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x},{\text{ }}v = - \frac{1}{x}\)

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

\(\therefore V = \pi \left[ { - \frac{{{{(\ln x)}^2}}}{x} + 2\left( { - \frac{{\ln x}}{x} - \frac{1}{x}} \right)} \right]_1^{\text{e}}\)

\( = 2\pi - \frac{{5\pi }}{e}\)     A1

[6 marks]

METHOD 2

\(V = \pi \int_1^{\text{e}} {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \)     M1

Let \(\ln x = u \Rightarrow x = {{\text{e}}^u},{\text{ }}\frac{{{\text{d}}x}}{x} = {\text{d}}u\)     (M1)

\(\int {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x = \int {\frac{{{u^2}}}{{{{\text{e}}^u}}}{\text{d}}u = \int {{{\text{e}}^{ - u}}} {u^2}{\text{d}}u = - {{\text{e}}^{ - u}}{u^2} + 2\int {{{\text{e}}^{ - u}}u{\text{d}}u} } } \)     A1

\( = - {{\text{e}}^{ - u}}{u^2} + 2\left( { - {{\text{e}}^{ - u}}u + \int {{{\text{e}}^{ - u}}{\text{d}}u} } \right) = - {{\text{e}}^{ - u}}{u^2} - 2{{\text{e}}^{ - u}}u - 2{{\text{e}}^{ - u}}\)

\( = - {{\text{e}}^{ - u}}({u^2} + 2u + 2)\)     A1

When x = e, u = 1. When x = 1, u = 0 .

\(\therefore {\text{ Volume}} = \pi \left[ { - {{\text{e}}^{ - u}}({u^2} + 2u + 2)} \right]_0^1\)     M1

\( = \pi ( - 5{{\text{e}}^{ - 1}} + 2){\text{ }}\left( { = 2\pi  - \frac{{5\pi }}{{\text{e}}}} \right)\)     A1

[6 marks]

Only the best candidates were able to make significant progress with this question. It was disappointing to see that many candidates could not state that the formula for the required volume was \(\pi \int_1^e {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \) . Of those who could, very few either attempted integration by parts or used an appropriate substitution.

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

Determine the equation of the normal to the curve at the point \(x = 3\) in the form \(ax + by + c = 0\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(1 - x)^{ - 2}}\;\;\;\left( { = \frac{1}{{{{(1 - x)}^2}}}} \right)\)     (M1)A1

[2 marks]

gradient of Tangent \( = \frac{1}{4}\)     (A1)

gradient of Normal \( =  - 4\)     (M1)

\(y + \frac{1}{2} =  - 4(x - 3)\) or attempt to find \(c\) in \(y = mx + c\)     M1

\(8x + 2y - 23 = 0\)     A1

[4 marks]

Total [6 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).

If \(f(x) = x - 3{x^{\frac{2}{3}}},{\text{ }}x > 0\) ,

(a)     find the x-coordinate of the point P where \(f'(x) = 0\) ;

(b)     determine whether P is a maximum or minimum point.

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

\( \Rightarrow 1 - \frac{2}{{{x^{\frac{1}{3}}}}} = 0 \Rightarrow {x^{\frac{1}{3}}} = 2 \Rightarrow x = 8\)     A1

(b)     \(f''(x) = \frac{2}{{3{x^{\frac{4}{3}}}}}\)     A1

\(f''(8) > 0 \Rightarrow {\text{ at }}x = 8,{\text{ }}f(x){\text{ has a minimum.}}\)     M1A1

[5 marks]

Most candidates were able to correctly differentiate the function and find the point where \(f'(x) = 0\) . They were less successful in determining the nature of the point.

The normal to the curve \(x{{\text{e}}^{ - y}} + {{\text{e}}^y} = 1 + x\), at the point (c, \(\ln c\)), has a y-intercept \({c^2} + 1\).

Determine the value of c.

EITHER

differentiating implicitly:

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

at the point (c, \(\ln c\))

\(\frac{1}{c} - c \times \frac{1}{c}\frac{{{\text{d}}y}}{{{\text{d}}x}} + c\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{c}\,\,\,\,\,(c \ne 1)\)     (A1)

OR

reasonable attempt to make expression explicit     (M1)

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

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

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

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

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

\(y = 0,{\text{ }}y = \ln x\)     A1

Note: Do not penalize if y = 0 not stated.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{x}\)

gradient of tangent \( = \frac{1}{c}\)     A1

Note: If candidate starts with \(y = \ln x\) with no justification, award (M0)(A0)A1A1.

THEN

the equation of the normal is

\(y - \ln c = - c(x - c)\)     M1

\(x = 0,{\text{ }}y = {c^2} + 1\)

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

\(\ln c = 1\)

\(c = {\text{e}}\)     A1

[7 marks]

This was the first question to cause the majority of candidates a problem and only the better candidates gained full marks. Weaker candidates made errors in the implicit differentiation and those who were able to do this often were unable to simplify the expression they gained for the gradient of the normal in terms of c; a significant number of candidates did not know how to simplify the logarithms appropriately.

Express \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.

Find the value of \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) at the point on C where y = 1 and \(x > 0\).

attempt at implicit differentiation     M1

EITHER

\(\frac{{2x}}{y} - \frac{{{x^2}}}{{{y^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - 2 = \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     A1A1

Note: Award A1 for each side.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\frac{{2x}}{y} - 2}}{{\frac{1}{y} + \frac{{{x^2}}}{{{y^2}}}}}{\text{ }}\left( { = \frac{{2xy - 2{y^2}}}{{{x^2} + y}}} \right)\)     A1

OR

after multiplication by y

\(2x - 2y - 2x\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}x}}\ln y + y\frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     A1A1

Note: Award A1 for each side.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2(x - y)}}{{1 + 2x + \ln y}}\)     A1

[4 marks]

for \(y = 1,{\text{ }}{x^2} - 2x = 0\)

\(x = (0{\text{ or) 2}}\)     A1

for \(x = 2\), \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{2}{5}\)     A1

[2 marks]

Most candidates were familiar with the concept of implicit differentiation and the majority found the correct derivative function. In part (b), a significant number of candidates didn’t realise that the value of x was required.

Most candidates were familiar with the concept of implicit differentiation and the majority found the correct derivative function. In part (b), a significant number of candidates didn’t realise that the value of x was required.

Find the coordinates of the points A and B.

Given that the graph of the function has exactly one point of inflexion, find its coordinates.

\(f'(x) = {(\ln x)^2} + \frac{{2x\ln x}}{x}\left( { = {{(\ln x)}^2} + 2\ln x = \ln x(\ln x + 2)} \right)\)     M1A1

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

Note: Award M1 for an attempt to solve \(f'(x) = 0\).

\(A({e^{ - 2}},\,4{e^{ - 2}})\) and B(1, 0)     A1A1

Note: The final A1 is independent of prior working.

[5 marks]

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

\(f''(x) = 0{\text{ }}\left( { \Rightarrow x = {e^{ - 1}}} \right)\)     (M1)

inflexion point \(({e^{ - 1}},{\text{ }}{e^{ - 1}})\)     A1 

Note: M1 for attempt to solve \(f''(x) = 0\).

[3 marks]

This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.

This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.

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]

Show that \(f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) .

Obtain a similar expression for \({f^{(4)}}(x)\) .

Suggest an expression for \({f^{(2n)}}(x)\), \(n \in {\mathbb{Z}^ + }\), and prove your conjecture using mathematical induction.

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

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

\( = 2{{\text{e}}^x}\cos x\)     A1

\( = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\)     AG

[3 marks]

\(f'''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\)     A1

\({f^{(4)}}(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) - 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\)     A1

\( = 4{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\)     A1

\( = 4{{\text{e}}^x}\sin (x + \pi )\)     A1

[4 marks]

the conjecture is that

\({f^{(2n)}}(x) = {2^n}{{\text{e}}^x}\sin \left( {x + \frac{{n\pi }}{2}} \right)\)     A1

for n  = 1, this formula gives

\(f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) which is correct     A1

let the result be true for n = k , \(\left( {i.e.{\text{ }}{f^{(2k)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)} \right)\)     M1

consider \({f^{(2k + 1)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\)     M1

\({f^{\left( {2(k + 1)} \right)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) - {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)\)     A1

\( = {2^{k + 1}}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\)     A1

\( = {2^{k + 1}}{{\text{e}}^x}\sin \left( {x + \frac{{(k + 1)\pi }}{2}} \right)\)     A1

therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 1\)

the result is proved by induction.     R1

Note: Award the final R1 only if the two M marks have been awarded.

[8 marks]

The function f is defined by \(f(x) = x{{\text{e}}^{2x}}\) .

It can be shown that \({f^{(n)}}(x) = ({2^n}x + n{2^{n - 1}}){{\text{e}}^{2x}}\) for all \(n \in {\mathbb{Z}^ + }\), where \({f^{(n)}}(x)\) represents the \({n^{{\text{th}}}}\) derivative of \(f(x)\) .

(a)     By considering \({f^{(n)}}(x){\text{ for }}n = 1{\text{ and }}n = 2\) , show that there is one minimum point P on the graph of f , and find the coordinates of P.

(b)     Show that f has a point of inflexion Q at x = −1.

(c)     Determine the intervals on the domain of f where f is

(i)     concave up;

(ii)     concave down.

(d)     Sketch f , clearly showing any intercepts, asymptotes and the points P and Q.

(e)     Use mathematical induction to prove that \({f^{(n)}}(x) = ({2^n}x + n{2^{n - 1}}){{\text{e}}^{2x}}{\text{ for all }}n \in {\mathbb{Z}^ + },{\text{ where }}{f^{(n)}}{\text{ represents the }}{n^{{\text{th}}}}{\text{ derivative of }}f(x)\) .

(a)     \(f'(x) = (1 + 2x){{\text{e}}^{2x}}\)     A1

\(f'(x) = 0\)     M1

\( \Rightarrow (1 + 2x){{\text{e}}^{2x}} = 0 \Rightarrow x = - \frac{1}{2}\)     A1

\(f''(x) = ({2^2}x + 2 \times {2^{2 - 1}}){{\text{e}}^{2x}} = (4x + 4){{\text{e}}^{2x}}\)     A1

\(f''\left( { - \frac{1}{2}} \right) = \frac{2}{{\text{e}}}\)     A1

\(\frac{2}{{\text{e}}} > 0 \Rightarrow {\text{at }}x = - \frac{1}{2},{\text{ }}f(x){\text{ has a minimum.}}\)     R1

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

[7 marks]

(b)     \(f''(x) = 0 \Rightarrow 4x + 4 = 0 \Rightarrow x = - 1\)     M1A1

\({\text{Using the }}{{\text{2}}^{{\text{nd}}}}{\text{ derivative }}f''\left( { - \frac{1}{2}} \right) = \frac{2}{{\text{e}}}{\text{ and }}f''( - 2) = - \frac{4}{{{{\text{e}}^4}}},\)     M1A1

the sign change indicates a point of inflexion.     R1

[5 marks]

(c)     (i)     f(x) is concave up for \(x > - 1\) .     A1

(ii)     f(x) is concave down for \(x < - 1\) .     A1

[2 marks]

(d)

     A1A1A1A1

Note: Award A1 for P and Q, with Q above P,

A1 for asymptote at y = 0 ,

A1 for (0, 0) ,

A1 for shape.

[4 marks]

(e)     Show true for n = 1     (M1)

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

\( = {{\text{e}}^{2x}}(1 + 2x) = (2x + {2^0}){{\text{e}}^{2x}}\)

Assume true for \(n = k{\text{ , }}i.e.{\text{ }}{f^{(k)}}x = ({2^k}x + k \times {2^{k - 1}}){{\text{e}}^{2x}}{\text{, }}k \geqslant 1\)     M1A1

Consider \(n = k + 1{\text{ , }}i.e.{\text{ an attempt to find }}\frac{{\text{d}}}{{{\text{d}}x}}\left( {{f^k}(x)} \right)\) .     M1

\({F^{(k + 1)}}(x) = {2^k}{{\text{e}}^{2x}} + 2{{\text{e}}^{2x}}({2^k}x + k \times {2^{k - 1}})\)     A1

\( = \left( {{2^k} + 2({2^k}x + k \times {2^{k - 1}})} \right){{\text{e}}^{2x}}\)

\( = (2 \times {2^k}x + {2^k} + k \times 2 \times {2^{k - 1}}){{\text{e}}^{2x}}\)

\( = ({2^{k + 1}}x + {2^k} + k \times {2^k}){{\text{e}}^{2x}}\)     A1
\( = \left( {{2^{k + 1}}x + \left( {k + 1} \right){2^k}} \right){{\text{e}}^{2x}}\)     A1

P(n) is true for \(k \Rightarrow P(n)\) is true for k + 1, and since true for n = 1, result proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\)

Note: Only award R1 if a reasonable attempt is made to prove the \({(k + 1)^{{\text{th}}}}\) step.

[9 marks]

Total [27 marks]

This was the most accessible question in section B for these candidates. A majority of candidates produced partially correct answers to part (a), but a significant number struggled with demonstrating that the point is a minimum, despite the hint being given in the question. Part (b) started quite successfully but many students were unable to prove it is a point of inflexion or, more commonly, did not attempt to justify it. Correct answers were often seen for part (c). Part (d) was dependent on the successful completion of the first three parts. If candidates made errors in earlier parts, this often became obvious when they came to sketch the curve. However, few candidates realised that this part was a good way of checking that the above answers were at least consistent. The quality of curve sketching was rather weak overall, with candidates not marking points appropriately and not making features such as asymptotes clear. It is not possible to tell to what extent this was an effect of candidates not having a calculator, but it should be noted that asking students to sketch curves without a calculator will continue to appear on non-calculator papers. In part (e) the basic idea of proof by induction had clearly been taught with a significant majority of students understanding this. However, many candidates did not understand that they had to differentiate again to find the result for (k + 1) .

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]

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]

Find \(\int {\arcsin x\,{\text{d}}x} \)

attempt at integration by parts with \(u = \arcsin x\) and \(v' = 1\)     M1

\(\int {\arcsin x\,{\text{d}}x}  = x\arcsin x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}{\text{d}}x} {\text{ }}\)    A1A1

Note:     Award A1 for \(x\arcsin x\) and A1 for \( - \int {\frac{x}{{\sqrt {1 - {x^2}} }}{\text{d}}x} \).

solving \(\int {\frac{x}{{\sqrt {1 - {x^2}} }}{\text{d}}x} \) by substitution with \(u = 1 - {x^2}\) or inspection     (M1)

\(\int {\arcsin x{\text{d}}x} = x\arcsin x + \sqrt {1 - {x^2}} + c\)   A1

[5 marks]

Show that \(\int_0^{\frac{\pi }{6}} {x\sin 2x{\text{d}}x = \frac{{\sqrt 3 }}{8} - \frac{\pi }{{24}}} \).

Using integration by parts     (M1)

\(u = x{\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1,{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \sin 2x{\text{ and }}v = - \frac{1}{2}\cos 2x\)     (A1)

\(\left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} - \int_0^{\frac{\pi }{6}} {\left( { - \frac{1}{2}\cos 2x} \right){\text{d}}x} \)     A1

\( = \left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} + \left[ {\frac{1}{4}\sin 2x} \right]_0^{\frac{\pi }{6}}\)     A1

Note: Award the A1A1 above if the limits are not included.

\(\left[ {x\left( { - \frac{1}{2}\cos 2x} \right)} \right]_0^{\frac{\pi }{6}} = - \frac{\pi }{{24}}\)     A1

\(\left[ {\frac{1}{4}\sin 2x} \right]_0^{\frac{\pi }{6}} = \frac{{\sqrt 3 }}{8}\)     A1

\(\int_0^{\frac{\pi }{6}} {x\sin 2x{\text{d}}x = \frac{{\sqrt 3 }}{8} - \frac{\pi }{{24}}} \)     AG     N0

Note: Allow FT on the last two A1 marks if the expressions are the negative of the correct ones.

[6 marks]

This question was reasonably well done, with few candidates making the inappropriate choice of u and \(\frac{{{\text{d}}v}}{{{\text{d}}x}}\). The main source of a loss of marks was in finding v by integration. A few candidates used the double angle formula for sine, with poor results.

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.

The function \(f\) is defined as \(f(x) = a{x^2} + bx + c\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{R}\).

Hayley conjectures that \(\frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = \frac{{f'({x_2}) + f'({x_1})}}{2},{\text{ }}x1 \ne x2\).

Show that Hayley’s conjecture is correct.

\(\frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = \frac{{ax_2^2 + b{x_2} + c - (ax_1^2 + b{x_1} + c)}}{{{x_2} - {x_1}}}\)     (M1)

\( = \frac{{a(x_2^2 - x_1^2) + b({x_2} - {x_1})}}{{{x_2} - {x_1}}}\)     A1

\( = \frac{{a({x_2} - {x_1})({x_2} + {x_1}) + b({x_2} - {x_1})}}{{{x_2} - {x_1}}}\)     (A1)

\( = a({x_2} + {x_1}) + b\,\,\,\,\,({x_1} \ne {x_2})\)     A1

\(\frac{{f'({x_2}) + f'({x_1})}}{2} = \frac{{(2a{x_2} + b) + (2a{x_1} + b)}}{2}\)     M1

\( = \frac{{2a({x_2} + {x_1}) + 2b}}{2}\)

\( = a({x_2} + {x_1}) + b\)     A1

so Hayley’s conjecture is correct     AG

[6 marks]

This was generally answered very well. A small minority attempted to ‘prove’ the result by substituting specific values into the identity and thus gained little or no credit. Some started by assuming the result to be correct, then manipulated both sides until they derived an obvious identity. Reluctantly, they gained credit for this, though such an approach should be discouraged.

Find the \(x\)-coordinates of all the points on the curve \(y = 2{x^4} + 6{x^3} + \frac{7}{2}{x^2} - 5x + \frac{3}{2}\) at which

the tangent to the curve is parallel to the tangent at \(( - 1,{\text{ }}6)\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 8{x^3} + 18{x^2} + 7x - 5\)    A1

when \(x =  - 1,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 2\)     A1

\(8{x^3} + 18{x^2} + 7x - 5 =  - 2\)    M1

\(8{x^3} + 18{x^2} + 7x - 3 = 0\)

\((x + 1)\) is a factor     A1

\(8{x^3} + 18{x^2} + 7x - 3 = (x + 1)(8{x^2} + 10x - 3)\)    (M1)

Note:     M1 is for attempting to find the quadratic factor.

\((x + 1)(4x - 1)(2x + 3) = 0\)

\((x =  - 1),{\text{ }}x = 0.25,{\text{ }}x =  - 1.5\)    (M1)A1

Note:     M1 is for an attempt to solve their quadratic factor.

[7 marks]

The first half of the question was accessible to all the candidates. Some though saw the word ‘tangent’ and lost time calculating the equation of this. It was a pity that so many failed to spot that \(x + 1\) was a factor of the cubic and so did not make much progress with the final part of this question.

Consider the curve \(y = \frac{1}{{1 - x}} + \frac{4}{{x - 4}}\).

Find the x-coordinates of the points on the curve where the gradient is zero.

valid attempt to find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{\left( {1 - x} \right)}^2}}} - \frac{4}{{{{\left( {x - 4} \right)}^2}}}\)      A1A1

attempt to solve \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1

\(x = 2,\,\,x =  - 2\)     A1A1

[6 marks]

Use the substitution \(x = a\sec \theta \) to show that \(\int_{a\sqrt 2 }^{2a} {\frac{{{\text{d}}x}}{{{x^3}\sqrt {{x^2} - {a^2}} }} = \frac{1}{{24{a^3}}}\left( {3\sqrt 3  + \pi  - 6} \right)} \).

\(x = a\sec \theta \)

\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = a\sec \theta \tan \theta \)     (A1)

new limits:

\(x = a\sqrt 2  \Rightarrow \theta  = \frac{\pi }{4}\) and \(x = 2a \Rightarrow \theta  = \frac{\pi }{3}\)     (A1)

\(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{a\sec \theta \tan \theta }}{{{a^3}{{\sec }^3}\theta \sqrt {{a^2}{{\sec }^2}\theta  - {a^2}} }}{\text{d}}\theta } \)     M1

\( = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}\theta }}{{{a^3}}}{\text{d}}\theta } \)     A1

using \({\cos ^2}\theta  = \frac{1}{2}(\cos 2\theta  + 1)\)     M1

\(\frac{1}{{2{a^3}}}\left[ {\frac{1}{2}\sin 2\theta  + \theta } \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}\) or equivalent     A1

\( = \frac{1}{{4{a^3}}}\left( {\frac{{\sqrt 3 }}{2} + \frac{{2\pi }}{3} - 1 - \frac{\pi }{2}} \right)\) or equivalent     A1

\( = \frac{1}{{24{a^3}}}\left( {3\sqrt 3  + \pi  - 6} \right)\)     AG

[7 marks]

Calculate \(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} \) .

Find \(\int {{{\tan }^3}x{\text{d}}x} \) .

EITHER

let \(u = \tan x;{\text{ d}}u = {\sec ^2}x{\text{d}}x\)     (M1)

consideration of change of limits     (M1)

\(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{1}{{{u^{\frac{1}{3}}}}}{\text{d}}u} \)     (A1)

Note: Do not penalize lack of limits.

\( = \left[ {\frac{{3{u^{\frac{2}{3}}}}}{2}} \right]_1^{\sqrt 3 }\)     A1

\( = \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} - \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} - 3}}{2}} \right)\)     A1A1     N0

OR

\(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \left[ {\frac{{3{{(\tan x)}^{\frac{2}{3}}}}}{2}} \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}\)     M2A2

\( = \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} - \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} - 3}}{2}} \right)\)     A1A1     N0

[6 marks]

\(\int {{{\tan }^3}x{\text{d}}x}  = \int {\tan x({{\sec }^2}x - 1){\text{d}}x} \)     M1

\( = \int {(\tan x \times {{\sec }^2}x - \tan x){\text{d}}x} \)

\( = \frac{1}{2}{\tan ^2}x - \ln \left| {\sec x} \right| + C\)     A1A1

Note: Do not penalize the absence of absolute value or C.

[3 marks]

Quite a variety of methods were successfully employed to solve part (a).

Many candidates did not attempt part (b).

Find the value of \(\int_0^1 {t\ln (t + 1){\text{d}}t} \).

EITHER

attempt at integration by substitution     (M1)

using \(u = t + 1{\text{, d}}u = {\text{d}}t\), the integral becomes

\(\int_1^2 {(u - 1)\ln u{\text{d}}u} \)     A1

then using integration by parts     M1

\(\int_1^2 {(u - 1)\ln u{\text{d}}u}  = \left[ {\left( {\frac{{{u^2}}}{2} - u} \right)\ln u} \right]_1^2 - \int_1^2 {\left( {\frac{{{u^2}}}{2} - u} \right) \times \frac{1}{u}{\text{d}}u} \)     A1

\( =  - \left[ {\frac{{{u^2}}}{4} - u} \right]_1^2\)     (A1)

\( = \frac{1}{4}\,\,\,\,\,\)(accept 0.25)     A1

OR

attempt to integrate by parts     (M1)

correct choice of variables to integrate and differentiate     M1

\(\int_0^1 {t\ln (t + 1){\text{d}}t}  = \left[ {\frac{{{t^2}}}{2}\ln (t + 1)} \right]_0^1 - \int_0^1 {\frac{{{t^2}}}{2} \times \frac{1}{{t + 1}}{\text{d}}t} \)     A1

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

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

\( = \frac{1}{4}\,\,\,\,\,\)(accept 0.25)     A1

[6 marks]

Again very few candidates gained full marks on this question. The most common approach was to begin by integrating by parts, which was done correctly, but very few candidates then knew how to integrate \(\frac{{{t^2}}}{{t + 1}}\). Those who began with a substitution often made more progress. Again a number of candidates were let down by their inability to simplify appropriately.

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

Find \(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} \).

\(y = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{1}{{2\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} }}\left( { =  - \frac{1}{{\sqrt {4 - {x^2}} }}} \right)\)    M1A1

Note: M1 is for use of the chain rule.

[2 marks]

attempt at integration by parts     M1

\(u = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} =  - \frac{1}{{\sqrt {4 - {x^2}} }}\)

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 \Rightarrow v = x\)     (A1)

\(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x}  = \left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \int_0^1 {\frac{1}{{\sqrt {4 - {x^2}} }}} dx\)      A1

using integration by substitution or inspection      (M1)

\(\left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \left[ { - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}} \right]_0^1\)      A1

Note: Award A1 for \({ - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}}\) or equivalent.

Note: Condone lack of limits to this point.

attempt to substitute limits into their integral     M1

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

[7 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]

A particle moves in a straight line such that at time \(t\) seconds \((t \geqslant 0)\), its velocity \(v\), in \({\text{m}}{{\text{s}}^{ - 1}}\), is given by \(v = 10t{{\text{e}}^{ - 2t}}\). Find the exact distance travelled by the particle in the first half-second.

\(s = \int\limits_0^{\frac{1}{2}} {10t{{\text{e}}^{ - 2t}}{\text{d}}t} \)

attempt at integration by parts     M1

\( = \left[ { - 5t{{\text{e}}^{ - 2t}}} \right]_0^{\frac{1}{2}} - \int\limits_0^{\frac{1}{2}} { - 5{{\text{e}}^{ - 2t}}{\text{d}}t} \)     A1

\( = \left[ { - 5t{{\text{e}}^{ - 2t}} - \frac{5}{2}{{\text{e}}^{ - 2t}}} \right]_0^{\frac{1}{2}}\)     (A1)

Note:     Condone absence of limits (or incorrect limits) and missing factor of 10 up to this point.

\(s = \int\limits_0^{\frac{1}{2}} {10t{{\text{e}}^{ - 2t}}{\text{d}}t} \)     (M1)

\( =  - 5{{\text{e}}^{ - 1}} + \frac{5}{2}{\text{ }}\left( { = \frac{{ - 5}}{{\text{e}}} + \frac{5}{2}} \right){\text{ }}\left( { = \frac{{5{\text{e}} - 10}}{{2{\text{e}}}}} \right)\)     A1

[5 marks]

(a)     Show that \(\frac{3}{{x + 1}} + \frac{2}{{x + 3}} = \frac{{5x + 11}}{{{x^2} + 4x + 3}}\).

(b)     Hence find the value of k such that \(\int_0^2 {\frac{{5x + 11}}{{{x^2} + 4x + 3}}{\text{d}}x = \ln k} \) .

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

\( = \frac{{3x + 9 + 2x + 2}}{{{x^2} + 4x + 3}}\)     A1

\( = \frac{{5x + 11}}{{{x^2} + 4x + 3}}\)     AG

(b)     \(\int_0^2 {\frac{{5x + 11}}{{{x^2} + 4x + 3}}{\text{d}}x} = \int_0^2 {\left( {\frac{3}{{x + 1}} + \frac{2}{{x + 3}}} \right){\text{d}}x} \)     M1

\( = \left[ {3\ln (x + 1) + 2\ln (x + 3)} \right]_0^2\)     A1

\( = 3\ln 3 + 2\ln 5 - 3\ln 1 - 2\ln 3\,\,\,\,\,( = 3\ln 3 + 2\ln 5 - 2\ln 3)\)     A1

\( = \ln 3 + 2\ln 5\)

\( = \ln 75\,\,\,\,\,(k = 75)\)     A1

[6 marks]

Many students did not ‘Show’ enough in a) in order to be convincing. The need for the steps of the simplification to be shown was not clear. Too many did not link a) to b) and seemed to not be aware of the Command Term ‘hence’ and its implication for marking ( no marks will be awarded to alternative methods). The simplifications of the log expressions were done poorly by many and the fact that \({3^3} = 9\) was noted by too many. There were very few elegant solutions to this question.

Use the substitution \(u = {x^{\frac{1}{2}}}\) to find \(\int {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \).

Hence find the value of \(\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \), expressing your answer in the form arctan \(q\), where \(q \in \mathbb{Q}\).

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{2}{x^{ - \frac{1}{2}}}\) (accept \({\text{d}}u = \frac{1}{2}{x^{ - \frac{1}{2}}}{\text{d}}x\) or equivalent)       A1

substitution, leading to an integrand in terms of \(u\)     M1

\(\int {\frac{{2u{\text{d}}u}}{{{u^3} + u}}} \) or equivalent      A1

= 2 arctan \(\left( {\sqrt x } \right)\left( { + c} \right)\)     A1

[4 marks]

\(\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \) = arctan 3 − arctan 1     A1

tan(arctan 3 − arctan 1) = \(\frac{{3 - 1}}{{1 + 3 \times 1}}\)      (M1)

tan(arctan 3 − arctan 1) = \(\frac{1}{2}\)

arctan 3 − arctan 1 = arctan \(\frac{1}{2}\)     A1

[3 marks]

Find all values of x for \(0.1 \leqslant x \leqslant 1\) such that \(\sin (\pi {x^{ - 1}}) = 0\).

Find \(\int_{\frac{1}{{n + 1}}}^{\frac{1}{n}} {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}}){\text{d}}x} \), showing that it takes different integer values when n is even and when n is odd.

Evaluate \(\int_{0.1}^1 {\left| {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}})} \right|{\text{d}}x} \).

\(\sin (\pi {x^{ - 1}}) = 0{\text{ }}\frac{\pi }{x} = \pi ,{\text{ }}2\pi ( \ldots )\)     (A1) 

\(x = 1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8},\frac{1}{9},\frac{1}{{10}}\)     A1

[2 marks]

\(\left[ {\cos (\pi {x^{ - 1}})} \right]_{\frac{1}{{n + 1}}}^{\frac{1}{n}}\)     M1

\( = \cos (\pi n) - \cos \left( {\pi (n + 1)} \right)\)     A1

= 2 when n is even and = –2 when n is odd     A1

[3 marks]

\(\int_{0.1}^1 {\left| {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}})} \right|{\text{d}}x}  = 2 + 2 +  \ldots  + 2 = 18\)     (M1)A1

[2 marks]

There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it.

There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it.

There were disappointingly few correct answers to part (c) with candidates not realising that it was necessary to combine the previous two parts in order to write down the answer.

The folium of Descartes is a curve defined by the equation \({x^3} + {y^3} - 3xy = 0\), shown in the following diagram.

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the \(y\)-axis.

\({x^3} + {y^3} - 3xy = 0\)

\(3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3x\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3y = 0\)     M1A1

Note:     Differentiation wrt \(y\) is also acceptable.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3y - 3{x^2}}}{{3{y^2} - 3x}}{\text{ }}\left( { = \frac{{y - {x^2}}}{{{y^2} - x}}} \right)\)     (A1)

Note:     All following marks may be awarded if the denominator is correct, but the numerator incorrect.

\({y^2} - x = 0\)     M1

EITHER

\(x = {y^2}\)

\({y^6} + {y^3} - 3{y^3} = 0\)     M1A1

\({y^6} - 2{y^3} = 0\)

\({y^3}({y^3} - 2) = 0\)

\((y \ne 0)\therefore y = \sqrt[3]{2}\)     A1

\(x = {\left( {\sqrt[3]{2}} \right)^2}{\text{ }}\left( { = \sqrt[3]{4}} \right)\)     A1

OR

\({x^3} + xy - 3xy = 0\)     M1

\(x({x^2} - 2y) = 0\)

\(x \ne 0 \Rightarrow y = \frac{{{x^2}}}{2}\)     A1

\({y^2} = \frac{{{x^4}}}{4}\)

\(x = \frac{{{x^4}}}{4}\)

\(x({x^3} - 4) = 0\)

\((x \ne 0)\therefore x = \sqrt[3]{4}\)     A1

\(y = \frac{{{{\left( {\sqrt[3]{4}} \right)}^2}}}{2} = \sqrt[3]{2}\)     A1

[8 marks]

A body is moving in a straight line. When it is \(s\) metres from a fixed point O on the line its velocity, \(v\), is given by \(v =  - \frac{1}{{{s^2}}},{\text{ }}s > 0\).

Find the acceleration of the body when it is 50 cm from O.

\(\frac{{{\text{d}}v}}{{{\text{d}}s}} = 2{s^{ - 3}}\)     M1A1

Note:     Award M1 for \(2{s^{ - 3}}\) and A1 for the whole expression.

\(a = v\frac{{{\text{d}}v}}{{{\text{d}}s}}\)     (M1)

\(a =  - \frac{1}{{{s^2}}} \times \frac{2}{{{s^3}}}\left( { =  - \frac{2}{{{s^5}}}} \right)\)     (A1)

when \(s = \frac{1}{2},{\text{ }}a =  - \frac{2}{{{{(0.5)}^5}}}{\text{ }}( =  - 64){\text{ (m}}{{\text{s}}^{ - 2}})\)     M1A1

Note:     M1 is for the substitution of 0.5 into their equation for acceleration.

Award M1A0 if \(s = 50\) is substituted into the correct equation.

[6 marks]

Consider the curve \(y = x{{\text{e}}^x}\) and the line \(y = kx,{\text{ }}k \in \mathbb{R}\).

(a)     Let k = 0.

(i)     Show that the curve and the line intersect once.

(ii)     Find the angle between the tangent to the curve and the line at the point of intersection.

(b)     Let k =1. Show that the line is a tangent to the curve.

(c)     (i)     Find the values of k for which the curve \(y = x{{\text{e}}^x}\) and the line \(y = kx\) meet in two distinct points.

(ii)     Write down the coordinates of the points of intersection.

(iii)     Write down an integral representing the area of the region A enclosed by the curve and the line.

(iv)     Hence, given that \(0 < k < 1\), show that \(A < 1\).

(a)     (i)     \(x{{\text{e}}^x} = 0 \Rightarrow x = 0\)     A1

so, they intersect only once at (0, 0)

(ii)     \(y' = {{\text{e}}^x} + x{{\text{e}}^x} = (1 + x){{\text{e}}^x}\)     M1A1

\(y'(0) = 1\)     A1

\(\theta = \arctan 1 = \frac{\pi }{4}{\text{ }}(\theta = 45^\circ )\)     A1

[5 marks]

(b)     when \(k = 1,{\text{ }}y = x\)

\(x{{\text{e}}^x} = x \Rightarrow x({{\text{e}}^x} - 1) = 0\)     M1

\( \Rightarrow x = 0\)     A1

\(y'(0) = 1\) which equals the gradient of the line \(y = x\)     R1

so, the line is tangent to the curve at origin     AG

Note: Award full credit to candidates who note that the equation \(x({{\text{e}}^x} - 1) = 0\) has a double root x = 0 so y = x is a tangent.

[3 marks]

(c)     (i)     \(x{{\text{e}}^x} = kx \Rightarrow x({{\text{e}}^x} - k) = 0\)     M1

\( \Rightarrow x = 0{\text{ or }}x = \ln k\)     A1

\(k > 0{\text{ and }}k \ne 1\)     A1

(ii)     (0, 0) and \((\ln k,{\text{ }}k\ln k)\)     A1A1

(iii)     \(A = \left| {\int_0^{\ln k} {kx - x{{\text{e}}^x}{\text{d}}x} } \right|\)     M1A1

Note: Do not penalize the omission of absolute value.

(iv)     attempt at integration by parts to find \(\int {x{{\text{e}}^x}{\text{d}}x} \)     M1

\(\int {x{{\text{e}}^x}{\text{d}}x}  = x{{\text{e}}^x} - \int {{{\text{e}}^x}{\text{d}}x = {{\text{e}}^x}(x - 1)} \)     A1

as \(0 < k < 1 \Rightarrow \ln k < 0\)     R1

\(A = \int_{\ln k}^0 {kx - x{{\text{e}}^x}{\text{d}}x = \left[ {\frac{k}{2}{x^2} - (x - 1){{\text{e}}^x}} \right]_{\ln k}^0} \)     A1

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

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

\( = 1 - \frac{k}{2}\left( {{{(\ln k - 1)}^2} + 1} \right)\)     M1A1

since \(\frac{k}{2}\left( {{{(\ln k - 1)}^2} + 1} \right) > 0\)     R1

\(A < 1\)     AG

[15 marks]

Total [23 marks]

Many candidates solved (a) and (b) correctly but in (c), many failed to realise that the equation \(x{{\text{e}}^x} = kx\) has two roots under certain conditions and that the point of the question was to identify those conditions. Most candidates made a reasonable attempt to write down the appropriate integral in (c)(iii) with the modulus signs and limits often omitted but no correct solution has yet been seen to (c)(iv).

Calculate the value of the volume generated.

(i)     Given that \(\frac{{{\text{d}}V}}{{{\text{d}}h}} = \pi {(3\cos 2h + 4)^2}\), find an expression for \(\frac{{{\text{d}}h}}{{{\text{d}}t}}\).

(ii)     Find the value of \(\frac{{{\text{d}}h}}{{{\text{d}}t}}\) when \(h = \frac{\pi }{4}\).

(i)     Find \(\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}}\).

(ii)     Find the values of \(h\) for which \(\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}} = 0\).

(iii)     By making specific reference to the shape of the container, interpret \(\frac{{{\text{d}}h}}{{{\text{d}}t}}\) at the values of \(h\) found in part (c)(ii).

use of \(\pi \int_a^b {{x^2}{\text{d}}y} \)     (M1)

Note:     Condone any or missing limits.

\(V = \pi \int_0^\pi  {{{(3\cos 2y + 4)}^2}{\text{d}}y} \)    (A1)

\( = \pi \int_0^\pi  {(9{{\cos }^2}2y + 24\cos 2y + 16){\text{d}}y} \)    A1

\(9{\cos ^2}2y = \frac{9}{2}(1 + \cos 4y)\)    (M1)

\( = \pi \left[ {\frac{{9y}}{2} + \frac{9}{8}\sin 4y + 12\sin 2y + 16y} \right]_0^\pi \)    M1A1

\( = \pi \left( {\frac{{9\pi }}{2} + 16\pi } \right)\)    (A1)

\( = \frac{{41{\pi ^2}}}{2}{\text{ (c}}{{\text{m}}^3})\)    A1

Note:     If the coefficient “\(\pi \)” is absent, or eg, “\(2\pi \)” is used, only M marks are available.

[8 marks]

(i)     attempting to use \(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{{\text{d}}V}}{{{\text{d}}t}} \times \frac{{{\text{d}}h}}{{{\text{d}}V}}\) with \(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 2\)     M1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{2}{{\pi {{(3\cos 2h + 4)}^2}}}\)    A1

(ii)     substituting \(h = \frac{\pi }{4}\) into \(\frac{{{\text{d}}h}}{{{\text{d}}t}}\)     (M1)

\(\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{1}{{8\pi }}{\text{ (cm}}\,{\text{mi}}{{\text{n}}^{ - 1}})\)    A1

Note:     Do not allow FT marks for (b)(ii).

[4 marks]

(i)     \(\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}} = \frac{{\text{d}}}{{{\text{d}}t}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right) = \frac{{{\text{d}}h}}{{{\text{d}}t}} \times \frac{{\text{d}}}{{{\text{d}}h}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right)\)     (M1)

\( = \frac{2}{{\pi {{(3\cos 2h + 4)}^2}}} \times \frac{{24\sin 2h}}{{\pi {{(3\cos 2h + 4)}^3}}}\)    M1A1

Note:     Award M1 for attempting to find \(\frac{{\text{d}}}{{{\text{d}}h}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right)\).

\( = \frac{{48\sin 2h}}{{{\pi ^2}{{(3\cos 2h + 4)}^5}}}\)    A1

(ii)     \(\sin 2h = 0 \Rightarrow h = 0,{\text{ }}\frac{\pi }{2},{\text{ }}\pi \)     A1

Note:     Award A1 for \(\sin 2h = 0 \Rightarrow h = 0,{\text{ }}\frac{\pi }{2},{\text{ }}\pi \) from an incorrect \(\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}}\).

(iii)     METHOD 1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}}\) is a minimum at \(h = 0,{\text{ }}\pi \) and the container is widest at these values     R1

\(\frac{{{\text{d}}h}}{{{\text{d}}t}}\) is a maximum at \(h = \frac{\pi }{2}\) and the container is narrowest at this value     R1

[7 marks]

Part (a) was often answered well, though for some reason a minority tended to use the incorrect \(\pi \int {{{(3\cos 2y)}^2}{\text{d}}y} \) and gained few marks thereafter. Incorrect limits were sometimes seen, which led to only method marks being available. A pleasing number were able to deal with the integration of \({\cos ^2}2y\) through the use of the correct identity.

Part (b) was well answered and did not pose too many problems.

Correct answers to part (c) were rarely seen. Only the very best candidates appreciated the correct use of the chain rule when trying to determine an expression for \(\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}}\).

Find the coordinates of the points on C at which \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) .

The tangent to C at the point P(1, 2) cuts the x-axis at the point T. Determine the coordinates of T.

The normal to C at the point P cuts the y-axis at the point N. Find the area of triangle PTN.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x - \frac{1}{2}{x^3}\)     A1

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

\(x = 0,{\text{ }} \pm 2\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) at \(\left( {0,\frac{9}{8}} \right),{\text{ }}\left( { - 2,\frac{{25}}{8}} \right),{\text{ }}\left( {2,\frac{{25}}{8}} \right)\)     A1A1A1

Note: Award A2 for all three x-values correct with errors/omissions in y-values.

[4 marks]

at x =1, gradient of tangent \( = \frac{3}{2}\)     (A1)

Note: In the following, allow FT on incorrect gradient.

equation of tangent is \(y - 2 = \frac{3}{2}(x - 1)\,\,\,\,\,\left( {y = \frac{3}{2}x + \frac{1}{2}} \right)\)     (A1)

meets x-axis when y = 0 , \( - 2 = \frac{3}{2}(x - 1)\)     (M1)

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

coordinates of T are \(\left( { - \frac{1}{3},0} \right)\)     A1

[4 marks]

gradient of normal \( = - \frac{2}{3}\)     (A1)

equation of normal is \(y - 2 = - \frac{2}{3}(x - 1)\,\,\,\,\,\left( {y = - \frac{2}{3}x + \frac{8}{3}} \right)\)     (M1)

at x = 0 , \(y = \frac{8}{3}\)     A1

Note: In the following, allow FT on incorrect coordinates of T and N.

lengths of \({\text{PN}} = \sqrt {\frac{{13}}{9}} \) , \({\text{PT}} = \sqrt {\frac{{52}}{9}} \)     A1A1

area of triangle \({\text{PTN}} = \frac{1}{2} \times \sqrt {\frac{{13}}{9}}  \times \sqrt {\frac{{52}}{9}} \)     M1

\( = \frac{{13}}{9}\) (or equivalent e.g. \(\frac{{\sqrt {676} }}{{18}}\))     A1

[7 marks]

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. 

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. 

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. 

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]

A curve is given by the equation \(y = \sin (\pi \cos x)\).

Find the coordinates of all the points on the curve for which \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0,{\text{ }}0 \leqslant x \leqslant \pi \).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \cos (\pi \cos x) \times \pi \sin x\)    M1A1

Note:     Award follow through marks below if their answer is a multiple of the correct answer.

considering either \(\sin x = 0\) or \(\cos (\pi \cos x) = 0\)     (M1)

\(x = 0,{\text{ }}\pi \)    A1

\(\pi \cos x = \frac{\pi }{2},{\text{ }} - \frac{\pi }{2}{\text{ }}\left( { \Rightarrow \cos x = \frac{1}{2}, - \frac{1}{2}} \right)\)    M1

Note:     Condone absence of \( - \frac{\pi }{2}\).

\( \Rightarrow x = \frac{\pi }{3},{\text{ }}\frac{{2\pi }}{3}\)

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

\(\left( {\frac{{2\pi }}{3},{\text{ }} - 1} \right)\)    A1

[7 marks]

This was not a straight-forward differentiation and it was pleasing to see how many candidates managed to do it correctly. Having done this they found two of the solutions, often three of the solutions, successfully. The final solution was found by only a few candidates. Again candidates lost marks unnecessarily by not close reading the question and realising that they needed both coordinates of the points, not just the \(x\)-coordinates.

Show that \({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x\) where \(r,\,x \in {\mathbb{R}^ + }\).

Express \(y\) in terms of \(x\). Give your answer in the form \(y = p{x^q}\), where p , q are constants.

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines \(x = 1\) and \(x = \alpha \) where \(\alpha  > 1\). The area of R is \(\sqrt 2 \).

Find the value of \(\alpha \).

METHOD 1

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)\)     M1A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}\)     M1

\( = \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}\)     A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y =  - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}\)

\({\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)\)     M1A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0\)     M1

\(\sqrt 2 xy = 1\)     A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

the area of R is \(\int\limits_1^\alpha  {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x\)     M1

\( = \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha \)     A1

\( = \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha \)     A1

\(\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha  = \sqrt 2 \)     M1

\(\alpha  = {{\text{e}}^2}\)     A1

Note: Only follow through from part (b) if \(y\) is in the form \(y = p{x^q}\)

[5 marks]

Given that \(y = \frac{1}{{1 - x}}\), use mathematical induction to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = \frac{{n!}}{{{{(1 - x)}^{n + 1}}}},{\text{ }}n \in {\mathbb{Z}^ + }\).

proposition is true for n = 1 since \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{(1 - x)}^2}}}\)     M1

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

Note: Must see the 1! for the A1.

assume true for n = k, \(k \in {\mathbb{Z}^ + }\), i.e. \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = \frac{{k!}}{{{{(1 - x)}^{k + 1}}}}\)     M1

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

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

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

hence, \({{\text{P}}_{k + 1}}\) is true whenever \({{\text{P}}_{k}}\) is true, and \({{\text{P}}_1}\) is true, and therefore the proposition is true for all positive integers     R1

Note: The final R1 is only available if at least 4 of the previous marks have been awarded.