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HL Paper 1

Prove by mathematical induction that $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ for $n \in ℤ^{+}$ .

[7]

Hence or otherwise, determine the Maclaurin series of $f (x) = x^{2} e^{x}$ in ascending powers of $x$ , up to and including the term in $x^{4}$ .

[3]

Hence or otherwise, determine the value of $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}]$ .

[4]

Markscheme

For $n = 1$

LHS: $\frac{d}{d x} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} (= e^{x} (x^{2} + 2 x))$ A1

RHS: $(x^{2} + 2 (1) x + 1 (1 - 1)) e^{x} (= e^{x} (x^{2} + 2 x))$ A1

so true for $n = 1$

now assume true for $n = k$ ; i.e. $\frac{d^{k}}{d x^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x}$ M1

Note: Do not award M1 for statements such as "let $n = k$ ". Subsequent marks can still be awarded.

attempt to differentiate the RHS M1

$\frac{d^{k + 1}}{d x^{k + 1}} (x^{2} e^{x}) = \frac{d}{d x} ([x^{2} + 2 k x + k (k - 1)] e^{x})$

$= (2 x + 2 k) e^{x} + (x^{2} + 2 k x + k (k - 1)) e^{x}$ A1

$= [x^{2} + 2 (k + 1) x + k (k + 1)] e^{x}$ A1

so true for $n = k$ implies true for $n = k + 1$

therefore $n = 1$ true and $n = k$ true $\Rightarrow n = k + 1$ true

therefore, true for all $n \in ℤ^{+}$ R1

Note: Award R1 only if three of the previous four marks have been awarded

[7 marks]

METHOD 1

attempt to use $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ (M1)

Note: For $x = 0$ , $\frac{d^{n}}{d x^{n}} {(x^{2} e^{x})}_{x = 0} = n (n - 1)$ may be seen.

$f (0) = 0, f' (0) = 0, f'' (0) = 2, f''' (0) = 6, f^{(4)} (0) = 12$

use of $f (x) = f (0) + x f' (0) + \frac{x^{2}}{2!} f'' (0) + \frac{x^{3}}{3!} f''' (0) + \frac{x^{4}}{4!} f^{(4)} (0) + \dots$ (M1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

METHOD 2

' $x^{2} \times$ Maclaurin series of $e^{x}$ ' (M1)

$x^{2} (1 + x + \frac{x^{2}}{2!} + \dots)$ (A1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

[3 marks]

METHOD 1

attempt to substitute $x^{2} e^{x} \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ into $\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}$ M1

$\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}} \approx \frac{{(x^{2} + x^{3} + \frac{1}{2} x^{4} (+ \dots) - x^{2})}^{3}}{x^{9}}$ (A1)

EITHER

$= \frac{{(x^{3} + \frac{1}{2} x^{4} + \dots)}^{3}}{x^{9}}$ A1

$= \frac{x^{9} (+ higher order terms)}{x^{9}}$

${(\frac{x^{3} + \frac{1}{2} x^{4} (+ \dots)}{x^{3}})}^{3}$ A1

${(1 + \frac{1}{2} x (+ \dots))}^{3}$

THEN

$= 1 (+ higher order terms)$

so $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = 1$ A1

METHOD 2

$\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = \lim_{x \to 0} {(\frac{x^{2} e^{x} - x^{2}}{x^{3}})}^{3}$ M1

$= \lim_{x \to 0} {(\frac{e^{x} - 1}{x})}^{3}$ (A1)

attempt to use L'Hôpital's rule M1

$= \lim_{x \to 0} {(\frac{e^{x} - 0}{1})}^{3}$

$= {[\lim_{x \to 0} e^{x}]}^{3}$

$= 1$ A1

[4 marks]

Examiners report

[N/A]

Consider the function ${f_n}(x) = (\cos 2x)(\cos 4x) \ldots (\cos {2^n}x),{\text{ }}n \in {\mathbb{Z}^ + }$ .

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

[2]

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 $m \in \mathbb{Z}$ .

[8]

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

[3]

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

[8]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

even function A1

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

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]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^x}\,{\text{cos}}{\,^2}x$ , where 0 ≤ $x$ ≤ 5. The curve $y = f\left( x \right)$ is shown on the following graph which has local maximum points at A and C and touches the $x$ -axis at B and D.

Use integration by parts to show that $\int {{{\text{e}}^x}\,{\text{cos}}\,2x{\text{d}}x = } \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x + c$ , $c \in \mathbb{R}$ .

[5]

Hence, show that $\int {{{\text{e}}^x}\,{\text{cos}}{\,^2}x{\text{d}}x = } \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2} + c$ , $c \in \mathbb{R}$ .

[3]

Find the $x$ -coordinates of A and of C , giving your answers in the form $a + {\text{arctan}}\,b$ , where $a$ , $b \in \mathbb{R}$ .

[6]

Find the area enclosed by the curve and the $x$ -axis between B and D, as shaded on the diagram.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempt at integration by parts with $u = {{\text{e}}^x}$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\,{\text{d}}x - \int {\frac{{{{\text{e}}^x}}}{2}} {\text{sin}}\,2x\,{\text{d}}x$ A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x - \frac{1}{2}\left( { - \frac{{{{\text{e}}^x}}}{2}{\text{cos}}\,2x + \int {\frac{{{{\text{e}}^x}}}{2}} {\text{cos}}\,2x} \right)$ M1A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x - \frac{1}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore \frac{5}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\, + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 2

attempt at integration by parts with $u = {\text{cos}}\,2x$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {{\text{e}}^x}$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } {{\text{e}}^x}\,{\text{cos}}\,2x + 2\int {{{\text{e}}^x}\,{\text{sin}}\,2x\,{\text{d}}x}$ A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2\left( {{{\text{e}}^x}\,{\text{sin}}\,2x - 2\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} } \right)$ M1A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 3

attempt at use of table M1

A1A1

Note: A1 for first 2 lines correct, A1 for third line correct.

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = \,} {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$ M1

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

[5 marks]

$\int {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^2}\,x{\text{d}}x = } \int {\frac{{{{\text{e}}^x}}}{2}} \left( {{\text{cos}}\,2x + 1} \right){\text{d}}x$ M1A1

$= \frac{1}{2}\left( {\frac{{{\text{2}}{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x} \right) + \frac{{{{\text{e}}^x}}}{2}$ A1

$= \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}\left( { + c} \right)$ AG

Note: Do not accept solutions where the RHS is differentiated.

[3 marks]

$f'\left( x \right) = {{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x - 2{{\text{e}}^x}\,{\text{sin}}\,x\,{\text{cos}}\,x$ M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

${{\text{e}}^x}\,{\text{cos}}\,x\left( {{\text{cos}}\,x - 2\,{\text{sin}}\,x} \right) = 0$ (M1)

Note: Award M1 for an attempt to factorise ${{\text{cos}}\,x}$ or divide by ${\text{cos}}\,x\left( {{\text{cos}}\,x \ne 0} \right)$ .

discount ${\text{cos}}\,x = 0$ (as this would also be a zero of the function)

$\Rightarrow {\text{cos}}\,x - 2\,{\text{sin}}\,x = 0$

$\Rightarrow {\text{tan}}\,x = \frac{1}{2}$ (M1)

$\Rightarrow x = {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at A) and $x = \pi + {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at C) A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

[6 marks]

${\text{cos}}\,x = 0 \Rightarrow x = \frac{\pi }{2}$ or $\frac{{3\pi }}{2}$ A1

Note: The A1may be awarded for work seen in part (c).

$\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x} \right)} \,{\text{d}}x = \left[ {\frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}$ M1

$= \left( { - \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{2}} \right) - \left( { - \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{2}} \right)\left( { = \frac{{{\text{2}}{{\text{e}}^{\frac{{3\pi }}{2}}}}}{5} - \frac{{{\text{2}}{{\text{e}}^{\frac{\pi }{2}}}}}{5}} \right)$ M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for ${\text{sin}}\,3\pi = {\text{sin}}\,\pi = 0$ and ${\text{cos}}\,3\pi = {\text{cos}}\,\pi = - 1$ and A1 for a completely correct answer.

[5 marks]

Examiners report

[N/A]

Show that ${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = 1 + {\text{sin}}\,2x$ .

[2]

Show that ${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ .

[4]

Hence or otherwise find $\int_0^{\frac{\pi }{6}} {\left( {{\text{sec}}\,2x + {\text{tan}}\,2x} \right)} {\text{d}}x$ in the form ${\text{ln}}\left( {a + \sqrt b } \right)$ where $a$ , $b \in \mathbb{Z}$ .

[9]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = {\text{si}}{{\text{n}}^2}\,x + 2{\text{sin}}\,x\,{\text{cos}}\,x + {\text{co}}{{\text{s}}^2}\,x$ M1A1

Note: Do not award the M1 for just ${\text{si}}{{\text{n}}^2}\,x + {\text{co}}{{\text{s}}^2}\,x$ .

Note: Do not award A1 if correct expression is followed by incorrect working.

$= 1 + {\text{sin}}\,2x$ AG

[2 marks]

${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{1}{{{\text{cos}}\,2x}} + \frac{{{\text{sin}}\,2x}}{{{\text{cos}}\,2x}}$ M1

Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of ${\text{tan}}\,x$ .

$= \frac{{{\text{1}} + {\text{sin}}\,2x}}{{{\text{cos}}\,2x}}$

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{{\text{co}}{{\text{s}}^2}\,x - {\text{si}}{{\text{n}}^2}\,x}}$ A1A1

Note: Award A1 for numerator, A1 for denominator.

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right)}}$ M1

$= \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ AG

Note: Apply MS in reverse if candidates have worked from RHS to LHS.

Note: Alternative method using ${\text{tan}}\,2x$ and ${\text{sec}}\,2x$ in terms of ${\text{tan}}\,x$ .

[4 marks]

METHOD 1

$\int_0^{\frac{\pi }{6}} {\left( {\frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}} \right)} {\text{d}}x$ A1

Note: Award A1 for correct expression with or without limits.

EITHER

$= \left[ { - {\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_0^{\frac{\pi }{6}}$ or $\left[ {{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_{\frac{\pi }{6}}^0$ (M1)A1A1

Note: Award M1 for integration by inspection or substitution, A1 for ${{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)}$ , A1 for completely correct expression including limits.

$= - {\text{ln}}\left( {{\text{cos}}\,\frac{\pi }{6} - {\text{sin}}\,\frac{\pi }{6}} \right) + {\text{ln}}\left( {{\text{cos}}\,0 - {\text{sin}}\,0} \right)$ M1

Note: Award M1 for substitution of limits into their integral and subtraction.

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

let $u = {\text{cos}}\,x - {\text{sin}}\,x$ M1

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = - {\text{sin}}\,x - {\text{cos}}\,x = - \left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)$

$- \int_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}} {\left( {\frac{1}{u}} \right)} {\text{d}}u$ A1A1

Note: Award A1 for correct limits even if seen later, A1 for integral.

$= \left[ { - {\text{ln}}\,u} \right]_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}$ or $\left[ {{\text{ln}}\,u} \right]_{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}^1$ A1

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

THEN

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

Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.

$= {\text{ln}}\left( {1 + \sqrt 3 } \right)$ (M1)A1

METHOD 2

$\left[ {\frac{1}{2}{\text{ln}}\left( {{\text{tan}}\,2x + {\text{sec}}\,2x} \right) - \frac{1}{2}{\text{ln}}\left( {{\text{cos}}\,2x} \right)} \right]_0^{\frac{\pi }{6}}$ A1A1

$= \frac{1}{2}{\text{ln}}\left( {\sqrt 3 + 2} \right) - \frac{1}{2}{\text{ln}}\left( {\frac{1}{2}} \right) - 0$ A1A1(A1)

$= \frac{1}{2}{\text{ln}}\left( {4 + 2\sqrt 3 } \right)$ M1

$= \frac{1}{2}{\text{ln}}\left( {{{\left( {1 + \sqrt 3 } \right)}^2}} \right)$ M1A1

$= {\text{ln}}\left( {1 + \sqrt 3 } \right)$ A1

[9 marks]

Examiners report

[N/A]

Consider the function $f$ defined by $f(x) = {x^2} - {a^2},{\text{ }}x \in \mathbb{R}$ where $a$ is a positive constant.

The function $g$ is defined by $g(x) = x\sqrt {f(x)}$ for $\left| x \right| > a$ .

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

$y = f(x)$ ;

[2]

a.i.

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

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

[4]

a.ii.

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

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

[2]

a.iii.

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

[5]

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

[4]

Markscheme

M17/5/MATHL/HP1/ENG/TZ2/09.a.i/M

A1 for correct shape

A1 for correct $x$ and $y$ intercepts and minimum point

[2 marks]

a.i.

M17/5/MATHL/HP1/ENG/TZ2/09.a.ii/M

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

a.ii.

M17/5/MATHL/HP1/ENG/TZ2/09.a.iii/M

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

A1 for correctly labelled minimum point

[2 marks]

a.iii.

EITHER

attempt at integration by parts (M1)

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

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

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

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

attempt at integration by parts (M1)

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

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

$= {x^2}\sin x + 2x\cos x - 2\sin x$

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

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

[5 marks]

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

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

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

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

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

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

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Consider the functions $f$ and $g$ defined on the domain $0 < x < 2\pi$ by $f\left( x \right) = 3\,{\text{cos}}\,2x$ and $g\left( x \right) = 4 - 11\,{\text{cos}}\,x$ .

The following diagram shows the graphs of $y = f\left( x \right)$ and $y = g\left( x \right)$

Find the $x$ -coordinates of the points of intersection of the two graphs.

[6]

Find the exact area of the shaded region, giving your answer in the form $p\pi + q\sqrt 3$ , where $p$ , $q \in \mathbb{Q}$ .

[5]

At the points A and B on the diagram, the gradients of the two graphs are equal.

Determine the $y$ -coordinate of A on the graph of $g$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$3\,{\text{cos}}\,2x = 4 - 11\,{\text{cos}}\,x$

attempt to form a quadratic in ${\text{cos}}\,x$ M1

$3\left( {2\,{\text{co}}{{\text{s}}^2}\,x - 1} \right) = 4 - 11\,{\text{cos}}\,x$ A1

$\left( {6\,{\text{co}}{{\text{s}}^2}\,x + 11\,{\text{cos}}\,x - 7 = 0} \right)$

valid attempt to solve their quadratic M1

$\left( {3\,{\text{cos}}\,x + 7} \right)\left( {2\,{\text{cos}}\,x - 1} \right) = 0$

${\text{cos}}\,x = \frac{1}{2}$ A1

$x = \frac{\pi }{3}{\text{,}}\,\,\frac{{5\pi }}{3}$ A1A1

Note: Ignore any “extra” solutions.

[6 marks]

consider (±) $\int\limits_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}} {\left( {4 - 11\,{\text{cos}}\,x - 3\,{\text{cos}}\,2x} \right)} \,{\text{d}}x$ M1

$= \left( \pm \right)\left[ {4x - 11\,{\text{sin}}\,x - \frac{3}{2}{\text{sin}}\,2x} \right]_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}}$ A1

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

attempt to substitute their limits into their integral M1

$= \frac{{20\pi }}{3} - 11\,{\text{sin}}\frac{{5\pi }}{3} - \frac{3}{2}{\text{sin}}\frac{{10\pi }}{3} - \left( {\frac{{4\pi }}{3} - 11\,{\text{sin}}\frac{\pi }{3} - \frac{3}{2}{\text{sin}}\frac{{2\pi }}{3}} \right)$

$= \frac{{16\pi }}{3} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4}$

$= \frac{{16\pi }}{3} + \frac{{25\sqrt 3 }}{2}$ A1A1

[5 marks]

attempt to differentiate both functions and equate M1

$- 6\,{\text{sin}}\,2x = 11\,{\text{sin}}\,x$ A1

attempt to solve for $x$ M1

$11\,{\text{sin}}\,x + 12\,{\text{sin}}\,x\,{\text{cos}}\,x = 0$

${\text{sin}}\,x\left( {11 + 12\,{\text{cos}}\,x} \right) = 0$

${\text{cos}}\,x = - \frac{{11}}{{12}}\,\,\left( {{\text{or}}\,\,{\text{sin}}\,x = 0\,} \right)$ A1

$\Rightarrow y = 4 - 11\left( { - \frac{{11}}{{12}}} \right)$ M1

$y = \frac{{169}}{{12}}\,\left( { = 14\frac{1}{{12}}} \right)$ A1

[6 marks]

Examiners report

[N/A]

A continuous random variable $X$ has the probability density function

$f (x) = \{\begin{matrix} \frac{2}{(b - a) (c - a)} (x - a), & a \leq x \leq c \\ \frac{2}{(b - a) (b - c)} (b - x), & c < x \leq b \\ 0, & otherwise \end{matrix}$ .

The following diagram shows the graph of $y = f (x)$ for $a \leq x \leq b$ .

Given that $c \geq \frac{a + b}{2}$ , find an expression for the median of $X$ in terms of $a, b$ and $c$ .

Markscheme

let $m$ be the median

EITHER

attempts to find the area of the required triangle M1

base is $(m - a)$ (A1)

and height is $\frac{2}{(b - a) (c - a)} (m - a)$

area $= \frac{1}{2} (m - a) \times \frac{2}{(b - a) (c - a)} (m - a) (= \frac{{(m - a)}^{2}}{(b - a) (c - a)})$ A1

attempts to integrate the correct function M1

$\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$

$= \frac{2}{(b - a) (c - a)} {[\frac{1}{2} {(x - a)}^{2}]}_{a}^{m}$ OR $\frac{2}{(b - a) (c - a)} {[\frac{x^{2}}{2} - a x]}_{a}^{m}$ A1A1

Note: Award A1 for correct integration and A1 for correct limits.

THEN

sets up (their) $\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$ or area $= \frac{1}{2}$ M1

Note: Award M0A0A0M1A0A0 if candidates conclude that $m > c$ and set up their area or sum of integrals $= \frac{1}{2}$ .

$\frac{{(m - a)}^{2}}{(b - a) (c - a)} = \frac{1}{2}$

$m = a \pm \sqrt{\frac{(b - a) (c - a)}{2}}$ (A1)

as $m > a$ , rejects $m = a - \sqrt{\frac{(b - a) (c - a)}{2}}$

so $m = a + \sqrt{\frac{(b - a) (c - a)}{2}}$ A1

[6 marks]

Examiners report

[N/A]

Consider the function $f$ defined by $f(x) = {{\text{e}}^x}\sin x,{\text{ }}0 \leqslant x \leqslant \pi$ .

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

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

[2]

Find the $x$ -coordinate of the point of inflexion of the graph of $f$ .

[2]

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

[3]

Find the area of the region enclosed by the graph of $f$ and the $x$ -axis.

[6]

Find the value of the curvature of the graph of $f$ at the local maximum point.

[3]

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

[2]

Markscheme

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

N16/5/MATHL/HP1/ENG/TZ0/11.e/M

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

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

Examiners report

[N/A]

Let $f\left( x \right) = \frac{{2 - 3{x^5}}}{{2{x^3}}},\,\,x \in \mathbb{R},\,\,x \ne 0$ .

The graph of $y = f\left( x \right)$ has a local maximum at A. Find the coordinates of A.

[5]

Show that there is exactly one point of inflexion, B, on the graph of $y = f\left( x \right)$ .

[5]

b.i.

The coordinates of B can be expressed in the form B $\left( {{2^a},\,b \times {2^{ - 3a}}} \right)$ where a, b $\in \mathbb{Q}$ . Find the value of a and the value of b.

[3]

b.ii.

Sketch the graph of $y = f\left( x \right)$ showing clearly the position of the points A and B.

[4]

Markscheme

attempt to differentiate (M1)

$f'\left( x \right) = - 3{x^{ - 4}} - 3x$ A1

Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example $f'\left( x \right) = \frac{{ - 15{x^4} \times 2{x^3} - 6{x^2}\left( {2 - 3{x^5}} \right)}}{{{{\left( {2{x^3}} \right)}^2}}}$ .

$- \frac{3}{{{x^4}}} - 3x = 0$ M1

$\Rightarrow {x^5} = - 1 \Rightarrow x = - 1$ A1

${\text{A}}\left( { - 1,\, - \frac{5}{2}} \right)$ A1

[5 marks]

$f''\left( x \right) = 0$ M1

$f''\left( x \right) = 12{x^{ - 5}} - 3\left( { = 0} \right)$ A1

Note: Award A1 for correct derivative seen even if not simplified.

$\Rightarrow x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)$ A1

hence (at most) one point of inflexion R1

Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.

$f''\left( x \right)$ changes sign at $x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)$ R1

so exactly one point of inflexion

[5 marks]

b.i.

$x = \sqrt[5]{4} = {2^{\frac{2}{5}}}\left( { \Rightarrow a = \frac{2}{5}} \right)$ A1

$f\left( {{2^{\frac{2}{5}}}} \right) = \frac{{2 - 3 \times {2^2}}}{{2 \times {2^{\frac{6}{5}}}}} = - 5 \times {2^{ - \frac{6}{5}}}\left( { \Rightarrow b = - 5} \right)$ (M1)A1

Note: Award M1 for the substitution of their value for $x$ into $f\left( x \right)$ .

[3 marks]

b.ii.

A1A1A1A1

A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.

Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

By using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} {sec}^{n} x \tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.

Markscheme

METHOD 1

$u = sec x \Rightarrow d u = sec x \tan x d x$ (A1)

attempts to express the integral in terms of $u$ M1

$\int_{1}^{2} u^{n - 1} d u$ A1

$= \frac{1}{n} {[u^{n}]}_{1}^{2} (= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}})$ A1

Note: Condone the absence of or incorrect limits up to this point.

$= \frac{2^{n} - 1^{n}}{n}$ M1

$= \frac{2^{n} - 1}{n}$ A1

Note: Award M1 for correct substitution of their limits for $u$ into their antiderivative for $u$ (or given limits for $x$ into their antiderivative for $x$ ).

METHOD 2

$\int {sec}^{n} x \tan x d x = \int {sec}^{n - 1} x sec x \tan x d x$ (A1)

applies integration by inspection (M1)

$= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}}$ A2

Note: Award A2 if the limits are not stated.

$= \frac{1}{n} ({sec}^{n} \frac{π}{3} - {sec}^{n} 0)$ M1

Note: Award M1 for correct substitution into their antiderivative.

$= \frac{2^{n} - 1}{n}$ A1

[6 marks]

Examiners report

[N/A]

The continuous random variable $X$ has probability density function

$f (x) = \{\begin{cases} \frac{k}{\sqrt{4 - 3 x^{2}}}, & 0 \leq x \leq 1 \\ 0, & otherwise . \end{cases}$

Find the value of $k$ .

[4]

Find $E (X)$ .

[4]

Markscheme

attempt to integrate $\frac{k}{\sqrt{4 - 3 x^{2}}}$ (M1)

$= k ⌊\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)⌋$ A1

Note: Award (M1)A0 for $arcsin (\frac{\sqrt{3}}{2} x)$ .
Condone absence of $k$ up to this stage.

equating their integrand to $1$ M1

$k {[\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)]}_{0}^{1} = 1$

$k = \frac{3 \sqrt{3}}{π}$ A1

[4 marks]

$E (X) = \frac{3 \sqrt{3}}{π} \int_{0}^{1} \frac{x}{\sqrt{4 - 3 x^{2}}} d x$ A1

Note: Condone absence of limits if seen at a later stage.

EITHER

attempt to integrate by inspection (M1)

$= \frac{3 \sqrt{3}}{π} \times - \frac{1}{6} \int - 6 x {(4 - 3 x^{2})}^{- \frac{1}{2}} d x$

$= \frac{3 \sqrt{3}}{π} {[- \frac{1}{3} \sqrt{4 - 3 x^{2}}]}_{0}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

for example, $u = 4 - 3 x^{2} \Rightarrow \frac{d u}{d x} = - 6 x$

Note: Other substitutions may be used. For example $u = - 3 x^{2}$ .

$= - \frac{\sqrt{3}}{2 π} \int_{4}^{1} u^{- \frac{1}{2}} d u$ M1

Note: Condone absence of limits up to this stage.

$= - \frac{\sqrt{3}}{2 π} {[2 \sqrt{u}]}_{4}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

THEN

$= \frac{\sqrt{3}}{π}$ A1

Note: Award A0M1A1A0 for their $k [- \frac{1}{3} \sqrt{4 - 3 x^{2}}]$ or $k [- 2 \sqrt{u}]$ for working with incorrect or no limits.

[4 marks]

Examiners report

Most candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x² was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.

[N/A]

A function $f$ is defined by $f (x) = x \sqrt{1 - x^{2}}$ where $- 1 \leq x \leq 1$ .

The graph of $y = f (x)$ is shown below.

Show that $f$ is an odd function.

[2]

The range of $f$ is $a \leq y \leq b$ , where $a, b \in ℝ$ .

Find the value of $a$ and the value of $b$ .

[6]

Markscheme

attempts to replace $x$ with $- x$ M1

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

$= - x \sqrt{1 - {(- x)}^{2}} (= - f (x))$ A1

Note: Award M1A1 for an attempt to calculate both $f (- x)$ and $- f (- x)$ independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by $180 °$ about the origin or the graph is invariant after a reflection in the $y$ -axis and then in the $x$ -axis (or vice versa).

so $f$ is an odd function AG

[2 marks]

attempts both product rule and chain rule differentiation to find $f' (x)$ M1

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

$= \frac{1 - 2 x^{2}}{\sqrt{1 - x^{2}}}$

sets their $f' (x) = 0$ M1

$\Rightarrow x = \pm \frac{1}{\sqrt{2}}$ A1

attempts to find at least one of $f (\pm \frac{1}{\sqrt{2}})$ (M1)

Note: Award M1 for an attempt to evaluate $f (x)$ at least at one of their $f' (x) = 0$ roots.

$a = - \frac{1}{2}$ and $b = \frac{1}{2}$ A1

Note: Award A1 for $- \frac{1}{2} \leq y \leq \frac{1}{2}$ .

[6 marks]

Examiners report

[N/A]

Consider the function defined by $f (x) = \frac{k x - 5}{x - k}$ , where $x \in ℝ \ \{k\}$ and $k^{2} \neq 5$ .

Consider the case where $k = 3$ .

State the equation of the vertical asymptote on the graph of $y = f (x)$ .

[1]

State the equation of the horizontal asymptote on the graph of $y = f (x)$ .

[1]

Use an algebraic method to determine whether $f$ is a self-inverse function.

[4]

Sketch the graph of $y = f (x)$ , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.

[3]

The region bounded by the $x$ -axis, the curve $y = f (x)$ , and the lines $x = 5$ and $x = 7$ is rotated through $2 π$ about the $x$ -axis. Find the volume of the solid generated, giving your answer in the form $π (a + b \ln 2)$ , where $a, b \in ℤ$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x = k$ A1

[1 mark]

$y = k$ A1

[1 mark]

METHOD 1

$(f \circ f) (x) = \frac{k (\frac{k x - 5}{x - k}) - 5}{(\frac{k x - 5}{x - k}) - k}$ M1

$= \frac{k (k x - 5) - 5 (x - k)}{k x - 5 - k (x - k)}$ A1

$= \frac{k^{2} x - 5 k - 5 x + 5 k}{k x - 5 - k x + k^{2}}$

$= \frac{k^{2} x - 5 x}{k^{2} - 5}$ A1

$= \frac{x (k^{2} - 5)}{k^{2} - 5}$

$= x$

$(f \circ f) (x) = x$ , (hence $f$ is self-inverse) R1

Note: The statement $f (f (x)) = x$ could be seen anywhere in the candidate’s working to award R1.

METHOD 2

$f (x) = \frac{k x - 5}{x - k}$

$x = \frac{k y - 5}{y - k}$ M1

Note: Interchanging $x$ and $y$ can be done at any stage.

$x (y - k) = k y - 5$ A1

$x y - x k = k y - 5$

$x y - k y = x k - 5$

$y (x - k) = k x - 5$ A1

$y = f^{- 1} (x) = \frac{k x - 5}{x - k}$ (hence $f$ is self-inverse) R1

[4 marks]

attempt to draw both branches of a rectangular hyperbola M1

$x = 3$ and $y = 3$ A1

$(0, \frac{5}{3})$ and $(\frac{5}{3}, 0)$ A1

[3 marks]

METHOD 1

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

EITHER

attempt to express $\frac{3 x - 5}{x - 3}$ in the form $p + \frac{q}{x - 3}$ M1

$\frac{3 x - 5}{x - 3} = 3 + \frac{4}{x - 3}$ A1

attempt to expand ${(\frac{3 x - 5}{x - 3})}^{2}$ or ${(3 x - 5)}^{2}$ and divide out M1

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24 x - 56}{{(x - 3)}^{2}}$ A1

THEN

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}$ A1

$volume = π \int_{5}^{7} (9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}) d x$

$= π {[9 x + 24 \ln (x - 3) - \frac{16}{x - 3}]}_{5}^{7}$ A1

$= π ⌊(63 + 24 \ln 4 - 4) - (45 + 24 \ln 2 - 8)⌋$

$= π (22 + 24 \ln 2)$ A1

METHOD 2

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

substituting $u = x - 3 \Rightarrow \frac{d u}{d x} = 1$ A1

$3 x - 5 = 3 (u + 3) - 5 = 3 u + 4$

$volume = π \int_{2}^{4} {(\frac{3 u + 4}{u})}^{2} d u$ M1

$= π \int_{2}^{4} 9 + \frac{16}{u^{2}} + \frac{24}{u} d u$ A1

$= π {[9 u - \frac{16}{u} + 24 \ln u]}_{2}^{4}$ A1

Note: Ignore absence of or incorrect limits seen up to this point.

$= π (22 + 24 \ln 2)$ A1

[6 marks]

Examiners report

[N/A]

The curve $C$ is given by the equation $y = x\,{\text{tan}}\left( {\frac{{\pi xy}}{4}} \right)$ .

At the point (1, 1) , show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 + \pi }}{{2 - \pi }}$ .

[5]

Hence find the equation of the normal to $C$ at the point (1, 1).

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to differentiate implicitly M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\,{\text{se}}{{\text{c}}^2}\left( {\frac{{\pi xy}}{4}} \right)\left[ {\left( {\frac{\pi }{4}x\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{\pi }{4}y} \right)} \right] + {\text{tan}}\left( {\frac{{\pi xy}}{4}} \right)$ A1A1

Note: Award A1 for each term.

attempt to substitute $x = 1$ , $y = 1$ into their equation for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{\pi }{2} + 1$

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

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 + \pi }}{{2 - \pi }}$ AG

[5 marks]

attempt to use gradient of normal $= \frac{{ - 1}}{{\frac{{{\text{d}}y}}{{{\text{d}}x}}}}$ (M1)

$= \frac{{\pi - 2}}{{\pi + 2}}$

so equation of normal is $y - 1 = \frac{{\pi - 2}}{{\pi + 2}}\left( {x - 1} \right)$ or $y = \frac{{\pi - 2}}{{\pi + 2}}x + \frac{4}{{\pi + 2}}$ A1

[2 marks]

Examiners report

[N/A]

Consider the function $f(x) = \frac{1}{{{x^2} + 3x + 2}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne - 2,{\text{ }}x \ne - 1$ .

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

[1]

a.i.

Factorize ${x^2} + 3x + 2$ .

[1]

a.ii.

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

[5]

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

[4]

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

[2]

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 $x = 1$ .

[3]

Markscheme

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

[1 mark]

a.i.

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

[1 mark]

a.ii.

M17/5/MATHL/HP1/ENG/TZ1/B11.b/M

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]

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

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

$= \ln 2 - \ln 3 - \ln 1 + \ln 2$ M1

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

$\therefore p = \frac{4}{3}$

[4 marks]

M17/5/MATHL/HP1/ENG/TZ1/B11.e/M

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]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Let $y = {\text{arccos}}\left( {\frac{x}{2}} \right)$

Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .

[2]

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

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

Examiners report

[N/A]

Let $f (x) = \sqrt{1 + x}$ for $x > - 1$ .

Show that $f'' (x) = - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ .

[3]

Use mathematical induction to prove that $f^{(n)} (x) = {(- \frac{1}{4})}^{n - 1} \frac{(2 n - 3)!}{(n - 2)!} {(1 + x)}^{\frac{1}{2} - n}$ for $n \in ℤ, n \geq 2$ .

[9]

Let $g (x) = e^{m x}, m \in ℚ$ .

Consider the function $h$ defined by $h (x) = f (x) \times g (x)$ for $x > - 1$ .

It is given that the $x^{2}$ term in the Maclaurin series for $h (x)$ has a coefficient of $\frac{7}{4}$ .

Find the possible values of $m$ .

[8]

Markscheme

attempt to use the chain rule M1

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

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

$= - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ AG

Note: Award M1A0A0 for $f' (x) = \frac{1}{\sqrt{1 + x}}$ or equivalent seen

[3 marks]

let $n = 2$

$f^{''} (x) = (- \frac{1}{4 \sqrt{{(1 + x)}^{3}}} =) {(- \frac{1}{4})}^{1} \frac{1!}{0!} {(1 + x)}^{\frac{1}{2} - 2}$ R1

Note: Award R0 for not starting at $n = 2$ . Award subsequent marks as appropriate.

assume true for $n = k$ , (so $f^{(k)} (x) = {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k}$ ) M1

Note: Do not award M1 for statements such as “let $n = k$ ” or “ $n = k$ is true”. Subsequent marks can still be awarded.

consider $n = k + 1$

$LHS = f^{(k + 1)} (x) = \frac{d (f^{(k)} (x))}{d x}$ M1

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1}{2} - k) {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

EITHER

$RHS = f^{(k + 1)} (x) = {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for $\frac{(2 k - 1)!}{(k - 1)!} = \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} (= \frac{2 (2 k - 1) (2 k - 3)!}{(k - 2)!})$

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$(= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1})$

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= (\frac{1}{2} - k) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: The following A marks can be awarded in any order.

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1 - 2 k}{2}) {(1 + x)}^{\frac{1}{2} - k - 1}$

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for isolating $(2 k - 1)$ correctly.

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(2 k - 2) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for multiplying top and bottom by $(k - 1)$ or $2 (k - 1)$ .

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$= {(- \frac{1}{4})}^{(k + 1) - 1} \frac{(2 (k + 1) - 3)!}{((k + 1) - 2)!} {(1 + x)}^{\frac{1}{2} - (k + 1)} = RHS$

THEN

since true for $n = 2$ , and true for $n = k + 1$ if true for $n = k$ , the statement is true for all, $n \in ℤ, n \geq 2$ by mathematical induction R1

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

[9 marks]

METHOD 1

$h (x) = \sqrt{1 + x} e^{m x}$

using product rule to find $h' (x)$ (M1)

$h' (x) = \sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}$ A1

$h'' (x) = m (\sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}) + \frac{1}{2 \sqrt{1 + x}} m e^{m x} - \frac{1}{4 \sqrt{{(1 + x)}^{3}}} e^{m x}$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = m^{2} + \frac{1}{2} m + \frac{1}{2} m - \frac{1}{4} (= m^{2} + m - \frac{1}{4})$ A1

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 2

EITHER

attempt to find $f (0), f' (0), f'' (0)$ (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} f (0) = 1$

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

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

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

attempt to apply binomial theorem for rational exponents (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} = 1 + \frac{1}{2} x + \frac{(\frac{1}{2}) (- \frac{1}{2})}{2!} x^{2} \dots$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

THEN

$g (x) = 1 + m x + \frac{m^{2}}{2} x^{2} + \dots$ (A1)

$h (x) = (1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots) (1 + m x + \frac{m^{2}}{2} x^{2} + \dots)$ (M1)

coefficient of $x^{2}$ is $\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8}$ A1

attempt to set equal to $\frac{7}{4}$ and solve M1

$\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8} = \frac{7}{4}$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 3

$g' (x) = m e^{m x}$ and $g'' (x) = m^{2} e^{m x}$ (A1)

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

using product rule to find $h' (x)$ and $h'' (x)$ (M1)

$h' (x) = f (x) g' (x) + f' (x) g (x)$

$h'' (x) = f (x) g'' (x) + 2 f' (x) g' (x) + f'' (x) g (x)$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = f (0) g'' (0) + 2 g' (0) f' (0) + g (0) f'' (0)$

$= 1 \times m^{2} + 2 m \times \frac{1}{2} + 1 \times (- \frac{1}{4}) (= m^{2} + m - \frac{1}{4})$ A1

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

[8 marks]

Examiners report

[N/A]

Consider $f\left( x \right) = \frac{{2x - 4}}{{{x^2} - 1}}{\text{, }} - 1 < x < 1$ .

For the graph of $y = f\left( x \right)$ ,

Find $f'\left( x \right)$ .

[2]

a.i.

Show that, if $f'\left( x \right) = 0$ , then $x = 2 - \sqrt 3$ .

[3]

a.ii.

find the coordinates of the $y$ -intercept.

[1]

b.i.

show that there are no $x$ -intercepts.

[2]

b.ii.

sketch the graph, showing clearly any asymptotic behaviour.

[2]

b.iii.

Show that $\frac{3}{{x + 1}} - \frac{1}{{x - 1}} = \frac{{2x - 4}}{{{x^2} - 1}}$ .

[2]

The area enclosed by the graph of $y = f\left( x \right)$ and the line $y = 4$ can be expressed as ${\text{ln}}\,v$ . Find the value of $v$ .

[7]

Markscheme

attempt to use quotient rule (or equivalent) (M1)

$f'\left( x \right) = \frac{{\left( {{x^2} - 1} \right)\left( 2 \right) - \left( {2x - 4} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ A1

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

[2 marks]

a.i.

$f'\left( x \right) = 0$

simplifying numerator (may be seen in part (i)) (M1)

$\Rightarrow {x^2} - 4x + 1 = 0$ or equivalent quadratic equation A1

EITHER

use of quadratic formula

$\Rightarrow x = \frac{{4 \pm \sqrt {12} }}{2}$ A1

use of completing the square

${\left( {x - 2} \right)^2} = 3$ A1

THEN

$x = 2 - \sqrt 3$ (since $2 + \sqrt 3$ is outside the domain) AG

Note: Do not condone verification that $x = 2 - \sqrt 3 \Rightarrow f'\left( x \right) = 0$ .

Do not award the final A1 as follow through from part (i).

[3 marks]

a.ii.

(0, 4) A1

[1 mark]

b.i.

$2x - 4 = 0 \Rightarrow x = 2$ A1

outside the domain R1

[2 marks]

b.ii.

A1A1

award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching $x = \pm 1$ asymptotically

[2 marks]

b.iii.

valid attempt to combine fractions (using common denominator) M1

$\frac{{3\left( {x - 1} \right) - \left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}$ A1

$= \frac{{3x - 3 - x - 1}}{{{x^2} - 1}}$

$= \frac{{2x - 4}}{{{x^2} - 1}}$ AG

[2 marks]

$f\left( x \right) = 4 \Rightarrow 2x - 4 = 4{x^2} - 4$ M1

( $x = 0$ or) $x = \frac{1}{2}$ A1

area under the curve is $\int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= \int_0^{\frac{1}{2}} {\frac{3}{{x + 1}} - \frac{1}{{x - 1}}{\text{d}}x}$

Note: Ignore absence of, or incorrect limits up to this point.

$= \left[ {3\,{\text{ln}}\,\left| {x + 1} \right| - {\text{ln}}\,\left| {x - 1} \right|} \right]_0^{\frac{1}{2}}$ A1

$= 3\,{\text{ln}}\frac{3}{2} - {\text{ln}}\frac{1}{2}\left( { - 0} \right)$

$= {\text{ln}}\frac{{27}}{4}$ A1

area is $2 - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ or $\int_0^{\frac{1}{2}} {4\,{\text{d}}x} - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= 2 - {\text{ln}}\frac{{27}}{4}$

$= {\text{ln}}\frac{{4\,{{\text{e}}^2}}}{{27}}$ A1

$\left( { \Rightarrow v = \frac{{4\,{{\text{e}}^2}}}{{27}}} \right)$

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

Given that $\int_{ - 2}^2 {f\left( x \right){\text{d}}x = 10}$ and $\int_0^2 {f\left( x \right){\text{d}}x = 12}$ , find

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

[4]

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

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

Examiners report

[N/A]

Using the substitution $u = {\text{sin}}\,x$ , find $\int {\frac{{{\text{co}}{{\text{s}}^3}x\,{\text{d}}x}}{{\sqrt {{\text{sin}}\,x} }}}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$u = {\text{sin}}\,x \Rightarrow {\text{d}}u = {\text{cos}}\,x{\text{d}}x$ (A1)

valid attempt to write integral in terms of $u$ and ${\text{d}}u$ M1

$\int {\frac{{{\text{co}}{{\text{s}}^3}x\,{\text{d}}x}}{{\sqrt {{\text{sin}}\,x} }}} = \int {\frac{{\left( {1 - {u^2}} \right){\text{d}}u}}{{\sqrt u }}}$ A1

$= \int {\left( {{u^{ - \frac{1}{2}}} - {u^{\frac{3}{2}}}} \right)} \,{\text{d}}u$

$= 2{u^{\frac{1}{2}}} - \frac{{2{u^{\frac{5}{2}}}}}{5}\left( { + c} \right)$ (A1)

$= 2\sqrt {{\text{sin}}\,x} - \frac{{2{{\left( {\sqrt {{\text{sin}}\,x} } \right)}^5}}}{5}\left( { + c} \right)$ or equivalent A1

[5 marks]

Examiners report

[N/A]

The function $f$ is defined by $f (x) = e^{x} \sin x$ , where $x \in ℝ$ .

The function $g$ is defined by $g (x) = e^{x} \cos x$ , where $x \in ℝ$ .

Find the Maclaurin series for $f (x)$ up to and including the $x^{3}$ term.

[4]

Hence, find an approximate value for $\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x$ .

[4]

Show that $g (x)$ satisfies the equation $g'' (x) = 2 (g' (x) - g (x))$ .

[4]

c.i.

Hence, deduce that $g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ .

[1]

c.ii.

Using the result from part (c), find the Maclaurin series for $g (x)$ up to and including the $x^{4}$ term.

[5]

Hence, or otherwise, determine the value of $\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}}$ .

[3]

Markscheme

METHOD 1

recognition of both known series (M1)

$e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots$ and $\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots$

attempt to multiply the two series up to and including $x^{3}$ term (M1)

$e^{x} \sin x = (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots)$

$= x - \frac{x^{3}}{3!} + x^{2} + \frac{x^{3}}{2!} + \dots$ (A1)

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

METHOD 2

$f (x) = e^{x} \sin x$

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

$f'' (x) = e^{x} \cos x - e^{x} \sin x + e^{x} \sin x + e^{x} \cos x (= 2 e^{x} \cos x)$

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

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

substitute $x = 0$ into $f$ or its derivatives to obtain Maclaurin series (M1)

$e^{x} \sin x = 0 + \frac{x}{1!} \times 1 + \frac{x^{2}}{2!} \times 2 + \frac{x^{3}}{3!} \times 2 + \dots$

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

[4 marks]

$e^{x^{2}} \sin (x^{2}) = x^{2} + x^{4} + \frac{1}{3} x^{6} + \dots$ (A1)

substituting their expression and attempt to integrate M1

$\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x \approx \int_{0}^{1} (x^{2} + x^{4} + \frac{1}{3} x^{6}) d x$

Note: Condone absence of limits up to this stage.

$= {[\frac{x^{3}}{3} + \frac{x^{5}}{5} + \frac{x^{7}}{21}]}_{0}^{1}$ A1

$= \frac{61}{105}$ A1

[4 marks]

attempt to use product rule at least once M1

$g' (x) = e^{x} \cos x - e^{x} \sin x$ A1

$g'' (x) = e^{x} \cos x - e^{x} \sin x - e^{x} \sin x - e^{x} \cos x (= - 2 e^{x} \sin x)$ A1

EITHER

$2 (g' (x) - g (x)) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x) = - 2 e^{x} \sin x$ A1

$g'' (x) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x)$ A1

THEN

$g'' (x) = 2 (g' (x) - g (x))$ AG

Note: Accept working with each side separately to obtain $- 2 e^{x} \sin x$ .

[4 marks]

c.i.

$g''' (x) = 2 (g'' (x) - g' (x))$ A1

$g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ AG

Note: Accept working with each side separately to obtain $- 4 e^{x} \cos x$ .

[1 mark]

c.ii.

attempt to substitute $x = 0$ into a derivative (M1)

$g (0) = 1, g' (0) = 1, g'' (0) = 0$ A1

$g''' (0) = - 2, g^{(4)} (0) = - 4$ (A1)

attempt to substitute into Maclaurin formula (M1)

$g (x) = 1 + x - \frac{2}{3!} x^{3} - \frac{4}{4!} x^{4} + \dots (= 1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots)$ A1

Note: Do not award any marks for approaches that do not use the part (c) result.

[5 marks]

METHOD 1

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \lim_{x \to 0} \frac{(1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots) - 1 - x}{x^{3}}$ M1

$= \lim_{x \to 0} (- \frac{1}{3} - \frac{1}{6} x + \dots)$ (A1)

$= - \frac{1}{3}$ A1

Note: Condone the omission of $+ \dots$ in their working.

METHOD 2

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \frac{0}{0}$ indeterminate form, attempt to apply l'Hôpital's rule M1

$= \lim_{x \to 0} \frac{e^{x} \cos x - e^{x} \sin x - 1}{3 x^{2}} (= \lim_{x \to 0} \frac{g' (x) - 1}{3 x^{2}})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x}{6 x} (= \lim_{x \to 0} \frac{g'' (x)}{6 x})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x - 2 e^{x} \cos x}{6} (= \lim_{x \to 0} \frac{g''' (x)}{6})$ A1

$= - \frac{1}{3}$ A1

[3 marks]

Examiners report

Part (a) was well answered using both methods. A number failed to see the connection between parts (a) and (b). Far too often, a candidate could not add three fractions together in part (b). There were many good responses to part (c) with candidates showing results on both sides are equal. A number of candidates failed to use the result from (c) in part (d). There were some good responses to part (e), with candidates working successfully with the series from (d) or applying l'Hôpital's rule. In particular, some responses were missing the appropriate limit notation and candidates following method 2 did not always show that the initial expression was of an indeterminate form before applying l'Hôpital's rule. Many candidates did not attempt parts (d) and (e).

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

The graph of $y = f'\left( x \right)$ , 0 ≤ $x$ ≤ 5 is shown in the following diagram. The curve intercepts the $x$ -axis at (1, 0) and (4, 0) and has a local minimum at (3, −1).

The shaded area enclosed by the curve $y = f'\left( x \right)$ , the $x$ -axis and the $y$ -axis is 0.5. Given that $f\left( 0 \right) = 3$ ,

The area enclosed by the curve $y = f'\left( x \right)$ and the $x$ -axis between $x = 1$ and $x = 4$ is 2.5 .

Write down the $x$ -coordinate of the point of inflexion on the graph of $y = f\left( x \right)$ .

[1]

find the value of $f\left( 1 \right)$ .

[3]

find the value of $f\left( 4 \right)$ .

[2]

Sketch the curve $y = f\left( x \right)$ , 0 ≤ $x$ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3 A1

[1 mark]

attempt to use definite integral of $f'\left( x \right)$ (M1)

$\int_0^1 {f'\left( x \right){\text{d}}x} = 0.5$

$f\left( 1 \right) - f\left( 0 \right) = 0.5$ (A1)

$f\left( 1 \right) = 0.5 + 3$

= 3.5 A1

[3 marks]

$\int_1^4 {f'\left( x \right){\text{d}}x} = - 2.5$ (A1)

Note: (A1) is for −2.5.

$f\left( 4 \right) - f\left( 1 \right) = - 2.5$

$f\left( 4 \right) = 3.5 - 2.5$

= 1 A1

[2 marks]

A1A1A1

A1 for correct shape over approximately the correct domain
A1 for maximum and minimum (coordinates or horizontal lines from 3.5 and 1 are required),
A1 for $y$ -intercept at 3

[3 marks]

Examiners report

[N/A]

A function $f$ is defined by $f (x) = \frac{3}{x^{2} + 2}, x \in ℝ$ .

The region $R$ is bounded by the curve $y = f (x)$ , the $x$ -axis and the lines $x = 0$ and $x = \sqrt{6}$ . Let $A$ be the area of $R$ .

The line $x = k$ divides $R$ into two regions of equal area.

Let $m$ be the gradient of a tangent to the curve $y = f (x)$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

[4]

Show that $A = \frac{\sqrt{2} π}{2}$ .

[4]

Find the value of $k$ .

[4]

Show that $m = - \frac{6 x}{{(x^{2} + 2)}^{2}}$ .

[2]

Show that the maximum value of $m$ is $\frac{27}{32} \sqrt{\frac{2}{3}}$ .

[7]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

a curve symmetrical about the $y$ -axis with correct concavity that has a local maximum point on the positive $y$ -axis A1

a curve clearly showing that $y \to 0$ as $x \to \pm \infty$ A1

$(0, \frac{3}{2})$ A1

horizontal asymptote $y = 0$ ( $x$ -axis) A1

[4 marks]

attempts to find $\int \frac{3}{x^{2} + 2} d x$ (M1)

$= [\frac{3}{\sqrt{2}} \arctan \frac{x}{\sqrt{2}}]$ A1

Note: Award M1A0 for obtaining $[k \arctan \frac{x}{\sqrt{2}}]$ where $k \neq \frac{3}{\sqrt{2}}$ .

Note: Condone the absence of or use of incorrect limits to this stage.

$= \frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan 0)$ (M1)

$= \frac{3}{\sqrt{2}} \times \frac{π}{3} (= \frac{π}{\sqrt{2}})$ A1

$A = \frac{\sqrt{2} π}{2}$ AG

[4 marks]

METHOD 1

EITHER

$\int_{0}^{k} \frac{3}{x^{2} + 2} d x = \frac{\sqrt{2} π}{4}$

$\frac{3}{\sqrt{2}} \arctan \frac{k}{\sqrt{2}} = \frac{\sqrt{2} π}{4}$ (M1)

$\int_{k}^{\sqrt{6}} \frac{3}{x^{2} + 2} d x = \frac{\sqrt{2} π}{4}$

$\frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}}) = \frac{\sqrt{2} π}{4}$ (M1)

$\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$

THEN

$\arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$ A1

$\frac{k}{\sqrt{2}} = \tan \frac{π}{6} (= \frac{1}{\sqrt{3}})$ A1

$k = \frac{\sqrt{6}}{3} (= \sqrt{\frac{2}{3}})$ A1

METHOD 2

$\int_{0}^{k} \frac{3}{x^{2} + 2} d x = \int_{k}^{\sqrt{6}} \frac{3}{x^{2} + 2} d x$

$\frac{3}{\sqrt{2}} \arctan \frac{k}{\sqrt{2}} = \frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}})$ (M1)

$\arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$ A1

$\frac{k}{\sqrt{2}} = \tan \frac{π}{6} (= \frac{1}{\sqrt{3}})$ A1

$k = \frac{\sqrt{6}}{3} (= \sqrt{\frac{2}{3}})$ A1

[4 marks]

attempts to find $\frac{d}{d x} (\frac{3}{x^{2} + 2})$ (M1)

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

so $m = - \frac{6 x}{{(x^{2} + 2)}^{2}}$ AG

[2 marks]

attempts product rule or quotient rule differentiation M1

EITHER

$\frac{d m}{d x} = (- 6 x) (- 2) (2 x) {(x^{2} + 2)}^{- 3} + {(x^{2} + 2)}^{- 2} (- 6)$ A1

$\frac{d m}{d x} = \frac{{(x^{2} + 2)}^{2} (- 6) - (- 6 x) (2) (2 x) (x^{2} + 2)}{{(x^{2} + 2)}^{4}}$ A1

Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.

THEN

attempts to express their $\frac{d m}{d x}$ as a rational fraction with a factorized numerator M1

$\frac{d m}{d x} = \frac{6 (x^{2} + 2) (3 x^{2} - 2)}{{(x^{2} + 2)}^{4}} (= \frac{6 (3 x^{2} - 2)}{{(x^{2} + 2)}^{3}})$

attempts to solve their $\frac{d m}{d x} = 0$ for $x$ M1

$x = \pm \sqrt{\frac{2}{3}}$ A1

from the curve, the maximum value of $m$ occurs at $x = - \sqrt{\frac{2}{3}}$ R1

(the minimum value of $m$ occurs at $x = \sqrt{\frac{2}{3}}$ )

Note: Award R1 for any equivalent valid reasoning.

maximum value of $m$ is $- \frac{6 (- \sqrt{\frac{2}{3}})}{{({(- \sqrt{\frac{2}{3}})}^{2} + 2)}^{2}}$ A1

leading to a maximum value of $\frac{27}{32} \sqrt{\frac{2}{3}}$ AG

[7 marks]

Examiners report

[N/A]

Show that ${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$ where $r,\,x \in {\mathbb{R}^ + }$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

The folium of Descartes is a curve defined by the equation ${x^3} + {y^3} - 3xy = 0$ , shown in the following diagram.

N17/5/MATHL/HP1/ENG/TZ0/07

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the $y$ -axis.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${x^3} + {y^3} - 3xy = 0$

$3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3x\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3y = 0$ 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

${x^3} + xy - 3xy = 0$ 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]

Examiners report

[N/A]

A particle moves along a straight line. Its displacement, $s$ metres, at time $t$ seconds is given by $s = t + \cos 2t,{\text{ }}t \geqslant 0$ . The first two times when the particle is at rest are denoted by ${t_1}$ and ${t_2}$ , where ${t_1} < {t_2}$ .

Find ${t_1}$ and ${t_2}$ .

[5]

Find the displacement of the particle when $t = {t_1}$

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$s = t + \cos 2t$

$\frac{{{\text{d}}s}}{{{\text{d}}t}} = 1 - 2\sin 2t$ 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]

Examiners report

[N/A]

Use l’Hôpital’s rule to determine the value of

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

using l’Hôpital’s rule,

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2 x}{3 x^{2}}$ M1A1

$= \lim_{x \to 0} \frac{- 2 \sin x + 4 \sin 2 x}{6 x}$ (M1)A1

$= \lim_{x \to 0} \frac{- 2 \cos x + 8 \cos 2 x}{6}$ A1

$= 1$ A1

[6 marks]

Examiners report

[N/A]

Write $2x - {x^2}$ in the form $a{\left( {x - h} \right)^2} + k$ , where $a{\text{, }}h{\text{, }}k \in \mathbb{R}$ .

[2]

Hence, find the value of $\int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{{\sqrt {2x - {x^2}} }}} {\text{d}}x$ .

[5]

Markscheme

attempt to complete the square or multiplication and equating coefficients (M1)

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

$a = - 1$ , $h = 1$ , $k = 1$

[2 marks]

use of their identity from part (a) $\left( {\int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} {\text{d}}x} \right)$ (M1)

$= \left[ {{\text{arc}}\,{\text{sin}}\left( {x - 1} \right)} \right]_{\frac{1}{2}}^{\frac{3}{2}}$ or $\left[ {{\text{arc}}\,{\text{sin}}\left( u \right)} \right]_{ - \frac{1}{2}}^{\frac{1}{2}}$ A1

Note: Condone lack of, or incorrect limits up to this point.

$= {\text{arc}}\,{\text{sin}}\left( {\frac{1}{2}} \right) - {\text{arc}}\,{\text{sin}}\left( { - \frac{1}{2}} \right)$ (M1)

$= \frac{\pi }{6} - \left( { - \frac{\pi }{6}} \right)$ (A1)

$= \frac{\pi }{3}$ A1

[5 marks]

Examiners report

[N/A]

Consider the expression $\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x}$ where $a \in ℚ, a \neq 0$ .

The binomial expansion of this expression, in ascending powers of $x$ , as far as the term in $x^{2}$ is $4 b x + b x^{2}$ , where $b \in ℚ$ .

Find the value of $a$ and the value of $b$ .

[6]

State the restriction which must be placed on $x$ for this expansion to be valid.

[1]

Markscheme

attempt to expand binomial with negative fractional power (M1)

$\frac{1}{\sqrt{1 + a x}} = {(1 + a x)}^{- \frac{1}{2}} = 1 - \frac{a x}{2} + \frac{3 a^{2} x^{2}}{8} + \dots$ A1

$\sqrt{1 - x} = {(1 - x)}^{\frac{1}{2}} = 1 - \frac{x}{2} - \frac{x^{2}}{8} + \dots$ A1

$\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x} = \frac{(1 - a)}{2} x + (\frac{3 a^{2} + 1}{8}) x^{2} + \dots$

attempt to equate coefficients of $x$ or $x^{2}$ (M1)

$x : \frac{1 - a}{2} = 4 b; x^{2} : \frac{3 a^{2} + 1}{8} = b$

attempt to solve simultaneously (M1)

$a = - \frac{1}{3}, b = \frac{1}{6}$ A1

[6 marks]

$|x| < 1$ A1

[1 mark]

Examiners report

[N/A]

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

[4]

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

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

A curve has equation $3x - 2{y^2}{{\text{e}}^{x - 1}} = 2$ .

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

[5]

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

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

Markscheme

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = \int (3 - 5 x^{- \frac{1}{2}}) d x$ (A1)

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[5 marks]

Examiners report

A mixed response was noted for this question. Candidates who simplified the algebraic fraction before integrating were far more successful in gaining full marks in this question. Many candidates used other valid approaches such as integration by substitution and integration by parts with varying degrees of success. A small number of candidates substituted the limits without integrating.

A function $f$ is defined by $f (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x \neq - 1, x \neq 3$ .

A function $g$ is defined by $g (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x > 3$ .

The inverse of $g$ is $g^{- 1}$ .

A function $h$ is defined by $h (x) = arctan \frac{x}{2}$ , where $x \in ℝ$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]

Show that $g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ .

[6]

b.i.

State the domain of $g^{- 1}$ .

[1]

b.ii.

Given that $(h \circ g) (a) = \frac{π}{4}$ , find the value of $a$ .

Give your answer in the form $p + \frac{q}{2} \sqrt{r}$ , where $p, q, r \in ℤ^{+}$ .

[7]

Markscheme

$y$ -intercept $(0, - \frac{1}{3})$ A1

Note: Accept an indication of $- \frac{1}{3}$ on the $y$ -axis.

vertical asymptotes $x = - 1$ and $x = 3$ A1

horizontal asymptote $y = 0$ A1

uses a valid method to find the $x$ -coordinate of the local maximum point (M1)

Note: For example, uses the axis of symmetry or attempts to solve $f' (x) = 0$ .

local maximum point $(1, - \frac{1}{4})$ A1

Note: Award (M1)A0 for a local maximum point at $x = 1$ and coordinates not given.

three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1

[6 marks]

$x = \frac{1}{y^{2} - 2 y - 3}$ M1

Note: Award M1 for interchanging $x$ and $y$ (this can be done at a later stage).

EITHER

attempts to complete the square M1

$y^{2} - 2 y - 3 = {(y - 1)}^{2} - 4$ A1

$x = \frac{1}{{(y - 1)}^{2} - 4}$

${(y - 1)}^{2} - 4 = \frac{1}{x} ({(y - 1)}^{2} = 4 + \frac{1}{x})$ A1

$y - 1 = \pm \sqrt{4 + \frac{1}{x}} (= \pm \sqrt{\frac{4 x + 1}{x}})$

attempts to solve $x y^{2} - 2 x y - 3 x - 1 = 0$ for $y$ M1

$y = \frac{- (- 2 x) \pm \sqrt{{(- 2 x)}^{2} + 4 x (3 x + 1)}}{2 x}$ A1

Note: Award A1 even if $-$ (in $\pm$ ) is missing

$= \frac{2 x \pm \sqrt{16 x^{2} + 4 x}}{2 x}$ A1

THEN

$= 1 \pm \frac{\sqrt{4 x^{2} + x}}{x}$ A1

$y > 3$ and hence $y = 1 - \frac{\sqrt{4 x^{2} + x}}{x}$ is rejected R1

Note: Award R1 for concluding that the expression for $y$ must have the ‘ $+$ ’ sign.
The R1 may be awarded earlier for using the condition $x > 3$ .

$y = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$

$g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ AG

[6 marks]

b.i.

domain of $g^{- 1}$ is $x > 0$ A1

[1 mark]

b.ii.

attempts to find $(h \circ g) (a)$ (M1)

$(h \circ g) (a) = arctan (\frac{g (a)}{2}) ((h \circ g) (a) = arctan (\frac{1}{2 (a^{2} - 2 a - 3)}))$ (A1)

$arctan (\frac{g (a)}{2}) = \frac{π}{4} (arctan (\frac{1}{2 (a^{2} - 2 a - 3)}) = \frac{π}{4})$

attempts to solve for $g (a)$ M1

$\Rightarrow g (a) = 2 (\frac{1}{(a^{2} - 2 a - 3)} = 2)$

EITHER

$\Rightarrow a = g^{- 1} (2)$ A1

attempts to find their $g^{- 1} (2)$ M1

$a = 1 + \frac{\sqrt{4 {(2)}^{2} + 2}}{2}$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

$\Rightarrow 2 a^{2} - 4 a - 7 = 0$ A1

attempts to solve their quadratic equation M1

$a = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} + 4 (2) (7)}}{4} (= \frac{4 \pm \sqrt{72}}{4})$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

THEN

$a = 1 + \frac{3}{2} \sqrt{2}$ (as $a > 3$ ) A1

$(p = 1, q = 3, r = 2)$

Note: Award A1 for $a = 1 + \frac{1}{2} \sqrt{18}$ $(p = 1, q = 1, r = 18)$

[7 marks]

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

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

[4]

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 $q \in \mathbb{Q}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{2}{x^{ - \frac{1}{2}}}$ (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]

Examiners report

[N/A]

A camera at point C is 3 m from the edge of a straight section of road as shown in the following diagram. The camera detects a car travelling along the road at $t$ = 0. It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.

A car travels along the road at a speed of 24 ms⁻¹. Let the position of the car be X and let OĈX = θ.

Find $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ , the rate of rotation of the camera, in radians per second, at the instant the car passes the point O .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

let OX = $x$

METHOD 1

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 24$ (or −24) (A1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}x}}{{{\text{d}}t}} \times \frac{{{\text{d}}\theta }}{{{\text{d}}x}}$ (M1)

$3\,{\text{tan}}\,\theta = x$ A1

EITHER

$3\,{\text{se}}{{\text{c}}^2}\,\theta = \frac{{{\text{d}}x}}{{{\text{d}}\theta }}$ A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{{3\,{\text{se}}{{\text{c}}^2}\,\theta }}$

attempt to substitute for $\theta = 0$ into their differential equation M1

$\theta = {\text{arctan}}\left( {\frac{x}{3}} \right)$

$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{1}{3} \times \frac{1}{{1 + \frac{{{x^2}}}{9}}}$ A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 24 \times \frac{1}{{3\left( {1 + \frac{{{x^2}}}{9}} \right)}}$

attempt to substitute for $x = 0$ into their differential equation M1

THEN

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

METHOD 2

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 24$ (or −24) (A1)

$3\,{\text{tan}}\,\theta = x$ A1

attempt to differentiate implicitly with respect to $t$ M1

$3\,{\text{se}}{{\text{c}}^2}\,\theta \times \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}x}}{{{\text{d}}t}}$ A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{{3\,{\text{se}}{{\text{c}}^2}\,\theta }}$

attempt to substitute for $\theta = 0$ into their differential equation M1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{24}}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

Note: Can be done by consideration of CX, use of Pythagoras.

METHOD 3

let the position of the car be at time $t$ be $d - 24t$ from O (A1)

${\text{tan}}\,\theta = \frac{{d - 24t}}{3}\left( { = \frac{d}{3} - 8t} \right)$ M1

Note: For ${\text{tan}}\,\theta = \frac{{24t}}{3}$ award A0M1 and follow through.

EITHER

attempt to differentiate implicitly with respect to $t$ M1

${\text{se}}{{\text{c}}^2}\,\theta \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = - 8$ A1

attempt to substitute for $\theta = 0$ into their differential equation M1

$\theta = {\text{arctan}}\left( {\frac{d}{3} - 8t} \right)$ M1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{8}{{1 + {{\left( {\frac{d}{3} - 8t} \right)}^2}}}$ A1

at O, $t = \frac{d}{{24}}$ A1

THEN

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = - 8$ A1

[6 marks]

Examiners report

[N/A]

Use l’Hôpital’s rule to determine the value of $\lim_{x \to 0} (\frac{2 x \cos (x^{2})}{5 \tan x})$ .

Markscheme

attempts to apply l’Hôpital’s rule on $\lim_{x \to 0} (\frac{2 x \cos (x^{2})}{5 \tan x})$ M1

$= \lim_{x \to 0} (\frac{2 \cos (x^{2}) - 4 x^{2} \sin (x^{2})}{5 \sec^{2} x})$ M1A1A1

Note: Award M1 for attempting to use product and chain rule differentiation on the numerator, A1 for a correct numerator and A1 for a correct denominator. The awarding of A1 for the denominator is independent of the M1.

$= \frac{2}{5}$ A1

[5 marks]

Examiners report

[N/A]

The acceleration, $a {ms}^{- 2}$ , of a particle moving in a horizontal line at time $t$ seconds, $t \geq 0$ , is given by $a = - (1 + v)$ where $v {ms}^{- 1}$ is the particle’s velocity and $v > - 1$ .

At $t = 0$ , the particle is at a fixed origin $O$ and has initial velocity $v_{0} {ms}^{- 1}$ .

Initially at $O$ , the particle moves in the positive direction until it reaches its maximum displacement from $O$ . The particle then returns to $O$ .

Let $s$ metres represent the particle’s displacement from $O$ and $s_{max}$ its maximum displacement from $O$ .

Let $v (T - k)$ represent the particle’s velocity $k$ seconds before it reaches $s_{max}$ , where

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$ .

Similarly, let $v (T + k)$ represent the particle’s velocity $k$ seconds after it reaches $s_{max}$ .

By solving an appropriate differential equation, show that the particle’s velocity at time $t$ is given by $v (t) = (1 + v_{0}) e^{- t} - 1$ .

[6]

Show that the time $T$ taken for the particle to reach $s_{max}$ satisfies the equation $e^{T} = 1 + v_{0}$ .

[2]

b.i.

By solving an appropriate differential equation and using the result from part (b) (i), find an expression for $s_{max}$ in terms of $v_{0}$ .

[5]

b.ii.

By using the result to part (b) (i), show that $v (T - k) = e^{k} - 1$ .

[2]

Deduce a similar expression for $v (T + k)$ in terms of $k$ .

[2]

Hence, show that $v (T - k) + v (T + k) \geq 0$ .

[3]

Markscheme

$\frac{d v}{d t} = - (1 + v)$ (A1)

$\int 1 d t = \int - \frac{1}{1 + v} d v$ (or equivalent / use of integrating factor) M1

$t = - \ln (1 + v) (+ C)$ A1

EITHER

attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$C = \ln (1 + v_{0})$

$t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{t - C} = \frac{1}{(1 + v)}$

Attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$e^{- C} = \frac{1}{(1 + v_{0})} \Rightarrow C = \ln (1 + v_{0})$

$t - \ln (1 + v_{0}) = - \ln (1 + v) \Rightarrow t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{- t + C} = 1 + v$ A1

$k e^{- t} - 1 = v$

Attempt to find $k$ with initial conditions $t = 0, v = v_{0}$ M1

$k = 1 + v_{0}$

$e^{- t} (1 + v_{0}) = 1 + v$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

Note: condone use of modulus within the ln function(s)

[6 marks]

recognition that when $t = T, v = 0$ M1

$(1 + v_{0}) e^{- T} - 1 = 0 \Rightarrow e^{- T} = \frac{1}{1 + v_{0}}$ A1

$e^{T} = 1 + v_{0}$ AG

Note: Award M1A0 for substituting $v_{0} = e^{T} - 1$ into $v$ and showing that $v = 0$ .

[6 marks]

b.i.

$s (t) = \int v (t) d t (= \int ((1 + v_{0}) e^{- t} - 1) d t)$ (M1)

$= - (1 + v_{0}) e^{- t} - t (+ D)$ A1

( $t = 0, s = 0$ so) $D = 1 + v_{0}$ A1

$s (t) = - (1 + v_{0}) e^{- t} - t + 1 + v_{0}$

at $s_{max}, e^{T} = 1 + v_{0} \Rightarrow T = \ln (1 + v_{0})$

Substituting into $s (t) (= - (1 + v_{0}) e^{- t} - t + 1 + v_{0})$ M1

$s_{max} = - (1 + v_{0}) (\frac{1}{1 + v_{0}}) - \ln (1 + v_{0}) + v_{0} + 1$ A1

$(s_{max} = v_{0} - \ln (1 + v_{0}))$

[5 marks]

b.ii.

METHOD 1

$v (T - k) = (1 + v_{0}) e^{- T} e^{k} - 1$ (M1)

$= (1 + v_{0}) (\frac{1}{1 + v_{0}}) e^{k} - 1$ A1

$= e^{k} - 1$ AG

METHOD 2

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$

$= e^{T} e^{- (T - k)} - 1$ M1

$= e^{T - T + k} - 1$ A1

$= e^{k} - 1$ AG

[2 marks]

METHOD 1

$v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1$ (A1)

$= e^{- k} - 1$ A1

METHOD 2

$v (T + k) = (1 + v_{0}) e^{- (T + k)} - 1$ (A1)

$= e^{T} e^{- (T + k)} - 1$

$= e^{T - T - k} - 1$

$= e^{- k} - 1$ A1

[2 marks]

METHOD 1

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

attempt to express as a square M1

$= {(e^{\frac{k}{2}} - e^{- \frac{k}{2}})}^{2} (\geq 0)$ A1

so $v (T - k) + v (T + k) \geq 0$ AG

METHOD 2

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

Attempt to solve $\frac{d}{d k} (e^{k} + e^{- k}) = 0 (\Rightarrow k = 0)$ M1

minimum value of $2$ , (when $k = 0$ ), hence $e^{k} + e^{- k} \geq 2$ R1

so $v (T - k) + v (T + k) \geq 0$ AG

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Use l’Hôpital’s rule to find $\lim_{x \to 0} (\frac{arctan 2 x}{\tan 3 x})$ .

Markscheme

attempt to differentiate numerator and denominator M1

$\lim_{x \to 0} (\frac{arctan 2 x}{\tan 3 x})$

$= \lim_{x \to 0} \frac{(\frac{2}{1 + 4 x^{2}})}{3 {sec}^{2} 3 x}$ A1A1

Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.

attempt to substitute $x = 0$ (M1)

$= \frac{2}{3}$ A1

Note: Award a maximum of M1A1A0M1A1 for absence of limits.

[5 marks]

Examiners report

[N/A]

Consider the curves ${C_1}$ and ${C_2}$ defined as follows

${C_1}\,{\text{:}}\,xy = 4$ , $x > 0$

${C_2}\,{\text{:}}\,{y^2} - {x^2} = 2$ , $x > 0$

Using implicit differentiation, or otherwise, find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ for each curve in terms of $x$ and $y$ .

[4]

Let P( $a$ , $b$ ) be the unique point where the curves ${C_1}$ and ${C_2}$ intersect.

Show that the tangent to ${C_1}$ at P is perpendicular to the tangent to ${C_2}$ at P.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${C_1}\,{\text{:}}\,y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ (M1)

Note: M1 is for use of both product rule and implicit differentiation.

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{y}{x}$ A1

Note: Accept $- \frac{4}{{{x^2}}}$

${C_2}\,{\text{:}}\,2y\frac{{{\text{d}}y}}{{{\text{d}}x}} - 2x = 0$ (M1)

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y}$ A1

Note: Accept $\pm \frac{x}{{\sqrt {2 + {x^2}} }}$

[4 marks]

substituting $a$ and $b$ for $x$ and $y$ M1

product of gradients at P is $\left( { - \frac{b}{a}} \right)\left( {\frac{a}{b}} \right) = - 1$ or equivalent reasoning R1

Note: The R1 is dependent on the previous M1.

so tangents are perpendicular AG

[2 marks]

Examiners report

[N/A]

Solve the differential equation $\frac{d y}{d x} = \frac{\ln 2 x}{x^{2}} - \frac{2 y}{x}, x > 0$ , given that $y = 4$ at $x = \frac{1}{2}$ .

Give your answer in the form $y = f (x)$ .

Markscheme

$\frac{d y}{d x} + \frac{2 y}{x} = \frac{\ln 2 x}{x^{2}}$ (M1)

attempt to find integrating factor (M1)

$(e^{\int \frac{2}{x} d x} = e^{2 \ln x}) = x^{2}$ (A1)

$x^{2} \frac{d y}{d x} + 2 x y = \ln 2 x$

$\frac{d}{d x} (x^{2} y) = \ln 2 x$

$x^{2} y = \int \ln 2 x d x$

attempt to use integration by parts (M1)

$x^{2} y = x \ln 2 x - x (+ c)$ A1

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{c}{x^{2}}$

substituting $x = \frac{1}{2}, y = 4$ into an integrated equation involving $c$ M1

$4 = 0 - 2 + 4 c$

$\Rightarrow c = \frac{3}{2}$

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{3}{2 x^{2}}$ A1

[7 marks]

Examiners report

[N/A]

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.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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]

Examiners report

[N/A]

Consider the function $q\left( x \right) = {x^5} - 10{x^2} + 15x - 6,{\text{ }}x \in \mathbb{R}$ .

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

[3]

e.i.

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

[3]

e.ii.

Markscheme

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

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

[3 marks]

e.i.

M17/5/MATHL/HP1/ENG/TZ1/B12.e.ii/M

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

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

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

[3 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

Consider the curve $C$ defined by $y^{2} = \sin (x y), y \neq 0$ .

Show that $\frac{d y}{d x} = \frac{y \cos (x y)}{2 y - x \cos (x y)}$ .

[5]

Prove that, when $\frac{d y}{d x} = 0, y = \pm 1$ .

[5]

Hence find the coordinates of all points on $C$ , for $0 < x < 4 π$ , where $\frac{d y}{d x} = 0$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at implicit differentiation M1

$2 y \frac{d y}{d x} = \cos (x y) ⌊x \frac{d y}{d x} + y⌋$ A1M1A1

Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.

$2 y \frac{d y}{d x} = x \frac{d y}{d x} \cos (x y) + y \cos (x y)$

$2 y \frac{d y}{d x} - x \frac{d y}{d x} \cos (x y) = y \cos (x y)$

$\frac{d y}{d x} (2 y - x \cos (x y)) = y \cos (x y)$ M1

Note: Award M1 for collecting derivatives and factorising.

$\frac{d y}{d x} = \frac{y \cos (x y)}{2 y - x \cos (x y)}$ AG

[5 marks]

setting $\frac{d y}{d x} = 0$

$y \cos (x y) = 0$ (M1)

$(y \neq 0) \Rightarrow \cos (x y) = 0$ A1

$\Rightarrow \sin (x y) (= \pm \sqrt{1 - \cos^{2} (x y)} = \pm \sqrt{1 - 0}) = \pm 1$ OR $x y = (2 n + 1) \frac{π}{2} (n \in ℤ)$ OR $x y = \frac{π}{2}, \frac{3 π}{2}, \dots$ A1

Note: If they offer values for $x y$ , award A1 for at least two correct values in two different ‘quadrants’ and no incorrect values.

$y^{2} (= \sin (x y)) > 0$ R1

$\Rightarrow y^{2} = 1$ A1

$\Rightarrow y = \pm 1$ AG

[5 marks]

$y = \pm 1 \Rightarrow 1 = \sin (\pm x) \Rightarrow \sin x = \pm 1$ OR $y = \pm 1 \Rightarrow 0 = \cos (\pm x) \Rightarrow \cos x = 0$ (M1)

$(\sin x = 1 \Rightarrow) (\frac{π}{2}, 1), (\frac{5 π}{2}, 1)$ A1A1

$(\sin x = - 1 \Rightarrow) (\frac{3 π}{2}, - 1), (\frac{7 π}{2}, - 1)$ A1A1

Note: Allow ‘coordinates’ expressed as $x = \frac{π}{2}, y = 1$ for example.
Note: Each of the A marks may be awarded independently and are not dependent on (M1) being awarded.

Note: Mark only the candidate’s first two attempts for each case of $\sin x$ .

[5 marks]

Examiners report

[N/A]

Find the coordinates of the points on the curve ${y^3} + 3x{y^2} - {x^3} = 27$ at which $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at implicit differentiation M1

$3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{y^2} + 6xy\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3{x^2} = 0$ A1A1

Note: Award A1 for the second & third terms, A1 for the first term, fourth term & RHS equal to zero.

substitution of $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ M1

$3{y^2} - 3{x^2} = 0$

$\Rightarrow y = \pm x$ A1

substitute either variable into original equation M1

$y = x \Rightarrow {x^3} = 9 \Rightarrow x = \sqrt[3]{9}$ (or ${y^3} = 9 \Rightarrow y = \sqrt[3]{9}$ ) A1

$y = - x \Rightarrow {x^3} = 27 \Rightarrow x = 3$ (or ${y^3} = - 27 \Rightarrow y = - 3$ ) A1

$\left( {\sqrt[3]{9}{\text{,}}\,\,\sqrt[3]{9}} \right)$ , (3, −3) A1

[9 marks]

Examiners report

[N/A]

Find $\int {\arcsin x\,{\text{d}}x}$

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^{{\text{sin}}\,x}}$ .

Find the first two derivatives of $f\left( x \right)$ and hence find the Maclaurin series for $f\left( x \right)$ up to and including the ${x^2}$ term.

[8]

Show that the coefficient of ${x^3}$ in the Maclaurin series for $f\left( x \right)$ is zero.

[4]

Using the Maclaurin series for ${\text{arctan}}\,x$ and ${{\text{e}}^{3x}} - 1$ , find the Maclaurin series for ${\text{arctan}}\left( {{{\text{e}}^{3x}} - 1} \right)$ up to and including the ${x^3}$ term.

[6]

Hence, or otherwise, find $\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{f\left( x \right) - 1}}{{{\text{arctan}}\left( {{{\text{e}}^{3x}} - 1} \right)}}$ .

[3]

Markscheme

attempting to use the chain rule to find the first derivative M1

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

attempting to use the product rule to find the second derivative M1

$f''\left( x \right) = {{\text{e}}^{{\text{sin}}\,x}}\left( {{\text{co}}{{\text{s}}^2}\,x - {\text{sin}}\,x} \right)$ (or equivalent) A1

attempting to find $f\left( 0 \right)$ , $f'\left( 0 \right)$ and $f''\left( 0 \right)$ M1

$f\left( 0 \right) = 1$ ; $f'\left( 0 \right) = \left( {{\text{cos}}\,0} \right){{\text{e}}^{{\text{sin}}\,0}} = 1$ ; $f''\left( 0 \right) = {{\text{e}}^{{\text{sin}}\,0}}\left( {{\text{co}}{{\text{s}}^2}\,0 - {\text{sin}}\,0} \right) = 1$ A1

substitution into the Maclaurin formula $f\left( x \right) = f\left( 0 \right) + xf'\left( 0 \right) + \frac{{{x^2}}}{{2{\text{!}}}}f''\left( 0 \right) + \ldots$ M1

so the Maclaurin series for $f\left( x \right)$ up to and including the ${x^2}$ term is $1 + x + \frac{{{x^2}}}{2}$ A1

[8 marks]

METHOD 1

attempting to differentiate $f''\left( x \right)$ M1

$f'''\left( x \right) = \left( {{\text{cos}}\,x} \right){{\text{e}}^{{\text{sin}}\,x}}\left( {{\text{co}}{{\text{s}}^2}\,x - {\text{sin}}\,x} \right) - \left( {{\text{cos}}\,x} \right){{\text{e}}^{{\text{sin}}\,x}}\left( {2\,{\text{sin}}\,x + 1} \right)$ (or equivalent) A2

substituting $x = 0$ into their $f'''\left( x \right)$ M1

$f'''\left( 0 \right) = 1\left( {1 - 0} \right) - 1\left( {0 + 1} \right) = 0$

so the coefficient of ${x^3}$ in the Maclaurin series for $f\left( x \right)$ is zero AG

METHOD 2

substituting ${{\text{sin}}\,x}$ into the Maclaurin series for ${{\text{e}}^x}$ (M1)

${{\text{e}}^{{\text{sin}}\,x}} = 1 + {\text{sin}}\,x + \frac{{{\text{si}}{{\text{n}}^2}\,x}}{{2{\text{!}}}} + \frac{{{\text{si}}{{\text{n}}^3}\,x}}{{3{\text{!}}}} + \ldots$

substituting Maclaurin series for ${{\text{sin}}\,x}$ M1

${{\text{e}}^{{\text{sin}}\,x}} = 1 + \left( {x - \frac{{{x^3}}}{{3{\text{!}}}} + \ldots } \right) + \frac{{{{\left( {x - \frac{{{x^3}}}{{3{\text{!}}}} + \ldots } \right)}^2}}}{{2{\text{!}}}} + \frac{{{{\left( {x - \frac{{{x^3}}}{{3{\text{!}}}} + \ldots } \right)}^3}}}{{3{\text{!}}}} + \ldots$ A1

coefficient of ${x^3}$ is $- \frac{1}{{3{\text{!}}}} + \frac{1}{{3{\text{!}}}} = 0$ A1

so the coefficient of ${x^3}$ in the Maclaurin series for $f\left( x \right)$ is zero AG

[4 marks]

substituting $3x$ into the Maclaurin series for ${{\text{e}}^x}$ M1

${{\text{e}}^{3x}} = 1 + 3x + \frac{{{{\left( {3x} \right)}^2}}}{{2{\text{!}}}} + \frac{{{{\left( {3x} \right)}^3}}}{{3{\text{!}}}} + \ldots$ A1

substituting $\left( {{{\text{e}}^{3x}} - 1} \right)$ into the Maclaurin series for ${\text{arctan}}\,x$ M1

${\text{arctan}}\left( {{{\text{e}}^{3x}} - 1} \right) = \left( {{{\text{e}}^{3x}} - 1} \right) - \frac{{{{\left( {{{\text{e}}^{3x}} - 1} \right)}^3}}}{3} + \frac{{{{\left( {{{\text{e}}^{3x}} - 1} \right)}^5}}}{5} - \ldots$

$= \left( {3x + \frac{{{{\left( {3x} \right)}^2}}}{{2{\text{!}}}} + \frac{{{{\left( {3x} \right)}^3}}}{{3{\text{!}}}} + \ldots } \right) - \frac{{{{\left( {3x + \frac{{{{\left( {3x} \right)}^2}}}{{2{\text{!}}}} + \frac{{{{\left( {3x} \right)}^3}}}{{3{\text{!}}}} + \ldots } \right)}^3}}}{3} + \ldots$ A1

selecting correct terms from above M1

$= \left( {3x + \frac{{{{\left( {3x} \right)}^2}}}{{2{\text{!}}}} + \frac{{{{\left( {3x} \right)}^3}}}{{3{\text{!}}}}} \right) - \frac{{{{\left( {3x} \right)}^3}}}{3}$

$= 3x + \frac{{9{x^2}}}{2} - \frac{{9{x^3}}}{2}$ A1

[6 marks]

METHOD 1

substitution of their series M1

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{x + \frac{{{x^2}}}{2} + \ldots }}{{3x + \frac{{9{x^2}}}{2} + \ldots }}$ A1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 + \frac{x}{2} + \ldots }}{{3 + \frac{{9x}}{2} + \ldots }}$

$= \frac{1}{3}$ A1

METHOD 2

use of l’Hôpital’s rule M1

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\left( {{\text{cos}}\,x} \right){{\text{e}}^{{\text{sin}}\,x}}}}{{\frac{{3{{\text{e}}^{3x}}}}{{1 + {{\left( {{{\text{e}}^{3x}} - 1} \right)}^2}}}}}$ (or equivalent) A1

$= \frac{1}{3}$ A1

[3 marks]

Examiners report

[N/A]

Find the equation of the tangent to the curve $y = e^{2 x} - 3 x$ at the point where $x = 0$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(x = 0 \Rightarrow) y = 1$ (A1)

appreciate the need to find $\frac{d y}{d x}$ (M1)

$(\frac{d y}{d x} =) 2 e^{2 x} - 3$ A1

$(x = 0 \Rightarrow) \frac{d y}{d x} = - 1$ A1

$\frac{y - 1}{x - 0} = - 1 (y = 1 - x)$ A1

[5 marks]

Examiners report

[N/A]

Given that $\int_0^{{\text{ln}}\,k} {{{\text{e}}^{2x}}} {\text{d}}x = 12$ , find the value of $k$ .

Markscheme

$\frac{1}{2}{{\text{e}}^{2x}}$ seen (A1)

attempt at using limits in an integrated expression $\left( {\left[ {\frac{1}{2}{{\text{e}}^{2x}}} \right]_0^{{\text{ln}}\,k} = \frac{1}{2}{{\text{e}}^{2\,{\text{ln}}\,k}} - \frac{1}{2}{{\text{e}}^0}} \right)$ (M1)

$= \frac{1}{2}{{\text{e}}^{{\text{ln}}\,{k^2}}} - \frac{1}{2}{{\text{e}}^0}$ (A1)

Setting their equation $= 12$ M1

Note: their equation must be an integrated expression with limits substituted.

$\frac{1}{2}{k^2} - \frac{1}{2} = 12$ A1

$\left( {{k^2} = 25 \Rightarrow } \right)k = 5$ A1

Note: Do not award final A1 for $k = \pm 5$ .

[6 marks]

Examiners report

[N/A]

The lines $l_{1}$ and $l_{2}$ have the following vector equations where $λ, μ \in ℝ$ .

$l_{1} : r_{1} = (\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) + λ (\begin{matrix} 2 \\ - 2 \\ 2 \end{matrix})$

$l_{2} : r_{2} = (\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$

By using the substitution $u = \sin x$ , find $\int \frac{\sin x \cos x}{\sin^{2} x - \sin x - 2} d x$ .

Markscheme

$u = \sin x \Rightarrow d u = \cos x d x$ (or equivalent) A1

$= \int \frac{u}{u^{2} - u - 2} d u$ A1

attempt to use partial fractions M1

$(\frac{u}{(u + 1) (u - 2)} \equiv \frac{A}{u + 1} + \frac{B}{u - 2} \Rightarrow u \equiv A (u - 2) + B (u + 1))$

Valid attempt to solve for $A$ and $B$ (M1)

$A = \frac{1}{3}$ and $B = \frac{2}{3}$ A1

$\frac{u}{(u + 1) (u - 2)} \equiv \frac{1}{3 (u + 1)} + \frac{2}{3 (u - 2)}$

$\int (\frac{1}{3 (u + 1)} + \frac{2}{3 (u - 2)}) d u = \frac{1}{3} \ln |u + 1| + \frac{2}{3} \ln |u - 2| (+ C)$ (or equivalent) A1

Note: Condone the absence of $+ C$ or lack of moduli here but not in the final answer.

$= \frac{1}{3} \ln |\sin x + 1| + \frac{2}{3} \ln |\sin x - 2| + C$ A1

Note: Condone further simplification of the correct answer.

[7 marks]

Examiners report

[N/A]

A right circular cone of radius $r$ is inscribed in a sphere with centre O and radius $R$ as shown in the following diagram. The perpendicular height of the cone is $h$ , X denotes the centre of its base and B a point where the cone touches the sphere.

Show that the volume of the cone may be expressed by $V = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ .

[4]

Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is $\frac{{32\pi {R^3}}}{{81}}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to use Pythagoras in triangle OXB M1

$\Rightarrow {r^2} = {R^2} - {\left( {h - R} \right)^2}$ A1

substitution of their ${r^2}$ into formula for volume of cone $V = \frac{{\pi {r^2}h}}{3}$ M1

$= \frac{{\pi h}}{3}\left( {{R^2} - {{\left( {h - R} \right)}^2}} \right)$

$= \frac{{\pi h}}{3}\left( {{R^2} - \left( {{h^2} + {R^2} - 2hR} \right)} \right)$ A1

Note: This A mark is independent and may be seen anywhere for the correct expansion of ${{{\left( {h - R} \right)}^2}}$ .

$= \frac{{\pi h}}{3}\left( {2hR - {h^2}} \right)$

$= \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ AG

[4 marks]

at max, $\frac{{{\text{d}}V}}{{{\text{d}}h}} = 0$ R1

$\frac{{{\text{d}}V}}{{{\text{d}}h}} = \frac{\pi }{3}\left( {4Rh - 3{h^2}} \right)$

$\Rightarrow 4Rh = 3{h^2}$

$\Rightarrow h = \frac{{4R}}{3}$ (since $h \ne 0$ ) A1

EITHER

${V_{{\text{max}}}} = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ from part (a)

$= \frac{\pi }{3}\left( {2R{{\left( {\frac{{4R}}{3}} \right)}^2} - {{\left( {\frac{{4R}}{3}} \right)}^3}} \right)$ A1

$= \frac{\pi }{3}\left( {2R\frac{{16{R^2}}}{9} - \left( {\frac{{64{R^3}}}{{27}}} \right)} \right)$ A1

${r^2} = {R^2} - {\left( {\frac{{4R}}{3} - R} \right)^2}$

${r^2} = {R^2} - \frac{{{R^2}}}{9} = \frac{{8{R^2}}}{9}$ A1

$\Rightarrow {V_{{\text{max}}}} = \frac{{\pi {r^2}}}{3}\left( {\frac{{4R}}{3}} \right)$

$= \frac{{4\pi R}}{9}\left( {\frac{{8{R^2}}}{9}} \right)$ A1

THEN

$= \frac{{32\pi {R^3}}}{{81}}$ AG

[4 marks]

Examiners report

[N/A]

Consider the functions $f,\,\,g,$ defined for $x \in \mathbb{R}$ , given by $f\left( x \right) = {{\text{e}}^{ - x}}\,{\text{sin}}\,x$ and $g\left( x \right) = {{\text{e}}^{ - x}}\,{\text{cos}}\,x$ .

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

Markscheme

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]

Examiners report

[N/A]