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HL Paper 1

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]

Consider the equation ${z^4} + a{z^3} + b{z^2} + cz + d = 0$ , where $a$ , $b$ , $c$ , $d \in \mathbb{R}$ and $z \in \mathbb{C}$ .

Two of the roots of the equation are log₂6 and $i\sqrt 3$ and the sum of all the roots is 3 + log₂3.

Show that 6 $a$ + $d$ + 12 = 0.

Markscheme

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

$- i\sqrt 3$ is a root (A1)

$3 + {\text{lo}}{{\text{g}}_2}3 - {\text{lo}}{{\text{g}}_2}6\left( { = 3 + {\text{lo}}{{\text{g}}_2}\frac{1}{2} = 3 - 1 = 2} \right)$ is a root (A1)

sum of roots: $- a = 3 + {\text{lo}}{{\text{g}}_2}3 \Rightarrow a = - 3 - {\text{lo}}{{\text{g}}_2}3$ M1

Note: Award M1 for use of $- a$ is equal to the sum of the roots, do not award if minus is missing.

Note: If expanding the factored form of the equation, award M1 for equating $a$ to the coefficient of ${z^3}$ .

product of roots: ${\left( { - 1} \right)^4}d$ $= 2\left( {{\text{lo}}{{\text{g}}_2}6} \right)\left( {i\sqrt 3 } \right)\left( { - i\sqrt 3 } \right)$ M1

$= 6\,{\text{lo}}{{\text{g}}_2}6$ A1

Note: Award M1A0 for $d = - 6\,{\text{lo}}{{\text{g}}_2}6$

$6a + d + 12 = - 18 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}6 + 12$

EITHER

$= - 6 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

$= - 6 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

[7 marks]

Examiners report

[N/A]

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

The graph of $y = f (x)$ has a vertical asymptote and a horizontal asymptote.

Write down the equation of the vertical asymptote.

[1]

a.i.

Write down the equation of the horizontal asymptote.

[1]

a.ii.

On the set of axes below, sketch the graph of $y = f (x)$ .

On your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.

[3]

Hence, solve the inequality $0 < \frac{2 x - 1}{x + 1} < 2$ .

[1]

Solve the inequality $0 < \frac{2 |x| - 1}{|x| + 1} < 2$ .

[2]

Markscheme

$x = - 1$ A1

[1 mark]

a.i.

$y = 2$ A1

[1 mark]

a.ii.

rational function shape with two branches in opposite quadrants, with two correctly positioned asymptotes and asymptotic behaviour shown A1

axes intercepts clearly shown at $x = \frac{1}{2}$ and $y = - 1$ A1A1

[3 marks]

$x > \frac{1}{2}$ A1

Note: Accept correct alternative correct notation, such as $(\frac{1}{2}, \infty)$ and $] \frac{1}{2}, \infty [$ .

[1 mark]

EITHER

attempts to sketch $y = \frac{2 |x| - 1}{|x| + 1}$ (M1)

attempts to solve $2 |x| - 1 = 0$ (M1)

Note: Award the (M1) if $x = \frac{1}{2}$ and $x = - \frac{1}{2}$ are identified.

THEN

$x < - \frac{1}{2}$ or $x > \frac{1}{2}$ A1

Note: Accept the use of a comma. Condone the use of ‘and’. Accept correct alternative notation.

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

In the following Argand diagram, the points $Z_{1}$ , $O$ and $Z_{2}$ are the vertices of triangle $Z_{1} {OZ}_{2}$ described anticlockwise.

The point $Z_{1}$ represents the complex number $z_{1} = r_{1} e^{i α}$ , where $r_{1} > 0$ . The point $Z_{2}$ represents the complex number $z_{2} = r_{2} e^{i θ}$ , where $r_{2} > 0$ .

Angles $α, θ$ are measured anticlockwise from the positive direction of the real axis such that $0 \leq α, θ < 2 π$ and $0 < α - θ < π$ .

In parts (c), (d) and (e), consider the case where $Z_{1} {OZ}_{2}$ is an equilateral triangle.

Let $z_{1}$ and $z_{2}$ be the distinct roots of the equation $z^{2} + a z + b = 0$ where $z \in ℂ$ and $a, b \in ℝ$ .

Show that $z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ where ${z_{2}}^{*}$ is the complex conjugate of $z_{2}$ .

[2]

Given that $Re (z_{1} {z_{2}}^{*}) = 0$ , show that $Z_{1} {OZ}_{2}$ is a right-angled triangle.

[2]

Express $z_{1}$ in terms of $z_{2}$ .

[2]

c.i.

Hence show that ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ .

[4]

c.ii.

Use the result from part (c)(ii) to show that $a^{2} - 3 b = 0$ .

[5]

Consider the equation $z^{2} + a z + 12 = 0$ , where $z \in ℂ$ and $a \in ℝ$ .

Given that $0 < α - θ < π$ , deduce that only one equilateral triangle $Z_{1} {OZ}_{2}$ can be formed from the point $O$ and the roots of this equation.

[3]

Markscheme

${z_{2}}^{*} = r_{2} e^{-i θ}$ (A1)

$z_{1} {z_{2}}^{*} = r_{1} e^{i α} r_{2} e^{-i θ}$ A1

$z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ AG

Note: Accept working in modulus-argument form

[2 marks]

$Re (z_{1} {z_{2}}^{*}) = r_{1} r_{2} \cos (α - θ) (= 0)$ A1

$α - θ = arcos 0 (r_{1}, r_{2} > 0)$

$α - θ = \frac{π}{2}$ (as $0 < α - θ < π$ ) A1

so $Z_{1} {OZ}_{2}$ is a right-angled triangle AG

[2 marks]

EITHER

$\frac{z_{1}}{z_{2}} (= \frac{r_{1}}{r_{2}} e^{i (α - θ)}) = e^{i \frac{π}{3}}$ (since $r_{1} = r_{2}$ ) (M1)

$z_{1} = r_{2} e^{i (θ + \frac{π}{3})} (= r_{2} e^{i θ} e^{i \frac{π}{3}})$ (M1)

THEN

$z_{1} = z_{2} e^{i \frac{π}{3}}$ A1

Note: Accept working in either modulus-argument form to obtain $z_{1} = z_{2} (\cos \frac{π}{3} + i \sin \frac{π}{3})$ or in Cartesian form to obtain $z_{1} = z_{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ .

[2 marks]

c.i.

substitutes $z_{1} = z_{2} e^{i \frac{π}{3}}$ into ${z_{1}}^{2} + {z_{2}}^{2}$ M1

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ A1

EITHER

$e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ A1

${z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1) = {z_{2}}^{2} (- \frac{1}{2} + \frac{\sqrt{3}}{2} i + 1)$

$= {z_{2}}^{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ A1

THEN

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$

$= z_{2} (z_{2} e^{i \frac{π}{3}})$ and $z_{2} e^{i \frac{π}{3}} = z_{1}$ A1

so ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ AG

Note: For candidates who work on the LHS and RHS separately to show equality, award M1A1 for ${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ , A1 for $z_{1} z_{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$ and A1 for $e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ . Accept working in either modulus-argument form or in Cartesian form.

[4 marks]

c.ii.

METHOD 1

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

$a^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

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

substitutes $b = z_{1} z_{2}$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

METHOD 2

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

${(z_{1} + z_{2})}^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

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

substitutes $b = z_{1} z_{2}$ and $z_{1} + z_{2} = - a$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

[5 marks]

$a^{2} - 3 \times 12 = 0$

$a = \pm 6 (\Rightarrow z^{2} \pm 6 z + 12 = 0)$ A1

for $a = - 6 :$

$z_{1} = 3 + \sqrt{3} i, z_{2} = 3 - \sqrt{3} i$ and $α - θ = - \frac{5 π}{3}$ which does not satisfy $0 < α - θ < π$ R1

for $a = 6 :$

$z_{1} = - 3 - \sqrt{3} i, z_{2} = - 3 + \sqrt{3} i$ and $α - θ = \frac{π}{3}$ A1

so (for $0 < α - θ < π$ ), only one equilateral triangle can be formed from point $O$ and the two roots of this equation AG

[3 marks]

Examiners report

The vast majority of candidates scored full marks in parts (a) and (b). If they did not, it was normally due to the lack of rigour in setting out of the answer to a "show that" question. Part (c) was, though, more often than not poorly done. Many candidates could not use the given condition (equilateral triangle) to find $z_{1}$ in terms of $z_{2}$ . Part (d) was well answered by a rather high number of candidates.

Only a handful of students made good progress in (e), not even finding the possible values for $a$ .

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Let the roots of the equation ${z^3} = - 3 + \sqrt 3 {\text{i}}$ be $u$ , $v$ and $w$ .

On an Argand diagram, $u$ , $v$ and $w$ are represented by the points U, V and W respectively.

Express $- 3 + \sqrt 3 {\text{i}}$ in the form $r{{\text{e}}^{{\text{i}}\theta }}$ , where $r > 0$ and $- \pi < \theta \leqslant \pi$ .

[5]

Find $u$ , $v$ and $w$ expressing your answers in the form $r{{\text{e}}^{{\text{i}}\theta }}$ , where $r > 0$ and $- \pi < \theta \leqslant \pi$ .

[5]

Find the area of triangle UVW.

[4]

By considering the sum of the roots $u$ , $v$ and $w$ , show that

${\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{7\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} = 0$ .

[4]

Markscheme

attempt to find modulus (M1)

$r = 2\sqrt 3 \left( { = \sqrt {12} } \right)$ A1

attempt to find argument in the correct quadrant (M1)

$\theta = \pi + {\text{arctan}}\left( { - \frac{{\sqrt 3 }}{3}} \right)$ A1

$= \frac{{5\pi }}{6}$ A1

$- 3 + \sqrt 3 {\text{i}} = \sqrt {12} {{\text{e}}^{\frac{{5\pi {\text{i}}}}{6}}}\left( { = 2\sqrt 3 {{\text{e}}^{\frac{{5\pi {\text{i}}}}{6}}}} \right)$

[5 marks]

attempt to find a root using de Moivre’s theorem M1

${12^{\frac{1}{6}}}{{\text{e}}^{\frac{{5\pi {\text{i}}}}{{18}}}}$ A1

attempt to find further two roots by adding and subtracting $\frac{{2\pi }}{3}$ to the argument M1

${12^{\frac{1}{6}}}{{\text{e}}^{ - \frac{{7\pi {\text{i}}}}{{18}}}}$ A1

${12^{\frac{1}{6}}}{{\text{e}}^{\frac{{17\pi {\text{i}}}}{{18}}}}$ A1

Note: Ignore labels for $u$ , $v$ and $w$ at this stage.

[5 marks]

METHOD 1
attempting to find the total area of (congruent) triangles UOV, VOW and UOW M1

Area $= 3\left( {\frac{1}{2}} \right)\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right){\text{sin}}\frac{{2\pi }}{3}$ A1A1

Note: Award A1 for $\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right)$ and A1 for ${\text{sin}}\frac{{2\pi }}{3}$

= $\frac{{3\sqrt 3 }}{4}\left( {{{12}^{\frac{1}{3}}}} \right)$ (or equivalent) A1

METHOD 2

UV² $= {\left( {{{12}^{\frac{1}{6}}}} \right)^2} + {\left( {{{12}^{\frac{1}{6}}}} \right)^2} - 2\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right){\text{cos}}\frac{{2\pi }}{3}$ (or equivalent) A1

UV $= \sqrt 3 \left( {{{12}^{\frac{1}{6}}}} \right)$ (or equivalent) A1

attempting to find the area of UVW using Area = ${\frac{1}{2}}$ × UV × VW × sin $\alpha$ for example M1

Area $= \frac{1}{2}\left( {\sqrt 3 \times {{12}^{\frac{1}{6}}}} \right)\left( {\sqrt 3 \times {{12}^{\frac{1}{6}}}} \right){\text{sin}}\frac{\pi }{3}$

= $\frac{{3\sqrt 3 }}{4}\left( {{{12}^{\frac{1}{3}}}} \right)$ (or equivalent) A1

[4 marks]

$u$ + $v$ + $w$ = 0 R1

${12^{\frac{1}{6}}}\left( {{\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{i}}\,{\text{sin}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{cos}}\frac{{5\pi }}{{18}} + {\text{i}}\,{\text{sin}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} + {\text{i}}\,{\text{sin}}\frac{{17\pi }}{{18}}} \right) = 0$ A1

consideration of real parts M1

${12^{\frac{1}{6}}}\left( {{\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}}} \right) = 0$

${\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) = {\text{cos}}\frac{{17\pi }}{{18}}$ explicitly stated A1

${\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{7\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} = 0$ AG

[4 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]

Consider the function $f (x) = a x^{3} + b x^{2} + c x + d$ , where $x \in ℝ$ and $a, b, c, d \in ℝ$ .

Consider the function $g (x) = \frac{1}{2} x^{3} - 3 x^{2} + 6 x - 8$ , where $x \in ℝ$ .

The graph of $y = g (x)$ may be obtained by transforming the graph of $y = x^{3}$ using a sequence of three transformations.

Write down an expression for $f' (x)$ .

[1]

a.i.

Hence, given that $f^{- 1}$ does not exist, show that $b^{2} - 3 a c > 0$ .

[3]

a.ii.

Show that $g^{- 1}$ exists.

[2]

b.i.

$g (x)$ can be written in the form $p (x - 2)^{3} + q$ , where $p, q \in ℝ$ .

Find the value of $p$ and the value of $q$ .

[3]

b.ii.

Hence find $g^{- 1} (x)$ .

[3]

b.iii.

State each of the transformations in the order in which they are applied.

[3]

Sketch the graphs of $y = g (x)$ and $y = g^{- 1} (x)$ on the same set of axes, indicating the points where each graph crosses the coordinate axes.

[5]

Markscheme

$f' (x) = 3 a x^{2} + 2 b x + c$ A1

[1 mark]

a.i.

since $f^{- 1}$ does not exist, there must be two turning points R1

( $\Rightarrow f' (x) = 0$ has more than one solution)

using the discriminant $Δ > 0$ M1

$4 b^{2} - 12 a c > 0$ A1

$b^{2} - 3 a c > 0$ AG

[4 marks]

a.ii.

METHOD 1

$b^{2} - 3 a c = {(- 3)}^{2} - 3 \times \frac{1}{2} \times 6$ M1

$= 9 - 9$

$= 0$ A1

hence $g^{- 1}$ exists AG

METHOD 2

$g' (x) = \frac{3}{2} x^{2} - 6 x + 6$ M1

$Δ = {(- 6)}^{2} - 4 \times \frac{3}{2} \times 6$

$Δ = 36 - 36 = 0 \Rightarrow$ there is (only) one point with gradient of $0$ and this must be a point of inflexion (since $g (x)$ is a cubic.) R1

hence $g^{- 1}$ exists AG

[2 marks]

b.i.

$p = \frac{1}{2}$ A1

${(x - 2)}^{3} = x^{3} - 6 x^{2} + 12 x - 8$ (M1)

$\frac{1}{2} (x^{3} - 6 x^{2} + 12 x - 8) = \frac{1}{2} x^{3} - 3 x^{2} + 6 x - 4$

$g (x) = \frac{1}{2} {(x - 2)}^{3} - 4 \Rightarrow q = - 4$ A1

[3 marks]

b.ii.

$x = \frac{1}{2} {(y - 2)}^{3} - 4$ (M1)

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

$2 (x + 4) = {(y - 2)}^{3}$ (M1)

$\sqrt[3]{2 (x + 4)} = y - 2$

$y = \sqrt[3]{2 (x + 4)} + 2$

$g^{- 1} (x) = \sqrt[3]{2 (x + 4)} + 2$ A1

Note: $g^{- 1} (x) = \dots$ must be seen for the final A mark.

[3 marks]

b.iii.

translation through $(\begin{matrix} 2 \\ 0 \end{matrix})$ , A1

Note: This can be seen anywhere.

EITHER
a stretch scale factor $\frac{1}{2}$ parallel to the $y$ -axis then a translation through $(\begin{matrix} 0 \\ - 4 \end{matrix})$ A2
OR
a translation through $(\begin{matrix} 0 \\ - 8 \end{matrix})$ then a stretch scale factor $\frac{1}{2}$ parallel to the $y$ -axis A2

Note: Accept ‘shift’ for translation, but do not accept ‘move’. Accept ‘scaling’ for ‘stretch’.

[3 marks]

A1A1A1M1A1

Note: Award A1 for correct ‘shape’ of $g$ (allow non-stationary point of inflexion)
Award A1 for each correct intercept of $g$
Award M1 for attempt to reflect their graph in $y = x$ , A1 for completely correct $g^{- 1}$ including intercepts

[5 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]

Consider the series $\ln x + p \ln x + \frac{1}{3} \ln x + \dots$ , where $x \in ℝ, x > 1$ and $p \in ℝ, p \neq 0$ .

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference $d$ .

Show that $p = \pm \frac{1}{\sqrt{3}}$ .

[2]

a.i.

Hence or otherwise, show that the series is convergent.

[1]

a.ii.

Given that $p > 0$ and $S_{\infty} = 3 + \sqrt{3}$ , find the value of $x$ .

[3]

a.iii.

Show that $p = \frac{2}{3}$ .

[3]

b.i.

Write down $d$ in the form $k \ln x$ , where $k \in ℚ$ .

[1]

b.ii.

The sum of the first $n$ terms of the series is $\ln (\frac{1}{x^{3}})$ .

Find the value of $n$ .

[8]

b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms M1

$\frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x}$ OR $\frac{1}{3} \ln x = (\ln x) r^{2}$ OR $p \ln x = \ln x (\frac{1}{3 p})$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in geometric sequence

Award M1 for $\frac{p}{1} = \frac{\frac{1}{3}}{p}$ .

$r = p$ and $r^{2} = \frac{1}{3}$ M1

THEN

$p^{2} = \frac{1}{3}$ OR $r = \pm \frac{1}{\sqrt{3}}$ A1

$p = \pm \frac{1}{\sqrt{3}}$ AG

Note: Award M0A0 for $r^{2} = \frac{1}{3}$ or $p^{2} = \frac{1}{3}$ with no other working seen.

[2 marks]

a.i.

EITHER

since, $|p| = \frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}} < 1$ R1

since, $|p| = \frac{1}{\sqrt{3}}$ and $- 1 < p < 1$ R1

THEN

$\Rightarrow$ the geometric series converges. AG

Note: Accept $r$ instead of $p$ .
Award R0 if both values of $p$ not considered.

[1 mark]

a.ii.

$\frac{\ln x}{1 - \frac{1}{\sqrt{3}}} (= 3 + \sqrt{3})$ (A1)

$\ln x = 3 - \frac{3}{\sqrt{3}} + \sqrt{3} - \frac{\sqrt{3}}{\sqrt{3}}$ OR $\ln x = 3 - \sqrt{3} + \sqrt{3} - 1 (\Rightarrow \ln x = 2)$ A1

$x = e^{2}$ A1

[3 marks]

a.iii.

METHOD 1

attempt to find a difference from consecutive terms or from $u_{2}$ M1

correct equation A1

$p \ln x - \ln x = \frac{1}{3} \ln x - p \ln x$ OR $\frac{1}{3} \ln x = \ln x + 2 (p \ln x - \ln x)$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in arithmetic sequence.

Award M1A1 for $p - 1 = \frac{1}{3} - p$

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 2

attempt to use arithmetic mean $u_{2} = \frac{u_{1} + u_{3}}{2}$ M1

$p \ln x = \frac{\ln x + \frac{1}{3} \ln x}{2}$ A1

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 3

attempt to find difference using $u_{3}$ M1

$\frac{1}{3} \ln x = \ln x + 2 d (\Rightarrow d = - \frac{1}{3} \ln x)$

$u_{2} = \ln x + \frac{1}{2} (\frac{1}{3} \ln x - \ln x)$ OR $p \ln x - \ln x = - \frac{1}{3} \ln x$ A1

$p \ln x = \frac{2}{3} \ln x$ A1

$p = \frac{2}{3}$ AG

[3 marks]

b.i.

$d = - \frac{1}{3} \ln x$ A1

[1 mark]

b.ii.

METHOD 1

$S_{n} = \frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋$

attempt to substitute into $S_{n}$ and equate to $\ln (\frac{1}{x^{3}})$ (M1)

$\frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋ = \ln (\frac{1}{x^{3}})$

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

correct working with $S_{n}$ (seen anywhere) (A1)

$\frac{n}{2} ⌊2 \ln x - \frac{n}{3} \ln x + \frac{1}{3} \ln x⌋$ OR $n \ln x - \frac{n (n - 1)}{6} \ln x$ OR $\frac{n}{2} (\ln x + (\frac{4 - n}{3}) \ln x)$

correct equation without $\ln x$ A1

$\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ OR $n - \frac{n (n - 1)}{6} = - 3$ or equivalent

Note: Award as above if the series $1 + p + \frac{1}{3} + \dots$ is considered leading to $\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ .

attempt to form a quadratic $= 0$ (M1)

$n^{2} - 7 n - 18 = 0$

attempt to solve their quadratic (M1)

$(n - 9) (n + 2) = 0$

$n = 9$ A1

METHOD 2

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

listing the first $7$ terms of the sequence (A1)

$\ln x + \frac{2}{3} \ln x + \frac{1}{3} \ln x + 0 - \frac{1}{3} \ln x - \frac{2}{3} \ln x - \ln x + \dots$

recognizing first $7$ terms sum to $0$ M1

$8$ ^th term is $- \frac{4}{3} \ln x$ (A1)

$9$ ^th term is $- \frac{5}{3} \ln x$ (A1)

sum of $8$ ^th and $9$ ^th term $= - 3 \ln x$ (A1)

$n = 9$ A1

[8 marks]

b.iii.

Examiners report

Part (a)(i) was well done with few candidates incorrectly using the value of p to verify rather than to 'show' the given result. In part (a)(ii) most did not consider both values of r and some did know the condition for convergence of a geometric series. Part (a)(iii) was generally well done but some had difficulty in simplifying the surd. Part (b) (i) and (ii) was generally well done. Although many completely correct answers to part b (iii) were noted, weaker candidates often made errors in properties of logarithms or algebraic manipulation leading to an incorrect quadratic equation.

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

The cubic equation $x^{3} - k x^{2} + 3 k = 0$ where $k > 0$ has roots $α, β$ and $α + β$ .

Given that $α β = - \frac{k^{2}}{4}$ , find the value of $k$ .

Markscheme

$α + β + α + β = k$ (A1)

$α + β = \frac{k}{2}$

$α β (α + β) = - 3 k$ (A1)

$(- \frac{k^{2}}{4}) (\frac{k}{2}) = - 3 k (- \frac{k^{3}}{8} = - 3 k)$ M1

attempting to solve $- \frac{k^{3}}{8} + 3 k = 0$ (or equivalent) for $k$ (M1)

$k = 2 \sqrt{6} (= \sqrt{24}) (k > 0)$ A1

Note: Award A0 for $k = \pm 2 \sqrt{6} (\pm \sqrt{24})$ .

[5 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]

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]

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

N17/5/MATHL/HP1/ENG/TZ0/06.a

Markscheme

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

N17/5/MATHL/HP1/ENG/TZ0/06.a/M

correct vertical asymptote A1

shape including correct horizontal asymptote A1

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

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

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

[4 marks]

Examiners report

[N/A]

The function $f$ is defined by $f (x) = \frac{2 x + 4}{3 - x}$ , where $x \in ℝ, x \neq 3$ .

Write down the equation of

Find the coordinates where the graph of $f$ crosses

the vertical asymptote of the graph of $f$ .

[1]

a.i.

the horizontal asymptote of the graph of $f$ .

[1]

a.ii.

the $x$ -axis.

[1]

b.i.

the $y$ -axis.

[1]

b.ii.

Sketch the graph of $f$ on the axes below.

[1]

The function $g$ is defined by $g (x) = \frac{a x + 4}{3 - x}$ , where $x \in ℝ, x \neq 3$ and $a \in ℝ$ .

Given that $g (x) = g^{- 1} (x)$ , determine the value of $a$ .

[4]

Markscheme

$x = 3$ A1

[1 mark]

a.i.

$y = - 2$ A1

[1 mark]

a.ii.

$(- 2, 0)$ (accept $x = - 2$ ) A1

[1 mark]

b.i.

$(0, \frac{4}{3})$ (accept $y = \frac{4}{3}$ and $f (0) = \frac{4}{3}$ ) A1

[1 mark]

b.ii.

Note: Award A1 for completely correct shape: two branches in correct quadrants with asymptotic behaviour.

[1 mark]

METHOD 1

$(g (x) =) y = \frac{a x + 4}{3 - x}$

attempt to find $x$ in terms of $y$ (M1)

OR exchange $x$ and $y$ and attempt to find $y$ in terms of $x$

$3 y - x y = a x + 4$ A1

$a x + x y = 3 y - 4$

$x (a + y) = 3 y - 4$

$x = \frac{3 y - 4}{y + a}$

$g^{- 1} (x) = \frac{3 x - 4}{x + a}$ A1

Note: Condone use of $y =$

$g (x) \equiv g^{- 1} (x)$

$\frac{a x + 4}{3 - x} \equiv \frac{3 x - 4}{x + a}$

$\Rightarrow a = - 3$ A1

METHOD 2

$g (x) = \frac{a x + 4}{3 - x}$

attempt to find an expression for $g (g (x))$ and equate to $x$ (M1)

$g g (x) = \frac{a (\frac{a x + 4}{3 - x}) + 4}{3 - (\frac{a x + 4}{3 - x})} = x$ A1

$\frac{a (a x + 4) + 4 (3 - x)}{(9 - 3 x) - (a x + 4)} = x$

$\frac{a (a x + 4) + 4 (3 - x)}{5 - (3 + a) x} = x$

$a (a x + 4) + 4 (3 - x) = x (5 - (3 + a) x)$ A1

equating coefficients of $x^{2}$ (or similar)

$a = - 3$ A1

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Let $f\left( x \right) = \frac{{{x^2} - 10x + 5}}{{x + 1}}{\text{,}}\,\,x \in \mathbb{R}{\text{,}}\,\,x \ne - 1$ .

Find the co-ordinates of all stationary points.

[4]

Write down the equation of the vertical asymptote.

[1]

With justification, state if each stationary point is a minimum, maximum or horizontal point of inflection.

[4]

Markscheme

$f'\left( x \right) = \frac{{\left( {2x - 10} \right)\left( {x + 1} \right) - \left( {{x^2} - 10x + 5} \right)1}}{{{{\left( {x + 1} \right)}^2}}}$ M1

$f'\left( x \right) = 0 \Rightarrow {x^2} + 2x - 15 = 0 \Rightarrow \left( {x + 5} \right)\left( {x - 3} \right) = 0$ M1

Stationary points are $\left( { - 5{\text{,}}\,\, - 20} \right)\,\,{\text{and}}\,\,\left( {3{\text{,}}\,\, - 4} \right)$ A1A1

[4 marks]

$x = - 1$ A1

[1 mark]

Looking at the nature table

M1A1

$\left( { - 5{\text{,}}\, - 20} \right)$ is a max and $\left( {3{\text{,}}\, - 4} \right)$ is a min A1A1

[4 marks]

Examiners report

[N/A]

Consider the functions $f$ and $g$ defined by $f\left( x \right) = {\text{ln}}\left| x \right|$ , $x \in \mathbb{R}$ \ $\left\{ 0 \right\}$ , and $g\left( x \right) = {\text{ln}}\left| {x + k} \right|$ , $x \in \mathbb{R}$ \ $\left\{ { - k} \right\}$ , where $k \in \mathbb{R}$ , $k > 2$ .

The graphs of $f$ and $g$ intersect at the point P .

Describe the transformation by which $f\left( x \right)$ is transformed to $g\left( x \right)$ .

[1]

State the range of $g$ .

[1]

Sketch the graphs of $y = f\left( x \right)$ and $y = g\left( x \right)$ on the same axes, clearly stating the points of intersection with any axes.

[6]

Find the coordinates of P.

[2]

Markscheme

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

translation $k$ units to the left (or equivalent) A1

[1 mark]

range is $\left( {g\left( x \right) \in } \right)\mathbb{R}$ A1

[1 mark]

correct shape of $y = f\left( x \right)$ A1

their $f\left( x \right)$ translated $k$ units to left (possibly shown by $x = - k$ marked on $x$ -axis) A1

asymptote included and marked as $x = - k$ A1

$f\left( x \right)$ intersects $x$ -axis at $x = - 1$ , $x = 1$ A1

$g\left( x \right)$ intersects $x$ -axis at $x = - k - 1$ , $x = - k + 1$ A1

$g\left( x \right)$ intersects $y$ -axis at $y = {\text{ln}}\,k$ A1

Note: Do not penalise candidates if their graphs “cross” as $x \to \pm \infty$ .

Note: Do not award FT marks from the candidate’s part (a) to part (c).

[6 marks]

at P ${\text{ln}}\left( {x + k} \right) = {\text{ln}}\left( { - x} \right)$

attempt to solve $x + k = - x$ (or equivalent) (M1)

$x = - \frac{k}{2} \Rightarrow y = {\text{ln}}\left( {\frac{k}{2}} \right)\,\,$ (or $y = {\text{ln}}\left| {\frac{k}{2}} \right|$ ) A1

P $\left( { - \frac{k}{2},\,\,{\text{ln}}\frac{k}{2}} \right)$ (or P $\left( { - \frac{k}{2},\,\,{\text{ln}}\left| {\frac{k}{2}} \right|} \right)$ )

[2 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]

Sketch the graphs of $y = \frac{x}{2} + 1$ and $y = \left| {x - 2} \right|$ on the following axes.

Markscheme

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

straight line graph with correct axis intercepts A1

modulus graph: V shape in upper half plane A1

modulus graph having correct vertex and y-intercept A1

[3 marks]

Examiners report

[N/A]

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]

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^{2x}} - 6{{\text{e}}^x} + 5{\text{,}}\,\,x \in \mathbb{R}{\text{,}}\,\,x \leqslant a$ . The graph of $y = f\left( x \right)$ is shown in the following diagram.

Find the largest value of $a$ such that $f$ has an inverse function.

[3]

For this value of $a$ , find an expression for ${f^{ - 1}}\left( x \right)$ , stating its domain.

[5]

Markscheme

attempt to differentiate and set equal to zero M1

$f'\left( x \right) = 2{{\text{e}}^{2x}} - 6{{\text{e}}^x} = 2{{\text{e}}^x}\left( {{{\text{e}}^x} - 3} \right) = 0$ A1

minimum at $x = {\text{ln}}\,3$

$a = {\text{ln}}\,3$ A1

[3 marks]

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

$y = {\left( {{{\text{e}}^x} - 3} \right)^2} - 4$ (M1)

${{\text{e}}^x} - 3 = \pm \sqrt {y + 4}$ A1

as $x \leqslant {\text{ln}}\,3$ , $x = {\text{ln}}\left( {3 - \sqrt {y + 4} } \right)$ R1

so ${f^{ - 1}}\left( x \right) = {\text{ln}}\left( {3 - \sqrt {x + 4} } \right)$ A1

domain of ${f^{ - 1}}$ is $x \in \mathbb{R}$ , $- 4 \leqslant x < 5$ A1

[5 marks]

Examiners report

[N/A]

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

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

Markscheme

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

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

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

attempt to solve their two equations M1

p = −3, q = 2 A1

[5 marks]

Examiners report

[N/A]

Consider the polynomial $q(x) = 3{x^3} - 11{x^2} + kx + 8$ .

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

[3]

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

[3]

Markscheme

$q(4) = 0$ (M1)

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

$k = - 6$ A1

[3 marks]

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

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

$- 12 + p = - 11$

$p = 1$

$(x - 4)(3{x^2} + x - 2)$ (A1)

$(x - 4)(3x - 2)(x + 1)$ A1

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

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

[3 marks]

Examiners report

[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]

Let $f\left( x \right) = \frac{{2x + 6}}{{{x^2} + 6x + 10}}{\text{,}}\,\,x \in \mathbb{R}.$

Show that $f\left( x \right)$ has no vertical asymptotes.

[3]

Find the equation of the horizontal asymptote.

[2]

Find the exact value of $\int\limits_0^1 {f\left( x \right)} \,dx$ , giving the answer in the form ${\text{ln}}\,q{\text{,}}\,\,q \in \mathbb{Q}$ .

[3]

Markscheme

${x^2} + 6x + 10 = {x^2} + 6x + 9 + 1 = {\left( {x + 3} \right)^2} + 1$ M1A1

So the denominator is never zero and thus there are no vertical asymptotes. (or use of discriminant is negative) R1

[3 marks]

$x \to \pm \infty {\text{,}}\,\,f\left( x \right) \to 0$ so the equation of the horizontal asymptote is $y =0$ M1A1

[2 marks]

$\int\limits_0^1 {\frac{{2x + 6}}{{{x^2} + 6x + 10}}} \,dx = \left[ {{\text{ln}}\left( {{x^2} + 6x + 10} \right)} \right]_0^1 = {\text{ln}}\,17 - {\text{ln}}\,10 = {\text{ln}}\,\frac{{17}}{{10}}$ M1A1A1

[3 marks]

Examiners report

[N/A]

The equation $3 p x^{2} + 2 p x + 1 = p$ has two real, distinct roots.

Find the possible values for $p$ .

[5]

Consider the case when $p = 4$ . The roots of the equation can be expressed in the form $x = \frac{a \pm \sqrt{13}}{6}$ , where $a \in ℤ$ . Find the value of $a$ .

[2]

Markscheme

attempt to use discriminant $b^{2} - 4 a c (> 0)$ M1

${(2 p)}^{2} - 4 (3 p) (1 - p) (> 0)$

$16 p^{2} - 12 p (> 0)$ (A1)

$p (4 p - 3) (> 0)$

attempt to find critical values $(p = 0, p = \frac{3}{4})$ M1

recognition that discriminant $> 0$ (M1)

$p < 0$ or $p > \frac{3}{4}$ A1

Note: Condone ‘or’ replaced with ‘and’, a comma, or no separator

[5 marks]

$p = 4 \Rightarrow 12 x^{2} + 8 x - 3 = 0$

valid attempt to use $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ (or equivalent) M1

$x = \frac{- 8 \pm \sqrt{208}}{24}$

$x = \frac{- 2 \pm \sqrt{13}}{6}$

$a = - 2$ A1

[2 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]

The following diagram shows the graph of $y = f\left( x \right)$ . The graph has a horizontal asymptote at $y = - 1$ . The graph crosses the $x$ -axis at $x = - 1$ and $x = 1$ , and the $y$ -axis at $y = 2$ .

On the following set of axes, sketch the graph of $y = {\left[ {f\left( x \right)} \right]^2} + 1$ , clearly showing any asymptotes with their equations and the coordinates of any local maxima or minima.

Markscheme

no $y$ values below 1 A1

horizontal asymptote at $y = 2$ with curve approaching from below as $x \to \pm \infty$ A1

(±1,1) local minima A1

(0,5) local maximum A1

smooth curve and smooth stationary points A1

[5 marks]

Examiners report

[N/A]

Consider the function $g\left( x \right) = 4\,{\text{cos}}\,x + 1$ , $a \leqslant x \leqslant \frac{\pi }{2}$ where $a < \frac{\pi }{2}$ .

For $a = - \frac{\pi }{2}$ , sketch the graph of $y = g\left( x \right)$ . Indicate clearly the maximum and minimum values of the function.

[3]

Write down the least value of $a$ such that $g$ has an inverse.

[1]

For the value of $a$ found in part (b), write down the domain of ${g^{ - 1}}$ .

[1]

c.i.

For the value of $a$ found in part (b), find an expression for ${g^{ - 1}}\left( x \right)$ .

[2]

c.ii.

Markscheme

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

concave down and symmetrical over correct domain A1

indication of maximum and minimum values of the function (correct range) A1A1

[3 marks]

$a$ = 0 A1

Note: Award A1 for $a$ = 0 only if consistent with their graph.

[1 mark]

$1 \leqslant x \leqslant 5$ A1

Note: Allow FT from their graph.

[1 mark]

c.i.

$y = 4\,{\text{cos}}\,x + 1$

$x = 4\,{\text{cos}}\,y + 1$

$\frac{{x - 1}}{4} = {\text{cos}}\,y$ (M1)

$\Rightarrow y = {\text{arccos}}\left( {\frac{{x - 1}}{4}} \right)$

$\Rightarrow {g^{ - 1}}\left( x \right) = {\text{arccos}}\left( {\frac{{x - 1}}{4}} \right)$ A1

[2 marks]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Consider the function $f\left( x \right) = {x^4} - 6{x^2} - 2x + 4$ , $x \in \mathbb{R}$ .

The graph of $f$ is translated two units to the left to form the function $g\left( x \right)$ .

Express $g\left( x \right)$ in the form $a{x^4} + b{x^3} + c{x^2} + dx + e$ where $a$ , $b$ , $c$ , $d$ , $e \in \mathbb{Z}$ .

Markscheme

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

$g\left( x \right) = f\left( {x + 2} \right)\left( { = {{\left( {x + 2} \right)}^4} - 6{{\left( {x + 2} \right)}^2} - 2\left( {x + 2} \right) + 4} \right)$ M1

attempt to expand ${{{\left( {x + 2} \right)}^4}}$ M1

${\left( {x + 2} \right)^4} = {x^4} + 4\left( {2{x^3}} \right) + 6\left( {{2^2}{x^2}} \right) + 4\left( {{2^3}x} \right) + {2^4}$ (A1)

$= {x^4} + 8{x^3} + 24{x^2} + 32x + 16$ A1

$g\left( x \right) = {x^4} + 8{x^3} + 24{x^2} + 32x + 16 - 6\left( {{x^2} + 4x + 4} \right) - 2x - 4 + 4$

$= {x^4} + 8{x^3} + 18{x^2} + 6x - 8$ A1

Note: For correct expansion of $f\left( {x - 2} \right) = {x^4} - 8{x^3} + 18{x^2} - 10x$ award max M0M1(A1)A0A1.

[5 marks]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = \frac{{ax + b}}{{cx + d}}$ , for $x \in \mathbb{R},\,\,x \ne - \frac{d}{c}$ .

The function $g$ is defined by $g\left( x \right) = \frac{{2x - 3}}{{x - 2}},\,\,x \in \mathbb{R},\,\,x \ne 2$

Express $g\left( x \right)$ in the form $A + \frac{B}{{x - 2}}$ where A, B are constants.

Markscheme

$g\left( x \right) = 2 + \frac{1}{{x - 2}}$ A1A1

[2 marks]

Examiners report

[N/A]

Sketch the graph of $y = \frac{{x - 4}}{{2x - 5}}$ , stating the equations of any asymptotes and the coordinates of any points of intersection with the axes.

Markscheme

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

correct shape: two branches in correct quadrants with asymptotic behaviour A1

crosses at (4, 0) and $\left( {0{\text{,}}\,\,\frac{4}{5}} \right)$ A1A1

asymptotes at $x = \frac{5}{2}$ and $y = \frac{1}{2}$ A1A1

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

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 $f\left( x \right) = \frac{{2{x^2} - 5x - 12}}{{x + 2}}{\text{,}}\,\,x \in \mathbb{R}{\text{,}}\,\,x \ne - 2$ .

Find all the intercepts of the graph of $f\left( x \right)$ with both the $x$ and $y$ axes.

[4]

Write down the equation of the vertical asymptote.

[1]

As $x \to \pm \infty$ the graph of $f\left( x \right)$ approaches an oblique straight line asymptote.

Divide $2{x^2} - 5x - 12$ by $x + 2$ to find the equation of this asymptote.

[4]

Markscheme

$x = 0 \Rightarrow y = - 6$ intercept on the $y$ axes is (0, −6) A1

$2{x^2} - 5x - 12 = 0 \Rightarrow \left( {2x + 3} \right)\left( {x - 4} \right) = 0 \Rightarrow x = \frac{{ - 3}}{2}\,\,{\text{or}}\,\,{\text{4}}$ M1

intercepts on the $x$ axes are $\left( {\frac{{ - 3}}{2}{\text{,}}\,\,0} \right)\,\,{\text{and}}\,\,\left( {4{\text{,}}\,\,0} \right)$ A1A1

[4 marks]

$x = - 2$ A1

[1 mark]

$f(x) = 2x - 9 + \frac{6}{{x + 2}}$ M1A1

So equation of asymptote is $y = 2x - 9$ M1A1

[4 marks]

Examiners report

[N/A]

The function $f$ is defined by $f(x) = 2{x^3} + 5,{\text{ }} - 2 \leqslant x \leqslant 2$ .

Write down the range of $f$ .

[2]

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

[2]

Write down the domain and range of ${f^{ - 1}}$ .

[2]

Markscheme

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

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

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

[2 marks]

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

[2 marks]

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

[2 marks]

Examiners report

[N/A]

Solve the equation $\log_{3} \sqrt{x} = \frac{1}{2 \log_{2} 3} + \log_{3} (4 x^{3})$ , where $x > 0$ .

Markscheme

attempt to use change the base (M1)

$\log_{3} \sqrt{x} = \frac{\log_{3} 2}{2} + \log_{3} (4 x^{3})$

attempt to use the power rule (M1)

$\log_{3} \sqrt{x} = \log_{3} \sqrt{2} + \log_{3} (4 x^{3})$

attempt to use product or quotient rule for logs, $\ln a + \ln b = \ln a b$ (M1)

$\log_{3} \sqrt{x} = \log_{3} (4 \sqrt{2} x^{3})$

Note: The M marks are for attempting to use the relevant log rule and may be applied in any order and at any time during the attempt seen.

$\sqrt{x} = 4 \sqrt{2} x^{3}$

$x = 32 x^{6}$

$x^{5} = \frac{1}{32}$ (A1)

$x = \frac{1}{2}$ A1

[5 marks]

Examiners report

[N/A]

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

Markscheme

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

$\alpha + \beta = 2k$ A1

$\alpha \beta = k - 1$ A1

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

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

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

attempt to solve quadratic (M1)

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

[6 marks]

Examiners report

[N/A]

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

Markscheme

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

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

EITHER

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

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

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

THEN

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

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

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

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

[6 marks]

Examiners report

[N/A]

The following diagram shows the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ , with asymptotes at $y = - \frac{π}{4}$ and $y = \frac{3 π}{4}$ .

Describe a sequence of transformations that transforms the graph of $y = arctan x$ to the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ .

[3]

Show that $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ .

[4]

Verify that $arctan (2 x + 1) = arctan (\frac{x}{x + 1}) + \frac{π}{4}$ for $x \in ℝ, x > 0$ .

[3]

Using mathematical induction and the result from part (b), prove that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$ .

[9]

Markscheme

EITHER
horizontal stretch/scaling with scale factor $\frac{1}{2}$

Note: Do not allow ‘shrink’ or ‘compression’

followed by a horizontal translation/shift $\frac{1}{2}$ units to the left A2

Note: Do not allow ‘move’

horizontal translation/shift $1$ unit to the left

followed by horizontal stretch/scaling with scale factor $\frac{1}{2}$ A2

THEN

vertical translation/shift up by $\frac{π}{4}$ (or translation through $(\begin{matrix} 0 \\ \frac{π}{4} \end{matrix})$ A1
(may be seen anywhere)

[3 marks]

let $α = arctan p$ and $β = arctan q$ M1

$p = tan α$ and $q = tan β$ (A1)

$\tan (α + β) = \frac{p + q}{1 - p q}$ A1

$α + β = arctan (\frac{p + q}{1 - p q})$ A1

so $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ . AG

[4 marks]

METHOD 1

$\frac{π}{4} = arctan 1$ (or equivalent) A1

$arctan (\frac{x}{x + 1}) + arctan 1 = arctan (\frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1} (1)})$ A1

$= arctan (\frac{\frac{x + x + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}})$ A1

$= arctan (2 x + 1)$ AG

METHOD 2

$\tan \frac{π}{4} = 1$ (or equivalent) A1

Consider $arctan (2 x + 1) - arctan (\frac{x}{x + 1}) = \frac{π}{4}$

$\tan (arctan (2 x + 1) - arctan (\frac{x}{x + 1}))$

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

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

$= arctan 1$ AG

METHOD 3

$tan (arctan (2 x + 1)) = \tan (arctan (\frac{x}{x + 1}) + \frac{π}{4})$

$\tan \frac{π}{4} = 1$ (or equivalent) A1

$LHS = 2 x + 1$ A1

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

[3 marks]

let $P (n)$ be the proposition that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$

consider $P (1)$

when $n = 1, Σ_{r = 1}^{1} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{1}{2}) = RHS$ and so $P (1)$ is true R1

assume $P (k)$ is true, ie. $Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{k}{k + 1}) (k \in ℤ^{+})$ M1

Note: Award M0 for statements such as “let $n = k$ ”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

consider $P (k + 1)$ :

$Σ_{r = 1}^{k + 1} arctan (\frac{1}{2 r^{2}}) = Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ (M1)

$= arctan (\frac{k}{k + 1}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ A1

$= arctan (\frac{\frac{k}{k + 1} + \frac{1}{2 {(k + 1)}^{2}}}{1 - (\frac{k}{k + 1}) (\frac{1}{2 {(k + 1)}^{2}})})$ M1

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

Note: Award A1 for correct numerator, with $(k + 1)$ factored. Denominator does not need to be simplified

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 k^{3} + 6 k^{2} + 5 k + 2})$ A1

Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{(k + 2) (2 k^{2} + 2 k + 1)}) = arctan (\frac{k + 1}{k + 2})$ A1

Note: The word ‘arctan’ must be present to be able to award the last three A marks

$P (k + 1)$ is true whenever $P (k)$ is true and $P (1)$ is true, so

$P (n)$ is true for for $n \in ℤ^{+}$ R1

Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of $P (k)$ must be mentioned. ‘ $P (k)$ implies $P (k + 1)$ ’ is insufficient to award the mark.

[9 marks]

Examiners report

[N/A]

Use the binomial theorem to expand ${(\cos θ + i \sin θ)}^{4}$ . Give your answer in the form $a + b i$ where $a$ and $b$ are expressed in terms of $\sin θ$ and $\cos θ$ .

[3]

Use de Moivre’s theorem and the result from part (a) to show that $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ .

[5]

Use the identity from part (b) to show that the quadratic equation $x^{2} - 6 x + 1 = 0$ has roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ .

[5]

Hence find the exact value of ${cot}^{2} \frac{3 π}{8}$ .

[4]

Deduce a quadratic equation with integer coefficients, having roots ${cosec}^{2} \frac{π}{8}$ and ${cosec}^{2} \frac{3 π}{8}$ .

[3]

Markscheme

uses the binomial theorem on ${(\cos θ + i \sin θ)}^{4}$ M1

$= {}_{4}C_{0} \cos^{4} θ + {}_{4}C_{1} \cos^{3} θ (i \sin θ) + {}_{4}C_{2} \cos^{2} θ (i^{2} \sin^{2} θ) + {}_{4}C_{3} \cos θ (i^{3} \sin^{3} θ) + {}_{4}C_{4} (i^{4} \sin^{4} θ)$ A1

$= (\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ A1

[3 marks]

(using de Moivre’s theorem with $n = 4$ gives) $\cos 4 θ + i \sin 4 θ$ (A1)

equates both the real and imaginary parts of $\cos 4 θ + i \sin 4 θ$ and $(\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ M1

$\cos 4 θ = \cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ$ and $\sin 4 θ = 4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ$

recognizes that $\cot 4 θ = \frac{\cos 4 θ}{\sin 4 θ}$ (A1)

substitutes for $\sin 4 θ$ and $\cos 4 θ$ into $\frac{\cos 4 θ}{\sin 4 θ}$ M1

$\cot 4 θ = \frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}$

divides the numerator and denominator by $\sin^{4} θ$ to obtain

$\cot 4 θ = \frac{\frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{\sin^{4} θ}}{\frac{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}{\sin^{4} θ}}$ A1

$\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ AG

[5 marks]

setting $\cot 4 θ = 0$ and putting $x = \cot^{2} θ$ in the numerator of $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ gives $x^{2} - 6 x + 1 = 0$ M1

attempts to solve $\cot 4 θ = 0$ for $θ$ M1

$4 θ = \frac{π}{2}, \frac{3 π}{2}, \dots (4 θ = \frac{1}{2} (2 n + 1) π, n = 0, 1, \dots)$ (A1)

$θ = \frac{π}{8}, \frac{3 π}{8}$ A1

Note: Do not award the final A1 if solutions other than $θ = \frac{π}{8}, \frac{3 π}{8}$ are listed.

finding the roots of $\cot 4 θ = 0 (θ = \frac{π}{8}, \frac{3 π}{8})$ corresponds to finding the roots of $x^{2} - 6 x + 1 = 0$ where $x = \cot^{2} θ$ R1

so the equation $x^{2} - 6 x + 1 = 0$ as roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ AG

[5 marks]

attempts to solve $x^{2} - 6 x + 1 = 0$ for $x$ M1

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

since ${cot}^{2} \frac{π}{8} > {cot}^{2} \frac{3 π}{8}, {cot}^{2} \frac{3 π}{8}$ has the smaller value of the two roots R1

Note: Award R1 for an alternative convincing valid reason.

so ${cot}^{2} \frac{3 π}{8} = 3 -2 \sqrt{2}$ A1

[4 marks]

let $y = {cosec}^{2} θ$

uses $\cot^{2} θ = {cosec}^{2} θ - 1$ where $x = \cot^{2} θ$ (M1)

$x^{2} - 6 x + 1 = 0 \Rightarrow {(y - 1)}^{2} - 6 (y - 1) + 1 = 0$ M1

$y^{2} - 8 y + 8 = 0$ A1

[3 marks]

Examiners report

[N/A]