IB Questionbank

User interface language: English | Español

HL Paper 2

The function $f$ is defined by $f\left( x \right) = \frac{{2\,{\text{ln}}\,x + 1}}{{x - 3}}$ , 0 < $x$ < 3.

Draw a set of axes showing $x$ and $y$ values between −3 and 3. On these axes

Hence, or otherwise, find the coordinates of the point of inflexion on the graph of $y = f\left( x \right)$ .

[4]

sketch the graph of $y = f\left( x \right)$ , showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]

c.i.

sketch the graph of $y = {f^{ - 1}}\left( x \right)$ , showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]

c.ii.

Hence, or otherwise, solve the inequality $f\left( x \right) > {f^{ - 1}}\left( x \right)$ .

[3]

Markscheme

finding turning point of $y = f'\left( x \right)$ or finding root of $y = f''\left( x \right)$ (M1)

$x = 0.899$ A1

$y = f\left( {0.899048 \ldots } \right) = - 0.375$ (M1)A1

(0.899, −0.375)

Note: Do not accept $x = 0.9$ . Accept y-coordinates rounding to −0.37 or −0.375 but not −0.38.

[4 marks]

smooth curve over the correct domain which does not cross the y-axis

and is concave down for $x$ > 1 A1

$x$ -intercept at 0.607 A1

equations of asymptotes given as $x$ = 0 and $x$ = 3 (the latter must be drawn) A1A1

[4 marks]

c.i.

attempt to reflect graph of $f$ in $y$ = $x$ (M1)

smooth curve over the correct domain which does not cross the $x$ -axis and is concave down for $y$ > 1 A1

$y$ -intercept at 0.607 A1

equations of asymptotes given as $y$ = 0 and $y$ = 3 (the latter must be drawn) A1

Note: For FT from (i) to (ii) award max M1A0A1A0.

[4 marks]

c.ii.

solve $f\left( x \right) = {f^{ - 1}}\left( x \right)$ or $f\left( x \right) = x$ to get $x$ = 0.372 (M1)A1

0 < $x$ < 0.372 A1

Note: Do not award FT marks.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Consider $f(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right)$

The function $f$ is defined by $f(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right),{\text{ }}x \in D$

The function $g$ is defined by $g(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right),{\text{ }}x \in \left] {1,{\text{ }}\infty } \right[$ .

Find the largest possible domain $D$ for $f$ to be a function.

[2]

Sketch the graph of $y = f(x)$ showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.

[3]

Explain why $f$ is an even function.

[1]

Explain why the inverse function ${f^{ - 1}}$ does not exist.

[1]

Find the inverse function ${g^{ - 1}}$ and state its domain.

[4]

Find $g'(x)$ .

[3]

Hence, show that there are no solutions to $g'(x) = 0$ ;

[2]

g.i.

Hence, show that there are no solutions to $({g^{ - 1}})'(x) = 0$ .

[2]

g.ii.

Markscheme

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

${x^2} - 1 > 0$ (M1)

$x < - 1$ or $x > 1$ A1

[2 marks]

M17/5/MATHL/HP2/ENG/TZ1/12.b/M

shape A1

$x = 1$ and $x = - 1$ A1

$x$ -intercepts A1

[3 marks]

EITHER

$f$ is symmetrical about the $y$ -axis R1

$f( - x) = f(x)$ R1

[1 mark]

EITHER

$f$ is not one-to-one function R1

horizontal line cuts twice R1

Note: Accept any equivalent correct statement.

[1 mark]

$x = - 1 + \ln \left( {\sqrt {{y^2} - 1} } \right)$ M1

${{\text{e}}^{2x + 2}} = {y^2} - 1$ M1

${g^{ - 1}}(x) = \sqrt {{{\text{e}}^{2x + 2}} + 1} ,{\text{ }}x \in \mathbb{R}$ A1A1

[4 marks]

$g'(x) = \frac{1}{{\sqrt {{x^2} - 1} }} \times \frac{{2x}}{{2\sqrt {{x^2} - 1} }}$ M1A1

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

[3 marks]

$g'(x) = \frac{x}{{{x^2} - 1}} = 0 \Rightarrow x = 0$ M1

which is not in the domain of $g$ (hence no solutions to $g'(x) = 0$ ) R1

[2 marks]

g.i.

$({g^{ - 1}})'(x) = \frac{{{{\text{e}}^{2x + 2}}}}{{\sqrt {{{\text{e}}^{2x + 2}} + 1} }}$ M1

as ${{\text{e}}^{2x + 2}} > 0 \Rightarrow ({g^{ - 1}})'(x) > 0$ so no solutions to $({g^{ - 1}})'(x) = 0$ R1

Note: Accept: equation ${{\text{e}}^{2x + 2}} = 0$ has no solutions.

[2 marks]

g.ii.

Examiners report

[N/A]

g.i.

[N/A]

g.ii.

A scientist conducted a nine-week experiment on two plants, $A$ and $B$ , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant $A$ was given fertilizer regularly, while Plant $B$ was not.

The scientist found that the height of Plant $A, h_{A} cm$ , at time $t$ weeks can be modelled by the function $h_{A} (t) = \sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ .

The scientist found that the height of Plant $B, h_{B} cm$ , at time $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

Use the scientist’s models to find the initial height of

Plant $B$ .

[1]

a.i.

Plant $A$ correct to three significant figures.

[2]

a.ii.

Find the values of $t$ when $h_{A} (t) = h_{B} (t)$ .

[3]

For $t > 6$ , prove that Plant $A$ was always taller than Plant $B$ .

[3]

For $0 \leq t \leq 9$ , find the total amount of time when the rate of growth of Plant $B$ was greater than the rate of growth of Plant $A$ .

[6]

Markscheme

$32$ (cm) A1

[1 mark]

a.i.

$h_{A} (0) = \sin (6) + 27$ (M1)

$= 26.7205 \dots$

$= 26.7$ (cm) A1

[2 marks]

a.ii.

attempts to solve $h_{A} (t) = h_{B} (t)$ for $t$ (M1)

$t = 4.0074 \dots, 4.7034 \dots, 5.88332 \dots$

$t = 4.01, 4.70, 5.88$ (weeks) A2

[3 marks]

$h_{A} (t) - h_{B} (t) = \sin (2 t + 6) + t - 5$ A1

EITHER

for $t > 6, t - 5 > 1$ A1

and as $\sin (2 t + 6) \geq - 1 \Rightarrow h_{A} (t) - h_{B} (t) > 0$ R1

the minimum value of $\sin (2 t + 6) = - 1$ R1

so for $t > 6, h_{A} (t) - h_{B} (t) = t - 6 > 0$ A1

THEN

hence for $t > 6$ , Plant $A$ was always taller than Plant $B$ AG

[3 marks]

recognises that $h_{A}' (t)$ and $h_{B}' (t)$ are required (M1)

attempts to solve $h_{A}' (t) = h_{B}' (t)$ for $t$ (M1)

$t = 1.18879 \dots$ and $2.23598 \dots$ OR $4.33038 \dots$ and $5.37758 \dots$ OR $7.47197 \dots$ and $8.51917 \dots$ (A1)

Note: Award full marks for $t = \frac{4 π}{3} - 3, \frac{5 π}{3} - 3, (\frac{7 π}{3} - 3, \frac{8 π}{3} - 3 \frac{10 π}{3} - 3, \frac{11 π}{3} - 3)$ .

Award subsequent marks for correct use of these exact values.

$1.18879 \dots < t < 2.23598 \dots$ OR $4.33038 \dots < t < 5.37758 \dots$ OR $7.47197 \dots < t < 8.51917 \dots$ (A1)

attempts to calculate the total amount of time (M1)

$3 (2.2359 \dots - 1.1887 \dots) (= 3 ((\frac{5 π}{3} - 3) - (\frac{4 π}{3} - 3)))$

$= 3.14 (= π)$ (weeks) A1

[6 marks]

Examiners report

Part (a) In general, very well done, most students scored full marks. Some though had an incorrect answer for part(a)(ii) because they had their GDC in degrees.

Part (b) Well attempted. Some accuracy errors and not all candidates listed all three values.

Part (c) Most students tried a graphical approach (but this would only get them one out of three marks) and only some provided a convincing algebraic justification. Many candidates tried to explain in words without a convincing mathematical justification or used numerical calculations with specific time values. Some arrived at the correct simplified equation for the difference in heights but could not do much with it. Then only a few provided a correct mathematical proof.

Part (d) In general, well attempted by many candidates. The common error was giving the answer as 3.15 due to the pre-mature rounding. Some candidates only provided the values of time when the rates are equal, some intervals rather than the total time.

a.i.

[N/A]

a.ii.

[N/A]

The voltage $v$ in a circuit is given by the equation

$v\left( t \right) = 3\,{\text{sin}}\left( {100\pi t} \right)$ , $t \geqslant 0$ where $t$ is measured in seconds.

The current $i$ in this circuit is given by the equation

$i\left( t \right) = 2\,{\text{sin}}\left( {100\pi \left( {t + 0.003} \right)} \right)$ .

The power $p$ in this circuit is given by $p\left( t \right) = v\left( t \right) \times i\left( t \right)$ .

The average power ${p_{av}}$ in this circuit from $t = 0$ to $t = T$ is given by the equation

${p_{av}}\left( T \right) = \frac{1}{T}\int_0^T {p\left( t \right){\text{d}}t}$ , where $T > 0$ .

Write down the maximum and minimum value of $v$ .

[2]

Write down two transformations that will transform the graph of $y = v\left( t \right)$ onto the graph of $y = i\left( t \right)$ .

[2]

Sketch the graph of $y = p\left( t \right)$ for 0 ≤ $t$ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.

[3]

Find the total time in the interval 0 ≤ $t$ ≤ 0.02 for which $p\left( t \right)$ ≥ 3.

[3]

Find ${p_{av}}$ (0.007).

[2]

With reference to your graph of $y = p\left( t \right)$ explain why ${p_{av}}\left( T \right)$ > 0 for all $T$ > 0.

[2]

Given that $p\left( t \right)$ can be written as $p\left( t \right) = a\,{\text{sin}}\left( {b\left( {t - c} \right)} \right) + d$ where $a$ , $b$ , $c$ , $d$ > 0, use your graph to find the values of $a$ , $b$ , $c$ and $d$ .

[6]

Markscheme

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

3, −3 A1A1

[2 marks]

stretch parallel to the $y$ -axis (with $x$ -axis invariant), scale factor $\frac{2}{3}$ A1

translation of $\left( {\begin{array}{*{20}{c}} { - 0.003} \\ 0 \end{array}} \right)$ (shift to the left by 0.003) A1

Note: Can be done in either order.

[2 marks]

correct shape over correct domain with correct endpoints    A1
first maximum at (0.0035, 4.76)    A1
first minimum at (0.0085, −1.24)    A1

[3 marks]

$p$ ≥ 3 between $t$ = 0.0016762 and 0.0053238 and $t$ = 0.011676 and 0.015324 (M1)(A1)

Note: Award M1A1 for either interval.

= 0.00730 A1

[3 marks]

${p_{av}} = \frac{1}{{0.007}}\int_0^{0.007} {6\,{\text{sin}}\left( {100\pi t} \right)} {\text{sin}}\left( {100\pi \left( {t + 0.003} \right)} \right){\text{d}}t$ (M1)

= 2.87 A1

[2 marks]

in each cycle the area under the $t$ axis is smaller than area above the $t$ axis R1

the curve begins with the positive part of the cycle R1

[2 marks]

$a = \frac{{4.76 - \left( { - 1.24} \right)}}{2}$ (M1)

$a = 3.00$ A1

$d = \frac{{4.76 + \left( { - 1.24} \right)}}{2}$

$d = 1.76$ A1

$b = \frac{{2\pi }}{{0.01}}$

$b = 628\left( { = 200\pi } \right)$ A1

$c = 0.0035 - \frac{{0.01}}{4}$ (M1)

$c = 0.00100$ A1

[6 marks]

Examiners report

[N/A]

Consider the function $f (x) = \frac{x^{2} - x - 12}{2 x - 15}, x \in ℝ, x \neq \frac{15}{2}$ .

Find the coordinates where the graph of $f$ crosses the

$x$ -axis.

[2]

a.i.

$y$ -axis.

[1]

a.ii.

Write down the equation of the vertical asymptote of the graph of $f$ .

[1]

The oblique asymptote of the graph of $f$ can be written as $y = a x + b$ where $a, b \in ℚ$ .

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

[4]

Sketch the graph of $f$ for $- 30 \leq x \leq 30$ , clearly indicating the points of intersection with each axis and any asymptotes.

[3]

Express $\frac{1}{f (x)}$ in partial fractions.

[3]

e.i.

Hence find the exact value of $\int_{0}^{3} \frac{1}{f (x)} d x$ , expressing your answer as a single logarithm.

[4]

e.ii.

Markscheme

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

attempts to solve $x^{2} - x - 12 = 0$ (M1)

$(- 3, 0)$ and $(4, 0)$ A1

[2 marks]

a.i.

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

$(0, \frac{4}{5})$ A1

[1 mark]

a.ii.

$x = \frac{15}{2}$ A1

Note: Award A0 for $x \neq \frac{15}{2}$ .
Award A1 in part (b), if $x = \frac{15}{2}$ is seen on their graph in part (d).

[1 mark]

METHOD 1

$(a x + b) (2 x - 15) \equiv x^{2} - x - 12$

attempts to expand $(a x + b) (2 x - 15)$ (M1)

$2 a x^{2} - 15 a x + 2 b x - 15 b \equiv x^{2} - x - 12$

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

equates coefficients of $x$ (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 2

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x}{2} + \frac{13}{4} + \dots$ M1

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

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 3

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

$\frac{x^{2} - x - 12}{2 x - 15} \equiv \frac{x}{2} + b + \frac{c}{2 x - 15}$ M1

$x^{2} - x - 12 \equiv \frac{(2 x - 15) x}{2} + (2 x - 15) b + c$

equates coefficients of $x$ : (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 4

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x^{2} - x - 12}{2 x - 15} = \frac{x}{2} + \frac{\frac{13 x}{2} - 12}{2 x - 15}$

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

$\frac{\frac{13 x}{2} - 12}{2 x - 15} = \frac{13}{4} + \dots$ M1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

[4 marks]

two branches with approximately correct shape (for $- 30 \leq x \leq 30$ ) A1

their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1

their axes intercepts in approximately the correct positions A1

Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.

[3 marks]

attempts to split into partial fractions: (M1)

$\frac{2 x - 15}{(x + 3) (x - 4)} \equiv \frac{A}{x + 3} + \frac{B}{x - 4}$

$2 x - 15 \equiv A (x - 4) + B (x + 3)$

$A = 3$ A1

$B = - 1$ A1

$(\frac{3}{x + 3} - \frac{1}{x - 4})$

[3 marks]

e.i.

$\int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x$

attempts to integrate and obtains two terms involving ‘ln’ (M1)

$= {[3 \ln |x + 3| - \ln |x - 4|]}_{0}^{3}$ A1

$= 3 \ln 6 - \ln 1 - 3 \ln 3 + \ln 4$ A1

$= 3 \ln 2 + \ln 4 (= \ln 8 + \ln 4)$

$= \ln 32 (= 5 \ln 2)$ A1

Note: The final A1 is dependent on the previous two A marks.

[4 marks]

e.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

e.i.

[N/A]

e.ii.

The function $f$ is defined by $f\left( x \right) = {\text{sec}}\,x + 2$ , $0 \leqslant x < \frac{\pi }{2}$ .

Write down the range of $f$ .

[1]

Find $f'\left( x \right)$ , stating its domain.

[4]

Markscheme

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

$f\left( x \right)$ ≥ 3 A1

[1 mark]

$x = {\text{sec}}\,y + 2$ (M1)

Note: Exchange of variables can take place at any point.

${\text{cos}}\,y = \frac{1}{{x - 2}}$ (A1)

$f'\left( x \right) = {\text{arccos}}\left( {\frac{1}{{x - 2}}} \right)$ , $x$ ≥ 3 A1A1

Note: Allow follow through from (a) for last A1 mark which is independent of earlier marks in (b).

[4 marks]

Examiners report

[N/A]

The population, $P$ , of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$

where $t$ is the time measured in years and $k, N$ are positive constants.

The constant $N$ represents the maximum population of this species of marsupial that the island can sustain indefinitely.

Let $P_{0}$ be the initial population of marsupials.

In the context of the population model, interpret the meaning of $\frac{d P}{d t}$ .

[1]

Show that $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

[4]

Hence show that the population of marsupials will increase at its maximum rate when $P = \frac{N}{2}$ . Justify your answer.

[5]

Hence determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .

[2]

By solving the logistic differential equation, show that its solution can be expressed in the form

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ .

[7]

After $10$ years, the population of marsupials is $3 P_{0}$ . It is known that $N = 4 P_{0}$ .

Find the value of $k$ for this population model.

[2]

Markscheme

rate of growth (change) of the (marsupial) population (with respect to time) A1

[1 mark]

Note: Do not accept growth (change) in the (marsupials) population per year.

METHOD 1

attempts implicit differentiation on $\frac{d P}{d t} = k P - \frac{k P^{2}}{N}$ be expanding $k P (1 - \frac{P}{N})$ (M1)

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} - 2 \frac{k P}{N} \frac{d P}{d t}$ A1A1

$= k \frac{d P}{d t} (1 - \frac{2 P}{N})$ A1

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$ and so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

METHOD 2

attempts implicit differentiation (product rule) on $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ M1

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) \frac{d P}{d t})$ A1

substitutes $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ into their $\frac{d^{2} P}{d t^{2}}$ M1

$\frac{d^{2} P}{d t^{2}} = k (k P (1 - \frac{P}{N})) (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) k P (1 - \frac{P}{N}))$

$= k^{2} P {(1 - \frac{P}{N})}^{2} - k^{2} P (1 - \frac{P}{N}) (\frac{P}{N})$

$= k^{2} P (1 - \frac{P}{N}) (1 - \frac{P}{N} - \frac{P}{N})$ A1

so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

[4 marks]

$\frac{d^{2} P}{d t^{2}} = 0 \Rightarrow k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N}) = 0$ (M1)

$P = 0, \frac{N}{2}, N$ A2

Note: Award A1 for $P = \frac{N}{2}$ only.

uses the second derivative to show that concavity changes at $P = \frac{N}{2}$ or the first derivative to show a local maximum at $P = \frac{N}{2}$ M1

EITHER

a clearly labelled correct sketch of $\frac{d^{2} P}{d t^{2}}$ versus $P$ showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

a correct and clearly labelled sign diagram (table) showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

for example, $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (> 0)$ with $P = \frac{N}{4}$ and $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (< 0)$ with $P = \frac{3 N}{4}$ showing $P = \frac{N}{2}$ corresponds to a local maximum point for $\frac{d P}{d t}$ R1

so the population is increasing at its maximum rate when $P = \frac{N}{2}$ AG

[5 marks]

substitutes $P = \frac{N}{2}$ into $\frac{d P}{d t}$ (M1)

$\frac{d P}{d t} = k (\frac{N}{2}) (1 - \frac{\frac{N}{2}}{N})$

the maximum value of $\frac{d P}{d t}$ is $\frac{k N}{4}$ A1

[2 marks]

METHOD 1

attempts to separate variables M1

$\int \frac{N}{P (N - P)} d P = \int k d t$

attempts to write $\frac{N}{P (N - P)}$ in partial fractions form M1

$\frac{N}{P (N - P)} \equiv \frac{A}{P} + \frac{B}{(N - P)} \Rightarrow N \equiv A (N - P) + B P$

$A = 1, B = 1$ A1

$\frac{N}{P (N - P)} \equiv \frac{1}{P} + \frac{1}{(N - P)}$

$\int (\frac{1}{P} + \frac{1}{(N - P)}) d P = \int k d t$

$\Rightarrow \ln P - \ln (N - P) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (N - P)$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln P_{0} - \ln (N - P_{0})$

$k t = \ln (\frac{P}{N - P}) - \ln (\frac{P_{0}}{N - P_{o}}) (= \ln (\frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}}))$ A1

so $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 2

attempts to separate variables M1

$\int \frac{1}{P (1 - \frac{P}{N})} d P = \int k d t$

attempts to write $\frac{1}{P (1 - \frac{P}{N})}$ in partial fractions form M1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{A}{P} + \frac{B}{1 - \frac{P}{N}} \Rightarrow 1 \equiv A (1 - \frac{P}{N}) + B P$

$A = 1, B = \frac{1}{N}$ A1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})}$

$\int \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})} d P = \int k d t$

$\Rightarrow \ln P - \ln (1 - \frac{P}{N}) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (1 - \frac{P}{N})$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

$\ln (\frac{P}{1 - \frac{P}{N}}) = k t + C \Rightarrow \ln (\frac{N P}{N - P}) = k t + C$

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln (\frac{N P_{0}}{N - P_{0}})$

$k t = \ln (\frac{N P}{N - P}) - \ln (\frac{N P_{0}}{N - P_{0}}) (= \ln \frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 3

lets $u = \frac{1}{P}$ and forms $\frac{d u}{d t} = - \frac{1}{P^{2}} \frac{d P}{d t}$ M1

multiplies both sides of the differential equation by $- \frac{1}{P^{2}}$ and makes the above substitutions M1

$- \frac{1}{P^{2}} \frac{d P}{d t} = k (\frac{1}{N} - \frac{1}{P}) \Rightarrow \frac{d u}{d t} = k (\frac{1}{N} - u)$

$\frac{d u}{d t} + k u = \frac{k}{N}$ (linear first-order DE) A1

$IF = e^{\int k d t} = e^{k t} \Rightarrow e^{k t} \frac{d u}{d t} + k e^{k t} u = \frac{k}{N} e^{k t}$ (M1)

$\frac{d}{d t} (u e^{k t}) = \frac{k}{N} e^{k t}$

$u e^{k t} = \frac{1}{N} e^{k t} (+ C) (\frac{1}{P} e^{k t} = \frac{1}{N} e^{k t} (+ C))$ A1

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}, u = \frac{1}{P_{0}}$ and so $C = \frac{1}{P_{0}} - \frac{1}{N} (= \frac{N - P_{0}}{N P_{0}})$

$e^{k t} (\frac{N - P}{N P}) = \frac{N - P_{0}}{N P_{0}}$

$e^{k t} = (\frac{P}{N - P}) (\frac{N - P_{0}}{P_{0}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

[7 marks]

substitutes $t = 10, P = 3 P_{0}$ and $N = 4 P_{0}$ into $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ M1

$10 k = \ln 3 (\frac{4 P_{0} - P_{0}}{4 P_{0} - 3 P_{0}}) (= \ln 9)$

$k = 0.220 (= \frac{1}{10} \ln 9, = \frac{1}{5} \ln 3)$ A1

[2 marks]

Examiners report

An extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.

Part (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.

Part (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.

Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer $P = \frac{N}{2}$ omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.

Part (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.

Part (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.

Part (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.

[N/A]

Consider the function $f(x) = 2{\sin ^2}x + 7\sin 2x + \tan x - 9,{\text{ }}0 \leqslant x < \frac{\pi }{2}$ .

Let $u = \tan x$ .

Determine an expression for $f’(x)$ in terms of $x$ .

[2]

a.i.

Sketch a graph of $y = f’(x)$ for $0 \leqslant x < \frac{\pi }{2}$ .

[4]

a.ii.

Find the $x$ -coordinate(s) of the point(s) of inflexion of the graph of $y = f(x)$ , labelling these clearly on the graph of $y = f’(x)$ .

[2]

a.iii.

Express $\sin x$ in terms of $\mu$ .

[2]

b.i.

Express $\sin 2x$ in terms of $u$ .

[3]

b.ii.

Hence show that $f(x) = 0$ can be expressed as ${u^3} - 7{u^2} + 15u - 9 = 0$ .

[2]

b.iii.

Solve the equation $f(x) = 0$ , giving your answers in the form $\arctan k$ where $k \in \mathbb{Z}$ .

[3]

Markscheme

$f’(x) = 4\sin x\cos x + 14\cos 2x + {\sec ^2}x$ (or equivalent) (M1)A1

[2 marks]

a.i.

N17/5/MATHL/HP2/ENG/TZ0/11.a.ii/M A1A1A1A1

Note: Award A1 for correct behaviour at $x = 0$ , A1 for correct domain and correct behaviour for $x \to \frac{\pi }{2}$ , A1 for two clear intersections with $x$ -axis and minimum point, A1 for clear maximum point.

[4 marks]

a.ii.

$x = 0.0736$ A1

$x = 1.13$ A1

[2 marks]

a.iii.

attempt to write $\sin x$ in terms of $u$ only (M1)

$\sin x = \frac{u}{{\sqrt {1 + {u^2}} }}$ A1

[2 marks]

b.i.

$\cos x = \frac{1}{{\sqrt {1 + {u^2}} }}$ (A1)

attempt to use $\sin 2x = 2\sin x\cos x{\text{ }}\left( { = 2\frac{u}{{\sqrt {1 + {u^2}} }}\frac{1}{{\sqrt {1 + {u^2}} }}} \right)$ (M1)

$\sin 2x = \frac{{2u}}{{1 + {u^2}}}$ A1

[3 marks]

b.ii.

$2{\sin ^2}x + 7\sin 2x + \tan x - 9 = 0$

$\frac{{2{u^2}}}{{1 + {u^2}}} + \frac{{14u}}{{1 + {u^2}}} + u - 9{\text{ }}( = 0)$ M1

$\frac{{2{u^2} + 14u + u(1 + {u^2}) - 9(1 + {u^2})}}{{1 + {u^2}}} = 0$ (or equivalent) A1

${u^3} - 7{u^2} + 15u - 9 = 0$ AG

[2 marks]

b.iii.

$u = 1$ or $u = 3$ (M1)

$x = \arctan (1)$ A1

$x = \arctan (3)$ A1

Note: Only accept answers given the required form.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

Consider the polynomial $P\left( z \right) \equiv {z^4} - 6{z^3} - 2{z^2} + 58z - 51,\,\,z \in \mathbb{C}$ .

Sketch the graph of $y = {x^4} - 6{x^3} - 2{x^2} + 58x - 51$ , stating clearly the coordinates of any maximum and minimum points and intersections with axes.

[6]

Hence, or otherwise, state the condition on $k \in \mathbb{R}$ such that all roots of the equation $P\left( z \right) = k$ are real.

[2]

Markscheme

shape A1

$x$ -axis intercepts at (−3, 0), (1, 0) and $y$ -axis intercept at (0, −51) A1A1

minimum points at (−1.62, −118) and (3.72, 19.7) A1A1

maximum point at (2.40, 26.9) A1

Note: Coordinates may be seen on the graph or elsewhere.

Note: Accept −3, 1 and −51 marked on the axes.

[6 marks]

from graph, 19.7 ≤ $k$ ≤ 26.9 A1A1

Note: Award A1 for correct endpoints and A1 for correct inequalities.

[2 marks]

Examiners report

[N/A]

A curve C is given by the implicit equation $x + y - {\text{cos}}\left( {xy} \right) = 0$ .

The curve $xy = - \frac{\pi }{2}$ intersects C at P and Q.

Show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$ .

[5]

Find the coordinates of P and Q.

[4]

b.i.

Given that the gradients of the tangents to C at P and Q are m₁ and m₂ respectively, show that m₁ × m₂ = 1.

[3]

b.ii.

Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line $y = - x$ .

[7]

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

$1 + \frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right){\text{sin}}\left( {xy} \right) = 0$ A1M1A1

Note: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.

$\left( {1 + x\,{\text{sin}}\left( {xy} \right)} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1 - y\,{\text{sin}}\left( {xy} \right)$ A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$ AG

[5 marks]

EITHER

when $xy = - \frac{\pi }{2},\,\,{\text{cos}}\,xy = 0$ M1

$\Rightarrow x + y = 0$ (A1)

$x - \frac{\pi }{{2x}} - {\text{cos}}\left( {\frac{{ - \pi }}{2}} \right) = 0$ or equivalent M1

$x - \frac{\pi }{{2x}} = 0$ (A1)

THEN

therefore ${x^2} = \frac{\pi }{2}\left( {x = \pm \sqrt {\frac{\pi }{2}} } \right)\left( {x = \pm 1.25} \right)$ A1

${\text{P}}\left( {\sqrt {\frac{\pi }{2}} ,\, - \sqrt {\frac{\pi }{2}} } \right),\,\,{\text{Q}}\left( { - \sqrt {\frac{\pi }{2}} ,\,\sqrt {\frac{\pi }{2}} } \right)$ or $P\left( {1.25,\, - 1.25} \right),\,Q\left( { - 1.25,\,1.25} \right)$ A1

[4 marks]

b.i.

m₁= $- \left( {\frac{{1 - \sqrt {\frac{\pi }{2}} \times - 1}}{{1 + \sqrt {\frac{\pi }{2}} \times - 1}}} \right)$ M1A1

m₂= $- \left( {\frac{{1 + \sqrt {\frac{\pi }{2}} \times - 1}}{{1 - \sqrt {\frac{\pi }{2}} \times - 1}}} \right)$ A1

m₁m₂= 1 AG

Note: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.

[3 marks]

b.ii.

equate derivative to −1 M1

$\left( {y - x} \right){\text{sin}}\left( {xy} \right) = 0$ (A1)

$y = x,\,{\text{sin}}\left( {xy} \right) = 0$ R1

in the first case, attempt to solve $2x = {\text{cos}}\left( {{x^2}} \right)$ M1

(0.486,0.486) A1

in the second case, ${\text{sin}}\left( {xy} \right) = 0 \Rightarrow xy = 0$ and $x + y = 1$ (M1)

(0,1), (1,0) A1

[7 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let $x$ grams represent the amount of salt in the tank and let $t$ minutes represent the time since the salt water began flowing into the tank.

The rate of change of the amount of salt in the tank, $\frac{{{\text{d}}x}}{{{\text{d}}t}}$ , is described by the differential equation $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 10{{\text{e}}^{- \frac{t}{4}}} - \frac{x}{{t + 1}}$ .

Show that $t$ + 1 is an integrating factor for this differential equation.

[2]

Hence, by solving this differential equation, show that $x\left( t \right) = \frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{t + 1}}$ .

[8]

Sketch the graph of $x$ versus $t$ for 0 ≤ $t$ ≤ 60 and hence find the maximum amount of salt in the tank and the value of $t$ at which this occurs.

[5]

Find the value of $t$ at which the amount of salt in the tank is decreasing most rapidly.

[2]

The rate of change of the amount of salt leaving the tank is equal to $\frac{x}{{t + 1}}$ .

Find the amount of salt that left the tank during the first 60 minutes.

[4]

Markscheme

METHOD 1

$I(t) = {{\text{e}}^{\int {P\left( t \right)} \,{\text{d}}t}}$ M1

${{\text{e}}^{\int {\frac{1}{{t + 1}}} \,{\text{d}}t}}$

= ${{\text{e}}^{{\text{ln}}\left( {t + 1} \right)}}$ A1

$= t + 1$ AG

METHOD 2

attempting product rule differentiation on $\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right)$ M1

$\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right) = \frac{{{\text{d}}x}}{{{\text{d}}t}}\left( {t + 1} \right) + x$

$= \left( {t + 1} \right)\left( {\frac{{{\text{d}}x}}{{{\text{d}}t}} + \frac{x}{{t + 1}}} \right)$ A1

so $t + 1$ is an integrating factor for this differential equation AG

[2 marks]

attempting to multiply through by $\left( {t + 1} \right)$ and rearrange to give (M1)

$\left( {t + 1} \right)\frac{{{\text{d}}x}}{{{\text{d}}t}} + x = 10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ A1

$\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right) = 10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$

$x\left( {t + 1} \right) = \int {10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}} {\text{d}}t$ A1

attempting to integrate the RHS by parts M1

$= - 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} + 40\int {{{\text{e}}^{ - \frac{t}{4}}}} \,{\text{d}}t$

$= - 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} - 160{{\text{e}}^{ - \frac{t}{4}}} + C$ A1

Note: Condone the absence of C.

EITHER

substituting $t = 0,\,\,x = 0 \Rightarrow C = 200$ M1

$x = \frac{{- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} - 160{{\text{e}}^{ - \frac{t}{4}}} + 200}}{{t + 1}}$ A1

using $- 40{{\text{e}}^{ - \frac{t}{4}}}$ as the highest common factor of $- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ and $- 160{{\text{e}}^{ - \frac{t}{4}}}$ M1

using $- 40{{\text{e}}^{ - \frac{t}{4}}}$ as the highest common factor of $- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ and $- 160{{\text{e}}^{ - \frac{t}{4}}}$ giving

$x\left( {t + 1} \right) = - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right) + C$ (or equivalent) M1A1

substituting $t = 0,\,\,x = 0 \Rightarrow C = 200$ M1

THEN

$x\left( t \right) = \frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{t + 1}}$ AG

[8 marks]

graph starts at the origin and has a local maximum (coordinates not required) A1

sketched for 0 ≤ $t$ ≤ 60 A1

correct concavity for 0 ≤ $t$ ≤ 60 A1

maximum amount of salt is 14.6 (grams) at $t$ = 6.60 (minutes) A1A1

[5 marks]

using an appropriate graph or equation (first or second derivative) M1

amount of salt is decreasing most rapidly at $t$ = 12.9 (minutes) A1

[2 marks]

EITHER

attempting to form an integral representing the amount of salt that left the tank M1

$\int\limits_0^{60} {\frac{{x\left( t \right)}}{{t + 1}}{\text{d}}t}$

$\int\limits_0^{60} {\frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{{{\left( {t + 1} \right)}^2}}}{\text{d}}t}$ A1

attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at $t$ = 60(minutes)

amount of salt that left the tank is $\int\limits_0^{60} {10} {{\text{e}}^{ - \frac{t}{4}}}\,{\text{d}}t - x\left( {60} \right)$ A1

THEN

= 36.7 (grams) A2

[4 marks]

Examiners report

[N/A]

Consider the differential equation $x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2}$ for $x > 0$ and $y > 2 x$ . It is given that $y = 3$ when $x = 1$ .

Use Euler’s method, with a step length of $0.1$ , to find an approximate value of $y$ when $x = 1.5$ .

[4]

Use the substitution $y = v x$ to show that $x \frac{d v}{d x} = v^{2} - v - 2$ .

[3]

By solving the differential equation, show that $y = \frac{8 x + x^{4}}{4 - x^{3}}$ .

[10]

c.i.

Find the actual value of $y$ when $x = 1.5$ .

[1]

c.ii.

Using the graph of $y = \frac{8 x + x^{4}}{4 - x^{3}}$ , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of $y$ at $x = 1.5$ .

[1]

c.iii.

Markscheme

attempt to use Euler’s method (M1)

$x_{n + 1} = x_{n} + 0.1; y_{n + 1} = y_{n} + 0.1 \times \frac{d y}{d x}$ , where $\frac{d y}{d x} = \frac{y^{2} - 2 x^{2}}{x^{2}}$

correct intermediate $y$ -values (A1)(A1)

$3.7, 4.63140 \dots, 5.92098, 7.79542 \dots$

Note: A1 for any two correct $y$ -values seen

$y = 10.6958 \dots$

$y = 10.7$ A1

Note: For the final A1, the value $10.7$ must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.

[4 marks]

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ (A1)

replacing $y$ with $v x$ and $\frac{d y}{d x}$ with $v + x \frac{d v}{d x}$ M1

$x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2} \Rightarrow x^{2} (v + x \frac{d v}{d x}) = v^{2} x^{2} - 2 x^{2}$ A1

$v + x \frac{d v}{d x} = v^{2} - 2$ (since $x > 0$ )

$x \frac{d v}{d x} = v^{2} - v - 2$ AG

[3 marks]

attempt to separate variables $v$ and $x$ (M1)

$\int \frac{d v}{v^{2} - v - 2} = \int \frac{d x}{x}$

$\int \frac{d v}{(v - 2) (v + 1)} = \int \frac{d x}{x}$ (A1)

attempt to express in partial fraction form M1

$\frac{1}{(v - 2) (v + 1)} \equiv \frac{A}{v - 2} + \frac{B}{v + 1}$

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

$\frac{1}{3} \int (\frac{1}{v - 2} - \frac{1}{v + 1}) d v = \int \frac{d x}{x}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| (+ c)$ A1

Note: Condone absence of modulus signs throughout.

EITHER

attempt to find $c$ using $x = 1, y = 3, v = 3$ M1

$c = \frac{1}{3} \ln \frac{1}{4}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| + \frac{1}{3} \ln \frac{1}{4}$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (\frac{{|x|}^{3}}{4})$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (A {|x|}^{3})$

attempt to find $A$ using $x = 1, y = 3, v = 3$ M1

$A = \frac{1}{4}$

THEN

$|\frac{v - 2}{v + 1}| = \frac{1}{4} x^{3}$ (since $x > 0$ )

substitute $v = \frac{y}{x}$ (seen anywhere) M1

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

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

attempt to make $y$ the subject M1

$y - \frac{x^{3} y}{4} = 2 x + \frac{x^{4}}{4}$ A1

$y = \frac{8 x + x^{4}}{4 - x^{3}}$ AG

[10 marks]

c.i.

actual value at $y (1.5) = 27.3$ A1

[1 mark]

c.ii.

gradient changes rapidly (during the interval considered) OR

the curve has a vertical asymptote at $x = \sqrt[3]{4} (= 1.5874 \dots)$ R1

[1 mark]

c.iii.

Examiners report

Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.

There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).

Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Consider the function $f (x) = \sqrt{x^{2} - 1}$ , where $1 \leq x \leq 2$ .

The curve $y = f (x)$ is rotated $2 π$ about the $y$ -axis to form a solid of revolution that is used to model a water container.

At $t = 0$ , the container is empty. Water is then added to the container at a constant rate of $0.4 m^{3} s^{- 1}$ .

Sketch the curve $y = f (x)$ , clearly indicating the coordinates of the endpoints.

[2]

Show that the inverse function of $f$ is given by $f^{- 1} (x) = \sqrt{x^{2} + 1}$ .

[3]

b.i.

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

[2]

b.ii.

Show that the volume, $V m^{3}$ , of water in the container when it is filled to a height of $h$ metres is given by $V = π (\frac{1}{3} h^{3} + h)$ .

[3]

c.i.

Hence, determine the maximum volume of the container.

[2]

c.ii.

Find the time it takes to fill the container to its maximum volume.

[2]

Find the rate of change of the height of the water when the container is filled to half its maximum volume.

[6]

Markscheme

correct shape (concave down) within the given domain $1 \leq x \leq 2$ A1

$(1, 0)$ and $(2, \sqrt{3}) (= (2, 1.73))$ A1

Note: The coordinates of endpoints may be seen on the graph or marked on the axes.

[2 marks]

interchanging $x$ and $y$ (seen anywhere) M1

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

$x^{2} = y^{2} - 1$ A1

$y = \sqrt{x^{2} + 1}$ A1

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

[3 marks]

b.i.

$0 \leq x \leq \sqrt{3}$ OR domain $[0, \sqrt{3}] (= [0, 1.73])$ A1

$1 \leq y \leq 2$ OR $1 \leq f^{- 1} (x) \leq 2$ OR range $[1, 2]$ A1

[2 marks]

b.ii.

attempt to substitute $x = \sqrt{y^{2} + 1}$ into the correct volume formula (M1)

$V = π \int_{0}^{h} {(\sqrt{y^{2} + 1})}^{2} d y (= π \int_{0}^{h} (y^{2} + 1) d y)$ A1

$= π {[\frac{1}{3} y^{3} + y]}_{0}^{h}$ A1

$= π (\frac{1}{3} h^{3} + h)$ AG

Note: Award marks as appropriate for correct work using a different variable e.g. $π \int_{0}^{h} {(\sqrt{x^{2} + 1})}^{2} d x$

[3 marks]

c.i.

attempt to substitute $h = \sqrt{3} (= 1.732 \dots)$ into $V$ (M1)

$V = 10.8828 \dots$

$V = 10.9 (m^{3}) (= 2 \sqrt{3} π) (m^{3})$ A1

[2 marks]

c.ii.

time $= \frac{10.8828 \dots}{0.4} (= \frac{2 \sqrt{3} π}{0.4})$ (M1)

$= 27.207 \dots$

$= 27.2 (= 5 \sqrt{3} π) (s)$ A1

[2 marks]

attempt to find the height of the tank when $V = 5.4414 \dots (= \sqrt{3} π)$ (M1)

$π (\frac{1}{3} h^{3} + h) = 5.4414 \dots (= \sqrt{3} π)$

$h = 1.1818 \dots$ (A1)

attempt to use the chain rule or differentiate $V = π (\frac{1}{3} h^{3} + h)$ with respect to $t$ (M1)

$\frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} = \frac{1}{π (h^{2} + 1)} \times \frac{d V}{d t}$ OR $\frac{d V}{d t} = π (h^{2} + 1) \frac{d h}{d t}$ (A1)

attempt to substitute their $h$ and $\frac{d V}{d t} = 0.4$ (M1)

$\frac{d h}{d t} = \frac{0.4}{π (1.1818 \dots^{2} + 1)} = 0.053124 \dots$

$= 0.0531 ({m s}^{- 1})$ A1

[6 marks]

Examiners report

Part a) was generally well done, with the most common errors being to use an incorrect domain or not to give the coordinates of the endpoints. Some graphs appeared to be straight lines; some candidates drew sketches which were too small which made it more difficult for them to show the curvature.

Most candidates were able to show the steps to find an inverse function in part b), although occasionally a candidate did not explicitly swop the x and y variables before writing down the inverse function, which was given in the question. Many candidates struggled to identify the domain and range of the inverse, despite having a correct graph.

Part c) required a rotation around the y-axis, but a number of candidates attempted to rotate around the x-axis or failed to include limits. In the same vein, many substituted 2 into the formula instead of the square root of 3 when answering the second part. Many subsequently gained follow through marks on part d).

There were a number of good attempts at related rates in part e), with the majority differentiating V with respect to t, using implicit differentiation. However, many did not find the value of h which corresponded to halving the volume, and a number did not differentiate with respect to t, only with respect to h.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

The following diagram shows the curve $\frac{x^{2}}{36} + \frac{{(y - 4)}^{2}}{16} = 1$ , where $h \leq y \leq 4$ .

The curve from point $Q$ to point $B$ is rotated $360 °$ about the $y$ -axis to form the interior surface of a bowl. The rectangle $OPQR$ , of height $h cm$ , is rotated $360 °$ about the $y$ -axis to form a solid base.

The bowl is assumed to have negligible thickness.

Given that the interior volume of the bowl is to be $285 {cm}^{3}$ , determine the height of the base.

Markscheme

attempts to express $x^{2}$ in terms of $y$ (M1)

$V = π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y$ A1

Note: Correct limits are required.

Attempts to solve $π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y = 285$ for $h$ (M1)

Note: Award M1 for attempting to solve $36 π (\frac{h^{3}}{48} - \frac{h^{2}}{4} + \frac{8}{3}) = 285$ or equivalent for $h$ .

$h = 0.7926 \dots$

$h = 0.793$ (cm) A2

[5 marks]

Examiners report

This question was a struggle for many candidates. To start with, many candidates had difficulty understanding the diagram. Some candidates tried to include the base in their equation.

Because of this confusion, the question was poorly attempted. Some only received one mark for rearranging the equation to make $x^{2}$ the subject but were unable to set the correct definite integral with correct terminals. Again, many candidates tried to solve by hand instead of using their GDC. The correct answer was not seen that often.

Those candidates who recognised that the volume was around the y -axis and used their GDC to solve, usually achieved full marks for this question.

The function $f$ is defined by $f\left( x \right) = {\text{sec}}\,x + 2$ , $0 \leqslant x < \frac{\pi }{2}$ .

Use integration by parts to find $\int {{{\left( {{\text{ln}}\,x} \right)}^2}} {\text{d}}x$ .

Markscheme

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

METHOD 1

write as $\int {1 \times {{\left( {{\text{ln}}\,x} \right)}^2}} {\text{d}}x$ (M1)

$= x{\left( {{\text{ln}}\,x} \right)^2} - \int {x \times \frac{{2\left( {{\text{ln}}\,x} \right)}}{x}} {\text{d}}x\left( { = x{{\left( {{\text{ln}}\,x} \right)}^2} - \int {2\,{\text{ln}}\,x} } \right)$ M1A1

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + \int 2 \,{\text{d}}x$ (M1)(A1)

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + 2x + c$ A1

METHOD 2

let $u = {\text{ln}}\,x$ M1

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$

$\int {{u^2}{{\text{e}}^u}} {\text{d}}u$ A1

$= {u^2}{{\text{e}}^u} - \int {2u{{\text{e}}^u}} {\text{d}}u$ M1

$= {u^2}{{\text{e}}^u} - 2u{{\text{e}}^u} + \int {2{{\text{e}}^u}} {\text{d}}u$ A1

$= {u^2}{{\text{e}}^u} - 2u{{\text{e}}^u} + 2{{\text{e}}^u} + c$

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

METHOD 3

Setting up $u = {\text{ln}}\,x$ and $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {\text{ln}}\,x$ M1

${\text{ln}}\,x\left( {x\,{\text{ln}}\,x - x} \right) - \int {\left( {{\text{ln}}\,x - 1} \right)} {\text{d}}x$ M1A1

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

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + 2x + c$ A1

[6 marks]

Examiners report

[N/A]

Two airplanes, $A$ and $B$ , have position vectors with respect to an origin $O$ given respectively by

$r_{A} = (\begin{matrix} 19 \\ - 1 \\ 1 \end{matrix}) + t (\begin{matrix} - 6 \\ 2 \\ 4 \end{matrix})$

$r_{B} = (\begin{matrix} 1 \\ 0 \\ 12 \end{matrix}) + t (\begin{matrix} 4 \\ 2 \\ - 2 \end{matrix})$

where $t$ represents the time in minutes and $0 \leq t \leq 2.5$ .

Entries in each column vector give the displacement east of $O$ , the displacement north of $O$ and the distance above sea level, all measured in kilometres.

The two airplanes’ lines of flight cross at point $P$ .

Find the three-figure bearing on which airplane $B$ is travelling.

[2]

Show that airplane $A$ travels at a greater speed than airplane $B$ .

[2]

Find the acute angle between the two airplanes’ lines of flight. Give your answer in degrees.

[4]

Find the coordinates of $P$ .

[5]

d.i.

Determine the length of time between the first airplane arriving at $P$ and the second airplane arriving at $P$ .

[2]

d.ii.

Let $D (t)$ represent the distance between airplane $A$ and airplane $B$ for $0 \leq t \leq 2.5$ .

Find the minimum value of $D (t)$ .

[5]

Markscheme

let $ϕ$ be the required angle (bearing)

EITHER

$ϕ = 90 ° - arctan \frac{1}{2} (= arctan 2)$ (M1)

Note: Award M1 for a labelled sketch.

$\cos ϕ = \frac{(\begin{matrix} 0 \\ 1 \end{matrix}) \cdot (\begin{matrix} 4 \\ 2 \end{matrix})}{\sqrt{1} \times \sqrt{20}} (= 0.4472 \dots, = \frac{1}{\sqrt{5}})$ (M1)

$ϕ = arccos (0.4472 \dots)$

THEN

$063 °$ A1

Note: Do not accept $063.6 °$ or $63.4 °$ or $1 . 10^{c}$ .

[2 marks]

METHOD 1

let $|b_{A}|$ be the speed of $A$ and let $|b_{B}|$ be the speed of $B$

attempts to find the speed of one of $A$ or $B$ (M1)

$|b_{A}| = \sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}}$ or $|b_{B}| = \sqrt{4^{2} + 2^{2} + {(- 2)}^{2}}$

Note: Award M0 for $|b_{A}| = \sqrt{19^{2} + {(- 1)}^{2} + 1^{2}}$ and $|b_{B}| = \sqrt{1^{2} + 0^{2} + 12^{2}}$ .

$|b_{A}| = 7.48 \dots (= \sqrt{56})$ (km min^-1) and $|b_{B}| = 4.89 \dots (= \sqrt{24})$ (km min^-1) A1

$|b_{A}| > |b_{B}|$ so $A$ travels at a greater speed than $B$ AG

METHOD 2

attempts to use $speed = \frac{distance}{time}$

${speed}_{A} = \frac{|r_{A} (t_{2}) - r_{A} (t_{1})|}{t_{2} - t_{1}}$ and ${speed}_{B} = \frac{|r_{B} (t_{2}) - r_{B} (t_{1})|}{t_{2} - t_{1}}$ (M1)

for example:

${speed}_{A} = \frac{|r_{A} (1) - r_{A} (0)|}{1}$ and ${speed}_{B} = \frac{|r_{B} (1) - r_{B} (0)|}{1}$

${speed}_{A} = \frac{\sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}}}{1}$ and ${speed}_{B} = \frac{\sqrt{4^{2} + 2^{2} + 2^{2}}}{1}$

${speed}_{A} = 7.48 \dots (2 \sqrt{14})$ and ${speed}_{B} = 4.89 \dots (\sqrt{24})$ A1

${speed}_{A} > {speed}_{B}$ so $A$ travels at a greater speed than $B$ AG

[2 marks]

attempts to use the angle between two direction vectors formula (M1)

$\cos θ = \frac{(- 6) (4) + (2) (2) + (4) (- 2)}{\sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}} \sqrt{4^{2} + 2^{2} + {(- 2)}^{2}}}$ (A1)

$\cos θ = - 0.7637 \dots (= - \frac{7}{\sqrt{84}})$ or $θ = arccos (- 0.7637 \dots) (= 2.4399 \dots)$

attempts to find the acute angle $180 ° - θ$ using their value of $θ$ (M1)

$= 40.2 °$ A1

[4 marks]

for example, sets $r_{A} (t_{1}) = r_{B} (t_{2})$ and forms at least two equations (M1)

$19 - 6 t_{1} = 1 + 4 t_{2}$

$- 1 + 2 t_{1} = 2 t_{2}$

$1 + 4 t_{1} = 12 - 2 t_{2}$

Note: Award M0 for equations involving $t$ only.

EITHER

attempts to solve the system of equations for one of $t_{1}$ or $t_{2}$ (M1)

$t_{1} = 2$ or $t_{2} = \frac{3}{2}$ A1

attempts to solve the system of equations for $t_{1}$ and $t_{2}$ (M1)

$t_{1} = 2$ or $t_{2} = \frac{3}{2}$ A1

THEN

substitutes their $t_{1}$ or $t_{2}$ value into the corresponding $r_{A}$ or $r_{B}$ (M1)

$P (7, 3, 9)$ A1

Note: Accept $\vec{OP} = (\begin{matrix} 7 \\ 3 \\ 9 \end{matrix})$ . Accept $7$ km east of $O$ , $3$ km north of $O$ and $9$ km above sea level.

[5 marks]

d.i.

attempts to find the value of $t_{1} - t_{2}$ (M1)

$t_{1} - t_{2} = 2 - \frac{3}{2}$

$0.5$ minutes ( $30$ seconds) A1

[2 marks]

d.ii.

EITHER

attempts to find $r_{B} - r_{A}$ (M1)

$r_{B} - r_{A} = (\begin{matrix} - 18 \\ 1 \\ 11 \end{matrix}) + t (\begin{matrix} 10 \\ 0 \\ - 6 \end{matrix})$

attempts to find their $D (t)$ (M1)

$D (t) = \sqrt{{(10 t - 18)}^{2} + 1 + {(11 - 6 t)}^{2}}$ A1

attempts to find $r_{A} - r_{B}$ (M1)

$r_{A} - r_{B} = (\begin{matrix} 18 \\ - 1 \\ - 11 \end{matrix}) + t (\begin{matrix} - 10 \\ 0 \\ 6 \end{matrix})$

attempts to find their $D (t)$ (M1)

$D (t) = \sqrt{{(18 - 10 t)}^{2} + {(- 1)}^{2} + {(- 11 + 6 t)}^{2}}$ A1

Note: Award M0M0A0 for expressions using two different time parameters.

THEN

either attempts to find the local minimum point of $D (t)$ or attempts to find the value of $t$ such that $D' (t) = 0$ (or equivalent) (M1)

$t = 1.8088 \dots (= \frac{123}{68})$

$D (t) = 1.01459 \dots$

minimum value of $D (t)$ is $1.01 (= \frac{\sqrt{1190}}{34})$ (km) A1

Note: Award M0 for attempts at the shortest distance between two lines.

[5 marks]

Examiners report

General comment about this question: many candidates were not exposed to this setting of vectors question and were rather lost.

Part (a) Probably the least answered question on the whole paper. Many candidates left it blank, others tried using 3D vectors. Out of those who calculated the angle correctly, only a small percentage were able to provide the correct true bearing as a 3-digit figure.

Part (b) Well done by many candidates who used the direction vectors to calculate and compare the speeds. A number of candidates tried to use the average rate of change but mostly unsuccessfully.

Part (c) Most candidates used the correct vectors and the formula to obtain the obtuse angle. Then only some read the question properly to give the acute angle in degrees, as requested.

Part (d) Well done by many candidates who used two different parameters. They were able to solve and obtain two values for time, the difference in minutes and the correct point of intersection. A number of candidates only had one parameter, thus scoring no marks in part (d) (i). The frequent error in part (d)(ii) was providing incorrect units.

Part (e) Many correct answers were seen with an efficient way of setting the question and using their GDC to obtain the answer, graphically or numerically. Some gave time only instead of actually giving the minimal distance. A number of candidates tried to find the distance between two skew lines ignoring the fact that the lines intersect.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Xavier, the parachutist, jumps out of a plane at a height of $h$ metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, $v\,{\text{m}}{{\text{s}}^{ - 1}}$ , $t$ seconds after jumping from the plane, can be modelled by the function

$v(t) = \left\{ {\begin{array}{*{20}{l}} {9.8t{\text{,}}}&{0 \leqslant t \leqslant 10} \\ {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }},}&{t > 10} \end{array}} \right.$

His velocity when he reaches the ground is $2.8{\text{ m}}{{\text{s}}^{ - 1}}$ .

Find his velocity when $t = 15$ .

[2]

Calculate the vertical distance Xavier travelled in the first 10 seconds.

[2]

Determine the value of $h$ .

[5]

Markscheme

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

$v(15) = \frac{{98}}{{\sqrt {1 + {{(15 - 10)}^2}} }}$ (M1)

$v(15) = 19.2{\text{ }}({\text{m}}{{\text{s}}^{ - 1}})$ A1

[2 marks]

$\int\limits_0^{10} {9.8t\,{\text{d}}t}$ (M1)

$= 490{\text{ }}({\text{m}})$ A1

[2 marks]

$\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }} = 2.8$ (M1)

$t = 44.985 \ldots {\text{ }}({\text{s}})$ A1

$h = 490 + \int\limits_{10}^{44.9...} {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }}{\text{d}}t}$ (M1)(A1)

$h = 906{\text{ (m}})$ A1

[5 marks]

Examiners report

[N/A]

Consider $\lim_{x \to 0} \frac{arctan (\cos x) - k}{x^{2}}$ , where $k \in ℝ$ .

Show that a finite limit only exists for $k = \frac{π}{4}$ .

[2]

Using l’Hôpital’s rule, show algebraically that the value of the limit is $- \frac{1}{4}$ .

[6]

Markscheme

(as $\lim_{x \to 0} x^{2} = 0$ , the indeterminate form $\frac{0}{0}$ is required for the limit to exist)

$\Rightarrow \lim_{x \to 0} (arctan (\cos x) - k) = 0$ M1

$arctan 1 - k = 0 (k = arctan 1)$ A1

so $k = \frac{π}{4}$ AG

Note: Award M1A0 for using $k = \frac{π}{4}$ to show the limit is $\frac{0}{0}$ .

[2 marks]

$\lim_{x \to 0} \frac{arctan (\cos x) - \frac{π}{4}}{x^{2}} (= \frac{0}{0})$

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x}$ A1A1

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

recognises to apply l’Hôpital’s rule again (M1)

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x} (= \frac{0}{0})$

Note: Award M0 if their limit is not the indeterminate form $\frac{0}{0}$ .

EITHER

$= \lim_{x \to 0} \frac{\frac{- \cos x (1 + \cos^{2} x) - 2 \sin^{2} x \cos x}{{(1 + \cos^{2} x)}^{2}}}{2}$ A1A1

Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.

$\lim_{x \to 0} \frac{- \cos x}{2 (1 + \cos^{2} x) - 4 x \sin x \cos x}$ A1A1

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

THEN

substitutes $x = 0$ into the correct expression to evaluate the limit A1

Note: The final A1 is dependent on all previous marks.

$= - \frac{1}{4}$ AG

[6 marks]

Examiners report

Part (a) Many candidates recognised the indeterminate form and provided a nice algebraic proof. Some verified by substituting the given value. Therefore, there is a need to teach the candidates the difference between proof and verification. Only a few candidates were able to give a complete 'show that' proof.

Part (b) Many candidates realised that they needed to apply the L'Hôpital's rule twice. There were many mistakes in differentiation using the chain rule. Not all candidates clearly showed the final substitution.

[N/A]

A particle $P$ moves in a straight line such that after time $t$ seconds, its velocity, $v$ in ${m s}^{- 1}$ , is given by $v = e^{- 3 t} \sin 6 t$ , where $0 < t < \frac{π}{2}$ .

At time $t$ , $P$ has displacement $s (t)$ ; at time $t = 0$ , $s (0) = 0$ .

At successive times when the acceleration of $P$ is $0 {m s}^{- 2}$ , the velocities of $P$ form a geometric sequence. The acceleration of $P$ is zero at times $t_{1}, t_{2}, t_{3}$ where $t_{1} < t_{2} < t_{3}$ and the respective velocities are $v_{1}, v_{2}, v_{3}$ .

Find the times when $P$ comes to instantaneous rest.

[2]

Find an expression for $s$ in terms of $t$ .

[7]

Find the maximum displacement of $P$ , in metres, from its initial position.

[2]

Find the total distance travelled by $P$ in the first $1.5$ seconds of its motion.

[2]

Show that, at these times, $\tan 6 t = 2$ .

[2]

e.i.

Hence show that $\frac{v_{2}}{v_{1}} = \frac{v_{3}}{v_{2}} = - e^{- \frac{π}{2}}$ .

[5]

e.ii.

Markscheme

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

$\frac{π}{6} (= 0.524)$ A1

$\frac{π}{3} (= 1.05)$ A1

[2 marks]

attempt to use integration by parts M1

$s = \int e^{- 3 t} \sin 6 t d t$

EITHER

$= - \frac{e^{- 3 t} \sin 6 t}{3} - \int - 2 e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} - \int - 4 e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} + 4 s)$

$5 s = \frac{-3 e^{- 3 t} \sin 6 t - 6 e^{- 3 t} cos 6 t}{9}$ M1

$= - \frac{e^{- 3 t} cos 6 t}{6} - \int \frac{1}{2} e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \int \frac{1}{4} e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \frac{1}{4} s)$

$\frac{5}{4} s = \frac{- 2 e^{- 3 t} cos 6 t - e^{- 3 t} sin 6 t}{12}$ M1

THEN

$s = - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15} (+ c)$ A1

at $t = 0, s = 0 \Rightarrow 0 = - \frac{2}{15} + c$ M1

$c = \frac{2}{15}$ A1

$s = \frac{2}{15} - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15}$

[7 marks]

EITHER

substituting $t = \frac{π}{6}$ into their equation for $s$ (M1)

$(s = \frac{2}{15} - \frac{e^{- \frac{π}{2}} (sin π + 2 cos π)}{15})$

using GDC to find maximum value (M1)

evaluating $\int_{0}^{\frac{π}{6}} v d t$ (M1)

THEN

$= 0.161 (= \frac{2}{15} (1 + e^{- \frac{π}{2}}))$ A1

[2 marks]

METHOD 1

EITHER

distance required $= \int_{0}^{1.5} |e^{- 3 t} \sin 6 t| d t$ (M1)

distance required $= \int_{0}^{\frac{π}{6}} e^{- 3 t} \sin 6 t d t + |\int_{\frac{π}{6}}^{\frac{π}{3}} e^{- 3 t} \sin 6 t d t| + \int_{\frac{π}{3}}^{1.5} e^{- 3 t} \sin 6 t d t$ (M1)

$(= 0.16105 \dots + 0.033479 \dots + 0.006806 \dots)$

THEN

$= 0.201 (m)$ A1

METHOD 2

using successive minimum and maximum values on the displacement graph (M1)

$0.16105 \dots + (0.16105 \dots - 0.12757 \dots) + (0.13453 \dots - 0.12757 \dots)$

$= 0.201 (m)$ A1

[2 marks]

valid attempt to find $\frac{d v}{d t}$ using product rule and set $\frac{d v}{d t} = 0$ M1

$\frac{d v}{d t} = e^{- 3 t} 6 \cos 6 t - 3 e^{- 3 t} \sin 6 t$ A1

$\frac{d v}{d t} = 0 \Rightarrow \tan 6 t = 2$ AG

[2 marks]

e.i.

attempt to evaluate $t_{1}, t_{2}, t_{3}$ in exact form M1

$6 t_{1} = arctan 2 (\Rightarrow t_{1} = \frac{1}{6} arctan 2)$

$6 t_{2} = π + arctan 2 (\Rightarrow t_{2} = \frac{π}{6} + \frac{1}{6} arctan 2)$

$6 t_{3} = 2 π + arctan 2 (\Rightarrow t_{3} = \frac{π}{3} + \frac{1}{6} arctan 2)$ A1

Note: The A1 is for any two consecutive correct, or showing that $6 t_{2} = π + 6 t_{1}$ or $6 t_{3} = π + 6 t_{2}$ .

showing that $\sin 6 t_{n + 1} = - \sin 6 t_{n}$

eg $\tan 6 t = 2 \Rightarrow \sin 6 t = \pm \frac{2}{\sqrt{5}}$ M1A1

showing that $\frac{e^{- 3 t_{n +1}}}{e^{- 3 t_{n}}} = e^{- \frac{π}{2}}$ M1

eg $e^{- 3 (\frac{π}{6} + k)} \div e^{- 3 k} = e^{- \frac{π}{2}}$

Note: Award the A1 for any two consecutive terms.

$\frac{v_{3}}{v_{2}} = \frac{v_{2}}{v_{1}} = - e^{- \frac{π}{2}}$ AG

[5 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

The following diagram shows part of the graph of $2{x^2} = {\text{si}}{{\text{n}}^3}\,y$ for $0 \leqslant y \leqslant \pi$ .

The shaded region $R$ is the area bounded by the curve, the $y$ -axis and the lines $y = 0$ and $y = \pi$ .

Using implicit differentiation, find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .

[4]

a.i.

Find the equation of the tangent to the curve at the point $\left( {\frac{1}{4}{\text{, }}\frac{{5\pi }}{6}} \right)$ .

[4]

a.ii.

Find the area of $R$ .

[3]

The region $R$ is now rotated about the $y$ -axis, through $2\pi$ radians, to form a solid.

By writing ${{\text{si}}{{\text{n}}^3}\,y}$ as $\left( {1 - {\text{co}}{{\text{s}}^2}\,y} \right){\text{sin}}\,y$ , show that the volume of the solid formed is $\frac{{2\pi }}{3}$ .

[6]

Markscheme

valid attempt to differentiate implicitly (M1)

$4x = 3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y\frac{{{\text{d}}y}}{{{\text{d}}x}}$ A1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4x}}{{3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y}}$ A1

[4 marks]

a.i.

at $\left( {\frac{1}{4}{\text{, }}\frac{{5\pi }}{6}} \right){\text{, }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4x}}{{3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y}} = \frac{1}{{3{{\left( {\frac{1}{2}} \right)}^2}\left( { - \frac{{\sqrt 3 }}{2}} \right)}}$ (M1)

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{8}{{3\sqrt 3 }}\left( { = - 1.54} \right)$ A1

hence equation of tangent is

$y - \frac{{5\pi }}{6} = - 1.54\left( {x - \frac{1}{4}} \right)$ OR $y = - 1.54x + 3.00$ (M1)A1

Note: Accept $y = - 1.54x + 3$ .

[4 marks]

a.ii.

$x = \sqrt {\frac{1}{2}{\text{si}}{{\text{n}}^3}\,y}$ (M1)

$\int_0^\pi {\sqrt {\frac{1}{2}{\text{si}}{{\text{n}}^3}\,y\,{\text{d}}y} }$ (A1)

$= 1.24$ A1

[3 marks]

use of volume $= \int {\pi {x^2}} \,{\text{d}}y$ (M1)

$= \int_0^\pi {\frac{1}{2}} \pi \,{\text{si}}{{\text{n}}^3}\,y\,{\text{d}}y$ A1

$= \frac{1}{2}\pi \int_0^\pi {\left( {{\text{sin}}\,y - {\text{sin}}\,y\,{\text{co}}{{\text{s}}^2}\,y} \right){\text{d}}y}$

Note: Condone absence of limits up to this point.

reasonable attempt to integrate (M1)

$= \frac{1}{2}\pi \left[ { - {\text{cos}}\,y + \frac{1}{3}{\text{co}}{{\text{s}}^3}\,y} \right]_0^\pi$ A1A1

Note: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.

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

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

Note: Do not accept decimal answer equivalent to $\frac{{2\pi }}{3}$ .

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The following diagram shows part of the graph of $y = p + q \sin (r x)$ . The graph has a local maximum point at $(- \frac{9 π}{4}, 5)$ and a local minimum point at $(- \frac{3 π}{4}, - 1)$ .

Determine the values of $p$ , $q$ and $r$ .

[4]

Hence find the area of the shaded region.

[4]

Markscheme

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

the principal axis is $\frac{5 + (- 1)}{2} (= 2)$

so $p = 2$ A1

the amplitude is $\frac{5 - (- 1)}{2} (= 3)$

so $q = 3$ A1

EITHER

one period is $2 (- \frac{3 π}{4} - (- \frac{9 π}{4}))$ (M1)

$= 3 π$

$\Rightarrow \frac{2 π}{r} = 3 π$

Substituting a point eg $- 1 = 2 + \sin (- \frac{3 π}{4} r)$

$\sin (- \frac{3 π}{4} r) = - 1 \Rightarrow - \frac{3 π}{4} r = \dots - \frac{5 π}{2}, - \frac{π}{2}, \frac{3 π}{2}, \dots$

Choice of correct solution $- \frac{3 π}{4} r = - \frac{π}{2}$ (M1)

THEN

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

$(\Rightarrow y = 2 + 3 \sin (\frac{2 x}{3}))$

Note: $q$ and $r$ can be both given as negatives for full marks

[4 marks]

roots are $x = - 1.09459 \dots, x = - 3.617797 \dots$ (A1)

$\int_{- 3.617797 \dots}^{- 1.09459 \dots} (2 + 3 \sin (\frac{2 x}{3})) d x$ (M1)

$= - 1.66 (= - 1.66179 \dots)$ (A1)

so area $= 1.66 ({units}^{2})$ A1

[4 marks]

Examiners report

[N/A]

A function $f$ satisfies the conditions $f\left( 0 \right) = - 4$ , $f\left( 1 \right) = 0$ and its second derivative is $f''\left( x \right) = 15\sqrt x + \frac{1}{{{{\left( {x + 1} \right)}^2}}}$ , $x$ ≥ 0.

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

Markscheme

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

$f'\left( x \right) = \int {\left( {15\sqrt x + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)} \,{\text{d}}x = 10{x^{\frac{3}{2}}} - \frac{1}{{x + 1}}\left( { + c} \right)$ (M1)A1A1

Note: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.

$f\left( x \right) = \int {\left( {10{x^{\frac{3}{2}}} - \frac{1}{{x + 1}} + c} \right)} \,{\text{d}}x = 4{x^{\frac{5}{2}}} - {\text{ln}}\left( {x + 1} \right) + cx + d$ A1

Note: Allow FT from incorrect $f'\left( x \right)$ if it is of the form $f'\left( x \right) = A{x^{\frac{3}{2}}} + \frac{B}{{x + 1}} + c$ .

Accept ${\text{ln}}\left| {x + 1} \right|$ .

attempt to use at least one boundary condition in their $f\left( x \right)$ (M1)

$x = 0$ , $y = - 4$

⇒ $d = - 4$ A1

$x = 1$ , $y = 0$

⇒ $0 = 4 - {\text{ln}}\,2 + c - 4$

⇒ $c = {\text{ln}}\,2\left( { = 0.693} \right)$ A1

$f\left( x \right) = 4{x^{\frac{5}{2}}} - {\text{ln}}\left( {x + 1} \right) + x\,{\text{ln}}\,2 - 4$

[7 marks]

Examiners report

[N/A]

A point P moves in a straight line with velocity $v$ ms⁻¹ given by $v\left( t \right) = {{\text{e}}^{ - t}} - 8{t^2}{{\text{e}}^{ - 2t}}$ at time t seconds, where t ≥ 0.

Determine the first time t₁ at which P has zero velocity.

[2]

Find an expression for the acceleration of P at time t.

[2]

b.i.

Find the value of the acceleration of P at time t₁.

[1]

b.ii.

Markscheme

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

attempt to solve $v\left( t \right) = 0$ for t or equivalent (M1)

t₁ = 0.441(s) A1

[2 marks]

$a\left( t \right) = \frac{{{\text{d}}v}}{{{\text{d}}t}} = - {{\text{e}}^{ - t}} - 16t{{\text{e}}^{ - 2t}} + 16{t^2}{{\text{e}}^{ - 2t}}$ M1A1

Note: Award M1 for attempting to differentiate using the product rule.

[2 marks]

b.i.

$a\left( {{t_1}} \right) = - 2.28$ (ms⁻²) A1

[1 mark]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Let $l$ be the tangent to the curve $y = x{{\text{e}}^{2x}}$ at the point (1, ${{\text{e}}^2}$ ).

Find the coordinates of the point where $l$ meets the $x$ -axis.

Markscheme

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

METHOD 1

equation of tangent is $y = 22.167 \ldots x - 14.778 \ldots$ OR $y = - 7.389 \ldots = 22.167 \ldots \left( {x - 1} \right)$ (M1)(A1)

meets the $x$ -axis when $y = 0$

$x = 0.667$

meets $x$ -axis at (0.667, 0) $\left( { = \left( {\frac{2}{3},\,\,0} \right)} \right)$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

METHOD 1

Attempt to differentiate (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{2x}} + 2x{{\text{e}}^{2x}}$

when $x = 1$ , $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{{\text{e}}^2}$ (M1)

equation of the tangent is $y - {{\text{e}}^2} = 3{{\text{e}}^2}\left( {x - 1} \right)$

$y = 3{{\text{e}}^2}x - 2{{\text{e}}^2}$

meets $x$ -axis at $x = \frac{2}{3}$

${\left( {\frac{2}{3},\,\,0} \right)}$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

[4 marks]

Examiners report

[N/A]

A body moves in a straight line such that its velocity, $v\,{\text{m}}{{\text{s}}^{ - 1}}$ , after $t$ seconds is given by $v = 2\,{\text{sin}}\left( {\frac{t}{{10}} + \frac{\pi }{5}} \right)\csc \left( {\frac{t}{{30}} + \frac{\pi }{4}} \right)$ for $0 \leqslant t \leqslant 60$ .

The following diagram shows the graph of $v$ against $t$ . Point ${\text{A}}$ is a local maximum and point ${\text{B}}$ is a local minimum.

The body first comes to rest at time $t = {t_1}$ . Find

Determine the coordinates of point ${\text{A}}$ and the coordinates of point ${\text{B}}$ .

[4]

a.i.

Hence, write down the maximum speed of the body.

[1]

a.ii.

the value of ${t_1}$ .

[2]

b.i.

the distance travelled between $t = 0$ and $t = {t_1}$ .

[2]

b.ii.

the acceleration when $t = {t_1}$ .

[2]

b.iii.

Find the distance travelled in the first 30 seconds.

[3]

Markscheme

${\text{A}}\left( {7.47{\text{, }}2.28} \right)$ and ${\text{B}}\left( {43.4{\text{,}} - 2.45} \right)$ A1A1A1A1

[4 marks]

a.i.

maximum speed is $2.45\,\left( {{\text{m}}{{\text{s}}^{ - 1}}} \right)$ A1

[1 mark]

a.ii.

$v = 0 \Rightarrow {t_1} = 25.1\,\left( {\text{s}} \right)$ (M1)A1

[2 marks]

b.i.

$\int_0^{{t_1}} {v\,{\text{d}}t}$ (M1)

$= 41.0\,\left( {\text{m}} \right)$ A1

[2 marks]

b.ii.

$a = \frac{{{\text{d}}v}}{{{\text{d}}t}}$ at $t = {t_1} = 25.1$ (M1)

$a = - 0.200\,\,\left( {{\text{m}}{{\text{s}}^{ - 2}}} \right)$ A1

Note: Accept $a = - 0.2$ .

[2 marks]

b.iii.

attempt to integrate between 0 and 30 (M1)

Note: An unsupported answer of 38.6 can imply integrating from 0 to 30.

EITHER

$\int_0^{30} {\left| v \right|} \,{\text{d}}t$ (A1)

$41.0 - \int_{{t_1}}^{30} {v\,{\text{d}}t}$ (A1)

THEN

$= 43.3\,\left( {\text{m}} \right)$ A1

[3 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]

A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is $\theta$ radians.

M17/5/MATHL/HP2/ENG/TZ1/08

The volume of water is increasing at a constant rate of $0.0008{\text{ }}{{\text{m}}^3}{{\text{s}}^{ - 1}}$ .

Find an expression for the volume of water $V{\text{ }}({{\text{m}}^3})$ in the trough in terms of $\theta$ .

[3]

Calculate $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ when $\theta = \frac{\pi }{3}$ .

[4]

Markscheme

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

area of segment $= \frac{1}{2} \times {0.5^2} \times (\theta - \sin \theta )$ M1A1

$V = {\text{area of segment}} \times 10$

$V = \frac{5}{4}(\theta - \sin \theta )$ A1

[3 marks]

METHOD 1

$\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{5}{4}(1 - \cos \theta )\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ M1A1

$0.0008 = \frac{5}{4}\left( {1 - \cos \frac{\pi }{3}} \right)\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128{\text{ }}({\text{rad}}\,{s^{ - 1}})$ A1

METHOD 2

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

$\frac{{{\text{d}}V}}{{{\text{d}}\theta }} = \frac{5}{4}(1 - \cos \theta )$ A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{4 \times 0.0008}}{{5\left( {1 - \cos \frac{\pi }{3}} \right)}}$ (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128\left( {\frac{4}{{3125}}} \right)({\text{rad }}{s^{ - 1}})$ A1

[4 marks]

Examiners report

[N/A]

The function $f$ is defined by $f (x) = \frac{3 x + 2}{4 x^{2} - 1}$ , for $x \in ℝ$ , $x \neq p$ , $x \neq q$ .

The graph of $y = f (x)$ has exactly one point of inflexion.

The function $g$ is defined by $g (x) = \frac{4 x^{2} - 1}{3 x + 2}$ , for $x \in ℝ, x \neq - \frac{2}{3}$ .

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

[2]

Find an expression for $f' (x)$ .

[3]

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

[2]

Sketch the graph of $y = f (x)$ for $- 3 \leq x \leq 3$ , showing the values of any axes intercepts, the coordinates of any local maxima and local minima, and giving the equations of any asymptotes.

[5]

Find the equations of all the asymptotes on the graph of $y = g (x)$ .

[4]

By considering the graph of $y = g (x) - f (x)$ , or otherwise, solve $f (x) < g (x)$ for $x \in ℝ$ .

[4]

Markscheme

attempt to solve $4 x^{2} - 1 = 0$ e.g. by factorising $4 x^{2} - 1$ (M1)

$p = \frac{1}{2}, q = - \frac{1}{2}$ or vice versa A1

[2 marks]

attempt to use quotient rule or product rule (M1)

EITHER

$f' (x) = \frac{3 (4 x^{2} - 1) - 8 x (3 x + 2)}{{(4 x^{2} - 1)}^{2}} (= \frac{- 12 x^{2} - 16 x - 3}{{(4 x^{2} - 1)}^{2}})$ A1A1

Note: Award A1 for each term in the numerator with correct signs, provided correct denominator is seen.

$f' (x) = - 8 x (3 x + 2) {(4 x^{2} - 1)}^{- 2} + 3 {(4 x^{2} - 1)}^{- 1}$ A1A1

Note: Award A1 for each term.

[3 marks]

attempt to find the local min point on $y = f' (x)$ OR solve $f'' (x) = 0$ (M1)

$x = - 1.60$ A1

[2 marks]

A1A1A1A1A1

Note: Award A1 for both vertical asymptotes with their equations, award A1 for horizontal asymptote with equation, award A1 for each correct branch including asymptotic behaviour, coordinates of minimum and maximum points (may be seen next to the graph) and values of axes intercepts.
If vertical asymptotes are absent (or not vertical) and the branches overlap as a consequence, award maximum A0A1A0A1A1.

[5 marks]

$x = - \frac{2}{3} (= - 0.667)$ A1

(oblique asymptote has) gradient $\frac{4}{3} (= 1.33)$ (A1)

appropriate method to find complete equation of oblique asymptote M1

$3 x + 2 \overset{\frac{4}{3} x - \frac{8}{9}}{4 x^{2} + 0 x - 1}$

$\frac{4 x^{2} + \frac{8}{3} x}{- \frac{8}{3} x - 1}$

$- \frac{8}{3} x - \frac{16}{\frac{9}{\frac{7}{9}}}$

$y = \frac{4}{3} x - \frac{8}{9} (= 1.33 x - 0.889)$ A1

Note: Do not award the final A1 if the answer is not given as an equation.

[4 marks]

attempting to find at least one critical value $(x = - 0.568729 \dots, x = 1.31872 \dots)$ (M1)

$- \frac{2}{3} < x < - 0.569$ OR $- 0.5 < x < 0.5$ OR $x > 1.32$ A1A1A1

Note: Only penalize once for use of $\leq$ rather than $<$ .

[4 marks]

Examiners report

[N/A]

The region $A$ is enclosed by the graph of $y = 2\arcsin (x - 1) - \frac{\pi }{4}$ , the $y$ -axis and the line $y = \frac{\pi }{4}$ .

Write down a definite integral to represent the area of $A$ .

[4]

Calculate the area of $A$ .

[2]

Markscheme

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

METHOD 1

$2\arcsin (x - 1) - \frac{\pi }{4} = \frac{\pi }{4}$ (M1)

$x = 1 + \frac{1}{{\sqrt 2 }}\,\,\,( = 1.707 \ldots )$ (A1)

$\int\limits_0^{1 + \frac{1}{{\sqrt 2 }}} {\frac{\pi }{4} - \left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx}$ M1A1

Note: Award M1 for an attempt to find the difference between two functions, A1 for all correct.

METHOD 2

when $x = 0,{\text{ }}y = \frac{{ - 5\pi }}{4}\,\,\,( = - 3.93)$ A1

$x = 1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)$ M1A1

Note: Award M1 for an attempt to find the inverse function.

$\int_{\frac{{ - 5\pi }}{4}}^{\frac{\pi }{4}} {\left( {1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)} \right){\text{d}}y}$ A1

METHOD 3

$\int_0^{1.38...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right){\text{d}}x} \left| + \right.\int\limits_0^{1.71...} {\frac{\pi }{4}{\text{d}}x - \int\limits_{1.38...}^{1.71...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx} }$ M1A1A1A1

Note: Award M1 for considering the area below the $x$ -axis and above the $x$ -axis and A1 for each correct integral.

[4 marks]

${\text{area}} = 3.30{\text{ (square units)}}$ A2

[2 marks]

Examiners report

[N/A]

The following graph shows the two parts of the curve defined by the equation ${x^2}y = 5 - {y^4}$ , and the normal to the curve at the point P(2 , 1).

Show that there are exactly two points on the curve where the gradient is zero.

[7]

Find the equation of the normal to the curve at the point P.

[5]

The normal at P cuts the curve again at the point Q. Find the $x$ -coordinate of Q.

[3]

The shaded region is rotated by 2 $\pi$ about the $y$ -axis. Find the volume of the solid formed.

[7]

Markscheme

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

differentiating implicitly: M1

$2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}$ A1A1

Note: Award A1 for each side.

if $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ then either $x = 0$ or $y = 0$ M1A1

$x = 0 \Rightarrow$ two solutions for $y\left( {y = \pm \sqrt[4]{5}} \right)$ R1

$y = 0$ not possible (as 0 ≠ 5) R1

hence exactly two points AG

Note: For a solution that only refers to the graph giving two solutions at $x = 0$ and no solutions for $y = 0$ award R1 only.

[7 marks]

at (2, 1) $4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 4\frac{{{\text{d}}y}}{{{\text{d}}x}}$ M1

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

gradient of normal is 2 M1

1 = 4 + c (M1)

equation of normal is $y = 2x - 3$ A1

[5 marks]

substituting (M1)

${x^2}\left( {2x - 3} \right) = 5 - {\left( {2x - 3} \right)^4}$ or ${\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 - {y^4}$ (A1)

$x = 0.724$ A1

[3 marks]

recognition of two volumes (M1)

volume $1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 - {y^4}}}{y}} {\text{d}}y\left( { = 101\pi = 3.178 \ldots } \right)$ M1A1A1

Note: Award M1 for attempt to use $\pi \int {{x^2}} {\text{d}}y$ , A1 for limits, A1 for ${\frac{{5 - {y^4}}}{y}}$ Condone omission of $\pi$ at this stage.

volume 2

EITHER

$= \frac{1}{3}\pi \times {2^2} \times 4\left( { = 16.75 \ldots } \right)$ (M1)(A1)

$= \pi \int_{ - 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)$ (M1)(A1)

THEN

total volume = 19.9 A1

[7 marks]

Examiners report

[N/A]

The function $f$ has a derivative given by $f' (x) = \frac{1}{x (k - x)}, x \in ℝ, x \neq o, x \neq k$ where $k$ is a positive constant.

Consider $P$ , the population of a colony of ants, which has an initial value of $1200$ .

The rate of change of the population can be modelled by the differential equation $\frac{d P}{d t} = \frac{P (k - P)}{5 k}$ , where $t$ is the time measured in days, $t \geq 0$ , and $k$ is the upper bound for the population.

At $t = 10$ the population of the colony has doubled in size from its initial value.

The expression for $f' (x)$ can be written in the form $\frac{a}{x} + \frac{b}{k - x}$ , where $a, b \in ℝ$ . Find $a$ and $b$ in terms of $k$ .

[3]

Hence, find an expression for $f (x)$ .

[3]

By solving the differential equation, show that $P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ .

[8]

Find the value of $k$ , giving your answer correct to four significant figures.

[3]

Find the value of $t$ when the rate of change of the population is at its maximum.

[3]

Markscheme

$\frac{1}{x (k - x)} \equiv \frac{a}{x} + \frac{b}{k - x}$

$a (k - x) + b x = 1$ (A1)

attempt to compare coefficients OR substitute $x = k$ and $x = 0$ and solve (M1)

$a = \frac{1}{k}$ and $b = \frac{1}{k}$ A1

$f' (x) = \frac{1}{k x} + \frac{1}{k (k - x)}$

[3 marks]

attempt to integrate their $\frac{a}{x} + \frac{b}{k - x}$ (M1)

$f (x) \frac{1}{k} \int (\frac{1}{x} + \frac{1}{k - x}) d x$

$= \frac{1}{k} (\ln |x| - \ln |k - x|) (+ c) (= \frac{1}{k} \ln |\frac{x}{k - x}| (+ c))$ A1A1

Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of $+ c$ .

[3 marks]

attempt to separate variables and integrate both sides M1

$5 k \int \frac{1}{P (k - P)} d P = \int 1 d t$

$5 (\ln P - \ln (k - P)) = t + c$ A1

Note: There are variations on this which should be accepted, such as $\frac{1}{k} (\ln P - \ln (k - P)) = \frac{1}{5 k} t + c$ . Subsequent marks for these variations should be awarded as appropriate.

EITHER

attempt to substitute $t = 0, P = 1200$ into an equation involving $c$ M1

$c = 5 (\ln 1200 - \ln (k - 1200)) (= 5 \ln (\frac{1200}{k - 1200}))$ A1

$5 (\ln P - \ln (k - P)) = t + 5 (\ln 1200 - \ln (k - 1200))$ A1

$\ln (\frac{P (k - 1200)}{1200 (k - P)}) = \frac{t}{5}$

$\frac{P (k - 1200)}{1200 (k - P)} = e^{\frac{t}{5}}$ A1

$\ln (\frac{P}{k - P}) = \frac{t + c}{5}$

$\frac{P}{k - P} = A e^{\frac{t}{5}}$ A1

attempt to substitute $t = 0, P = 1200$ M1

$\frac{1200}{k - 1200} = A$ A1

$\frac{P}{k - P} = \frac{1200 e^{\frac{t}{5}}}{k - 1200}$ A1

THEN

attempt to rearrange and isolate $P$ M1

$P k - 1200 P = 1200 k e^{\frac{t}{5}} - 1200 P e^{\frac{t}{5}}$ OR $P k e^{- \frac{t}{5}} - 1200 P e^{- \frac{t}{5}} = 1200 k - 1200 P$ OR $\frac{k}{P} - 1 = \frac{k - 1200}{1200 e^{\frac{t}{5}}}$

$P (k - 1200 + 1200 e^{\frac{t}{5}}) = 1200 k e^{\frac{t}{5}}$ OR $P (k e^{- \frac{t}{5}} - 1200 e^{- \frac{t}{5}} + 1200) = 1200 k$ A1

$P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ AG

[8 marks]

attempt to substitute $t = 10, P = 2400$ (M1)

$2400 = \frac{1200 k}{(k - 1200) e^{- 2} + 1200}$ (A1)

$k = 2845.34 \dots$

$k = 2845$ A1

Note: Award (M1)(A1)A0 for any other value of $k$ which rounds to $2850$

[3 marks]

attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that $P = \frac{k}{2} (= 1422.67 \dots)$ (M1)

$t = 1.57814 \dots$

$= 1.58$ (days) A2

Note: Accept any value which rounds to $1.6$ .

[3 marks]

Examiners report

[N/A]

Write down the first three terms of the binomial expansion of $(1 + t)^{- 1}$ in ascending powers of $t$ .

[1]

By using the Maclaurin series for $\cos x$ and the result from part (a), show that the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ .

[4]

By using the Maclaurin series for $arctan x$ and the result from part (b), find $\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1})$ .

[3]

Markscheme

$1 - t + t^{2}$ A1

Note: Accept $1, - t$ and $t^{2}$ .

[1 mark]

$sec x = \frac{1}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)} (= {(1 - \frac{x^{2}}{2!} + (\frac{x^{4}}{4!} (- \dots)))}^{- 1})$ (M1)

$t = \cos x - 1$ or $sec x = 1 - (\cos x - 1) + {(\cos x - 1)}^{2}$ (M1)

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

$= 1 + \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{4}}{4}$ A1

so the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ AG

Note: Condone the absence of ‘…’

[4 marks]

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

$\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1}) = \lim_{x \to 0} (\frac{x (2 x - \frac{{(2 x)}^{3}}{3} + \dots)}{(1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}) - 1})$ M1

$= \lim_{x \to 0} (\frac{2 x^{2} - \frac{8 x^{4}}{3} + \dots}{\frac{x^{2}}{2} + \frac{5 x^{4}}{24}})$ A1

$= \lim_{x \to 0} (\frac{2 x^{2} (1 - \frac{4 x^{2}}{3})}{\frac{x^{2}}{2} (1 + \frac{5 x^{2}}{12})})$

$= 4$ A1

Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless $x$ is replaced by $2 x$ in $arctan$ .

[3 marks]

Examiners report

[N/A]

A function $f$ is defined by $f (x) = \frac{k e^{\frac{x}{2}}}{1 + e^{x}}$ , where $x \in ℝ, x \geq 0$ and $k \in ℝ^{+}$ .

The region enclosed by the graph of $y = f (x)$ , the $x$ -axis, the $y$ -axis and the line $x = \ln 16$ is rotated $360 °$ about the $x$ -axis to form a solid of revolution.

Pedro wants to make a small bowl with a volume of $300 {cm}^{3}$ based on the result from part (a). Pedro’s design is shown in the following diagrams.

The vertical height of the bowl, $BO$ , is measured along the $x$ -axis. The radius of the bowl’s top is $OA$ and the radius of the bowl’s base is $BC$ . All lengths are measured in $cm$ .

For design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.

Show that the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units.

[6]

Find the value of $k$ that satisfies the requirements of Pedro’s design.

[2]

Find $OA$ .

[2]

c.i.

Find $BC$ .

[2]

c.ii.

By sketching the graph of a suitable derivative of $f$ , find where the cross-sectional radius of the bowl is decreasing most rapidly.

[4]

d.i.

State the cross-sectional radius of the bowl at this point.

[2]

d.ii.

Markscheme

attempt to use $V = π \int_{a}^{b} {(f (x))}^{2} d x$ (M1)

$V = π \int_{0}^{\ln 16} {(\frac{k e^{\frac{x}{2}}}{1 + e^{x}})}^{2} d x (V = k^{2} π \int_{0}^{\ln 16} \frac{e^{x}}{{(1 + e^{x})}^{2}} d x)$

EITHER

applying integration by recognition (M1)

$= k^{2} π {[- \frac{1}{1 + e^{x}}]}_{0}^{\ln 16}$ A3

$u = 1 + e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 2$ and when $x = \ln 16, u = 17$

$V = k^{2} π \int_{2}^{17} \frac{1}{u^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{u}]}_{2}^{17}$ A1

$u = e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 1$ and when $x = \ln 16, u = 16$

$V = k^{2} π \int_{1}^{16} \frac{1}{{(1 + u)}^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{1 + u}]}_{1}^{16}$ A1

Note: Accept equivalent working with indefinite integrals and original limits for $x$ .

THEN

$= k^{2} π (\frac{1}{2} - \frac{1}{17})$ A1

so the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units AG

Note: Award (M1)(A0)(M0)(A0)(A0)(A1) when $\frac{15}{34}$ is obtained from GDC

[6 marks]

a valid algebraic or graphical attempt to find $k$ (M1)

$k^{2} = \frac{300 \times 34}{15 π}$

$k = 14.7 (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (as $k \in ℝ^{+}$ ) A1

Note: Candidates may use their GDC numerical solve feature.

[2 marks]

attempting to find $OA = f (0) = \frac{k}{2}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$OA = 7.36 (= \sqrt{\frac{170}{π}})$ A1

[2 marks]

c.i.

attempting to find $BC = f (\ln 16) = \frac{4 k}{17}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$BC = 3.46 (= \frac{8}{17} \sqrt{\frac{170}{π}} = \frac{8 \sqrt{10}}{\sqrt{17 π}})$ A1

[2 marks]

c.ii.

EITHER

recognising to graph $y = f' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f' (x) = \frac{k e^{\frac{x}{2}} (1 - e^{x})}{2 {(1 + e^{x})}^{2}}$

for $x > 0$ graph decreasing to the local minimum A1

before increasing towards the $x$ -axis A1

recognising to graph $y = f'' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f'' (x) = \frac{k e^{\frac{x}{2}} (e^{2 x} - 6 e^{x} +1)}{4 {(1 + e^{x})}^{3}}$

for $x > 0$ , graph increasing towards and beyond the $x$ -intercept A1

recognising $f'' (x) = 0$ for maximum rate (A1)

THEN

$x = 1.76 (= \ln (2 \sqrt{2} + 3))$ A1

Note: Only award A marks if either graph is seen.

[4 marks]

d.i.

attempting to find $f (1.76 \dots)$ (M1)

the cross-sectional radius at this point is $5.20 (\sqrt{\frac{85}{π}}) (cm)$ A1

[2 marks]

d.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

The function $f$ is defined by $f\left( x \right) = {\left( {x - 1} \right)^2}$ , $x$ ≥ 1 and the function $g$ is defined by $g\left( x \right) = {x^2} + 1$ , $x$ ≥ 0.

The region $R$ is bounded by the curves $y = f\left( x \right)$ , $y = g\left( x \right)$ and the lines $y = 0$ , $x = 0$ and $y = 9$ as shown on the following diagram.

The shape of a clay vase can be modelled by rotating the region $R$ through 360˚ about the $y$ -axis.

Find the volume of clay used to make the vase.

Markscheme

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

volume $= \pi {\int_0^9 {\left( {{y^{\frac{1}{2}}} + 1} \right)} ^2}{\text{d}}y - \pi \int_1^9 {\left( {y - 1} \right)} {\text{d}}y$ (M1)(M1)(M1)(A1)(A1)

Note: Award (M1) for use of formula for rotating about $y$ -axis, (M1) for finding at least one inverse, (M1) for subtracting volumes, (A1)(A1)for each correct expression, including limits.

$= 268.6 \ldots - 100.5 \ldots \left( {85.5\pi - 32\pi } \right)$

$= 168\left( { = 53.5\pi } \right)$ A2

[7 marks]

Examiners report

[N/A]

Consider the function $f(x) = \frac{{\sqrt x }}{{\sin x}},{\text{ }}0 < x < \pi$ .

Consider the region bounded by the curve $y = f(x)$ , the $x$ -axis and the lines $x = \frac{\pi }{6},{\text{ }}x = \frac{\pi }{3}$ .

Show that the $x$ -coordinate of the minimum point on the curve $y = f(x)$ satisfies the equation $\tan x = 2x$ .

[5]

a.i.

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

[2]

a.ii.

Sketch the graph of $y = f(x)$ showing clearly the minimum point and any asymptotic behaviour.

[3]

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y = - x$ .

[4]

This region is now rotated through $2\pi$ radians about the $x$ -axis. Find the volume of revolution.

[3]

Markscheme

attempt to use quotient rule or product rule M1

$f’(x) = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}{\text{ }}\left( { = \frac{1}{{2\sqrt x \sin x}} - \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}} \right)$ A1A1

Note: Award A1 for $\frac{1}{{2\sqrt x \sin x}}$ or equivalent and A1 for $- \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}$ or equivalent.

setting $f’(x) = 0$ M1

$\frac{{\sin x}}{{2\sqrt x }} - \sqrt x \cos x = 0$

$\frac{{\sin x}}{{2\sqrt x }} = \sqrt x \cos x$ or equivalent A1

$\tan x = 2x$ AG

[5 marks]

a.i.

$x = 1.17$

$0 < x \leqslant 1.17$ A1A1

Note: Award A1 for $0 < x$ and A1 for $x \leqslant 1.17$ . Accept $x < 1.17$ .

[2 marks]

a.ii.

N17/5/MATHL/HP2/ENG/TZ0/10.b/M

concave up curve over correct domain with one minimum point above the $x$ -axis. A1

approaches $x = 0$ asymptotically A1

approaches $x = \pi$ asymptotically A1

Note: For the final A1 an asymptote must be seen, and $\pi$ must be seen on the $x$ -axis or in an equation.

[3 marks]

$f’(x){\text{ }}\left( { = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}} \right) = 1$ (A1)

attempt to solve for $x$ (M1)

$x = 1.96$ A1

$y = f(1.96 \ldots )$

$= 1.51$ A1

[4 marks]

$V = \pi \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{x{\text{d}}x}}{{{{\sin }^2}x}}}$ (M1)(A1)

Note: M1 is for an integral of the correct squared function (with or without limits and/or $\pi$ ).

$= 2.68{\text{ }}( = 0.852\pi )$ A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the differential equation

$\frac{d y}{d x} = f (\frac{y}{x}), x > 0$

The curve $y = f (x)$ for $x > 0$ has a gradient function given by

$\frac{d y}{d x} = \frac{y^{2} + 3 x y + 2 x^{2}}{x^{2}}$ .

The curve passes through the point $(1, - 1)$ .

Use the substitution $y = v x$ to show that $\int \frac{d v}{f (v) - v} = \ln x + C$ where $C$ is an arbitrary constant.

[3]

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation $y = x (\tan (\ln x) - 1)$ .

[9]

The curve has a point of inflexion at $(x_{1}, y_{1})$ where $e^{- \frac{π}{2}} < x_{1} < e^{\frac{π}{2}}$ . Determine the coordinates of this point of inflexion.

[6]

Use the differential equation $\frac{d y}{d x} = \frac{y^{2} + 3 x y + 2 x^{2}}{x^{2}}$ to show that the points of zero gradient on the curve lie on two straight lines of the form $y = m x$ where the values of $m$ are to be determined.

[4]

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.

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ M1

$v + x \frac{d v}{d x} = f (v)$ A1

$\int \frac{d v}{f (v) - v} = \int \frac{d x}{x}$ A1

integrating the RHS, $\int \frac{d v}{f (v) - v} = \ln x + C$ AG

[3 marks]

EITHER

attempts to find $f (v)$ M1

$f (v) = v^{2} + 3 v + 2$ (A1)

substitutes their $f (v)$ into $\int \frac{d v}{f (v) - v}$ M1

$\int \frac{d v}{f (v) - v} = \int \frac{d v}{v^{2} + 2 v + 2}$

attempts to complete the square (M1)

$\int \frac{d v}{{(v + 1)}^{2} + 1}$ A1

$arctan (v + 1) (= \ln x + C)$ A1

attempts to find $f (v)$ M1

$v + x \frac{d v}{d x} = v^{2} + 3 v + 2$ A1

$\int \frac{d v}{v^{2} + 2 v + 2} = \frac{d x}{x}$ M1

attempts to complete the square (M1)

$\int \frac{d v}{{(v + 1)}^{2} + 1} (= \int \frac{d x}{x})$ A1

$arctan (v + 1) (= \ln x + C)$ A1

THEN

when $x = 1$ , $v = - 1$ (or $y = - 1$ ) and so $C = 0$ M1

substitutes for $v$ into their expression M1

$arctan (\frac{y}{x} + 1) = \ln x$

$\frac{y}{x} + 1 = \tan (\ln x)$ A1

so $y = x (\tan (\ln x) - 1)$ AG

[9 marks]

METHOD 1

EITHER

a correct graph of $y = f' (x)$ (for approximately $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ ) with a local minimum point below the $x$ -axis A2

Note: Award M1A1 for $\frac{d y}{d x} = \tan (\ln x) + {sec}^{2} (\ln x) - 1$ .

attempts to find the $x$ -coordinate of the local minimum point on the graph of $y = f' (x)$ (M1)

a correct graph of $y = f'' (x)$ (for approximately $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ ) showing the location of the $x$ -intercept A2

Note: Award M1A1 for $\frac{d^{2} y}{d x^{2}} = \frac{{sec}^{2} (\ln x)}{x} + \frac{2 {sec}^{2} (\ln x) \tan (\ln x)}{x}$ .

attempts to find the $x$ -intercept (M1)

THEN

$x = 0.629 (= e^{- arctan \frac{1}{2}})$ A1

attempts to find $f (0.629) (f (e^{- arctan \frac{1}{2}}))$ (M1)

the coordinates are $(0.629, - 0.943) (e^{- arctan \frac{1}{2}}, - \frac{3}{2} e^{- arctan \frac{1}{2}})$ A1

METHOD 2

attempts implicit differentiation on $\frac{d y}{d x}$ to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = \frac{(2 y + 3 x) (x \frac{d y}{d x} - y)}{x^{3}}$ (or equivalent)

$\frac{d^{2} y}{d x^{2}} = 0 \Rightarrow y = - \frac{3 x}{2}$ ( $\frac{d y}{d x} \neq \frac{y}{x}$ ) A1

attempts to solve $- \frac{3 x}{2} = x (\tan (\ln x) - 1)$ for $x$ where $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ M1

$x = 0.629 (= e^{- arctan \frac{1}{2}})$ A1

attempts to find $f (0.629) (f (= e^{- arctan \frac{1}{2}}))$ (M1)

the coordinates are $(0.629, - 0.943) (e^{- arctan \frac{1}{2}}, - \frac{3}{2} e^{- arctan \frac{1}{2}})$ A1

[6 marks]

$\frac{d y}{d x} = 0 \Rightarrow y^{2} + 3 x y + 2 x^{2} = 0$ M1

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

$(y + 2 x) (y + x) = 0$ or $y = \frac{- 3 x \pm \sqrt{{(3 x)}^{2} - 4 (2 x^{2})}}{2} (= \frac{- 3 x \pm x}{2}, (x > 0))$ A1

$y = - 2 x$ and $y = - x (m = - 2, - 1)$ A1

Note: Award M1 for stating $\frac{d y}{d x} = 0$ , M1 for substituting $y = m x$ into $\frac{d y}{d x} (= 0)$ , A1 for $(m + 2) (m + 1) = 0$ and A1 for $m = - 2, - 1 \Rightarrow y = - 2 x$ and $y = - x$ .

[4 marks]

Examiners report

[N/A]

Consider the curve defined by the equation $4{x^2} + {y^2} = 7$ .

Find the volume of the solid formed when the region bounded by the curve, the $x$ -axis for $x \geqslant 0$ and the $y$ -axis for $y \geqslant 0$ is rotated through $2\pi$ about the $x$ -axis.

Markscheme

Use of $V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {{y^2}{\text{d}}x}$

$V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {\left( {7 - 4{x^2}} \right){\text{d}}x}$ (M1)(A1)

Note: Condone absence of limits or incorrect limits for M mark.

Do not condone absence of or multiples of $\pi$ .

$= 19.4\,\,\,\left( { = \frac{{7\sqrt 7 \pi }}{3}} \right)$ A1

[3 marks]

Examiners report

[N/A]

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

A function $g$ is defined by $g (x) = arcsin (\frac{x^{2} - 1}{x^{2} + 1}), x \in ℝ, x \geq 0$ .

Show that $f$ is an even function.

[1]

By considering limits, show that the graph of $y = f (x)$ has a horizontal asymptote and state its equation.

[2]

Show that $f' (x) = \frac{2 x}{\sqrt{x^{2}} (x^{2} + 1)}$ for $x \in ℝ, x \neq 0$ .

[6]

c.i.

By using the expression for $f' (x)$ and the result $\sqrt{x^{2}} = |x|$ , show that $f$ is decreasing for $x < 0$ .

[3]

c.ii.

Find an expression for $g^{- 1} (x)$ , justifying your answer.

[5]

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

[1]

Sketch the graph of $y = g^{- 1} (x)$ , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.

[3]

Markscheme

EITHER

$f (- x) = arcsin (\frac{{(- x)}^{2} - 1}{{(- x)}^{2} + 1}) = arcsin (\frac{x^{2} - 1}{x^{2} + 1}) = f (x)$ R1

a sketch graph of $y = f (x)$ with line symmetry in the $y$ -axis indicated R1

THEN

so $f (x)$ is an even function. AG

[1 mark]

as $x \to \pm \infty, f (x) \to arcsin 1 (\to \frac{π}{2})$ A1

so the horizontal asymptote is $y = \frac{π}{2}$ A1

[2 marks]

attempting to use the quotient rule to find $\frac{d}{d x} (\frac{x^{2} - 1}{x^{2} + 1})$ M1

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

attempting to use the chain rule to find $\frac{d}{d x} (arcsin (\frac{x^{2} - 1}{x^{2} + 1}))$ M1

let $u = \frac{x^{2} - 1}{x^{2} + 1}$ and so $y = arcsin u$ and $\frac{d y}{d u} = \frac{1}{\sqrt{1 - u^{2}}}$

$f' (x) = \frac{1}{\sqrt{1 - {(\frac{x^{2} - 1}{x^{2} + 1})}^{2}}} \times \frac{4 x}{{(x^{2} + 1)}^{2}}$ M1

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

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

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

[6 marks]

c.i.

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

EITHER

for $x < 0, |x| = - x$ (A1)

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

$|x| > 0$ and $x^{2} + 1 > 0$ A1

$2 x < 0, x < 0$ A1

THEN

$f' (x) < 0$ R1

Note: Award R1 for stating that in $f' (x)$ , the numerator is negative, and the denominator is positive.

so $f$ is decreasing for $x < 0$ AG

Note: Do not accept a graphical solution

[3 marks]

c.ii.

$x = arcsin (\frac{y^{2} - 1}{y^{2} + 1})$ M1

$sin x = \frac{y^{2} - 1}{y^{2} + 1} \Rightarrow y^{2} sin x + sin x = y^{2} - 1$ A1

$y^{2} = \frac{1 + sin x}{1 - sin x}$ A1

domain of $g$ is $x \in ℝ, x \geq 0$ and so the range of $g^{- 1}$ must be $y \in ℝ, y \geq 0$

hence the positive root is taken (or the negative root is rejected) R1

Note: The R1 is dependent on the above A1.

so $(g^{- 1} (x) =) \sqrt{\frac{1 + sin x}{1 - sin x}}$ A1

Note: The final A1 is not dependent on R1 mark.

[5 marks]

domain is $- \frac{π}{2} \leq x < \frac{π}{2}$ A1

Note: Accept correct alternative notations, for example, $⌊ - \frac{π}{2}, \frac{π}{2} ⌊$ or $⌊ - \frac{π}{2}, \frac{π}{2})$ .
Accept $[- 1.57, 1.57 [$ if correct to $3$ s.f.

[1 mark]

A1A1A1

Note: A1 for correct domain and correct range and $y$ -intercept at $y = 1$
A1 for asymptotic behaviour $x \to \frac{π}{2}$
A1 for $x = \frac{π}{2}$
Coordinates are not required.
Do not accept $x = 1.57$ or other inexact values.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

The curve $C$ is defined by equation $xy - \ln y = 1,{\text{ }}y > 0$ .

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

[4]

Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$

[3]

Markscheme

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

$y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ M1A1A1

Note: Award A1 for the first two terms, A1 for the third term and the 0.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 - xy}}$ A1

Note: Accept $\frac{{ - {y^2}}}{{\ln y}}$ .

Note: Accept $\frac{{ - y}}{{x - \frac{1}{y}}}$ .

[4 marks]

${m_T} = \frac{{{{\text{e}}^2}}}{{1 - {\text{e}} \times \frac{2}{{\text{e}}}}}$ (M1)

${m_T} = - {{\text{e}}^2}$ (A1)

$y - {\text{e}} = - {{\text{e}}^2}x + 2{\text{e}}$

$- {{\text{e}}^2}x - y + 3{\text{e}} = 0$ or equivalent A1

Note: Accept $y = - 7.39x + 8.15$ .

[3 marks]

Examiners report

[N/A]

Differentiate from first principles the function $f\left( x \right) = 3{x^3} - x$ .

Markscheme

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

METHOD 1

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$= \frac{{\left( {3{{\left( {x + h} \right)}^3} - \left( {x + h} \right)} \right) - \left( {3{x^3} - x} \right)}}{h}$ M1

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

$= \frac{{9{x^2}h + 9x{h^2} + 3{h^3} - h}}{h}$ A1

cancelling $h$ M1

$= 9{x^2} + 9xh + 3{h^2} - 1$

then $\mathop {{\text{lim}}}\limits_{h \to 0} \left( {9{x^2} + 9xh + 3{h^2} - 1} \right)$

$= 9{x^2} - 1$ A1

Note: Final A1 dependent on all previous marks.

METHOD 2

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$= \frac{{\left( {3{{\left( {x + h} \right)}^3} - \left( {x + h} \right)} \right) - \left( {3{x^3} - x} \right)}}{h}$ M1

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

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

cancelling $h$ M1

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

then $\mathop {{\text{lim}}}\limits_{h \to 0} \left( {3\left( {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right) - 1} \right)$

$= 9{x^2} - 1$ A1

Note: Final A1 dependent on all previous marks.

[5 marks]

Examiners report

[N/A]

By using the substitution ${x^2} = 2\sec \theta$ , show that $\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }} = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c}$ .

Markscheme

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

EITHER

${x^2} = 2\sec \theta$

$2x\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sec \theta \tan \theta$ M1A1

$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}}$

$= \int {\frac{{\sec \theta \tan \theta {\text{d}}\theta }}{{2\sec \theta \sqrt {4{{\sec }^2}\theta - 4} }}}$ M1A1

$x = \sqrt 2 {(\sec \theta )^{\frac{1}{2}}}{\text{ }}\left( { = \sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}} \right)$

$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = \frac{{\sqrt 2 }}{2}{(\sec \theta )^{\frac{1}{2}}}\tan \theta {\text{ }}\left( { = \frac{{\sqrt 2 }}{2}{{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta } \right)$ M1A1

$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}}$

$= \int {\frac{{\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\tan \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\sqrt {4{{\sec }^2}\theta - 4} }}{\text{ }}\left( { = \int {\frac{{\sqrt 2 {{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}\sqrt {4{{\sec }^2}\theta - 4} }}} } \right)}$ M1A1

THEN

$= \frac{1}{2}\int {\frac{{\tan \theta {\text{d}}\theta }}{{2\tan \theta }}}$ (M1)

$= \frac{1}{4}\int {{\text{d}}\theta }$

$= \frac{\theta }{4} + c$ A1

${x^2} = 2\sec \theta \Rightarrow \cos \theta = \frac{2}{{{x^2}}}$ M1

Note: This M1 may be seen anywhere, including a sketch of an appropriate triangle.

so $\frac{\theta }{4} + c = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c$ AG

[7 marks]

Examiners report

[N/A]

A continuous random variable $X$ has the probability density function $f$ given by

$f (x) = \{\begin{cases} \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} 0 \leq x \leq 4 \\ 0 otherwise \end{cases}$

where $k \in ℝ^{+}$ .

Show that $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ .

[5]

Find the value of $k$ .

[2]

Markscheme

recognition of the need to integrate $\frac{x}{\sqrt{{(x^{2} + k)}^{3}}}$ (M1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x (= 1)$

EITHER

$u = x^{2} + k \Rightarrow \frac{d u}{d x} = 2 x$ (or equivalent) (A1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x = \frac{1}{2} \int u^{- \frac{3}{2}} d u$

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

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

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

THEN

attempt to use correct limits for their integrand and set equal to $1$ M1

${[- u^{- \frac{1}{2}}]}_{k}^{16 + k} = 1$ OR ${[- {(x^{2} + k)}^{- \frac{1}{2}}]}_{0}^{4} = 1$

$- {(16 + k)}^{- \frac{1}{2}} + k^{- \frac{1}{2}} = 1 (\Rightarrow \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{16 + k}} = 1)$ A1

$\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ AG

[5 marks]

attempt to solve $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ (M1)

$k = 0.645038 \dots$

$= 0.645$ A1

[2 marks]

Examiners report

[N/A]

The curve $C$ has equation $e^{2 y} = x^{3} + y$ .

Show that $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ .

[3]

The tangent to $C$ at the point Ρ is parallel to the $y$ -axis.

Find the $x$ -coordinate of Ρ.

[4]

Markscheme

attempts implicit differentiation on both sides of the equation M1

$2 e^{2 y} \frac{d y}{d x} = 3 x^{2} + \frac{d y}{d x}$ A1

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

so $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ AG

[3 marks]

attempts to solve $2 e^{2 y} - 1 = 0$ for $y$ (M1)

$y = - 0.346 \dots (= \frac{1}{2} \ln \frac{1}{2})$ A1

attempts to solve $e^{2 y} = x^{3} + y$ for $x$ given their value of $y$ (M1)

$x = 0.946 (= {(\frac{1}{2} (1 - \ln \frac{1}{2}))}^{\frac{1}{3}})$ A1

[4 marks]

Examiners report

[N/A]

Given that $2{x^3} - 3x + 1$ can be expressed in the form $Ax\left( {{x^2} + 1} \right) + Bx + C$ , find the values of the constants $A$ , $B$ and $C$ .

[2]

Hence find $\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x$ .

[5]

Markscheme

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

$2{x^3} - 3x + 1 = Ax\left( {{x^2} + 1} \right) + Bx + C$

$A = 2,\,\,C = 1,$ A1

$A + B = - 3 \Rightarrow B = - 5$ A1

[2 marks]

$\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x = \int {\left( {2x - \frac{{5x}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} {\text{d}}x$ M1M1

Note: Award M1 for dividing by $\left( {{x^2} + 1} \right)$ to get $2x$ , M1 for separating the $5x$ and 1.

$= {x^2} - \frac{5}{2}{\text{ln}}\left( {{x^2} + 1} \right) + {\text{arctan}}\,x\left( { + c} \right)$ (M1)A1A1

Note: Award (M1)A1 for integrating ${\frac{{5x}}{{{x^2} + 1}}}$ , A1 for the other two terms.

[5 marks]

Examiners report

[N/A]

A particle moves along a horizontal line such that at time $t$ seconds, $t$ ≥ 0, its acceleration $a$ is given by $a$ = 2 $t$ − 1. When $t$ = 6 , its displacement $s$ from a fixed origin O is 18.25 m. When $t$ = 15, its displacement from O is 922.75 m. Find an expression for $s$ in terms of $t$ .

Markscheme

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

attempt to integrate $a$ to find $v$ M1

$v = \int {a\,{\text{d}}t = \int {\left( {2t - 1} \right)} } \,{\text{d}}t$

$= {t^2} - t + c$ A1

$s = \int {v\,{\text{d}}t = \int {\left( {{t^2} - t + c} \right)} } \,{\text{d}}t$

$= \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + ct + d$ A1

attempt at substitution of given values (M1)

at $t = 6{\text{,}}\,\,\,18.25 = 72 - 18 + 6c + d$

at $t = 15{\text{,}}\,\,\,922.75 = 1125 - 112.5 + 15c + d$

solve simultaneously: (M1)

$c = - 6{\text{,}}\,\,d = 0.25$ A1

$\Rightarrow s = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + - 6t + \frac{1}{4}$

[6 marks]

Examiners report

[N/A]

Consider the curve $C$ given by $y = x - x y \ln (x y)$ where $x > 0, y > 0$ .

Show that $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ .

[3]

Hence find the equation of the tangent to $C$ at the point where $x = 1$ .

[5]

Markscheme

METHOD 1

attempts to differentiate implicitly including at least one application of the product rule (M1)

$u = x y, v = \ln (x y), \frac{d u}{d x} = x \frac{d y}{d x} + y, \frac{d v}{d x} = (x \frac{d y}{d x} + y) \frac{1}{x y}$

$\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$ A1

Note: Award (M1)A1 for implicitly differentiating $y = x (1 - y \ln (x y))$ and obtaining $\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + x \frac{d y}{d x} \ln (x y) + y \ln (x y)]$ .

$\frac{d y}{d x} = 1 - [(x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$

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

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 2

$y = x - x y \ln x - x y \ln y$

attempts to differentiate implicitly including at least one application of the product rule (M1)

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

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - (x \ln x \frac{d y}{d x} + (1 + \ln x) y) - (y \ln y + x (\ln y \frac{d y}{d x} + \frac{d y}{d x}))$

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

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln (x y) + 1) - y (\ln (x y) + 1)$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 3

attempt to differentiate implicitly including at least one application of the product rule M1

$u = x \ln (x y), v = y, \frac{d u}{d x} = \ln (x y) + (x \frac{d y}{d x} + y) \frac{x}{x y}, \frac{d v}{d x} = \frac{d y}{d x}$

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

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

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 4

lets $w = x y$ and attempts to find $\frac{d y}{d x}$ where $y = x - w \ln w$ M1

$\frac{d y}{d x} = 1 - (\frac{d w}{d x} + \frac{d w}{d x} \ln w) (= 1 - \frac{d w}{d x} (1 + \ln w))$ A1

$\frac{d w}{d x} = x \frac{d y}{d x} + y$ A1

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y + (x \frac{d y}{d x} + y) \ln (x y)) (= 1 - (x \frac{d y}{d x} + y) (1 + \ln (x y)))$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

[3 marks]

METHOD 1

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$ A1

substitutes $x = 1$ and their non-zero value of $y$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$2 \frac{d y}{d x} = 0 (\frac{d y}{d x} = 0)$ A1

equation of the tangent is $y = 1$ A1

METHOD 2

substitutes $x = 1$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (y)) = 1$

EITHER

correctly substitutes $\ln y = \frac{1 - y}{y}$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (1 + \frac{1}{y}) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

correctly substitutes $y + y \ln y = 1$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (2 + \ln y) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

THEN

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$

equation of the tangent is $y = 1$ A1

[5 marks]

Examiners report

[N/A]

Consider the differential equation $\frac{d y}{d x} = \frac{y - x}{y + x}$ , where $x, y > 0$ .

It is given that $y = 2$ when $x = 1$ .

Solve the differential equation, giving your answer in the form $f (x, y) = 0$ .

[9]

The graph of $y$ against $x$ has a local maximum between $x = 2$ and $x = 3$ . Determine the coordinates of this local maximum.

[4]

Show that there are no points of inflexion on the graph of $y$ against $x$ .

[4]

Markscheme

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

puts $y = v x$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ M1

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

attempts to express $x \frac{d v}{d x}$ as a single rational fraction in $v$

$x \frac{d v}{d x} = - \frac{v^{2} + 1}{v + 1}$ M1

attempts to separate variables M1

$\int \frac{v + 1}{v^{2} + 1} d v = - \int \frac{1}{x} d x$

$\frac{1}{2} \ln (v^{2} + 1) + arctan v = - \ln x (+ C)$ A1A1

substitutes $y = 2, x = 1$ and attempts to find the value of $C$ M1

$C = \frac{1}{2} \ln 5 + arctan 2$ A1

the solution is

$\frac{1}{2} \ln (\frac{y^{2}}{x^{2}} + 1) + arctan (\frac{y}{x}) + \ln x - \frac{1}{2} \ln 5 - arctan 2 = 0$ A1

[9 marks]

at a maximum, $\frac{d y}{d x} = 0$ M1

attempts to substitute $y = x$ into their solution M1

$\frac{1}{2} \ln 2 + arctan 1 + \ln x = \frac{1}{2} \ln 5 + arctan 2$

attempts to solve for $x, y$ (M1)

$(2.18, 2.18) (\frac{\sqrt{10}}{2} e^{arctan 2 - \frac{π}{4}}, \frac{\sqrt{10}}{2} e^{arctan 2 - \frac{π}{4}})$ A1

Note: Accept all answers that round to the correct $2 sf$ answer.
Accept $x = 2.18, y = 2.18$ .

[4 marks]

METHOD 1

attempts (quotient rule) implicit differentiation M1

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

correctly substitutes $\frac{d y}{d x} = \frac{y - x}{y + x}$ into $\frac{d^{2} y}{d x^{2}}$

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

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

this expression can never be zero therefore no points of inflexion R1

METHOD 2

attempts implicit differentiation on $(y + x) \frac{d y}{d x} = y - x$ M1

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

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

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

$- 1 - {(\frac{d y}{d x})}^{2} < 0$ and $x + y > 0, \frac{d^{2} y}{d x^{2}} \neq 0$ therefore no points of inflexion R1

Note: Accept putting $\frac{d^{2} y}{d x^{2}} = 0$ and obtaining contradiction.

[4 marks]

Examiners report

[N/A]

Assuming the Maclaurin series for $\cos x$ and $\ln (1 + x)$ , show that the Maclaurin series for $\cos (\ln (1 + x))$ is

$1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$

[4]

By differentiating the series in part (a), show that the Maclaurin series for $\sin (\ln (1 + x))$ is $x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$ .

[4]

Hence determine the Maclaurin series for $\tan (\ln (1 + x))$ as far as the term in $x^{3}$ .

[5]

Markscheme

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

METHOD 1

attempts to substitute $\ln (x + 1) = x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots$ into

$\cos x = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4} - \dots$ M1

$\cos x = (\ln (1 + x)) = 1 - \frac{1}{2} {(x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots)}^{2} + \frac{1}{24} {(x + \dots)}^{4} + \dots$ A1

attempts to expand the $RHS$ up to and including the $x^{4}$ term M1

$= 1 - \frac{1}{2} (x^{2} - x^{3} + \frac{1}{4} x^{4} + \frac{2}{3} x^{4} \dots) + \frac{1}{24} x^{4} + \dots$ A1

$= 1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$ AG

METHOD 2

attempts to substitute $\ln (x + 1)$ into $\cos x = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4} - \dots$ M1

$\cos (\ln (1 + x)) = 1 - \frac{1}{2} {(\ln (1 + x))}^{2} + \frac{1}{24} {(\ln (1 + x))}^{4} - \dots$

attempts to find the Maclaurin series for ${(\ln (1 + x))}^{2}$ up to and including the $x^{4}$ term M1

${(\ln (1 + x))}^{2} = x^{2} - x^{3} + \frac{11}{12} x^{4} - \dots$ A1

${(\ln (1 + x))}^{2} = x^{4} - \dots$

$= 1 - \frac{1}{2} (x^{2} - x^{3} + \frac{11}{12} x^{4} + \dots) + \frac{1}{24} x^{4} + \dots$ A1

$= 1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$ AG

[4 marks]

$- \sin (\ln (1 + x)) \times \frac{1}{1 + x} = - x + \frac{3}{2} x^{2} - \frac{5}{3} x^{3} + \dots$ A1A1

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

attempts to expand the $RHS$ up to and including the $x^{3}$ term M1

$= x - \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + x^{2} - \frac{3}{2} x^{3} + \dots$ A1

$= x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$ AG

[4 marks]

METHOD 1

let $\tan (\ln (1 + x)) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$

uses $\sin (\ln (1 + x)) = \cos (\ln (1 + x)) \times \tan (\ln (1 + x))$ to form M1

$x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots = (1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \dots) (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots)$ A1

$= a_{0} + a_{1} x + (a_{2} - \frac{1}{2} a_{0}) x^{2} + (a_{3} - \frac{1}{2} a_{1} + \frac{1}{2} a_{0}) x^{3} + \dots$ (A1)

attempts to equate coefficients,

$a_{0} = 0, a_{1} = 1, a_{2} - \frac{1}{2} a_{0} = - \frac{1}{2}, a_{3} - \frac{1}{2} a_{1} + \frac{1}{2} a_{0} = \frac{1}{6}$ M1

$a_{0} = 0, a_{1} = 1, a_{2} = - \frac{1}{2}, a_{3} = \frac{2}{3}$ A1

so $\tan (\ln (1 + x)) = x - \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \dots$

METHOD 2

uses $\tan (\ln (1 + x)) = \frac{\sin (\ln (1 + x))}{\cos (\ln (1 + x))}$ to form M1

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

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

attempts to expand the $RHS$ up to and including the $x^{3}$ term M1

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

$= x + \frac{1}{2} x^{3} - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$

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

Note: Accept use of long division.

[5 marks]

Examiners report

[N/A]

Two boats $A$ and $B$ travel due north.

Initially, boat $B$ is positioned $50$ metres due east of boat $A$ .

The distances travelled by boat $A$ and boat $B$ , after $t$ seconds, are $x$ metres and $y$ metres respectively. The angle $θ$ is the radian measure of the bearing of boat $B$ from boat $A$ . This information is shown on the following diagram.

Show that $y = x + 50 cot θ$ .

[1]

At time $T$ , the following conditions are true.

Boat $B$ has travelled $10$ metres further than boat $A$ .
Boat $B$ is travelling at double the speed of boat $A$ .
The rate of change of the angle $θ$ is $- 0.1$ radians per second.

Find the speed of boat $A$ at time $T$ .

[6]

Markscheme

$\tan θ = \frac{50}{y - x}$ OR $cot θ = \frac{y - x}{50}$ A1

$y = x + 50 cot θ$ AG

Note: $y - x$ may be identified as a length on a diagram, and not written explicitly.

[1 mark]

attempt to differentiate with respect to $t$ (M1)

$\frac{d y}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

attempt to set speed of $B$ equal to double the speed of $A$ (M1)

$2 \frac{d x}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$

$\frac{d x}{d t} = - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

$θ = arctan 5 (= 1.373 \dots = 78.69 \dots °)$ OR ${cosec}^{2} θ = 1 + {cot}^{2} θ = 1 + {(\frac{1}{5})}^{2} = \frac{26}{25}$ (A1)

Note: This A1 can be awarded independently of previous marks.

$\frac{d x}{d t} = - 50 (\frac{26}{25}) \times - 0.1$

So the speed of boat $A$ is $5.2 ({ms}^{- 1})$ A1

Note: Accept $5.20$ from the use of inexact values.

[6 marks]

Examiners report

[N/A]

A small bead is free to move along a smooth wire in the shape of the curve $y = \frac{10}{3 - 2 e^{- 0.5 x}} (x \geq 0)$ .

Find an expression for $\frac{d y}{d x}$ .

[3]

At the point on the curve where $x = 4$ , it is given that $\frac{d y}{d t} = - 0.1 {m s}^{- 1}$

Find the value of $\frac{d x}{d t}$ at this exact same instant.

[3]

Markscheme

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

valid attempt to use chain rule or quotient rule (M1)

$\frac{d y}{d x} = \frac{- 10 e^{- 0.5 x}}{{(3 - 2 e^{- 0.5 x})}^{2}}$ OR $\frac{d y}{d x} = - 10 e^{- 0.5 x} {(3 - 2 e^{- 0.5 x})}^{- 2}$ A1A1

[3 marks]

Note: Award A1 for numerator and A1 for denominator, or A1 for each part if the second alternative given.

valid attempt to use chain rule $(eg \frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t})$ (M1)

$\frac{d x}{d t} = - 0.1 \div \frac{- 10 e^{- 2}}{{(3 - 2 e^{- 2})}^{2}} (= - 0.1 \div - 0.181676 \dots)$ or equivalent (A1)

$= 0.550428 \dots$

$\frac{d x}{d t} = 0.550 ({ms}^{- 1})$ A1

[3 marks]

Examiners report

[N/A]

An earth satellite moves in a path that can be described by the curve $72.5{x^2} + 71.5{y^2} = 1$ where $x = x(t)$ and $y = y(t)$ are in thousands of kilometres and $t$ is time in seconds.

Given that $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 7.75 \times {10^{ - 5}}$ when $x = 3.2 \times {10^{ - 3}}$ , find the possible values of $\frac{{{\text{d}}y}}{{{\text{d}}t}}$ .

Give your answers in standard form.

Markscheme

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

METHOD 1

substituting for $x$ and attempting to solve for $y$ (or vice versa) (M1)

$y = ( \pm )0.11821 \ldots$ (A1)

EITHER

$145x + 143y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{145x}}{{143y}}} \right)$ M1A1

$145x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 143y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$ M1A1

THEN

attempting to find $\frac{{{\text{d}}x}}{{{\text{d}}t}}{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}t}} = - \frac{{145(3.2 \times {{10}^{ - 3}})}}{{143\left( {( \pm )0.11821 \ldots } \right)}} \times (7.75 \times {{10}^{ - 5}})} \right)$ (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ - 6}}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

METHOD 2

$y = ( \pm )\sqrt {\frac{{1 - 72.5{x^2}}}{{71.5}}}$ M1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = ( \pm )0.0274 \ldots$ (M1)(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = ( \pm )0.0274 \ldots \times 7.75 \times {10^{ - 5}}$ (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ - 6}}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

[6 marks]

Examiners report

[N/A]