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HL Paper 3

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{{{\text{d}}P}}{{{\text{d}}t}} = kP$ , where $k \in \mathbb{R}$ , $t$ is the time (in years) and $P$ is the population

The initial population is 1000.

Given that $k = 0.003$ , use your answer from part (a) to find

Consider now the situation when $k$ is not a constant, but a function of time.

Given that $k = 0.003 + 0.002t$ , find

Another model for population growth assumes

there is a maximum value for the population, $L$ .
that $k$ is not a constant, but is proportional to $\left( {1 - \frac{P}{L}} \right)$ .

Show that the general solution of this differential equation is $P = A{{\text{e}}^{kt}}$ , where $A \in \mathbb{R}$ .

[5]

the population after 10 years

[2]

b.i.

the number of years it will take for the population to triple.

[2]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P$

[1]

b.iii.

the solution of the differential equation, giving your answer in the form $P = f\left( t \right)$ .

[5]

c.i.

the number of years it will take for the population to triple.

[4]

c.ii.

Show that $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , where $m \in \mathbb{R}$ .

[2]

Solve the differential equation $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , giving your answer in the form $P = g\left( t \right)$ .

[10]

Given that the initial population is 1000, $L = 10000$ and $m = 0.003$ , find the number of years it will take for the population to triple.

[4]

Markscheme

$\int {\frac{1}{P}} {\text{d}}P = \int {k{\text{d}}t}$ M1A1

${\text{ln}}\,P = kt + c$ A1A1

$P = {e^{kt + c}}$ A1

$P = A{e^{kt}}$ , where $A = {e^c}$ AG

[5 marks]

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow A = 1000$ A1

$P\left( {10} \right) = 1000{e^{0.003\left( {10} \right)}} = 1030$ A1

[2 marks]

b.i.

$3000 = 1000{e^{0.003t}}$ M1

$t = \frac{{{\text{ln}}\,3}}{{0.003}} = 366$ years A1

[2 marks]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P = \infty$ A1

[1 mark]

b.iii.

$\int {\frac{1}{P}} {\text{d}}P = \int {\left( {0.003 + 0.002t} \right){\text{d}}t}$ M1

${\text{ln}}\,P = 0.003t + 0.001{t^2} + c$ A1A1

$P = {e^{0.003t + 0.001{t^2} + c}}$ A1

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow {e^c} = 1000$ M1

$P = 1000{e^{0.003t + 0.001{t^2}}}$

[5 marks]

c.i.

$3000 = 1000{e^{0.003t + 0.001{t^2}}}$ M1

${\text{ln}}\,3 = 0.003t + 0.001{t^2}$ A1

Use of quadratic formula or GDC graph or GDC polysmlt M1

$t = 31.7$ years A1

[4 marks]

c.ii.

$k = m\left( {1 - \frac{P}{L}} \right)$ , where $m$ is the constant of proportionality A1

So $\frac{{{\text{d}}P}}{{{\text{d}}t}} = m\left( {1 - \frac{P}{L}} \right)P$ A1

$\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ AG

[2 marks]

$\int {\frac{1}{{P\left( {L - P} \right)}}} {\text{d}}P = \int {\frac{m}{L}{\text{d}}t}$ M1

$\frac{1}{{P\left( {L - P} \right)}} = \frac{A}{P} + \frac{B}{{L - P}}$ M1

$1 \equiv A\left( {L - P} \right) + BP$ A1

$A = \frac{1}{L}{\text{,}}\,\,B = \frac{1}{L}$ A1

$\frac{1}{L}\int {\left( {\frac{1}{P} + \frac{1}{{L - P}}} \right){\text{d}}P} = \int {\frac{m}{L}{\text{d}}t}$

$\frac{1}{L}\left( {{\text{ln}}\,P - {\text{ln}}\left( {L - P} \right)} \right) = \frac{m}{L}t + c$ A1A1

${\text{ln}}\left( {\frac{P}{{L - P}}} \right) = mt + d$ , where $d = cL$ M1

$\frac{P}{{L - P}} = C{e^{mt}}$ , where $C = {e^d}$ A1

$P\left( {1 + C{e^{mt}}} \right) = CL{e^{mt}}$ M1

$P = \frac{{CL{e^{mt}}}}{{\left( {1 + C{e^{mt}}} \right)}}{\text{ }}\left( { = \frac{L}{{\left( {D{e^{ - mt}} + 1} \right)}}{\text{,}}\,\,{\text{where}}\;D = \frac{1}{C}} \right)$ A1

[10 marks]

$1000 = \frac{{10000}}{{D + 1}}$ M1

$D = 9$ A1

$3000 = \frac{{10000}}{{9{e^{ - 0.003t}} + 1}}$ M1

$t = 450$ years A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

In this question you will explore some of the properties of special functions $f$ and $g$ and their relationship with the trigonometric functions, sine and cosine.

Functions $f$ and $g$ are defined as $f (z) = \frac{e^{z} + e^{- z}}{2}$ and $g (z) = \frac{e^{z} - e^{- z}}{2}$ , where $z \in ℂ$ .

Consider $t$ and $u$ , such that $t, u \in ℝ$ .

Using $e^{i u} = \cos u + i \sin u$ , find expressions, in terms of $\sin u$ and $\cos u$ , for

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point ( $\cos θ, \sin θ$ ) defines points on the unit circle with equation $x^{2} + y^{2} = 1$ .

The functions $f (x)$ and $g (x)$ are known as hyperbolic functions, as the general point ( $f (θ), g (θ)$ ) defines points on a curve known as a hyperbola with equation $x^{2} - y^{2} = 1$ . This hyperbola has two asymptotes.

Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .

[2]

Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .

[3]

$f (i u)$ .

[3]

c.i.

$g (i u)$ .

[2]

c.ii.

Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

[2]

Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .

[4]

Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.

[4]

The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .

[5]

Markscheme

$f' (t) = \frac{e^{t} - e^{- t}}{2}$ A1

$f'' (t) = \frac{e^{t} + e^{- t}}{2}$ A1

$= f (t)$ AG

[2 marks]

METHOD 1

${(f (t))}^{2} + {(g (t))}^{2}$

substituting $f$ and $g$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$

$= \frac{{(e^{t})}^{2} + 2 + {(e^{- t})}^{2} + {(e^{t})}^{2} - 2 + {(e^{- t})}^{2}}{4}$ (M1)

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2} (= \frac{e^{2 t} + e^{- 2 t}}{2})$ A1

$= f (2 t)$ AG

METHOD 2

$f (2 t) = \frac{e^{2 t} + e^{- 2 t}}{2}$

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2}$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$ M1A1

$= {(f (t))}^{2} + {(g (t))}^{2}$ AG

Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.

[3 marks]

substituting $e^{i u} = \cos u + i \sin u$ into the expression for $f$ (M1)

obtaining $e^{-i u} = \cos u - i \sin u$ (A1)

$f (i u) = \frac{\cos u + i \sin u + \cos u - i \sin u}{2}$

Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.

$= \frac{2 \cos u}{2}$

$= \cos u$ A1

[3 marks]

c.i.

$g (i u) = \frac{\cos u + i \sin u - \cos u + i \sin u}{2}$

substituting and attempt to simplify (M1)

$= \frac{2 i \sin u}{2}$

$= i \sin u$ A1

[2 marks]

c.ii.

METHOD 1

${(f (i u))}^{2} + {(g (i u))}^{2}$

substituting expressions found in part (c) (M1)

$= \cos^{2} u - \sin^{2} u (= \cos 2 u)$ A1

METHOD 2

$f (2 i u) = \frac{e^{2 i u} + e^{- 2 i u}}{2}$

$= \frac{\cos 2 u + i \sin 2 u + \cos 2 u - i \sin 2 u}{2}$ M1

$= \cos 2 u$ A1

Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg $2 \cos^{2} u - 1$ etc

[2 marks]

${(f (t))}^{2} - {(g (t))}^{2} = \frac{{(e^{t} + e^{- t})}^{2} - {(e^{t} - e^{- t})}^{2}}{4}$ M1

$= \frac{(e^{2 t} + e^{- 2 t} + 2) - (e^{2 t} + e^{- 2 t} - 2)}{4}$ A1

$= \frac{4}{4} = 1$ A1

Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.

${(f (i u))}^{2} - {(g (i u))}^{2} = \cos^{2} u + \sin^{2} u$ M1

$= 1$ (hence ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ ) AG

Note: Award full marks for showing that ${(f (z))}^{2} - {(g (z))}^{2} = 1, \forall z \in ℂ$ .

[4 marks]

A1A1A1A1

Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct $x$ -intercepts of $(- 1, 0)$ and $(1, 0)$ (condone $x = - 1$ and $1$ ), A1 for $y = x$ and $y = - x$ .

[4 marks]

attempt to rotate by $45 °$ in either direction (M1)

Note: Evidence of an attempt to relate to a sketch of $x y = k$ would be sufficient for this (M1).

attempting to rotate a particular point, eg $(1, 0)$ (M1)

$(1, 0)$ rotates to $(\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}})$ (or similar) (A1)

hence $k = \pm \frac{1}{2}$ A1A1

[5 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

This question asks you to investigate some properties of the sequence of functions of the form ${f_n}(x) = {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right)$ , −1 ≤ $x$ ≤ 1 and $n \in {\mathbb{Z}^ + }$ .

Important: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.

For odd values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for odd values of $n$ describing, in terms of $n$ , the number of

For even values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for even values of $n$ describing, in terms of $n$ , the number of

The sequence of functions, ${f_n}(x)$ , defined above can be expressed as a sequence of polynomials of degree $n$ .

Consider ${f_{n + 1}}(x) = {\text{cos}}\left( {\left( {n + 1} \right)\,{\text{arccos}}\,x} \right)$ .

On the same set of axes, sketch the graphs of $y = {f_1}(x)$ and $y = {f_3}(x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

b.i.

local minimum points;

[1]

b.ii.

On a new set of axes, sketch the graphs of $y = {f_2}(x)$ and $y = {f_4}(x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

d.i.

local minimum points.

[1]

d.ii.

Solve the equation ${f_n}^\prime (x) = 0$ and hence show that the stationary points on the graph of $y = {f_n}(x)$ occur at $x = {\text{cos}}\frac{{k\pi }}{n}$ where $k \in {\mathbb{Z}^ + }$ and 0 < $k$ < $n$ .

[4]

Use an appropriate trigonometric identity to show that ${f_2}(x) = 2{x^2} - 1$ .

[2]

Use an appropriate trigonometric identity to show that ${f_{n + 1}}(x) = {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right){\text{cos}}\left( {{\text{arccos}}\,x} \right) - {\text{sin}}\left( {n\,{\text{arccos}}\,x} \right){\text{sin}}\left( {{\text{arccos}}\,x} \right)$ .

[2]

Hence show that ${f_{n + 1}}(x) + {f_{n - 1}}(x) = 2x{f_n}\left( x \right)$ , $n \in {\mathbb{Z}^ + }$ .

[3]

h.i.

Hence express ${f_3}(x)$ as a cubic polynomial.

[2]

h.ii.

Markscheme

correct graph of $y = {f_1}(x)$ A1

correct graph of $y = {f_3}(x)$ A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied M1

eg $n$ = 3, 1 local maximum point and 1 local minimum point

$n$ = 5, 2 local maximum points and 2 local minimum points

$n$ = 7, 3 local maximum points and 3 local minimum points (A1)

$\frac{{n - 1}}{2}$ local maximum points A1

[3 marks]

b.i.

$\frac{{n - 1}}{2}$ local minimum points A1

Note: Allow follow through from an incorrect local maximum formula expression.

[1 mark]

b.ii.

correct graph of $y = {f_2}(x)$ A1

correct graph of $y = {f_4}(x)$ A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied M1

eg $n$ = 2, 0 local maximum point and 1 local minimum point

$n$ = 4, 1 local maximum points and 2 local minimum points

$n$ = 6, 2 local maximum points and 3 local minimum points (A1)

$\frac{{n - 2}}{2}$ local maximum points A1

[3 marks]

d.i.

$\frac{n}{2}$ local minimum points A1

[1 mark]

d.ii.

${f_n}(x) = {\text{cos}}\left( {n\,{\text{arccos}}\left( x \right)} \right)$

${f_n}^\prime (x) = \frac{{n\,{\text{sin}}\left( {n\,{\text{arccos}}\left( x \right)} \right)}}{{\sqrt {1 - {x^2}} }}$ M1A1

Note: Award M1 for attempting to use the chain rule.

${f_n}^\prime (x) = 0 \Rightarrow n\,{\text{sin}}\left( {n\,{\text{arccos}}\left( x \right)} \right) = 0$ M1

$n\,{\text{arccos}}\left( x \right) = k\pi \,\,\,\left( {k \in {\mathbb{Z}^ + }} \right)$ A1

leading to

$x = {\text{cos}}\frac{{k\pi }}{n}$ ( $k \in {\mathbb{Z}^ + }$ and 0 < $k$ < $n$ ) AG

[4 marks]

${f_2}(x) = {\text{cos}}\left( {2\,{\text{arccos}}\,x} \right)$

$= 2{\left( {{\text{cos}}\left( {{\text{arccos}}\,x} \right)} \right)^2} - 1$ M1

stating that $\left( {{\text{cos}}\left( {{\text{arccos}}\,x} \right)} \right) = x$ A1

so ${f_2}(x) = 2{x^2} - 1$ AG

[2 marks]

${f_{n + 1}}(x) = {\text{cos}}\left( {\left( {n + 1} \right)\,{\text{arccos}}\,x} \right)$

$= {\text{cos}}\left( {n\,{\text{arccos}}\,x + {\text{arccos}}\,x} \right)$ A1

use of cos(A + B) = cos A cos B − sin A sin B leading to M1

$= {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right){\text{cos}}\left( {{\text{arccos}}\,x} \right) - {\text{sin}}\left( {n\,{\text{arccos}}\,x} \right){\text{sin}}\left( {{\text{arccos}}\,x} \right)$ AG

[2 marks]

${f_{n - 1}}(x) = {\text{cos}}\left( {\left( {n - 1} \right)\,{\text{arccos}}\,x} \right)$ A1

$= {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right){\text{cos}}\left( {{\text{arccos}}\,x} \right) + {\text{sin}}\left( {n\,{\text{arccos}}\,x} \right){\text{sin}}\left( {{\text{arccos}}\,x} \right)$ M1

${f_{n + 1}}(x) + {f_{n - 1}}(x) = 2\,{\text{cos}}\left( {n\,{\text{arccos}}\,x} \right){\text{cos}}\left( {{\text{arccos}}\,x} \right)$ A1

$= 2x{f_n}\left( x \right)$ AG

[3 marks]

h.i.

${f_3}(x) = 2x{f_2}\left( x \right) - {f_1}(x)$ (M1)

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

$= 4{x^3} - 3x$ A1

[2 marks]

h.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

h.i.

[N/A]

h.ii.

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ , where $x \ne 0$ .

Consider the family of curves which satisfy the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ , where $x \ne 0$ .

Given that $y\left( 1 \right) = 1$ , use Euler’s method with step length $h$ = 0.25 to find an approximation for $y\left( 2 \right)$ . Give your answer to two significant figures.

[4]

Solve the equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ for $y\left( 1 \right) = 1$ .

[6]

Find the percentage error when $y\left( 2 \right)$ is approximated by the final rounded value found in part (a). Give your answer to two significant figures.

[3]

Find the equation of the isocline corresponding to $\frac{{{\text{d}}y}}{{{\text{d}}x}} = k$ , where $k \ne 0$ , $k \in \mathbb{R}$ .

[1]

d.i.

Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

[4]

d.ii.

Markscheme

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

attempt to apply Euler’s method (M1)

${x_{n + 1}} = {x_n} + 0.25{\text{;}}\,\,{y_{n + 1}} = {y_n} + 0.25 \times \left( {1 + \frac{{{y_n}}}{{{x_n}}}} \right)$

(A1)(A1)

Note: Award A1 for correct $x$ values, A1 for first three correct $y$ values.

$y$ = 3.3 A1

[4 marks]

METHOD 1

$I\left( x \right) = {{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }}$ (M1)

$= {{\text{e}}^{ - {\text{ln}}\,x}}$

$= \frac{1}{x}$ (A1)

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

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

$\frac{y}{x} = {\text{ln}}\left| x \right| + C$ A1

$y\left( 1 \right) = 1 \Rightarrow C = 1$ M1

$y = x\,{\text{ln}}\left| x \right| + x$ A1

METHOD 2

$v = \frac{y}{x}$ M1

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

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 + v$ M1

$\int 1 \,{\text{d}}v = \int {\frac{1}{x}} \,{\text{d}}x$

$v = {\text{ln}}\left| x \right| + C$

$\frac{y}{x} = {\text{ln}}\left| x \right| + C$ A1

$y\left( 1 \right) = 1 \Rightarrow C = 1$ M1

$y = x\,{\text{ln}}\left| x \right| + x$ A1

[6 marks]

$y\left( 2 \right) = 2\,{\text{ln}}\,2 + 2 = 3.38629 \ldots$

percentage error $= \frac{{3.38629 \ldots - 3.3}}{{3.38629 \ldots }} \times 100{\text{% }}$ (M1)(A1)

= 2.5% A1

[3 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = k \Rightarrow 1 + \frac{y}{x} = k$ A1

$y = \left( {k - 1} \right)x$

[1 mark]

d.i.

gradient of isocline equals gradient of normal (M1)

$k - 1 = - \frac{1}{k}$ or $k\left( {k - 1} \right) = - 1$ A1

${k^2} - k + 1 = 0$ A1

$\Delta = 1 - 4 < 0$ R1

$\therefore$ no solution AG

Note: Accept alternative reasons for no solutions.

[4 marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.

A power series in $x$ is defined as a function of the form $f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ where the ${a_i} \in \mathbb{R}$ .

It can be considered as an infinite polynomial.

This is an example of a power series, but is only a finite power series, since only a finite number of the ${a_i}$ are non-zero.

We will now attempt to generalise further.

Suppose ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ can be written as the power series ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ .

Expand ${\left( {1 + x} \right)^5}$ using the Binomial Theorem.

[2]

Consider the power series $1 - x + {x^2} - {x^3} + {x^4} - ...$

By considering the ratio of consecutive terms, explain why this series is equal to ${\left( {1 + x} \right)^{ - 1}}$ and state the values of $x$ for which this equality is true.

[4]

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 2}}$ .

[2]

Repeat this process to find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 3}}$ .

[2]

Hence, by recognising the pattern, deduce the first four terms in a power series for ${\left( {1 + x} \right)^{ - n}}$ , $n \in {\mathbb{Z}^ + }$ .

[3]

By substituting $x = 0$ , find the value of ${a_0}$ .

[1]

By differentiating both sides of the expression and then substituting $x = 0$ , find the value of ${a_1}$ .

[2]

Repeat this procedure to find ${a_2}$ and ${a_3}$ .

[4]

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ .

[1]

Write down the power series for $\frac{1}{{1 + {x^2}}}$ .

[2]

Hence, using integration, find the power series for ${\text{arctan}}\,x$ , giving the first four non-zero terms.

[4]

Markscheme

$1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5}$ M1A1

[2 marks]

It is an infinite GP with $a = 1{\text{,}}\,\,r = - x$ R1A1

${S_\infty } = \frac{1}{{1 - \left( { - x} \right)}} = \frac{1}{{1 + x}} = {\left( {1 + x} \right)^{ - 1}}$ M1A1AG

[4 marks]

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

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

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

[2 marks]

$- 2{\left( {1 + x} \right)^{ - 3}} = - 2 + 6x - 12{x^2} + 20{x^3}...$ A1

${\left( {1 + x} \right)^{ - 3}} = 1 - 3x + 6{x^2} - 10{x^3}...$ A1

[2 marks]

${\left( {1 + x} \right)^{ - n}} = 1 - nx + \frac{{n\left( {n + 1} \right)}}{{2{\text{!}}}}{x^2} - \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1A1A1

[3 marks]

${1^q} = {a_0} \Rightarrow {a_0} = 1$ A1

[1 mark]

$q{\left( {1 + x} \right)^{q - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ...$ A1

${a_1} = q$ A1

[2 marks]

$q\left( {q - 1} \right){\left( {1 + x} \right)^{q - 2}} = 1 \times 2{a_2} + 2 \times 3{a_3}x + ...$ A1

${a_2} = \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}$ A1

$q\left( {q - 1} \right)\left( {q - 2} \right){\left( {1 + x} \right)^{q - 3}} = 1 \times 2 \times 3{a_3} + ...$ A1

${a_3} = \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}$ A1

[4 marks]

${\left( {1 + x} \right)^q} = 1 + qx + \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}{x^2} + \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1

[1 mark]

$\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} + ...$ M1A1

[2 marks]

${\text{arctan}}\,x + c = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + ...$ M1A1

Putting $x = 0 \Rightarrow c = 0$ R1

So ${\text{arctan}}\,x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + ...$ A1

[4 marks]

Examiners report

[N/A]

The function $f$ is defined by $f(x){\text{ }}={\text{ }}{(\arcsin{\text{ }}x)^2},{\text{ }} - 1 \leqslant x \leqslant 1$ .

The function $f$ satisfies the equation $\left( {1 - {x^2}} \right)f''\left( x \right) - xf'\left( x \right) - 2 = 0$ .

Show that $f'\left( 0 \right) = 0$ .

[2]

By differentiating the above equation twice, show that

$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$

where ${f^{\left( 3 \right)}}\left( x \right)$ and ${f^{\left( 4 \right)}}\left( x \right)$ denote the 3rd and 4th derivative of $f\left( x \right)$ respectively.

[4]

Hence show that the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$ .

[3]

Use this series approximation for $f\left( x \right)$ with $x = \frac{1}{2}$ to find an approximate value for ${\pi ^2}$ .

[2]

Markscheme

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

$f'\left( x \right) = \frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}$ M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for $f'\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)$ .

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

[2 marks]

differentiating gives $\left( {1 - {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) - 2xf''\left( x \right) - f'\left( x \right) - xf''\left( x \right)\left( { = 0} \right)$ M1A1

differentiating again gives $\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 2x{f^{\left( 3 \right)}}\left( x \right) - 3f''\left( x \right) - 3x{f^{\left( 3 \right)}}\left( x \right) - f''\left( x \right)\left( { = 0} \right)$ M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$ AG

[4 marks]

attempting to find one of $f''\left( 0 \right)$ , ${f^{\left( 3 \right)}}\left( 0 \right)$ or ${f^{\left( 4 \right)}}\left( 0 \right)$ by substituting $x = 0$ into relevant differential equation(s) (M1)

Note: Condone $f''\left( 0 \right)$ found by calculating $\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}} \right)$ at $x = 0$ .

$\left( {f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0} \right)$

$f''\left( 0 \right) = 2$ and ${f^{\left( 4 \right)}}\left( 0 \right) - 4f''\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8$ A1

${f^{\left( 3 \right)}}\left( 0 \right) = 0$ and so $\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}$ A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to ${f^{\left( 3 \right)}}\left( 0 \right) = 0$ stated or ${f^{\left( 3 \right)}}\left( 0 \right) = 0$ seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if ${f^{\left( 4 \right)}}\left( 0 \right) = 8$ is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$ AG

[3 marks]

substituting $x = \frac{1}{2}$ into ${x^2} + \frac{1}{3}{x^4}$ M1

the series approximation gives a value of $\frac{{13}}{{48}}$

so ${\pi ^2} \simeq \frac{{13}}{{48}} \times 36$

$\simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)$ A1

Note: Accept 9.76.

[2 marks]

Examiners report

[N/A]

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4{x^2} + {y^2} - xy}}{{{x^2}}}$ , with $y = 2$ when $x = 1$ .

Use Euler’s method, with step length $h = 0.1$ , to find an approximate value of $y$ when $x = 1.4$ .

[5]

Sketch the isoclines for $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4$ .

[3]

Express ${m^2} - 2m + 4$ in the form ${\left( {m - a} \right)^2} + b$ , where $a{\text{, }}b \in \mathbb{Z}$ .

[1]

c.i.

Solve the differential equation, for $x > 0$ , giving your answer in the form $y = f\left( x \right)$ .

[10]

c.ii.

Sketch the graph of $y = f\left( x \right)$ for $1 \leqslant x \leqslant 1.4$ .

[1]

c.iii.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f\left( {1.4} \right)$ will be less than, equal to, or greater than your answer in part (a).

[2]

c.iv.

Markscheme

(M1)(A1)(A1)(A1)A1

$y\left( {1.4} \right) \approx 5.34$

Note: Award A1 for each correct $y$ value.
For the intermediate $y$ values, accept answers that are accurate to 2 significant figures.
The final $y$ value must be accurate to 3 significant figures or better.

[5 marks]

attempt to solve $\frac{{4{x^2} + {y^2} - xy}}{{{x^2}}} = 4$ (M1)

$\Rightarrow {y^2} - xy = 0$

$y\left( {y - x} \right) = 0$

$y = 0$ or $y = x$

A1A1

[3 marks]

${m^2} - 2m + 4 = {\left( {m - 1} \right)^2} + 3\,\,\,\,\,\left( {a = 1,\,b = 3} \right)$ A1

[1 mark]

c.i.

recognition of homogeneous equation,
let $y = vx$ M1

the equation can be written as

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

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2} - 2v + 4$

$\int {\frac{1}{{{v^2} - 2v + 4}}} {\text{d}}v = \int {\frac{1}{x}} {\text{d}}x$ M1

Note: Award M1 for attempt to separate the variables.

$\int {\frac{1}{{{{\left( {v - 1} \right)}^2} + 4}}} {\text{d}}v = \int {\frac{1}{x}} {\text{d}}x$ from part (c)(i) M1

$\frac{1}{{\sqrt 3 }}{\text{arctan}}\left( {\frac{{v - 1}}{{\sqrt 3 }}} \right) = {\text{ln}}\,x\,\,\left( { + c} \right)$ A1A1

$x = 1{\text{,}}\,y = 2 \Rightarrow v = 2$

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

Note: Award M1 for using initial conditions to find $c$ .

$\Rightarrow c = \frac{\pi }{{6\sqrt 3 }}\,\,\left( { = 0.302} \right)$ A1

${\text{arctan}}\left( {\frac{{v - 1}}{{\sqrt 3 }}} \right) = \sqrt 3 \,{\text{ln}}\,x + \frac{\pi }{6}$

substituting $v = \frac{y}{x}$ M1

Note: This M1 may be awarded earlier.

$y = x\left( {\sqrt 3 {\text{tan}}\left( {\sqrt 3 \,{\text{ln}}\,x + \frac{\pi }{6}} \right) + 1} \right)$ A1

[10 marks]

c.ii.

curve drawn over correct domain A1

[1 mark]

c.iii.

the sketch shows that $f$ is concave up A1

Note: Accept ${f'}$ is increasing.

this means the tangent drawn using Euler’s method will give an underestimate of the real value, so $f\left( {1.4} \right)$ > estimate in part (a) R1

Note: The R1 is dependent on the A1.

[2 marks]

c.iv.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

c.iv.

Consider the differential equation $2xy\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} - {x^2}$ , where $x > 0$ .

Solve the differential equation and show that a general solution is ${x^2} + {y^2} = cx$ where $c$ is a positive constant.

[11]

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$ .

[5]

Markscheme

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

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2} - {x^2}}}{{2xy}}$

let $y = vx$ M1

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

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} - {x^2}}}{{2v{x^2}}}$ (M1)

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

Note: Or equivalent attempt at simplification.

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ - {v^2} - 1}}{{2v}}\,\,\,\,\,\left( { = - \frac{v}{2} - \frac{1}{{2v}}} \right)$ A1

$\frac{{2v}}{{1 + {v^2}}}\frac{{{\text{d}}v}}{{{\text{d}}x}} = - \frac{1}{x}$ (M1)

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

${\text{ln}}\left( {1 + {v^2}} \right) = - {\text{ln}}\,x + {\text{ln}}\,c$ A1A1

Note: Award A1 for LHS and A1 for RHS and a constant.

${\text{ln}}\left( {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right) = - {\text{ln}}\,x + {\text{ln}}\,c$ M1

Note: Award M1 for substituting $v = \frac{y}{x}$ . May be seen at a later stage.

$1 + {\left( {\frac{y}{x}} \right)^2} = \frac{c}{x}$ A1

Note: Award A1 for any correct equivalent equation without logarithms.

${x^2} + {y^2} = cx$ AG

[11 marks]

METHOD 1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2} - {x^2}}}{{2xy}}$

(for horizontal tangents) $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ M1

$\left( { \Rightarrow {y^2} = {x^2}} \right) \Rightarrow y = \pm x$

EITHER

using ${x^2} + {y^2} = cx \Rightarrow 2{x^2} = cx$ M1

$2{x^2} - cx = 0 \Rightarrow x = \frac{c}{2}$ A1

Note: Award M1A1 for $2{y^2} = \pm cy$ .

using implicit differentiation of ${x^2} + {y^2} = cx$

$2x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = c$ M1

Note: Accept differentiation of $y = \sqrt {cx - {x^2}}$ .

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow x = \frac{c}{2}$ A1

THEN

tangents at $y = \frac{c}{2},\,\,y = - \frac{c}{2}$ A1A1

hence there are two tangents AG

METHOD 2

${x^2} + {y^2} = cx$

${\left( {x - \frac{c}{2}} \right)^2} + {y^2} = \frac{{{c^2}}}{4}$ M1A1

this is a circle radius ${\frac{c}{2}}$ centre $\left( {\frac{c}{2},\,\,0} \right)$ A1

hence there are two tangents AG

tangents at $y = \frac{c}{2},\,\,y = - \frac{c}{2}$ A1A1

[5 marks]

Examiners report

[N/A]

Find the value of $\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x}$ .

[3]

Illustrate graphically the inequality $\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[4]

Hence write down a lower bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[1]

Find an upper bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[3]

Markscheme

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

$\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} = \mathop {{\text{lim}}}\limits_{R \to \infty } \int\limits_4^R {\frac{1}{{{x^3}}}{\text{d}}x}$ (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of $\mathop {{\text{lim}}}\limits_{x \to \infty }$ .

Do not award this mark to candidates who use $\infty$ as the upper limit throughout.

= $\mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_4^R\left( { = \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_4^\infty } \right)$ M1

$= \mathop {{\text{lim}}}\limits_{R \to \infty } \left( { - \frac{1}{2}\left( {{R^{ - 2}} - {4^{ - 2}}} \right)} \right)$

$= \frac{1}{{32}}$ A1

[3 marks]

A1A1A1A1

A1 for the curve
A1 for rectangles starting at $x = 4$
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

$\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ AG

[4 marks]

a lower bound is $\frac{1}{{32}}$ A1

Note: Allow FT from part (a).

[1 mark]

METHOD 1

$\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{32}}$ (M1)

$\frac{1}{{64}} + \sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} = \frac{1}{{32}} + \frac{1}{{64}}$ (M1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{3}{{64}}$ , an upper bound A1

Note: Allow FT from part (a).

METHOD 2

changing the lower limit in the inequality in part (b) gives

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \int\limits_3^\infty {\frac{1}{{{x^3}}}{\text{d}}x} \left( { < \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}}}} } \right)$ (A1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_3^R$ (M1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{18}}$ , an upper bound A1

Note: Condone candidates who do not use a limit.

[3 marks]

Examiners report

[N/A]

This question asks you to explore cubic polynomials of the form $(x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ and corresponding cubic equations with one real root and two complex roots of the form $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ .

In parts (a), (b) and (c), let $r = 1, a = 4$ and $b = 1$ .

Consider the equation $(z - 1) (z^{2} - 8 z + 17) = 0$ for $z \in ℂ$ .

Consider the function $f (x) = (x - 1) (x^{2} - 8 x + 17)$ for $x \in ℝ$ .

Consider the function $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

The equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ has roots $r$ and $a \pm b i$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

On the Cartesian plane, the points $C_{1} (a, \sqrt{g' (a)})$ and $C_{2} (a, - \sqrt{g' (a)})$ represent the real and imaginary parts of the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ .

The following diagram shows a particular curve of the form $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ and the tangent to the curve at the point $A (a, 80)$ . The curve and the tangent both intersect the $x$ -axis at the point $R (- 2, 0)$ . The points $C_{1}$ and $C_{2}$ are also shown.

Consider the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a \neq r, b > 0$ . The points $A (a, g (a))$ and $R (r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$ .

Consider the special case where $a = r$ and $b > 0$ .

Given that $1$ and $4 + i$ are roots of the equation, write down the third root.

[1]

a.i.

Verify that the mean of the two complex roots is $4$ .

[1]

a.ii.

Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .

[4]

Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.

[2]

Show that $g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ .

[2]

d.i.

Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .

[6]

d.ii.

Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .

[1]

Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .

[4]

f.i.

State the coordinates of $C_{2}$ .

[1]

f.ii.

Show that the $x$ -coordinate of $P$ is $\frac{1}{3} (2 a + r)$ .

You are not required to demonstrate a change in concavity.

[2]

g.i.

Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$ .

[1]

g.ii.

Sketch the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a = r = 1$ and $b = 2$ .

[2]

h.i.

For $a = r$ and $b > 0$ , state in terms of $r$ , the coordinates of points $P$ and $A$ .

[1]

h.ii.

Markscheme

$4 - i$ A1

[1 mark]

a.i.

mean $= \frac{1}{2} (4 + i + 4 - i)$ A1

$= 4$ AG

[1 mark]

a.ii.

METHOD 1

attempts product rule differentiation (M1)

Note: Award (M1) for attempting to express $f (x)$ as $f (x) = x^{3} - 9 x^{2} + 25 x - 17$

$f' (x) = (x - 1) (2 x - 8) + x^{2} - 8 x + 17 (f' (x) = 3 x^{2} - 18 x + 25)$ A1

$f' (4) = 1$ A1

Note: Where $f' (x)$ is correct, award A1 for solving $f' (x) = 1$ and obtaining $x = 4$ .

EITHER

$y - 3 = 1 (x - 4)$ A1

$y = x + c$

$3 = 4 + c \Rightarrow c = - 1$ A1

states the gradient of $y = x - 1$ is also $1$ and verifies that $(4, 3)$ lies on the line $y = x - 1$ A1

THEN

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f' (x)$ .

METHOD 2

sets $f (x) = x - 1$ to form $x - 1 = (x - 1) (x^{2} - 8 x + 17)$ (M1)

EITHER

$(x - 1) (x^{2} - 8 x + 16) = 0 (x^{3} - 9 x^{2} + 24 x - 16 = 0)$ A1

attempts to solve a correct cubic equation (M1)

$(x - 1) {(x - 4)}^{2} = 0 \Rightarrow x = 1, 4$

recognises that $x \neq 1$ and forms $x^{2} - 8 x + 17 = 1 (x^{2} - 8 x + 16 = 0)$ A1

attempts to solve a correct quadratic equation (M1)

${(x - 4)}^{2} = 0 \Rightarrow x = 4$

THEN

$x = 4$ is a double root R1

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Candidates using this method are not required to verify that $y = 3$ .

[4 marks]

a positive cubic with an $x$ -intercept $(x = 1)$ , and a local maximum and local minimum in the first quadrant both positioned to the left of $A$ A1

Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as $A, (4, 3)$ or the point labelled from both axes. Coordinates are not required.

a correct sketch of the tangent passing through $A$ and crossing the $x$ -axis at the same point $(x = 1)$ as the curve A1

Note: Award A1A0 if both graphs cross the $x$ -axis at distinctly different points.

[2 marks]

EITHER

$g' (x) = (x - r) (2 x - 2 a) + x^{2} - 2 a x + a^{2} + b^{2}$ (M1)A1

$g (x) = x^{3} - (2 a + r) x^{2} + (a^{2} + b^{2} + 2 a r) x - (a^{2} + b^{2}) r$

attempts to find $g' (x)$ M1

$g' (x) = 3 x^{2} - 2 (2 a + r) x + a^{2} + b^{2} + 2 a r$

$= 2 x^{2} - 2 (a + r) x + 2 a r + x^{2} - 2 a x + a^{2} + b^{2}$ A1

$(= 2 (x^{2} - a x - r x + a r) + x^{2} - 2 a x + a^{2} + b^{2})$

THEN

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ AG

[2 marks]

d.i.

METHOD 1

$g (a) = b^{2} (a - r)$ (A1)

$g' (a) = b^{2}$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y - g (a) = g' (a) (x - a)$ M1

$y - b^{2} (a - r) = b^{2} (x - a)$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $- b^{2} (a - r) = b^{2} (x - a)$ M1

$b > 0$ OR $b \neq 0 \Rightarrow - (a - r) = x - a$ R1

$x = r$ A1

THEN

so the tangent intersects the $x$ -axis at the point $R (r, 0)$ AG

METHOD 2

$g' (a) = b^{2}$ (A1)

$g (a) = b^{2} (a - r)$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y = g' (a) x + c$ and attempts to find $c$ M1

$c = - b^{2} r$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0$ OR $b \neq 0 \Rightarrow x - r = 0$ R1

$x = r$ A1

METHOD 3

$g' (a) = b^{2}$ (A1)

the line through $R (r, 0)$ parallel to the tangent at $A$ has equation
$y = b^{2} (x - r)$ A1

sets $g (x) = b^{2} (x - r)$ to form $b^{2} (x - r) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ M1

$b^{2} = x^{2} - 2 a x + a^{2} + b^{2}, (x \neq r)$ A1

${(x - a)}^{2} = 0$ A1

since there is a double root $(x = a)$ , this parallel line through $R (r, 0)$ is the required tangent at $A$ R1

[6 marks]

d.ii.

EITHER

$g' (a) = b^{2} \Rightarrow b = \sqrt{g' (a)}$ (since $b > 0$ ) R1

Note: Accept $b = \pm \sqrt{g' (a)}$ .

$(a \pm b i=) a \pm i \sqrt{b^{2}}$ and $g' (a) = b^{2}$ R1

THEN

hence the complex roots can be expressed as $a \pm i \sqrt{g' (a)}$ AG

[1 mark]

$b = 4$ (seen anywhere) A1

EITHER

attempts to find the gradient of the tangent in terms of $a$ and equates to $16$ (M1)

OR

substitutes $r = - 2, x = a$ and $y = 80$ to form $80 = (a - (- 2)) (a^{2} - 2 a^{2} + a^{2} + 16)$ (M1)

substitutes $r = - 2, x = a$ and $y = 80$ into $y = 16 (x - r)$ (M1)

THEN

$\frac{80}{a + 2} = 16 \Rightarrow a = 3$

roots are $- 2$ (seen anywhere) and $3 \pm 4 i$ A1A1

Note: Award A1 for $- 2$ and A1 for $3 \pm 4 i$ . Do not accept coordinates.

[4 marks]

f.i.

$(3, - 4)$ A1

Note: Accept “ $x = 3$ and $y = - 4$ ”.
Do not award A1FT for $(a, - 4)$ .

[1 mark]

f.ii.

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$

attempts to find $g'' (x)$ M1

$g'' (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a (= 6 x - 2 r - 4 a)$

sets $g'' (x) = 0$ and correctly solves for $x$ A1

for example, obtaining $x - r + 2 (x - a) = 0$ leading to $3 x = 2 a + r$

so $x = \frac{1}{3} (2 a + r)$ AG

Note: Do not award A1 if the answer does not lead to the AG.

[2 marks]

g.i.

point $P$ is $\frac{2}{3}$ of the horizontal distance (way) from point $R$ to point $A$ A1

Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ $P$ is between $R$ and $A$ , closer to $A$ ”.

[1 mark]

g.ii.

$y = (x - 1) (x^{2} - 2 x + 5)$ (A1)

a positive cubic with no stationary points and a non-stationary point of inflexion at $x = 1$ A1

Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of $x = 1$ is apparent.
Coordinates are not required and the $y$ -intercept need not be indicated.

[2 marks]

h.i.

$(r, 0)$ A1

[1 mark]

h.ii.

Examiners report

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4 - i$ as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of $4 + i$ and $4 - i$ .

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f' (x)$ and a good number of those candidates were able to determine that $f' (4) = 1$ . Candidates that did not determine the equation of the tangent had to state that the gradient of $y = x - 1$ is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find $f' (x)$ .

Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.

Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding $g (x)$ , then differentiating to find $g' (x)$ and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that $g' (a) = b^{2}$ and/or $g (a) = b^{2} (a - r)$ . Only the very best candidates obtained full marks by concluding that as $b > 0$ or $b \neq 0$ , then $x = r$ when $y = 0$ .

In general, only the best candidates were able to use the result $g' (a) = b^{2}$ to deduce that the complex roots of the equation can be expressed as $a \pm i \sqrt{g' (a)}$ . Although given the complex roots $a \pm b i$ , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation $z^{2} - 2 a z + a^{2} + b^{2} = 0$ .

In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state $- 2$ as a root. Some candidates determined that $b = 4$ but were unable to use the diagram to determine that $a = 3$ . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C₂ . The most frequent error was to give the y-coordinate as $3 - 4 i$ .

Of the candidates who attempted part (g) (i), most were able to find an expression for $g'' (x)$ and a reasonable number of these were then able to convincingly show that $x = \frac{1}{3} (2 a + r)$ . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated $x = \frac{1}{3} (2 a + r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y = (x - 1) (x^{2} - 2 x + 5)$ . However, most graphs were poorly drawn with many showing a change in concavity at $x = 0$ rather than at $x = 1$ . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).

a.i.

[N/A]

a.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

h.i.

[N/A]

h.ii.

This question will investigate methods for finding definite integrals of powers of trigonometrical functions.

Let ${I_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Let ${J_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{co}}{{\text{s}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}.$

Let ${T_n} = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Find the exact values of ${I_0}$ , ${I_1}$ and ${I_2}$ .

[6]

Use integration by parts to show that ${I_n} = \frac{{n - 1}}{n}{I_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[5]

b.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

b.ii.

Hence, find the exact values of ${I_3}$ and ${I_4}$ .

[2]

Use the substitution $x = \frac{\pi }{2} - u$ to show that ${J_n} = {I_n}$ .

[4]

Hence, find the exact values of ${J_{5}}$ and ${J_{6}}$

[2]

Find the exact values of ${T_{0}}$ and ${T_{1}}$ .

[3]

Use the fact that ${\text{ta}}{{\text{n}}^2}x = {\text{se}}{{\text{c}}^2}x - 1$ to show that ${T_n} = \frac{1}{{n - 1}} - {T_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[3]

g.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

g.ii.

Hence, find the exact values of ${T_{2}}$ and ${T_{3}}$ .

[2]

Markscheme

${I_0} = \int\limits_0^{\tfrac{\pi }{2}} 1 \,dx = \left[ x \right]_0^{\tfrac{\pi }{2}} = \frac{\pi }{2}$ M1A1

${I_1} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{sin}}\,x} \,dx = \left[ { - {\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}} = 1$ M1A1

${I_2} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^2}x} \,dx = \int\limits_0^{\tfrac{\pi }{2}} {\frac{{1 - {\text{cos}}\,2x}}{2}} \,dx = \left[ {\frac{x}{2} - \frac{{{\text{sin}}\,2x}}{4}} \right]_0^{\tfrac{\pi }{2}} = \frac{\pi }{4}$ M1A1

[6 marks]

${u = {\text{si}}{{\text{n}}^{n - 1}}x}$ ${v = - \cos x}$

${\frac{{du}}{{dx}} = \left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\,{\text{cos}}\,x}$ ${\frac{{dv}}{{dx}} = {\text{sin}}\,x}$

${I_n} = \left[ { - {\text{si}}{{\text{n}}^{n - 1}}x\,{\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}} + \int\limits_0^{\tfrac{\pi }{2}} {\left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\,{\text{cos}}{\,^2}x} \,dx$ M1A1A1

$= 0 + \int\limits_0^{\tfrac{\pi }{2}} {\left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\left( {1 - {\text{si}}{{\text{n}}^2}x} \right)} \,dx = \left( {n - 1} \right)\left( {{I_{n - 2}} - {I_n}} \right)$ M1A1

$\Rightarrow n{I_n} = \left( {n - 1} \right){I_{n - 2}} \Rightarrow {I_n} = \frac{{\left( {n - 1} \right)}}{n}{I_{n - 2}}$ AG

[6 marks]

b.i.

need $n \geqslant 2$ so that ${\text{si}}{{\text{n}}^{n - 1}}\tfrac{\pi }{2} = 0$ in $\left[ { - {\text{si}}{{\text{n}}^{n - 1}}\,x\,{\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}}$ R1

[1 mark]

b.ii.

${I_3} = \frac{2}{3}{I_1} = \frac{2}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,{I_4} = \frac{3}{4}{I_2} = \frac{{3\pi }}{{16}}$ A1A1

[2 marks]

$x = \frac{\pi }{2} - u \Rightarrow \frac{{dx}}{{du}} = - 1$ A1

${J_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{co}}{{\text{s}}^n}} x\,dx\,\, = \int\limits_{\tfrac{\pi }{2}}^0 { - {\text{co}}{{\text{s}}^n}} \left( {\frac{\pi }{2} - u} \right)\,du = - \int\limits_{\tfrac{\pi }{2}}^0 {{\text{si}}{{\text{n}}^n}} u\,du\, = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^n}} u\,du\, = {I_n}$ M1A1A1AG

[4 marks]

${J_5} = {I_5} = \frac{4}{5}{I_3} = \frac{4}{5} \times \frac{2}{3} = \frac{8}{{15}}\,\,\,\,\,\,\,\,\,\,{J_6} = {I_6} = \frac{5}{6}{I_4} = \frac{5}{6} \times \frac{{3\pi }}{{16}} = \frac{{5\pi }}{{32}}$ A1A1

[2 marks]

${T_{0\,}}{\text{ = }}\int\limits_0^{\tfrac{\pi }{4}} {1\,dx} = \left[ x \right]_0^{\tfrac{\pi }{4}} = \frac{\pi }{4}$ A1

${T_{1\,}}{\text{ = }}\int\limits_0^{\tfrac{\pi }{4}} {{\text{tan}}\,dx} = \left[ { - \ln \left| {{\text{cos}}\,x} \right|} \right]_0^{\tfrac{\pi }{4}} = - {\text{ln}}\frac{1}{{\sqrt 2 }} = {\text{ln}}\sqrt 2$ M1A1

[3 marks]

${T_n} = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^n}} x\,dx\, = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{ta}}{{\text{n}}^2}x\,dx\, = \,\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\left( {{\text{se}}{{\text{c}}^2}x - 1} \right)\,dx$ M1

$\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{se}}{{\text{c}}^2}x\,dx - \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,dx = \left[ {\frac{{{\text{ta}}{{\text{n}}^{n - 1}}x}}{{n - 1}}} \right]_0^{\tfrac{\pi }{4}} - {T_{n - 2}} = \frac{1}{{n - 1}} - {T_{n - 2}}\,\,\,$ A1A1AG

[3 marks]

g.i.

need $n \geqslant 2$ so that the powers of tan in $\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{se}}{{\text{c}}^2}x\,dx - \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,dx$ are not negative R1

[1 mark]

g.ii.

${T_2} = 1 - {T_0} = 1 - \frac{\pi }{4}$ A1

${T_3} = \frac{1}{2} - {T_1} = \frac{1}{2} - \ln \sqrt 2$ A1

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

This question asks you to investigate regular $n$ -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as $n$ tends to infinity, to make an approximation for $\pi$ .

Let ${P_i}\left( n \right)$ represent the perimeter of any $n$ -sided regular polygon inscribed in a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.

Let ${P_c}\left( n \right)$ represent the perimeter of any $n$ -sided regular polygon circumscribed about a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.

The equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of $\frac{{2\pi }}{3}$ at O, as shown in the following diagram.

Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to $3\sqrt 3$ units.

[3]

Consider a square of side length, $x$ units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.

[3]

Find the perimeter of a regular hexagon, of side length, $x$ units, inscribed in a circle of radius 1 unit.

[2]

Show that ${P_i}\left( n \right) = 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ .

[3]

Use an appropriate Maclaurin series expansion to find $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_i}\left( n \right)$ and interpret this result geometrically.

[5]

Show that ${P_c}\left( n \right) = 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ .

[4]

By writing ${P_c}\left( n \right)$ in the form $\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}$ , find $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_c}\left( n \right)$ .

[5]

Use the results from part (d) and part (f) to determine an inequality for the value of $\pi$ in terms of $n$ .

[2]

The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $\pi$ .

Determine the least value for $n$ such that the lower bound and upper bound approximations are both within 0.005 of $\pi$ .

[3]

Markscheme

METHOD 1

consider right-angled triangle OCX where CX $= \frac{x}{2}$

${\text{sin}}\frac{\pi }{3} = \frac{{\frac{x}{2}}}{1}$ M1A1

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

${P_i} = 3 \times x = 3\sqrt 3$ AG

METHOD 2

eg use of the cosine rule ${x^2} = {1^2} + {1^2} - 2\left( 1 \right)\left( 1 \right){\text{cos}}\frac{{2\pi }}{3}$ M1A1

$x = \sqrt 3$ A1

${P_i} = 3 \times x = 3\sqrt 3$ AG

Note: Accept use of sine rule.

[3 marks]

${\text{sin}}\frac{\pi }{4} = \frac{1}{x}$ where $x$ = side of square M1

$x = \sqrt 2$ A1

${P_i} = 4\sqrt 2$ A1

[3 marks]

6 equilateral triangles ⇒ $x$ = 1 A1

${P_i} = 6$ A1

[2 marks]

in right-angled triangle ${\text{sin}}\left( {\frac{\pi }{n}} \right) = \frac{{\frac{x}{2}}}{1}$ M1

$\Rightarrow x = 2\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ A1

${P_i} = n \times x$

${P_i} = n \times 2\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ M1

${P_i} = 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ AG

[3 marks]

consider $\mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$

use of ${\text{sin}}\,x = x - \frac{{{x^3}}}{{3{\text{!}}}} + \frac{{{x^5}}}{{5{\text{!}}}} - \ldots$ M1

$2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) = 2n\left( {\frac{\pi }{n} - \frac{{{\pi ^3}}}{{6{n^3}}} + \frac{{{\pi ^5}}}{{120{n^5}}} - \ldots } \right)$ (A1)

$= 2\left( {\pi - \frac{{{\pi ^3}}}{{6{n^2}}} + \frac{{{\pi ^5}}}{{120{n^4}}} - \ldots } \right)$ A1

$\Rightarrow \mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) = 2\pi$ A1

as $n \to \infty$ polygon becomes a circle of radius 1 and ${P_i} = 2\pi$ R1

[5 marks]

consider an $n$ -sided polygon of side length $x$

2 $n$ right-angled triangles with angle $\frac{{2\pi }}{{2n}} = \frac{\pi }{n}$ at centre M1A1

opposite side $\frac{x}{2} = {\text{tan}}\left( {\frac{\pi }{n}} \right) \Rightarrow x = 2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ M1A1

Perimeter ${P_c} = 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ AG

[4 marks]

consider $\mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right) = \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}} \right)$

$= \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}} \right) = \frac{0}{0}$ R1

attempt to use L’Hopital’s rule M1

$= \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{ - \frac{{2\pi }}{{{n^2}}}{\text{se}}{{\text{c}}^2}\left( {\frac{\pi }{n}} \right)}}{{ - \frac{1}{{{n^2}}}}}} \right)$ A1A1

$= 2\pi$ A1

[5 marks]

${P_i} < 2\pi < {P_c}$

$2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) < 2\pi < 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ M1

$n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) < \pi < n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ A1

[2 marks]

attempt to find the lower bound and upper bound approximations within 0.005 of $\pi$ (M1)

$n$ = 46 A2

[3 marks]

Examiners report

[N/A]

Consider the differential equation $x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y = {x^p} + 1$ where $x \in \mathbb{R},\,x \ne 0$ and $p$ is a positive integer, $p > 1$ .

Solve the differential equation given that $y = - 1$ when $x = 1$ . Give your answer in the form $y = f\left( x \right)$ .

[8]

Show that the $x$ -coordinate(s) of the points on the curve $y = f\left( x \right)$ where $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ satisfy the equation ${x^{p - 1}} = \frac{1}{p}$ .

[2]

b.i.

Deduce the set of values for $p$ such that there are two points on the curve $y = f\left( x \right)$ where $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ . Give a reason for your answer.

[2]

b.ii.

Markscheme

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

METHOD 1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} = {x^{p - 1}} + \frac{1}{x}$ (M1)

integrating factor $= {{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }}$ M1

${\text{ = }}{{\text{e}}^{ - {\text{ln}}\,x}}$ (A1)

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

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

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

$\frac{y}{x} = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C$ A1

Note: Condone the absence of C.

$y = \frac{1}{{p - 1}}{x^p} + Cx - 1$

substituting $x = 1$ , $y = - 1 \Rightarrow C = - \frac{1}{{p - 1}}$ M1

Note: Award M1 for attempting to find their value of C.

$y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1$ A1

[8 marks]

METHOD 2

put $y = vx$ so that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$ M1(A1)

substituting, M1

$x\left( {v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}} \right) - vx = {x^p} + 1$ (A1)

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 1}} + \frac{1}{x}$ M1

$\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 2}} + \frac{1}{{{x^2}}}$

$v = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C$ A1

Note: Condone the absence of C.

$y = \frac{1}{{p - 1}}{x^p} + Cx - 1$

substituting $x = 1$ , $y = - 1 \Rightarrow C = - \frac{1}{{p - 1}}$ M1

Note: Award M1 for attempting to find their value of C.

$y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1$ A1

[8 marks]

METHOD 1

find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ and solve $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ for $x$

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

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow p{x^{p - 1}} - 1 = 0$ A1

$p{x^{p - 1}} = 1$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

${x^{p - 1}} = \frac{1}{p}$ AG

METHOD 2

substitute $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ and their $y$ into the differential equation and solve for $x$

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow - \left( {\frac{{{x^p} - x}}{{p - 1}}} \right) + 1 = {x^p} + 1$ M1

${x^p} - x = {x^p} - p{x^p}$ A1

$p{x^{p - 1}} = 1$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

${x^{p - 1}} = \frac{1}{p}$ AG

[2 marks]

b.i.

there are two solutions for $x$ when $p$ is odd (and $p > 1$ A1

if $p - 1$ is even there are two solutions (to ${x^{p - 1}} = \frac{1}{p}$ )

and if $p - 1$ is odd there is only one solution (to ${x^{p - 1}} = \frac{1}{p}$ ) R1

Note: Only award the R1 if both cases are considered.

[4 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

This question asks you to explore properties of a family of curves of the type $y^{2} = x^{3} + a x + b$ for various values of $a$ and $b$ , where $a, b \in ℕ$ .

On the same set of axes, sketch the following curves for $- 2 \leq x \leq 2$ and $- 2 \leq y \leq 2$ , clearly indicating any points of intersection with the coordinate axes.

Now, consider curves of the form $y^{2} = x^{3} + b$ , for $x \geq - \sqrt[3]{b}$ , where $b \in ℤ^{+}$ .

Next, consider the curve $y^{2} = x^{3} + x, x \geq 0$ .

The curve $y^{2} = x^{3} + x$ has two points of inflexion. Due to the symmetry of the curve these points have the same $x$ -coordinate.

$P (x, y)$ is defined to be a rational point on a curve if $x$ and $y$ are rational numbers.

The tangent to the curve $y^{2} = x^{3} + a x + b$ at a rational point $P$ intersects the curve at another rational point $Q$ .

Let $C$ be the curve $y^{2} = x^{3} + 2$ , for $x \geq - \sqrt[3]{2}$ . The rational point $P (- 1, - 1)$ lies on $C$ .

$y^{2} = x^{3}, x \geq 0$

[2]

a.i.

$y^{2} = x^{3} + 1, x \geq - 1$

[2]

a.ii.

Write down the coordinates of the two points of inflexion on the curve $y^{2} = x^{3} + 1$ .

[1]

b.i.

By considering each curve from part (a), identify two key features that would distinguish one curve from the other.

[1]

b.ii.

By varying the value of $b$ , suggest two key features common to these curves.

[2]

Show that $\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ , for $x > 0$ .

[3]

d.i.

Hence deduce that the curve $y^{2} = x^{3} + x$ has no local minimum or maximum points.

[1]

d.ii.

Find the value of this $x$ -coordinate, giving your answer in the form $x = \sqrt{\frac{p \sqrt{3} + q}{r}}$ , where $p, q, r \in ℤ$ .

[7]

Find the equation of the tangent to $C$ at $P$ .

[2]

f.i.

Hence, find the coordinates of the rational point $Q$ where this tangent intersects $C$ , expressing each coordinate as a fraction.

[2]

f.ii.

The point $S (- 1, 1)$ also lies on $C$ . The line $[QS]$ intersects $C$ at a further point. Determine the coordinates of this point.

[5]

Markscheme

approximately symmetric about the $x$ -axis graph of $y^{2} = x^{3}$ A1

including cusp/sharp point at $(0, 0)$ A1

[2 marks]

Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at $x$ -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.i.

approximately symmetric about the $x$ -axis graph of $y^{2} = x^{3} + 1$ with approximately correct gradient at axes intercepts A1
some indication of position of intersections at $x = - 1$ , $y = \pm 1$ A1

[2 marks]

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.ii.

$(0, 1)$ and $(0, - 1)$ A1

[1 mark]

b.i.

Any two from:

$y^{2} = x^{3}$ has a cusp/sharp point, (the other does not)

graphs have different domains

$y^{2} = x^{3} + 1$ has points of inflexion, (the other does not)

graphs have different $x$ -axis intercepts (one goes through the origin, and the other does not)

graphs have different $y$ -axis intercepts A1

Note: Follow through from their sketch in part (a)(i). In accordance with marking rules, mark their first two responses and ignore any subsequent.

[1 mark]

b.ii.

Any two from:

as , $x \to \infty, y \to \pm \infty$

as $x \to \infty, y^{2} = x^{3} + b$ is approximated by $y^{2} = x^{3}$ (or similar)

they have $x$ intercepts at $x = - \sqrt[3]{b}$

they have $y$ intercepts at $y = (\pm) \sqrt{b}$

they all have the same range

$y = 0$ (or $x$ -axis) is a line of symmetry

they all have the same line of symmetry $(y = 0)$

they have one $x$ -axis intercept

they have two $y$ -axis intercepts

they have two points of inflexion

at $x$ -axis intercepts, curve is vertical/infinite gradient

there is no cusp/sharp point at $x$ -axis intercepts A1A1

Note: The last example is the only valid answer for things “not” present. Do not credit an answer of “they are all symmetrical” without some reference to the line of symmetry.

Note: Do not allow same/ similar shape or equivalent.

Note: In accordance with marking rules, mark their first two responses and ignore any subsequent.

[2 marks]

METHOD 1

attempt to differentiate implicitly M1

$2 y \frac{d y}{d x} = 3 x^{2} + 1$ A1

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

$\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ AG

METHOD 2

attempt to use chain rule $y = (\pm) \sqrt{x^{3} + x}$ M1

$\frac{d y}{d x} = (\pm) \frac{1}{2} {(x^{3} + x)}^{- \frac{1}{2}} (3 x^{2} + 1)$ A1A1

Note: Award A1 for $(\pm) \frac{1}{2} {(x^{3} + x)}^{- \frac{1}{2}}$ , A1 for $(3 x^{2} + 1)$

$\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ AG

[3 marks]

d.i.

EITHER

local minima/maxima occur when $\frac{d y}{d x} = 0$

$1 + 3 x^{2} = 0$ has no (real) solutions (or equivalent) R1

$(x^{2} \geq 0 \Rightarrow) 3 x^{2} + 1 > 0$ , so $\frac{d y}{d x} \neq 0$ R1

THEN

so, no local minima/maxima exist AG

[1 mark]

d.ii.

EITHER

attempt to use quotient rule to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = (\pm) \frac{12 x \sqrt{x + x^{3}} - (1 + 3 x^{2}) {(x + x^{3})}^{- \frac{1}{2}} (1 + 3 x^{2})}{4 (x + x^{3})}$ A1A1

Note: Award A1 for correct $12 x \sqrt{x + x^{3}}$ and correct denominator, A1 for correct $- (1 + 3 x^{2}) {(x + x^{3})}^{- \frac{1}{2}} (1 + 3 x^{2})$ .

Note: Future A marks may be awarded if the denominator is missing or incorrect.

stating or using $\frac{d^{2} y}{d x^{2}} = 0$ (may be seen anywhere) (M1)

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

attempt to use product rule to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = \frac{1}{2} (3 x^{2} + 1) (- \frac{1}{2}) (3 x^{2} + 1) {(x^{3} + x)}^{- \frac{3}{2}} + 3 x {(x^{3} + x)}^{- \frac{1}{2}}$ A1A1

Note: Award A1 for correct first term, A1 for correct second term.

setting $\frac{d^{2} y}{d x^{2}} = 0$ (M1)

attempts implicit differentiation on $2 y \frac{d y}{d x} = 3 x^{2} + 1$ M1

$2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 6 x$ A1

recognizes that $\frac{d^{2} y}{d x^{2}} = 0$ (M1)

$\frac{d y}{d x} = \pm \sqrt{3 x}$

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

THEN

$12 x (x + x^{3}) = {(1 + 3 x^{2})}^{2}$

$12 x^{2} + 12 x^{4} = 9 x^{4} + 6 x^{2} + 1$

$3 x^{4} + 6 x^{2} - 1 = 0$ A1

attempt to use quadratic formula or equivalent (M1)

$x^{2} = \frac{- 6 \pm \sqrt{48}}{6}$

$(x > 0 \Rightarrow) x = \sqrt{\frac{2 \sqrt{3} - 3}{3}} (p = 2, q = - 3, r = 3)$ A1

Note: Accept any integer multiple of $p, q$ and $r$ (e.g. $4, - 6$ and $6$ ).

[7 marks]

attempt to find tangent line through $(- 1, - 1)$ (M1)

$y + 1 = - \frac{3}{2} (x + 1)$ OR $y = - 1.5 x - 2.5$ A1

[2 marks]

f.i.

attempt to solve simultaneously with $y^{2} = x^{3} + 2$ (M1)

Note: The M1 mark can be awarded for an unsupported correct answer in an incorrect format (e.g. $(4.25, - 8.875)$ ).

obtain $(\frac{17}{4}, - \frac{71}{8})$ A1

[2 marks]

f.ii.

attempt to find equation of $[QS]$ (M1)

$\frac{y - 1}{x + 1} = - \frac{79}{42} (= - 1.88095 \dots)$ (A1)

solve simultaneously with $y^{2} = x^{3} + 2$ (M1)

$x = 0.28798 \dots (= \frac{127}{441})$ A1

$y = - 1.4226 \dots (= \frac{13175}{9261})$ A1

$(0.228, - 1.42)$

attempt to find vector equation of $[QS]$ (M1)

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 1 \\ 1 \end{matrix}) + λ (\begin{matrix} \frac{21}{4} \\ - \frac{79}{8} \end{matrix})$ (A1)

$x = - 1 + \frac{21}{4} λ$

$y = 1 - \frac{79}{8} λ$

attempt to solve ${(1 - \frac{79}{8} λ)}^{2} = {(- 1 + \frac{21}{4} λ)}^{3} + 2$ (M1)

$λ = 0.2453 \dots$

$x = 0.28798 \dots (= \frac{127}{441})$ A1

$y = - 1.4226 \dots (= \frac{13175}{9261})$ A1

$(0.228, - 1.42)$

[5 marks]

Examiners report

This was a relatively straightforward start, though it was disappointing to see so many candidates sketch their graphs on two separate axes, despite the question stating they should be sketched on the same axes.

a.i.

[N/A]

a.ii.

Of those candidates producing clear sketches, the vast majority were able to recognise the points of inflexion and write down their coordinates. A small number embarked on a mostly fruitless algebraic approach rather than use their graphs as intended. The distinguishing features between curves tended to focus on points of intersection with the axes, which was accepted. Only a small number offered ideas such as $y \to \infty$ on both curves. A number of (incorrect) suggestions were seen, stating that both curves tended towards a linear asymptote.

b.i.

[N/A]

b.ii.

A majority of candidates' suggestions related to the number of intersection points with the coordinate axes, while the idea of the x-axis acting as a line of symmetry was also often seen.

The required differentiation was straightforward for the majority of candidates.

d.i.

[N/A]

d.ii.

The majority employed the quotient rule here, often doing so successfully to find a correct expression for $\frac{d^{2} y}{d x^{2}}$ . Despite realising that $\frac{d^{2} y}{d x^{2}} = 0$ , the resulting algebra to find the required solution proved a step too far for most. A number of slips were seen in candidates' working, though better candidates were able to answer the question confidently.

Mistakes proved to be increasingly common by this stage of the paper. Various equations of lines were suggested, with the incorrect $y = 1.5 x + 2.5$ appearing more than once. Only the better candidates were able to tackle the final part of the question with any success; it was pleasing to see a number of clear algebraic (only) approaches, though this was not necessary to obtain full marks.

f.i.

[N/A]

f.ii.

Significant work on this question part was rarely seen, and it may have been the case that many candidates chose to spend their remaining time on the second question, especially if they felt they were making little progress with part f. Having said that, correct final answers were seen from better candidates, though these were few and far between.

The function $f$ is defined by $f (x) = \ln (1 + x^{2})$ where $- 1 < x < 1$ .

The seventh derivative of $f$ is given by $f^{(7)} (x) = \frac{1440 x (x^{6} - 21 x^{4} + 35 x^{2} - 7)}{{(1 + x^{2})}^{7}}$ .

Use the Maclaurin series for $\ln (1 + x)$ to write down the first three non-zero terms of the Maclaurin series for $f (x)$ .

[2]

a.i.

Hence find the first three non-zero terms of the Maclaurin series for $\frac{x}{1 + x^{2}}$ .

[4]

a.ii.

Use your answer to part (a)(i) to write down an estimate for $f (0.4)$ .

[1]

Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating $f (0.4)$ , using the first three non-zero terms of the Maclaurin series for $f (x)$ .

[6]

c.i.

With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for $f (0.4)$ .

[2]

c.ii.

Markscheme

substitution of $x^{2}$ in $\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots$ (M1)

$x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{3}$ A1

[2 marks]

a.i.

$\frac{d}{d x} (\ln (1 + x^{2})) = \frac{2 x}{1 + x^{2}}$ (M1)

Note: Award (M1) if this is seen in part (a)(i).

attempt to differentiate their answer in part (a) (M1)

$\frac{2 x}{1 + x^{2}} = 2 x - \frac{4 x^{3}}{2} + \frac{6 x^{5}}{3}$ M1

Note: Award M1 for equating their derivatives.

$\frac{x}{1 + x^{2}} = x - x^{3} + x^{5}$ A1

[4 marks]

a.ii.

$f (0.4) \approx 0.149$ A1

Note: Accept an answer that rounds correct to $2 s.f.$ or better.

[1 mark]

attempt to find the maximum of $|f^{(7)} (c)|$ for $c \in [0, 0.4]$ (M1)

maximum of $|f^{(7)} (c)|$ occurs at $c = 0.199$ (A1)

$|f^{(7)} (c)| < 1232.97 \dots$ (for all $c \in] 0, 0.4 [$ ) (A1)

use of $x = 0.4$ (M1)

substitution of $n = 6$ and $a = 0$ and their value of $x$ and their value of $f^{(7)} (c)$ into Lagrange error term (M1)

Note: Award (M1) for substitution of $n = 3$ and $a = 0$ and their value of $x$ and their value of $f^{(4)} (c)$ into Lagrange error term.

$|R_{6} (0.4)| < \frac{1232.97 {(0.4)}^{7}}{7!}$

upper bound $= 0.000401$ A1

Note: Accept an answer that rounds correct to $1 s.f$ or better.

[6 marks]

c.i.

$f^{(7)} (c) < 0$ (for all $c \in] 0, 0.4 [$ ) R1

Note: Accept $R_{6} (c) < 0$ or “the error term is negative”.

the answer in (b) is an overestimate A1

Note: The A1 is dependent on the R1.

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

The curve $y = f (x)$ has a gradient function given by

$\frac{d y}{d x} = x - y$ .

The curve passes through the point $(1, 1)$ .

On the same set of axes, sketch and label isoclines for $\frac{d y}{d x} = - 1, 0$ and $1$ , and clearly indicate the value of each $y$ -intercept.

[3]

a.i.

Hence or otherwise, explain why the point $(1, 1)$ is a local minimum.

[3]

a.ii.

Find the solution of the differential equation $\frac{d y}{d x} = x - y$ , which passes through the point $(1, 1)$ . Give your answer in the form $y = f (x)$ .

[8]

Explain why the graph of $y = f (x)$ does not intersect the isocline $\frac{d y}{d x} = 1$ .

[2]

c.i.

Sketch the graph of $y = f (x)$ on the same set of axes as part (a)(i).

[2]

c.ii.

Markscheme

attempt to find equation of isoclines by setting $x - y = - 1, 0, 1$ M1

$3$ parallel lines with positive gradient A1

$y$ -intercept $= - c$ for $\frac{d y}{d x} = c$ A1

Note: To award A1, each $y$ -intercept should be clear, but condone a missing label (eg. $(0, 0)$ ).

If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.

[3 marks]

a.i.

at point $(1, 1), \frac{d y}{d x} = 0$ A1

EITHER

to the left of $(1, 1)$ , the gradient is negative R1

to the right of $(1, 1)$ , the gradient is positive R1

Note: Accept any correct reasoning using gradient, isoclines or slope field.

If a candidate uses left/right or $x < 1 / x > 1$ without explicitly referring to the point $(1, 1)$ or a correct region on the diagram, award R0R1.

$\frac{d^{2} y}{d x^{2}} = 1 - \frac{d y}{d x}$ A1

$\frac{d^{2} y}{d x^{2}} = 1 (> 0)$ A1

Note: accept correct reasoning $\frac{d y}{d x}$ that is increasing as $x$ increases.

THEN

hence $(1, 1)$ is a local minimum AG

[3 marks]

a.ii.

integrating factor $= e^{\int d x}$ (M1)

$= e^{x}$ (A1)

$\frac{d y}{d x} e^{x} + y e^{x} = x e^{x}$ (M1)

$y e^{x} = \int x e^{x} d x$ A1

$= x e^{x} - \int e^{x} d x$ (M1)

$= x e^{x} - e^{x} (+ c)$ A1

Note: Award A1 for the correct RHS.

substituting $(1, 1)$ gives

$e = e - e + c$ M1

$c = e$

$y = x - 1 + e^{1 - x}$ A1

[8 marks]

METHOD 1

EITHER

attempt to solve for the intersection $x - 1 + e^{1 - x} = x - 1$ (M1)

attempt to find the difference $x - 1 + e^{1 - x} - (x - 1)$ (M1)

THEN

$e^{1 - x} > 0$ for all $x$ R1

Note: Accept $e^{1 - x} \neq 0$ or equivalent reasoning.

therefore the curve does not intersect the isocline AG

METHOD 2

$y = x - 1$ is an (oblique) asymptote to the curve R1

Note: Do not accept “the curve is parallel to $y = x - 1$ "

$y = x - 1$ is the isocline for $\frac{d y}{d x} = 1$ R1

therefore the curve does not intersect the isocline AG

METHOD 3

The initial point is above $y = x - 1$ , so $\frac{d y}{d x} < 1$ R1

$\Rightarrow x - y < 1$

$\Rightarrow y > x - 1$ R1

therefore the curve does not intersect the isocline AG

[2 marks]

c.i.

concave up curve with minimum at approximately $(1, 1)$ A1

asymptote of curve is isocline $y = x - 1$ A1

Note: Only award FT from (b) if the above conditions are satisfied.

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

Consider the differential equation

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {\frac{y}{x}} \right),{\text{ }}x > 0.$

Use the substitution $y = vx$ to show that the general solution of this differential equation is

$\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant.}}$

[3]

Hence, or otherwise, solve the differential equation

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{x^2} + 3xy + {y^2}}}{{{x^2}}},{\text{ }}x > 0,$

given that $y = 1$ when $x = 1$ . Give your answer in the form $y = g(x)$ .

[10]

Markscheme

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

$y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$ M1

the differential equation becomes

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = f(v)$ A1

$\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \int {\frac{{{\text{d}}v}}{x}} }$ A1

integrating, Constant $\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant}}$ AG

[3 marks]

EITHER

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

$\left( {\int {\frac{{{\text{d}}v}}{{f(v) - v}} = } } \right)\,\,\,\int {\frac{{{\text{d}}v}}{{1 + 3v + {v^2} - v}} = \ln x + C}$ M1A1

$\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}} = (\ln x + C)}$ A1

Note: A1 is for correct factorization.

$- \frac{1}{{1 + v}}\,\,\,( = \ln x + C)$ A1

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 + 3v + {v^2}$ A1

$\int {\frac{{{\text{d}}v}}{{1 + 2v + {v^2}}} = \int {\frac{1}{x}{\text{d}}x} }$ M1

$\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}}\,\,\,\left( { = \int {\frac{1}{x}{\text{d}}x} } \right)}$ (A1)

Note: A1 is for correct factorization.

$- \frac{1}{{1 + v}} = \ln x( + C)$ A1A1

THEN

substitute $y = 1$ or $v = 1$ when $x = 1$ (M1)

therefore $C = - \frac{1}{2}$ A1

Note: This A1 can be awarded anywhere in their solution.

substituting for $v$ ,

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

Note: Award for correct substitution of $\frac{y}{x}$ into their expression.

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

Note: Award for any rearrangement of a correct expression that has $y$ in the numerator.

$y = x\left( {\frac{1}{{\left( {\frac{1}{2} - \ln x} \right)}} - 1} \right)\,\,\,{\text{(or equivalent)}}$ A1

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

[10 marks]

Examiners report

[N/A]

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

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\left( {2x} \right) - 4}}{{3{x^2}}}} \right)$

[5]

Hence find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\int_0^x {\left( {{{\text{e}}^{ - 3t}}^{^2} + 3\,{\text{cos}}\left( {2t} \right) - 4} \right)} \,{\text{d}}t}}{{\int_0^x {3{t^2}} \,{\text{d}}t}}} \right)$ .

[3]

Markscheme

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

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\,2x - 4}}{{3{x^2}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - {\text{6}}x{{\text{e}}^{ - 3x}}^{^2} - 6\,{\text{sin}}\,2x}}{{6x}} = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$ M1A1A1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - {\text{6}}{{\text{e}}^{ - 3x}}^{^2} + 36{x^2}{{\text{e}}^{ - 3x}}^{^2} - 12\,{\text{cos}}\,2x}}{6}$ A1

= −3 A1

[5 marks]

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\int_0^x {\left( {{{\text{e}}^{ - 3t}}^{^2} + 3\,{\text{cos}}\,2t - 4} \right)} \,{\text{d}}t}}{{\int_0^x {3{t^2}} \,{\text{d}}t}}} \right)$ is of the form $\frac{0}{0}$

applying l’Hôpital´s rule (M1)

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\,2x - 4}}{{3{x^2}}}$ (A1)

= −3 A1

[3 marks]

Examiners report

[N/A]

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y = x$ where $y = 1$ when $x = 0$ .

Show that $\sqrt {{x^2} + 1}$ is an integrating factor for this differential equation.

[4]

Solve the differential equation giving your answer in the form $y = f(x)$ .

[6]

Markscheme

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

METHOD 1

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

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

Note: Award M1 for use of $u = {x^2} + 1$ for example or $\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \ln f(x)}$ .

integrating factor $= {{\text{e}}^{\frac{1}{2}\ln ({x^2} + 1)}}$ A1

$= {{\text{e}}^{\ln \left( {\sqrt {{x^2} + 1} } \right)}}$ A1

Note: Award A1 for ${{\text{e}}^{\ln \sqrt u }}$ where $u = {x^2} + 1$ .

$= \sqrt {{x^2} + 1}$ AG

METHOD 2

$\frac{{\text{d}}}{{{\text{d}}x}}\left( {y\sqrt {{x^2} + 1} } \right) = \frac{{{\text{d}}y}}{{{\text{d}}x}}\sqrt {{x^2} + 1} + \frac{x}{{\sqrt {{x^2} + 1} }}y$ M1A1

$\sqrt {{x^2} + 1} \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y} \right)$ M1A1

Note: Award M1 for attempting to express in the form $\sqrt {{x^2} + 1} \times {\text{(LHS of de)}}$ .

so $\sqrt {{x^2} + 1}$ is an integrating factor for this differential equation AG

[4 marks]

$\sqrt {{x^2} + 1} \frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{\sqrt {{x^2} + 1} }}y = x\sqrt {{x^2} + 1}$ (or equivalent) (M1)

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

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

$= \frac{1}{3}{({x^2} + 1)^{\frac{3}{2}}} + C$ (M1)A1

Note: Award M1 for using an appropriate substitution.

Note: Condone the absence of $C$ .

substituting $x = 0,{\text{ }}y = 1 \Rightarrow C = \frac{2}{3}$ M1

Note: Award M1 for attempting to find their value of $C$ .

$y = \frac{1}{3}({x^2} + 1) + \frac{2}{{3\sqrt {{x^2} + 1} }}{\text{ }}\left( {y = \frac{{{{({x^2} + 1)}^{\frac{3}{2}}} + 2}}{{3\sqrt {{x^2} + 1} }}} \right)$ A1

[6 marks]

Examiners report

[N/A]

This question asks you to explore the behaviour and key features of cubic polynomials of the form $x^{3} - 3 c x + d$ .

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

The graphs of $y = f (x)$ for $c = - 1$ and $c = 0$ are shown in the following diagrams.

$c = - 1$ $c = 0$

On separate axes, sketch the graph of $y = f (x)$ showing the value of the $y$ -intercept and the coordinates of any points with zero gradient, for

Hence, or otherwise, find the set of values of $c$ such that the graph of $y = f (x)$ has

Given that the graph of $y = f (x)$ has one local maximum point and one local minimum point, show that

Hence, for $c > 0$ , find the set of values of $c$ such that the graph of $y = f (x)$ has

$c = 1$ .

[3]

a.i.

$c = 2$ .

[3]

a.ii.

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

[1]

a point of inflexion with zero gradient.

[1]

c.i.

one local maximum point and one local minimum point.

[2]

c.ii.

no points where the gradient is equal to zero.

[1]

c.iii.

the $y$ -coordinate of the local maximum point is $2 c^{\frac{3}{2}} + 2$ .

[3]

d.i.

the $y$ -coordinate of the local minimum point is $- 2 c^{\frac{3}{2}} + 2$ .

[1]

d.ii.

exactly one $x$ -axis intercept.

[2]

e.i.

exactly two $x$ -axis intercepts.

[2]

e.ii.

exactly three $x$ -axis intercepts.

[2]

e.iii.

Consider the function $g (x) = x^{3} - 3 c x + d$ for $x \in ℝ$ and where $c, d \in ℝ$ .

Find all conditions on $c$ and $d$ such that the graph of $y = g (x)$ has exactly one $x$ -axis intercept, explaining your reasoning.

[6]

Markscheme

$c = 1$ : positive cubic with correct $y$ -intercept labelled A1

local maximum point correctly labelled A1

local minimum point correctly labelled A1

[3 marks]

a.i.

$c = 2$ : positive cubic with correct $y$ -intercept labelled A1

local maximum point correctly labelled A1

local minimum point correctly labelled A1

Note: Accept the following exact answers:
Local maximum point coordinates $(- \sqrt{2}, 2 + 4 \sqrt{2})$ .
Local minimum point coordinates $(\sqrt{2}, 2 - 4 \sqrt{2})$ .

[3 marks]

a.ii.

$f' (x) = 3 x^{2} - 3 c$ A1

Note: Accept $3 x^{2} - 3 c$ (an expression).

[1 mark]

$c = 0$ A1

[1 mark]

c.i.

considers the number of solutions to their $f' (x) = 0$ (M1)

$3 x^{2} - 3 c = 0$

$c > 0$ A1

[2 marks]

c.ii.

$c < 0$ A1

Note: The (M1) in part (c)(ii) can be awarded for work shown in either (ii) or (iii).

[1 mark]

c.iii.

attempts to solve their $f' (x) = 0$ for $x$ (M1)

$x \pm \sqrt{c}$ (A1)

Note: Award (A1) if either $x = - \sqrt{c}$ or $x = \sqrt{c}$ is subsequently considered.
Award the above (M1)(A1) if this work is seen in part (c).

correctly evaluates $f (- \sqrt{c})$ A1

$f (- \sqrt{c}) = - c^{\frac{3}{2}} + 3 c^{\frac{3}{2}} + 2 (= - c \sqrt{c} + 3 c \sqrt{c} + 2)$

the $y$ -coordinate of the local maximum point is $2 c^{\frac{3}{2}} + 2$ AG

[3 marks]

d.i.

correctly evaluates $f (\sqrt{c})$ A1

$f (\sqrt{c}) = c^{\frac{3}{2}} - 3 c^{\frac{3}{2}} + 2 (= c \sqrt{c} - 3 c \sqrt{c} + 2)$

the $y$ -coordinate of the local minimum point is $- 2 c^{\frac{3}{2}} + 2$ AG

[1 mark]

d.ii.

the graph of $y = f (x)$ will have one $x$ -axis intercept if

EITHER

$- 2 c^{\frac{3}{2}} + 2 > 0$ (or equivalent reasoning) R1

the minimum point is above the $x$ -axis R1

Note: Award R1 for a rigorous approach that does not (only) refer to sketched graphs.

THEN

$0 < c < 1$ A1

Note: Condone $c < 1$ . The A1 is independent of the R1.

[2 marks]

e.i.

the graph of $y = f (x)$ will have two $x$ -axis intercepts if

EITHER

$- 2 c^{\frac{3}{2}} + 2 = 0$ (or equivalent reasoning) (M1)

evidence from the graph in part(a)(i) (M1)

THEN

$c = 1$ A1

[2 marks]

e.ii.

the graph of $y = f (x)$ will have three $x$ -axis intercepts if

EITHER

$- 2 c^{\frac{3}{2}} + 2 < 0$ (or equivalent reasoning) (M1)

reasoning from the results in both parts (e)(i) and (e)(ii) (M1)

THEN

$c > 1$ A1

[2 marks]

e.iii.

case 1:

$c \leq 0$ (independent of the value of $d$ ) A1

EITHER

$g' (x) = 0$ does not have two solutions (has no solutions or $1$ solution) R1

$\Rightarrow g' (x) \geq 0$ for $x \in^{~}$ R1

the graph of $y = f (x)$ has no local maximum or local minimum points, hence any vertical translation of this graph ( $y = g (x)$ ) will also have no local maximum or local minimum points R1

THEN

therefore there is only one $x$ -axis intercept AG

Note: Award at most A0R1 if only $c < 0$ is considered.

case 2

$c > 0$

$(- \sqrt{c}, 2 c^{\frac{3}{2}} + d)$ is a local maximum point and $(\sqrt{c}, - 2 c^{\frac{3}{2}} + d)$ is a local minimum point (A1)

Note: Award (A1) for a correct $y$ -coordinate seen for either the maximum or the minimum.

considers the positions of the local maximum point and/or the local minimum point (M1)

EITHER
considers both points above the $x$ -axis or both points below the $x$ -axis

OR

considers either the local minimum point only above the $x$ -axis OR the local maximum point only below the $x$ -axis

THEN

$d > 2 c^{\frac{3}{2}}$ (both points above the $x$ -axis) A1

$d < - 2 c^{\frac{3}{2}}$ (both points above the $x$ -axis) A1

Note: Award at most (A1)(M1)A0A0 for case 2 if $c > 0$ is not clearly stated.

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

[N/A]

This question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a $3$ by $6$ rectangle has an area of $18$ and a perimeter of $18$ .

For each polygon in this question, let the numerical value of its area be $A$ and let the numerical value of its perimeter be $P$ .

An $n$ -sided regular polygon can be divided into $n$ congruent isosceles triangles. Let $x$ be the length of each of the two equal sides of one such isosceles triangle and let $y$ be the length of the third side. The included angle between the two equal sides has magnitude $\frac{2 π}{n}$ .

Part of such an $n$ -sided regular polygon is shown in the following diagram.

Consider a $n$ -sided regular polygon such that $A = P$ .

The Maclaurin series for $\tan x$ is $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots$

Consider a right-angled triangle with side lengths $a, b$ and $\sqrt{a^{2} + b^{2}}$ , where $a \geq b$ , such that $A = P$ .

Find the side length, $s$ , where $s > 0$ , of a square such that $A = P$ .

[3]

Write down, in terms of $x$ and $n$ , an expression for the area, $A_{T}$ , of one of these isosceles triangles.

[1]

Show that $y = 2 x \sin \frac{π}{n}$ .

[2]

Use the results from parts (b) and (c) to show that $A = P = 4 n \tan \frac{π}{n}$ .

[7]

Use the Maclaurin series for $\tan x$ to find $\lim_{n \to \infty} (4 n \tan \frac{π}{n})$ .

[3]

e.i.

Interpret your answer to part (e)(i) geometrically.

[1]

e.ii.

Show that $a = \frac{8}{b - 4} + 4$ .

[7]

By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which $a, b, A, P \in ℤ$ .

[3]

g.i.

Determine the area and perimeter of these two right-angled triangles.

[1]

g.ii.

Markscheme

$A = s^{2}$ and $P = 4 s$ (A1)

$A = P \Rightarrow s^{2} = 4 s$ (M1)

$s (s - 4) = 0$

$\Rightarrow s = 4 (s > 0)$ A1

Note: Award A1M1A0 if both $s = 4$ and $s = 0$ are stated as final answers.

[3 marks]

$A_{T} = \frac{1}{2} x^{2} \sin \frac{2 π}{n}$ A1

Note: Award A1 for a correct alternative form expressed in terms of $x$ and $n$ only.

For example, using Pythagoras’ theorem, $A_{T} = x \sin \frac{π}{n} \sqrt{x^{2} - x^{2} \sin^{2} \frac{π}{n}}$ or $A_{T} = 2 (\frac{1}{2} (x \sin \frac{π}{n}) (x \cos \frac{π}{n}))$ or $A_{T} = x^{2} \sin \frac{π}{n} \cos \frac{π}{n}$ .

[1 mark]

METHOD 1

uses $\sin θ = \frac{opp}{hyp}$ (M1)

$\frac{\frac{y}{2}}{x} = \sin \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 2

uses Pythagoras’ theorem ${(\frac{y}{2})}^{2} + h^{2} = x^{2}$ and $h = x \cos \frac{π}{n}$ (M1)

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

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 3

uses the cosine rule (M1)

$y^{2} = 2 x^{2} - 2 x^{2} \cos \frac{2 π}{n} (= 2 x^{2} (1 - \cos \frac{2 π}{n}))$

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 4

uses the sine rule (M1)

$\frac{y}{\sin \frac{2 π}{n}} = \frac{x}{\sin (\frac{π}{2} - \frac{π}{n})}$

$y \cos \frac{π}{n} = 2 x \sin \frac{π}{n} \cos \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

[2 marks]

$A = P \Rightarrow n A_{T} = n y$ (M1)

Note: Award M1 for equating correct expressions for $A$ and $P$ .

$\frac{1}{2} n x^{2} \sin \frac{2 π}{n} = 2 n x \sin \frac{π}{n} (n x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 n x \sin \frac{π}{n})$

$\frac{1}{2} x^{2} \sin \frac{2 π}{n} = 2 x \sin \frac{π}{n} (x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n})$ A1

uses $\sin \frac{2 π}{n} = 2 \sin \frac{π}{n} \cos \frac{π}{n}$ (seen anywhere in part (d) or in part (b)) (M1)

$x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n}$

attempts to either factorise or divide their expression (M1)

$x \sin \frac{π}{n} (x \cos \frac{π}{n} - 2) = 0$

$x = \frac{2}{\cos \frac{π}{n}}, (x \sin \frac{π}{n} \neq 0)$ (or equivalent) A1

EITHER

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $P = n y$ (M1)

$P = 2 n (\frac{2}{\cos \frac{π}{n}}) (\sin \frac{π}{n})$ A1

Note: Other approaches are possible. For example, award A1 for $P = 2 n x \cos \frac{π}{n} \tan \frac{π}{n}$ and M1 for substituting $x = \frac{2}{\cos \frac{π}{n}}$ into $P$ .

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $A = n A_{T}$ (M1)

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (\sin \frac{2 π}{n})$

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (2 \sin \frac{π}{n} \cos \frac{π}{n})$ A1

THEN

$A = P = 4 n \tan \frac{π}{n}$ AG

[7 marks]

attempts to use the Maclaurin series for $\tan x$ with $x = \frac{π}{n}$ (M1)

$\tan \frac{π}{n} = \frac{π}{n} + \frac{{(\frac{π}{n})}^{3}}{3} + \frac{2 {(\frac{π}{n})}^{5}}{15} (+ \dots)$

$4 n \tan \frac{π}{n} = 4 n (\frac{π}{n} + \frac{π^{3}}{3 n^{3}} + \frac{2 π^{5}}{15 n^{5}} (+ \dots))$ (or equivalent) A1

$= 4 (π + \frac{π^{3}}{3 n^{2}} + \frac{2 π^{5}}{15 n^{4}} + \dots)$

$\Rightarrow \lim_{n \to \infty} (4 n \tan \frac{π}{n}) = 4 π$ A1

Note: Award a maximum of M1A1A0 if $\lim_{n \to \infty}$ is not stated anywhere.

[3 marks]

e.i.

(as $n \to \infty, P \to 4 π$ and $A \to 4 π$ )

the polygon becomes a circle of radius $2$ R1

Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area $4 π$ OR
the polygon becomes a circle of perimeter $4 π$ OR
the polygon becomes a circle with $A = P = 4 π$ .
Award R0 for polygon becomes a circle.

[1 mark]

e.ii.

$A = \frac{1}{2} a b$ and $P = a + b + \sqrt{a^{2} + b^{2}}$ (A1)(A1)

equates their expressions for $A$ and $P$ M1

$A = P \Rightarrow a + b + \sqrt{a^{2} + b^{2}} = \frac{1}{2} a b$

$\sqrt{a^{2} + b^{2}} = \frac{1}{2} a b - (a + b)$ M1

Note: Award M1 for isolating $\sqrt{a^{2} + b^{2}}$ or $\pm 2 \sqrt{a^{2} + b^{2}}$ . This step may be seen later.

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

$a^{2} + b^{2} = \frac{1}{4} a^{2} b^{2} - 2 (\frac{1}{2} a b) (a + b) + {(a + b)}^{2}$ M1

$(= \frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + a^{2} + 2 a b + b^{2})$

Note: Award M1 for attempting to expand their RHS of either $a^{2} + b^{2} = \dots$ or $4 (a^{2} + b^{2}) = \dots$ .

EITHER

$a b (\frac{1}{4} a b - a - b + 2) = 0 (a b \neq 0)$ A1

$\frac{1}{4} a b - a - b + 2 = 0$

$a b - 4 a = 4 b - 8$

$\frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + 2 a b = 0$

$a (\frac{1}{4} b^{2} - b) + (2 b - b^{2}) = 0 (a (b^{2} - 4 b) + (8 b - 4 b^{2}) = 0)$ A1

$a = \frac{4 b^{2} - 8 b}{b^{2} - 4 b}$

THEN

$\Rightarrow a = \frac{4 b - 8}{b - 4}$ A1

$a = \frac{4 b - 16 + 8}{b - 4}$

$a = \frac{8}{b - 4} + 4$ AG

Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that $A = P = \frac{16}{b - 4} + 2 b + 4$ gains $4$ of the $7$ marks.

[7 marks]

using an appropriate method (M1)

eg substituting values for $b$ or using divisibility properties

$(5, 12, 13)$ and $(6, 8, 10)$ A1A1

Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.

[3 marks]

g.i.

$A = P = 30$ and $A = P = 24$ A1

Note: Do not award A1FT.

[1 mark]

g.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

g.i.

[N/A]

g.ii.

This question asks you to explore the behaviour and some key features of the function $f_{n} (x) = x^{n} (a - x)^{n}$ , where $a \in ℝ^{+}$ and $n \in ℤ^{+}$ .

In parts (a) and (b), only consider the case where $a = 2$ .

Consider $f_{1} (x) = x (2 - x)$ .

Consider $f_{n} (x) = x^{n} {(2 - x)}^{n}$ , where $n \in ℤ^{+}, n > 1$ .

Now consider $f_{n} (x) = x^{n} {(a - x)}^{n}$ where $a \in ℝ^{+}$ and $n \in ℤ^{+}, n > 1$ .

By using the result from part (f) and considering the sign of ${f_{n}}^{'} (- 1)$ , show that the point $(0, 0)$ on the graph of $y = f_{n} (x)$ is

Sketch the graph of $y = f_{1} (x)$ , stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.

[3]

Use your graphic display calculator to explore the graph of $y = f_{n} (x)$ for

• the odd values $n = 3$ and $n = 5$ ;

• the even values $n = 2$ and $n = 4$ .

Hence, copy and complete the following table.

[6]

Show that ${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ .

[5]

State the three solutions to the equation ${f_{n}}^{'} (x) = 0$ .

[2]

Show that the point $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ on the graph of $y = f_{n} (x)$ is always above the horizontal axis.

[3]

Hence, or otherwise, show that ${f_{n}}^{'} (\frac{a}{4}) > 0$ , for $n \in ℤ^{+}$ .

[2]

a local minimum point for even values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[3]

g.i.

a point of inflexion with zero gradient for odd values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[2]

g.ii.

Consider the graph of $y = x^{n} {(a - x)}^{n} - k$ , where $n \in ℤ^{+}$ , $a \in ℝ^{+}$ and $k \in ℝ$ .

State the conditions on $n$ and $k$ such that the equation $x^{n} {(a - x)}^{n} = k$ has four solutions for $x$ .

[5]

Markscheme

inverted parabola extended below the $x$ -axis A1

$x$ -axis intercept values $x = 0, 2$ A1

Note: Accept a graph passing through the origin as an indication of $x = 0$ .

local maximum at $(1, 1)$ A1

Note: Coordinates must be stated to gain the final A1.
Do not accept decimal approximations.

[3 marks]

A1A1A1A1A1A1

Note: Award A1 for each correct value.

For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.

[6 marks]

METHOD 1

attempts to use the product rule (M1)

${f_{n}}^{'} (x) = - n x^{n} {(a - x)}^{n - 1} + n x^{n - 1} {(a - x)}^{n}$ A1A1

Note: Award A1 for a correct $u \frac{d v}{d x}$ and A1 for a correct $v \frac{d u}{d x}$ .

EITHER

attempts to factorise ${f_{n}}^{'} (x)$ (involving at least one of $n x^{n - 1}$ or ${(a - x)}^{n - 1}$ ) (M1)

$= n x^{n - 1} {(a - x)}^{n - 1} ((a - x) - x)$ A1

attempts to express ${f_{n}}^{'} (x)$ as the difference of two products with each product containing at least one of $n x^{n - 1}$ or ${(a - x)}^{n - 1}$ (M1)

$= (- x) (n x^{n - 1}) {(a - x)}^{n - 1} + (a - x) (n x^{n - 1}) {(a - x)}^{n - 1}$ A1

THEN

${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ AG

Note: Award the final (M1)A1 for obtaining any of the following forms:

${f_{n}}^{'} (x) = n x^{n} {(a - x)}^{n} (\frac{a - x - x}{x (a - x)}); {f_{n}}^{'} (x) = \frac{n x^{n} {(a - x)}^{n}}{x (a - x)} (a - x - x);$

${f_{n}}^{'} (x) = n x^{n - 1} ({(a - x)}^{n} - x {(a - x)}^{n - 1});$

${f_{n}}^{'} (x) = {(a - x)}^{n - 1} (n x^{n - 1} {(a - x)}^{n} - n x^{n})$

METHOD 2

$f_{n} (x) = {(x (a - x))}^{n}$ (M1)

$= {(a x - x^{2})}^{n}$ A1

attempts to use the chain rule (M1)

${f_{n}}^{'} (x) = n (a - 2 x) {(a x - x^{2})}^{n - 1}$ A1A1

Note: Award A1 for $n (a - 2 x)$ and A1 for ${(a x - x^{2})}^{n - 1}$ .

${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ AG

[5 marks]

$x = 0, x = \frac{a}{2}, x = a$ A2

Note: Award A1 for either two correct solutions or for obtaining $x = 0, x = - a, x = - \frac{a}{2}$
Award A0 otherwise.

[2 marks]

attempts to find an expression for $f_{n} (\frac{a}{2})$ (M1)

$f_{n} (\frac{a}{2}) = {(\frac{a}{2})}^{n} {(a - \frac{a}{2})}^{n}$

$= {(\frac{a}{2})}^{n} {(\frac{a}{2})}^{n} (= {(\frac{a}{2})}^{2 n}), (= {({(\frac{a}{2})}^{n})}^{2})$ A1

EITHER

since $a \in ℝ^{+}, {(\frac{a}{2})}^{2 n} > 0$ (for $n \in ℤ^{+}, n > 1$ and so $f_{n} (\frac{a}{2}) > 0$ ) R1

Note: Accept any logically equivalent conditions/statements on $a$ and $n$ .
Award R0 if any conditions/statements specified involving $a$ , $n$ or both are incorrect.

(since $a \in ℝ^{+}$ ), $\frac{a}{2}$ raised to an even power ( $2 n$ ) (or equivalent reasoning) is always positive (and so $f_{n} (\frac{a}{2}) > 0$ ) R1

Note: The condition $a \in ℝ^{+}$ is given in the question. Hence some candidates will assume $a \in ℝ^{+}$ and not state it. In these instances, award R1 for a convincing argument.
Accept any logically equivalent conditions/statements on on $a$ and $n$ .
Award R0 if any conditions/statements specified involving $a$ , $n$ or both are incorrect.

THEN

so $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ is always above the horizontal axis AG

Note: Do not award (M1)A0R1.

[3 marks]

METHOD 1

${f_{n}}^{'} (\frac{a}{4}) = n {(\frac{a}{4})}^{n - 1} (a - \frac{a}{2}) {(a - \frac{a}{4})}^{n - 1} (= n {(\frac{a}{4})}^{n - 1} (\frac{a}{2}) {(\frac{3 a}{4})}^{n - 1})$ A1

EITHER

$n {(\frac{a}{4})}^{n - 1} (\frac{a}{2}) {(\frac{3 a}{4})}^{n - 1} > 0$ as $a \in ℝ^{+}$ and $n \in ℤ^{+}$ R1

$n {(\frac{a}{4})}^{n - 1}, (a - \frac{a}{2})$ and ${(a - \frac{a}{4})}^{n - 1}$ are all $> 0$ R1

Note: Do not award A0R1.
Accept equivalent reasoning on correct alternative expressions for ${f_{n}}^{'} (\frac{a}{4})$ and accept any logically equivalent conditions/statements on $a$ and $n$ .

Exceptions to the above are condone $n > 1$ and condone $n > 0$ .

An alternative form for ${f_{n}}^{'} (\frac{a}{4})$ is $(2 n) {(3)}^{n - 1} {(\frac{a}{4})}^{2 n - 1}$ .

THEN

hence ${f_{n}}^{'} (\frac{a}{4}) > 0$ AG

METHOD 2

$f_{n} (0) = 0$ and $f_{n} (\frac{a}{2}) > 0$ A1

(since $f_{n}$ is continuous and there are no stationary points between $x = 0$ and $x = \frac{a}{2}$ )

the gradient (of the curve) must be positive between $x = 0$ and $x = \frac{a}{2}$ R1

Note: Do not award A0R1.

hence ${f_{n}}^{'} (\frac{a}{4}) > 0$ AG

[2 marks]

${f_{n}}^{'} (- 1) = n {(- 1)}^{n - 1} (a + 2) {(a + 1)}^{n - 1}$

for $n$ even:

$n {(- 1)}^{n - 1} (= - n) < 0$ (and $(a + 2), {(a + 1)}^{n - 1}$ are both $> 0$ ) R1

${f_{n}}^{'} (- 1) < 0$ A1

${f_{n}}^{'} (0) = 0$ and ${f_{n}}^{'} (\frac{a}{4}) > 0$ (seen anywhere) A1

Note: Candidates can give arguments based on the sign of ${(- 1)}^{n - 1}$ to obtain the R mark.
For example, award R1 for the following:
If $n$ is even, then $n - 1$ is odd and hence ${(- 1)}^{n - 1} < 0 (= - 1)$ .
Do not award R0A1.
The second A1 is independent of the other two marks.
The A marks can be awarded for correct descriptions expressed in words.
Candidates can state $(0, 0)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_{n}}^{'} (0) = 0$ . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. $n = 2$ ).

hence $(0, 0)$ is a local minimum point AG

[3 marks]

g.i.

for $n$ odd:

$n {(- 1)}^{n - 1} (= n) < 0$ , (and $(a + 2), {(a + 1)}^{n - 1}$ are both $> 0$ ) so ${f_{n}}^{'} (- 1) > 0$ R1

Note: Candidates can give arguments based on the sign of ${(- 1)}^{n - 1}$ to obtain the R mark.
For example, award R1 for the following:
If $n$ is odd, then $n - 1$ is even and hence ${(- 1)}^{n - 1} > 0 (= 1)$ .

${f_{n}}^{'} (0) = 0$ and ${f_{n}}^{'} (\frac{a}{4}) > 0$ (seen anywhere) A1

Note: The A1 is independent of the R1.
Candidates can state $(0, 0)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_{n}}^{'} (0) = 0$ . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. $n = 3$ ).

hence $(0, 0)$ is a point of inflexion with zero gradient AG

[2 marks]

g.ii.

considers the parity of $n$ (M1)

Note: Award M1 for stating at least one specific even value of $n$ .

$n$ must be even (for four solutions) A1

Note: The above 2 marks are independent of the 3 marks below.

$0 < k < {(\frac{a}{2})}^{2 n}$ A1A1A1

Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.

The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
For example, award A1A1A0 for $0 \leq k \leq {(\frac{a}{2})}^{2 n}$ . Award A1A0A0 for $k > 0$ .

Award A1A0A0 for $0 < k < f_{n} (\frac{a}{2})$ .

[5 marks]

Examiners report

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

In this question you will be exploring the strategies required to solve a system of linear differential equations.

Consider the system of linear differential equations of the form:

$\frac{d x}{d t} = x - y$ and $\frac{d y}{d t} = a x + y$ ,

where $x, y, t \in ℝ^{+}$ and $a$ is a parameter.

First consider the case where $a = 0$ .

Now consider the case where $a = - 1$ .

Now consider the case where $a = - 4$ .

From previous cases, we might conjecture that a solution to this differential equation is $y = F e^{λ t}$ , $λ \in ℝ$ and $F$ is a constant.

By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.

[3]

a.i.

Show that $\frac{d x}{d t} - x = - A e^{t}$ .

[1]

a.ii.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$ .

[4]

a.iii.

By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .

[3]

b.i.

By substituting $Y = \frac{d y}{d t}$ , show that $Y = B e^{2 t}$ where $B$ is a constant.

[3]

b.ii.

Hence find $y$ as a function of $t$ .

[2]

b.iii.

Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.

[3]

b.iv.

Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[3]

c.i.

Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[4]

c.ii.

Let the two values found in part (c)(ii) be $λ_{1}$ and $λ_{2}$ .

Verify that $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

[4]

c.iii.

Markscheme

METHOD 1

$\frac{d y}{d t} = y$

$\int \frac{d y}{y} = \int d t$ (M1)

$\ln y = t + c$ OR $\ln |y| = t + c$ A1A1

Note: Award A1 for $\ln y$ and A1 for $t$ and $c$ .

$y = A e^{t}$ AG

METHOD 2

rearranging to $\frac{d y}{d t} - y = 0$ AND multiplying by integrating factor $e^{- t}$ M1

$y e^{- t} = A$ A1A1

$y = A e^{t}$ AG

[3 marks]

a.i.

substituting $y = A e^{t}$ into differential equation in $x$ M1

$\frac{d x}{d t} = x - A e^{t}$

$\frac{d x}{d t} - x = - A e^{t}$ AG

[1 mark]

a.ii.

integrating factor (IF) is $e^{\int - 1 d t}$ (M1)

$= e^{- t}$ (A1)

$e^{- t} \frac{d x}{d t} - x e^{- t} = - A$

$x e^{- t} = - A t + D$ (A1)

$x = (- A t + D) e^{t}$ A1

Note: The first constant must be $A$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[4 marks]

a.iii.

$\frac{d^{2} y}{d t^{2}} = - \frac{d x}{d t} + \frac{d y}{d t}$ A1

EITHER

$= - x + y + \frac{d y}{d t}$ (M1)

$= \frac{d y}{d t} + \frac{d y}{d t}$ A1

$= - x + y + (- x + y)$ (M1)

$= 2 (- x + y)$ A1

THEN

$= 2 \frac{d y}{d t}$ AG

[3 marks]

b.i.

$\frac{d Y}{d t} = 2 Y$ A1

$\int \frac{d Y}{Y} = \int 2 d t$ M1

$\ln |Y| = 2 t + c$ OR $\ln Y = 2 t + c$ A1

$Y = B e^{2 t}$ AG

[3 marks]

b.ii.

$\frac{d y}{d t} = B e^{2 t}$

$y = \int B e^{2 t} d t$ M1

$y = \frac{B}{2} e^{2 t} + C$ A1

Note: The first constant must be $B$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[2 marks]

b.iii.

METHOD 1

substituting $\frac{d y}{d t} = B e^{2 t}$ and their (iii) into $\frac{d y}{d t} = - x + y$ M1(M1)

$B e^{2 t} = - x + \frac{B}{2} e^{2 t} + C$ A1

$x = - \frac{B}{2} e^{2 t} + C$ AG

Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.

METHOD 2

$\frac{d x}{d t} = x - \frac{B}{2} e^{2 t} - C$

$\frac{d x}{d t} - x = - \frac{B}{2} e^{2 t} - C$

$\frac{d (x e^{- t})}{d t} = - \frac{B}{2} e^{t} - C e^{- t}$ M1

$x e^{- t} = \int - \frac{B}{2} e^{t} - C e^{- t} d t$

$x e^{- t} = - \frac{B}{2} e^{t} - C e^{- t} + D$ A1

$x = - \frac{B}{2} e^{2 t} + C + D e^{t}$

$\frac{d y}{d t} = - x + y \Rightarrow B e^{2 t} = \frac{B}{2} e^{2 t} - C - D e^{t} + \frac{B}{2} e^{2 t} + C \Rightarrow D = 0$ M1

$x = - \frac{B}{2} e^{2 t} + C$ AG

[3 marks]

b.iv.

$\frac{d y}{d t} = - 4 x + y$

$\frac{d^{2} y}{d t^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t}$ seen anywhere M1

METHOD 1

$\frac{d^{2} y}{d t^{2}} = - 4 (x - y) + \frac{d y}{d t}$

attempt to eliminate $x$ M1

$= - 4 (\frac{1}{4} (y - \frac{d y}{d t}) - y) + \frac{d y}{d t}$

$= 2 \frac{d y}{d t} + 3 y$ A1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ AG

METHOD 2

rewriting LHS in terms of $x$ and $y$ M1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = (- 8 x + 5 y) - 2 (- 4 x + y) - 3 y$ A1

$= 0$ AG

[3 marks]

c.i.

$\frac{d y}{d t} = F λ e^{λ t}, \frac{d^{2} y}{d t^{2}} = F λ^{2} e^{λ t}$ (A1)

$F λ^{2} e^{λ t} - 2 F λ e^{λ t} - 3 F e^{λ t} = 0$ (M1)

$λ^{2} - 2 λ - 3 = 0$ (since $e^{λ t} \neq 0$ ) A1

$λ_{1}$ and $λ_{2}$ are $3$ and $- 1$ (either order) A1

[4 marks]

c.ii.

METHOD 1

$y = F e^{3 t} + G e^{- t}$

$\frac{d y}{d t} = 3 F e^{3 t} - G e^{- t}, \frac{d^{2} y}{d t^{2}} = 9 F e^{3 t} - G e^{- t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 9 F e^{3 t} + G e^{- t} - 2 (3 F e^{3 t} - G e^{- t}) - 3 (F e^{3 t} - G e^{- t})$ M1

$= 9 F e^{3 t} + G e^{- t} - 6 F e^{3 t} + 2 G e^{- t} - 3 F e^{3 t} - 3 G e^{- t}$ A1

$= 0$ AG

METHOD 2

$y = F e^{λ_{1} t} + G e^{λ_{2} t}$

$\frac{d y}{d t} = F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}, \frac{d^{2} y}{d t^{2}} = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t} - 2 (F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}) - 3 (F e^{λ_{1} t} + G e^{λ_{2} t})$ M1

$= F e^{λ_{1} t} (λ^{2} - 2 λ - 3) + G e^{λ_{2} t} (λ^{2} - 2 λ - 3)$ A1

$= 0$ AG

[4 marks]

c.iii.

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]

b.iv.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

The function $f$ is defined by $f\left( x \right) = {\text{arcsin}}\left( {2x} \right)$ , where $- \frac{1}{2} \leqslant x \leqslant \frac{1}{2}$ .

By finding a suitable number of derivatives of $f$ , find the first two non-zero terms in the Maclaurin series for $f$ .

[8]

Hence or otherwise, find $\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}}$ .

[3]

Markscheme

$f\left( x \right) = {\text{arcsin}}\left( {2x} \right)$

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

Note: Award M1A0 for $f'\left( x \right) = \frac{1}{{\sqrt {1 - 4{x^2}} }}$

$f''\left( x \right) = \frac{{8x}}{{{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}$ A1

EITHER

$f'''\left( x \right) = \frac{{8{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}} - 8x\left( {\frac{3}{2}\left( { - 8x} \right){{\left( {1 - 4{x^2}} \right)}^{\frac{1}{2}}}} \right)}}{{{{\left( {1 - 4{x^2}} \right)}^3}}}\,\,\,\left( { = \frac{{8{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}} + 96{x^2}{{\left( {1 - 4{x^2}} \right)}^{\frac{1}{2}}}}}{{{{\left( {1 - 4{x^2}} \right)}^3}}}} \right)$ A1

$f'''\left( x \right) = 8{\left( {1 - 4{x^2}} \right)^{ - \frac{3}{2}}} + 8x\left( { - \frac{3}{2}{{\left( {1 - 4{x^2}} \right)}^{ - \frac{5}{2}}}} \right)\left( { - 8x} \right)\,\,\,\left( { = 8{{\left( {1 - 4{x^2}} \right)}^{ - \frac{3}{2}}} + 96{x^2}{{\left( {1 - 4{x^2}} \right)}^{ - \frac{5}{2}}}} \right)$ A1

THEN

substitute $x = 0$ into $f$ or any of its derivatives (M1)

$f\left( 0 \right) = 0$ , $f'\left( 0 \right) = 2$ and $f''\left( 0 \right) = 0$ A1

$f'''\left( 0 \right) = 8$

the Maclaurin series is

$f\left( x \right) = 2x + \frac{{8{x^3}}}{6} + \ldots \,\left( { = 2x + \frac{{4{x^3}}}{3} + \ldots } \right)$ (M1)A1

[8 marks]

METHOD 1

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{2x + \frac{{4{x^3}}}{3} + \ldots - 2x}}{{8{x^3}}}$ M1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{4}{3} + \ldots {\text{ terms with }}x}}{8}$ (M1)

$= \frac{1}{6}$ A1

Note: Condone the omission of +… in their working.

METHOD 2

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}} = \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{2}{{\sqrt {1 - 4{x^2}} }} - 2}}{{24{x^2}}}$ M1

$= \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule again

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{{8x}}{{{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}}}{{48x}}\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{1}{{6{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}} \right)$ M1

Note: Award M1 only if their previous expression is in indeterminate form.

$= \frac{1}{6}$ A1

Note: Award FT for use of their derivatives from part (a).

[3 marks]

Examiners report

[N/A]

Using L’Hôpital’s rule, find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)$ .

Markscheme

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

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)$

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{3}}\,{\text{se}}{{\text{c}}^2}\,3x - 3\,{\text{se}}{{\text{c}}^2}\,x}}{{{\text{3}}\,{\text{cos}}\,3x - 3\,{\text{cos}}\,x}}} \right)\,\,\,\,\,\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{se}}{{\text{c}}^2}\,3x - {\text{se}}{{\text{c}}^2}\,x}}{{{\text{cos}}\,3x - {\text{cos}}\,x}}} \right)\,} \right)$ M1A1A1

Note: Award M1 for attempt at differentiation using l'Hopital's rule, A1 for numerator, A1 for denominator.

METHOD 1

using l’Hopital’s rule again

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{18}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 6\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x}}{{ - 9\,{\text{sin}}\,3x + 3\,{\text{sin}}\,x}}} \right)\,\,\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{6}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 2\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x}}{{ - 3\,{\text{sin}}\,3x + {\text{sin}}\,x}}} \right)} \right)$ A1A1

EITHER

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{108}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{ta}}{{\text{n}}^2}\,3x + 54\,{\text{se}}{{\text{c}}^4}\,3x - 12\,{\text{se}}{{\text{c}}^2}\,x\,{\text{ta}}{{\text{n}}^2}\,x - 6\,{\text{se}}{{\text{c}}^4}\,x}}{{{\text{ - 27}}\,{\text{cos}}\,3x + 3\,{\text{cos}}\,x}}} \right)$ A1A1

Note: Not all terms in numerator need to be written in final fraction. Award A1 for $54\,{\text{se}}{{\text{c}}^4}\,3x + \ldots - 6\,{\text{se}}{{\text{c}}^4}\,x \ldots -$ . However, if the terms are written, they
must be correct to award A1.

attempt to substitute $x = 0$ M1

$= \frac{{48}}{{ - 24}}$

$\frac{{\text{d}}}{{{\text{d}}x}}\left( {{\text{18}}\,{\text{se}}{{\text{c}}^2}\,3x\,{\text{tan}}\,3x - 6\,{\text{se}}{{\text{c}}^2}\,x\,{\text{tan}}\,x} \right)\left| {_{x = 0} = 48} \right.$ (M1)A1

$\frac{{\text{d}}}{{{\text{d}}x}}\left( { - 9\,{\text{sin}}\,3x + 3\,{\text{sin}}\,x} \right)\left| {_{x = 0} = - 24} \right.$ A1

THEN

$\left( {\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)} \right) = - 2$ A1

METHOD 2

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\frac{3}{{{\text{co}}{{\text{s}}^2}\,3x}} - \frac{3}{{{\text{co}}{{\text{s}}^2}\,x}}}}{{{\text{3}}\,{\text{cos}}\,3x - 3\,{\text{cos}}\,x}}} \right)$ M1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{co}}{{\text{s}}^2}\,x - {\text{co}}{{\text{s}}^2}\,3x}}{{{\text{co}}{{\text{s}}^2}\,3x\,{\text{co}}{{\text{s}}^2}\,x\left( {{\text{cos}}\,3x - {\text{cos}}\,x} \right)}}} \right)$ A1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{cos}}\,x + {\text{cos}}\,3x}}{{ - {\text{co}}{{\text{s}}^2}\,3x\,{\text{co}}{{\text{s}}^2}\,x}}} \right)$ M1A1

attempt to substitute $x = 0$ M1

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

$= - 2$ A1

[9 marks]

Examiners report

[N/A]

A simple model to predict the population of the world is set up as follows. At time $t$ years the population of the world is $x$ , which can be assumed to be a continuous variable. The rate of increase of $x$ due to births is 0.056 $x$ and the rate of decrease of $x$ due to deaths is 0.035 $x$ .

Show that $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.021x$ .

[1]

Find a prediction for the number of years it will take for the population of the world to double.

[6]

Markscheme

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

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.056x - 0.035x$ A1

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.021x$ AG

[1 mark]

METHOD 1

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.021x$

attempt to separate variables M1

$\int {\frac{1}{x}} {\text{d}}x = \int {0.021} \,{\text{d}}t$ A1

${\text{ln}}\,x = 0.021t( + c)$ A1

EITHER

$x = A{{\text{e}}^{0.021t}}$

$\Rightarrow 2A = A{{\text{e}}^{0.021t}}$ A1

Note: This A1 is independent of the following marks.

$t = 0{\text{,}}\,\,x = {x_0} \Rightarrow c = {\text{ln}}\,{x_0}$

$\Rightarrow {\text{ln}}\,2{x_0} = 0.021t + {\text{ln}}\,{x_0}$ A1

Note: This A1 is independent of the following marks.

THEN

$\Rightarrow {\text{ln}}\,2 = 0.021t$ (M1)

$\Rightarrow t = 33$ years A1

Note: If a candidate writes $t = 33.007$ , so $t = 34$ then award the final A1.

METHOD 2

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.021x$

attempt to separate variables M1

$\int_A^{2A} {\frac{1}{x}{\text{d}}x = \int_0^t {0.021\,{\text{d}}u} }$ A1A1

Note: Award A1 for correct integrals and A1 for correct limits seen anywhere. Do not penalize use of $t$ in place of $u$ .

$\left[ {{\text{ln}}\,x} \right]_A^{2A} = \left[ {0.021u} \right]_0^t$ A1

$\Rightarrow {\text{ln}}\,2 = 0.021t$ (M1)

$\Rightarrow t = 33$ A1

[6 marks]

Examiners report

[N/A]

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{x\ln (1 + x)}}.$

Markscheme

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

attempt to use l’Hôpital’s rule, M1

${\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin x\cos x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}$ $\,\,\,$ or $\,\,\,$ $\frac{{\sin 2x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

this gives 0/0 so use the rule again (M1)

$= \mathop {\lim }\limits_{x \to 0} \frac{{2{{\cos }^2}x - 2{{\sin }^2}x}}{{\frac{1}{{1 + x}} + \frac{{1 + x - x}}{{{{(1 + x)}^2}}}}}$ $\,\,\,$ or $\,\,\,$ $\frac{{2\cos 2x}}{{\frac{{2 + x}}{{{{(1 + x)}^2}}}}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

$= 1$ A1

Note: This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.

[7 marks]

Examiners report

[N/A]

Use l’Hôpital’s rule to find

$\lim_{x \to 1} \frac{\cos (x^{2} - 1) - 1}{e^{x - 1} - x}$ .

Markscheme

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

attempt to use l’Hôpital’s rule M1

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

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

substitution of $1$ into their expression (M1)

$= \frac{0}{0}$ hence use l’Hôpital’s rule again

Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.

attempt to use product rule in numerator M1

$= \lim_{x \to 1} \frac{- 4 x^{2} \cos (x^{2} - 1) - 2 \sin (x^{2} - 1)}{e^{x - 1}}$ A1

$= - 4$ A1

[7 marks]

Examiners report

[N/A]

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{2x}}{{1 + {x^2}}}} \right)y = {x^2}$ , given that $y = 2$ when $x = 0$ .

Show that $1 + {x^2}$ is an integrating factor for this differential equation.

[5]

Hence solve this differential equation. Give the answer in the form $y = f(x)$ .

[6]

Markscheme

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

METHOD 1

attempting to find an integrating factor (M1)

$\int {\frac{{2x}}{{1 + {x^2}}}{\text{d}}x = \ln (1 + {x^2})}$ (M1)A1

IF is ${{\text{e}}^{\ln (1 + {x^2})}}$ (M1)A1

$= 1 + {x^2}$ AG

METHOD 2

multiply by the integrating factor

$(1 + {x^2})\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = {x^2}(1 + {x^2})$ M1A1

left hand side is equal to the derivative of $(1 + {x^2})y$

[5 marks]

$(1 + {x^2})\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = (1 + {x^2}){x^2}$ (M1)

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

$(1 + {x^2})y = \left( {\int {{x^2} + {x^4}{\text{d}}x = } } \right){\text{ }}\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5}( + c)$ A1A1

$y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c} \right)$

$x = 0,{\text{ }}y = 2 \Rightarrow c = 2$ M1A1

$y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + 2} \right)$ A1

[6 marks]

Examiners report

[N/A]