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HL Paper 2

An ice-skater is skating such that her position vector when viewed from above at time $t$ seconds can be modelled by

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} a e^{b t} \cos t \\ a e^{b t} sin t \end{matrix})$

with respect to a rectangular coordinate system from a point $O$ , where the non-zero constants $a$ and $b$ can be determined. All distances are in metres.

At time $t = 0$ , the displacement of the ice-skater is given by $(\begin{matrix} 5 \\ 0 \end{matrix})$ and the velocity of the ice‑skater is given by $(\begin{matrix} - 3.5 \\ 5 \end{matrix})$ .

Find the velocity vector at time $t$ .

[3]

Show that the magnitude of the velocity of the ice-skater at time $t$ is given by

$a e^{b t} \sqrt{(1 + b^{2})}$ .

[4]

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

[3]

Find the magnitude of the velocity of the ice-skater when $t = 2$ .

[2]

At a point $P$ , the ice-skater is skating parallel to the $y$ -axis for the first time.

Find $OP$ .

[6]

Markscheme

use of product rule (M1)

$(\begin{matrix} \dot{x} \\ \dot{y} \end{matrix}) = (\begin{matrix} a b e^{b t} \cos t - a e^{b t} sin t \\ a b e^{b t} sin t + a e^{b t} \cos t \end{matrix})$ A1A1

[3 marks]

${|v|}^{2} = {\dot{x}}^{2} + {\dot{y}}^{2} = {[a b e^{b t} \cos t - a e^{b t} sin t]}^{2} + {[a b e^{b t} sin t + a e^{b t} \cos t]}^{2}$ M1

Note: It is more likely that an expression for $|v|$ is seen.
$\sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}}$ is not sufficient to award the M1, their part (a) must be substituted.

$= [a^{2} \sin^{2} t - 2 a^{2} b \sin t \cos t + a^{2} b^{2} \cos^{2} t + a^{2} \cos^{2} t + 2 a^{2} b \sin t \cos t + a^{2} b^{2} \sin^{2} t] e^{2 b t}$ A1

use of $\sin^{2} t + \cos^{2} t = 1$ within a factorized expression that leads to the final answer M1

$= a^{2} (b^{2} + 1) e^{2 b t}$ A1

magnitude of velocity is $a e^{b t} \sqrt{(1 + b^{2})}$ AG

[4 marks]

when $t = 0, a e^{b t} \cos t = 5$

$a = 5$ A1

$a b e^{b t} \cos t - a e^{b t} \sin t = - 3.5$ (M1)

$b = - 0.7$ A1

Note: Use of $a e^{b t} \sqrt{(1 + b^{2})}$ result from part (b) is an alternative approach.

[3 marks]

$5 e^{- 0.7 \times 2} \sqrt{(1 + {(- 0.7)}^{2})}$ (M1)

$1.51 (1.50504 \dots)$ A1

[2 marks]

$\dot{x} = 0$ (M1)

$a e^{b t} (b \cos t - \sin t) = 0$

$\tan t = b$

$t = 2.53 (2.53086 \dots)$ (A1)

correct substitution of their $t$ to find $x$ or $y$ (M1)
$x = - 0.697 (- 0.696591 \dots)$ and $y = 0.488 (0.487614 \dots)$ (A1)

use of Pythagoras / distance formula (M1)
$OP = 0.850 m (0.850297 \dots)$ A1

[6 marks]

Examiners report

[N/A]

Consider the following system of coupled differential equations.

$\frac{d x}{d t} = - 4 x$

$\frac{d y}{d t} = 3 x - 2 y$

Find the value of $\frac{d y}{d x}$

Find the eigenvalues and corresponding eigenvectors of the matrix $(\begin{matrix} - 4 & 0 \\ 3 & - 2 \end{matrix})$ .

[6]

Hence, write down the general solution of the system.

[2]

Determine, with justification, whether the equilibrium point $(0, 0)$ is stable or unstable.

[2]

(i) at $(4, 0)$ .

(ii) at $(- 4, 0)$ .

[3]

Sketch a phase portrait for the general solution to the system of coupled differential equations for $- 6 \leq x \leq 6$ , $- 6 \leq y \leq 6$ .

[4]

Markscheme

$|\begin{matrix} - 4 - λ & 0 \\ 3 & - 2 - λ \end{matrix}| = 0$ (M1)

$(- 4 - λ) (- 2 - λ) = 0$ (A1)

$λ = - 4$ OR $λ = - 2$ A1

$λ = - 4$

$(\begin{matrix} - 4 & 0 \\ 3 & - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 4 x \\ - 4 y \end{matrix})$ (M1)

Note: This M1 can be awarded for attempting to find either eigenvector.

$3 x - 2 y = - 4 y$

$3 x = - 2 y$

possible eigenvector is $(\begin{matrix} - 2 \\ 3 \end{matrix})$ (or any real multiple) A1

$λ = - 2$

$(\begin{matrix} - 4 & 0 \\ 3 & - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 2 x \\ - 2 y \end{matrix})$

$x = 0, y = 1$

possible eigenvector is $(\begin{matrix} 0 \\ 1 \end{matrix})$ (or any real multiple) A1

[6 marks]

$(\begin{matrix} x \\ y \end{matrix}) = A e^{- 4 t} (\begin{matrix} - 2 \\ 3 \end{matrix}) + B e^{- 2 t} (\begin{matrix} 0 \\ 1 \end{matrix})$ (M1)A1

Note: Award M1A1 for $x = - 2 A e^{- 4 t}, y = 3 A e^{- 4 t} + B e^{- 2 t}$ , M1A0 if LHS is missing or incorrect.

[2 marks]

two (distinct) real negative eigenvalues R1

(or equivalent (eg both $e^{- 4 t} \to 0, e^{- 2 t} \to 0$ as $t \to \infty$ ))

⇒ stable equilibrium point A1

Note: Do not award R0A1.

[2 marks]

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

(i) $(4, 0) \Rightarrow \frac{d y}{d x} = - \frac{3}{4}$ A1

(ii) $(- 4, 0) \Rightarrow \frac{d y}{d x} = - \frac{3}{4}$ A1

[3 marks]

A1A1A1A1

Note: Award A1 for a phase plane, with correct axes (condone omission of labels) and at least three non-overlapping trajectories. Award A1 for all trajectories leading to a stable node at $(0, 0)$ . Award A1 for showing gradient is negative at $x = 4$ and $- 4$ . Award A1 for both eigenvectors on diagram.

[4 marks]

Examiners report

[N/A]

A ball is attached to the end of a string and spun horizontally. Its position relative to a given point, $O$ , at time $t$ seconds, $t \geq 0$ , is given by the equation

$r = (\begin{matrix} 1.5 \cos (0.1 t^{2}) \\ 1.5 \sin (0.1 t^{2}) \end{matrix})$ where all displacements are in metres.

The string breaks when the magnitude of the ball’s acceleration exceeds $20 {ms}^{- 2}$ .

Show that the ball is moving in a circle with its centre at $O$ and state the radius of the circle.

[4]

Find an expression for the velocity of the ball at time $t$ .

[2]

b.i.

Hence show that the velocity of the ball is always perpendicular to the position vector of the ball.

[2]

b.ii.

Find an expression for the acceleration of the ball at time $t$ .

[3]

c.i.

Find the value of $t$ at the instant the string breaks.

[3]

c.ii.

How many complete revolutions has the ball completed from $t = 0$ to the instant at which the string breaks?

[3]

c.iii.

Markscheme

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

$|r| = \sqrt{1 . 5^{2} \cos^{2} (0.1 t^{2}) + 1 . 5^{2} \sin^{2} (0.1 t^{2})}$ M1

$= 1.5$ as $\sin^{2} θ + \cos^{2} θ = 1$ R1

Note: use of the identity needs to be explicitly stated.

Hence moves in a circle as displacement from a fixed point is constant. R1

Radius $= 1.5 m$ A1

[4 marks]

$v = (\begin{matrix} - 0.3 t \sin (0.1 t^{2}) \\ 0.3 t \cos (0.1 t^{2}) \end{matrix})$ M1A1

Note: M1 is for an attempt to differentiate each term

[2 marks]

b.i.

$v ∙ r = (\begin{matrix} 1.5 \cos (0.1 t^{2}) \\ 1.5 \sin (0.1 t^{2}) \end{matrix}) ∙ (\begin{matrix} - 0.3 t \sin (0.1 t^{2}) \\ 0.3 t \cos (0.1 t^{2}) \end{matrix})$ M1

Note: M1 is for an attempt to find $v ∙ r$

$= 1.5 \cos (0.1 t^{2}) \times (- 0.3 t \sin (0.1 t^{2})) + 1.5 \sin (0.1 t^{2}) \times 0.3 t \sin (0.1 t^{2}) = 0$ A1

Hence velocity and position vector are perpendicular. AG

[2 marks]

b.ii.

$a = (\begin{matrix} - 0.3 \sin (0.1 t^{2}) - 0.06 t^{2} \cos (0.1 t^{2}) \\ 0.3 \cos (0.1 t^{2}) - 0.06 t^{2} \sin (0.1 t^{2}) \end{matrix})$ M1A1A1

[3 marks]

c.i.

${(- 0.3 \sin (0.1 t^{2}) - 0.06 t^{2} \cos (0.1 t^{2}))}^{2} + {(0.3 \cos (0.1 t^{2}) - 0.06 t^{2} \sin (0.1 t^{2}))}^{2} = 400$ (M1)(A1)

Note: M1 is for an attempt to equate the magnitude of the acceleration to $20$ .

$t = 18.3 (18.256 \dots) (s)$ A1

[3 marks]

c.ii.

Angle turned through is $0.1 \times 18 . 256^{2} =$ M1

$= 33.329 \dots$ A1

$\frac{33.329}{2 π}$ M1

$\frac{33.329}{2 π} = 5.30 \dots$

$5$ complete revolutions A1

[4 marks]

c.iii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

A change in grazing habits has resulted in two species of herbivore, $X$ and $Y$ , competing for food on the same grasslands. At time $t = 0$ environmentalists begin to record the sizes of both populations. Let the size of the population of $X$ be $x$ , and the size of the population $Y$ be $y$ . The following model is proposed for predicting the change in the sizes of the two populations:

$\dot{x} = 0.3 x - 0.1 y$

$\dot{y} = - 0.2 x + 0.4 y$

for $x, y > 0$

For this system of coupled differential equations find

When $t = 0$ $X$ has a population of $2000$ .

It is known that $Y$ has an initial population of $2900$ .

the eigenvalues.

[3]

a.i.

the eigenvectors.

[3]

a.ii.

Hence write down the general solution of the system of equations.

[1]

Sketch the phase portrait for this system, for $x, y > 0$ .

On your sketch show

the equation of the line defined by the eigenvector in the first quadrant
at least two trajectories either side of this line using arrows on those trajectories to represent the change in populations as t increases

[3]

Write down a condition on the size of the initial population of $Y$ if it is to avoid its population reducing to zero.

[1]

Find the value of $t$ at which $x = 0$ .

[6]

e.i.

Find the population of $Y$ at this value of $t$ . Give your answer to the nearest $10$ herbivores.

[2]

e.ii.

Markscheme

$|\begin{matrix} 0.3 - λ & - 0.1 \\ - 0.2 & 0.4 - λ \end{matrix}| = 0$ (M1)(A1)

$λ = 0.5$ and $0.2$ A1

[3 marks]

a.i.

Attempt to solve either

$(\begin{matrix} - 0.2 & - 0.1 \\ - 0.2 & - 0.1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ or $(\begin{matrix} 0.1 & - 0.1 \\ - 0.2 & 0.2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

or equivalent (M1)

$(\begin{matrix} 1 \\ - 2 \end{matrix})$ or $(\begin{matrix} 1 \\ 1 \end{matrix})$ A1A1

Note: accept equivalent forms

[3 marks]

a.ii.

$(\begin{matrix} x \\ y \end{matrix}) = A e^{0.5 t} (\begin{matrix} 1 \\ - 2 \end{matrix}) + B e^{0.2 t} (\begin{matrix} 1 \\ 1 \end{matrix})$ A1

[1 mark]

A1A1A1

Note: A1 for $y = x$ correctly labelled, A1 for at least two trajectories above $y = x$ and A1 for at least two trajectories below $y = x$ , including arrows.

[3 marks]

$y > 2000$ A1

[1 mark]

$(\begin{matrix} x \\ y \end{matrix}) = A e^{0.5 t} (\begin{matrix} 1 \\ - 2 \end{matrix}) + B e^{0.2 t} (\begin{matrix} 1 \\ 1 \end{matrix})$

At $t = 0$ $2000 = A + B, 2900 = - 2 A + B$ M1A1

Note: Award M1 for the substitution of $2000$ and $2900$

Hence $A = - 300, B = 2300$ A1A1

$0 = - 300 e^{0.5 t} + 2300 e^{0.2 t}$ M1

$t = 6.79 (6.7896 \dots)$ (years) A1

[6 marks]

e.i.

$y = 600 e^{0.5 \times 6.79} + 2300 e^{0.2 \times 6.79}$ (M1)

$= 26827.9 \dots$

$= 26830$ (to the nearest $10$ animals) A1

[2 marks]

e.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

e.i.

[N/A]

e.ii.

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v {m s}^{- 1}$ at time $t$ seconds, where $v = - 2 t^{2} + 16 t - 24$ for $t \geq 0$ .

Find the times when $P$ is at instantaneous rest.

[2]

Find the magnitude of the particle’s acceleration at $6$ seconds.

[4]

Find the greatest speed of $P$ in the interval $0 \leq t \leq 6$ .

[2]

The particle starts from the origin $O$ . Find an expression for the displacement of $P$ from $O$ at time $t$ seconds.

[4]

Find the total distance travelled by $P$ in the interval $0 \leq t \leq 4$ .

[3]

Markscheme

solving $v = 0$ M1

$t = 2, t = 6$ A1

[2 marks]

use of power rule (M1)

$\frac{d v}{d t} = - 4 t + 16$ (A1)

$(t = 6)$

$\Rightarrow a = - 8$ (A1)

magnitude $= 8 {m s}^{- 2}$ A1

[4 marks]

using a sketch graph of $v$ (M1)

$24 {m s}^{- 1}$ A1

[2 marks]

METHOD ONE

$x = \int v d t$

attempt at integration of $v$ (M1)

$- \frac{2 t^{3}}{3} + 8 t^{2} - 24 t (+ c)$ A1

attempt to find $c$ (use of $t = 0, x = 0$ ) (M1)

$c = 0$ A1

$(x = - \frac{2 t^{3}}{3} + 8 t^{2} - 24 t)$

METHOD TWO

$x = \int_{0}^{t} v d t$

attempt at integration of $v$ (M1)

${[- \frac{2 t^{3}}{3} + 8 t^{2} - 24 t]}_{0}^{t}$ A1

attempt to substituted limits into their integral (M1)

$x = - \frac{2 t^{3}}{3} + 8 t^{2} - 24 t$ A1

[4 marks]

$\int_{0}^{4} |v| d t$ (M1)(A1)

Note: Award M1 for using the absolute value of $v$ , or separating into two integrals, A1 for the correct expression.

$= 32 m$ A1

[3 marks]

Examiners report

[N/A]

A shock absorber on a car contains a spring surrounded by a fluid. When the car travels over uneven ground the spring is compressed and then returns to an equilibrium position.

The displacement, $x$ , of the spring is measured, in centimetres, from the equilibrium position of $x = 0$ . The value of $x$ can be modelled by the following second order differential equation, where $t$ is the time, measured in seconds, after the initial displacement.

$\ddot{x} + 3 \dot{x} + 1.25 x = 0$

The differential equation can be expressed in the form $(\begin{matrix} \dot{x} \\ \dot{y} \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix})$ , where $A$ is a $2 \times 2$ matrix.

Given that $y = \dot{x}$ , show that $\dot{y} = - 1.25 x - 3 y$ .

[2]

Write down the matrix $A$ .

[1]

Find the eigenvalues of matrix $A$ .

[3]

c.i.

Find the eigenvectors of matrix $A$ .

[3]

c.ii.

Given that when $t = 0$ the shock absorber is displaced $8 cm$ and its velocity is zero, find an expression for $x$ in terms of $t$ .

[6]

Markscheme

$y = \dot{x} \Rightarrow \dot{y} = \ddot{x}$ A1

$\dot{y} + 3 (y) + 1.25 x = 0$ R1

Note: If no explicit reference is made to $\dot{y} = \ddot{x}$ , or equivalent, award A0R1 if second line is seen. If $\frac{d y}{d x}$ used instead of $\frac{d y}{d t}$ , award A0R0.

$\dot{y} = - 3 y - 1.25 x$ AG

[2 marks]

$A = (\begin{matrix} 0 & 1 \\ - 1.25 & - 3 \end{matrix})$ A1

[1 mark]

$|\begin{matrix} - λ & 1 \\ - 1.25 & - 3 - λ \end{matrix}| = 0$ (M1)

$λ (λ + 3) + 1.25 = 0$ (A1)

$λ = - 2.5; λ = - 0.5$ A1

[3 marks]

c.i.

$(\begin{matrix} 2.5 & 1 \\ - 1.25 & - 0.5 \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ (M1)

$2.5 a + b = 0$

$v_{1} = (\begin{matrix} - 2 \\ 5 \end{matrix})$ A1

$(\begin{matrix} 0.5 & 1 \\ - 1.25 & - 2.5 \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

$0.5 a + b = 0$

$v_{2} = (\begin{matrix} - 2 \\ 1 \end{matrix})$ A1

Note: Award M1 for a valid attempt to find either eigenvector. Accept equivalent forms of the eigenvectors.
Do not award FT for eigenvectors that do not satisfy both rows of the matrix.

[3 marks]

c.ii.

$(\begin{matrix} x \\ y \end{matrix}) = A e^{- 2.5 t} (\begin{matrix} - 2 \\ 5 \end{matrix}) + B e^{- 0.5 t} (\begin{matrix} - 2 \\ 1 \end{matrix})$ M1A1

$t = 0 \Rightarrow x = 8, \dot{x} = y = 0$ (M1)

$- 2 A - 2 B = 8$

$5 A + B = 0$ (M1)

$A = 1; B = - 5$ A1

$x = - 2 e^{- 2.5 t} + 10 e^{- 0.5 t}$ A1

Note: Do not award the final A1 if the answer is given in the form $(\begin{matrix} x \\ y \end{matrix}) = A e^{- 2.5 t} (\begin{matrix} - 2 \\ 5 \end{matrix}) + B e^{- 0.5 t} (\begin{matrix} - 2 \\ 1 \end{matrix})$ .

[6 marks]

Examiners report

There were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of $x$ and the first derivative of $y$ was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ $x =$ ….” as asked for in the question.

c.i.

c.ii.

The cross-sectional view of a tunnel is shown on the axes below. The line $[AB]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y = - 0.1 x^{3} + 0.8 x^{2}, 2 \leq x \leq 8$ , relative to an origin $O$ .

Point $A$ has coordinates $(2, 0)$ , point $B$ has coordinates $(2, 2.4)$ , and point $C$ has coordinates $(8, 0)$ .

Find the height of the tunnel when

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

[2]

a.i.

Hence find the maximum height of the tunnel.

[4]

a.ii.

$x = 4$ .

[2]

b.i.

$x = 6$ .

[1]

b.ii.

Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.

[3]

Write down the integral which can be used to find the cross-sectional area of the tunnel.

[2]

d.i.

Hence find the cross-sectional area of the tunnel.

[2]

d.ii.

Markscheme

evidence of power rule (at least one correct term seen) (M1)

$\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x$ A1

[2 marks]

a.i.

$- 0.3 x^{2} + 1.6 x = 0$ M1

$x = 5.33 (5.33333 \dots, \frac{16}{3})$ A1

$y = - 0.1 \times 5.33333 \dots^{3} + 0.8 \times 5.33333 \dots^{2}$ (M1)

Note: Award M1 for substituting their zero for $\frac{d y}{d x} (5.333 \dots)$ into $y$ .

$7.59 m (7.58519 \dots)$ A1

Note: Award M0A0M0A0 for an unsupported $7.59$ .
Award at most M0A0M1A0 if only the last two lines in the solution are seen.
Award at most M1A0M1A1 if their $x = 5.33$ is not seen.

[6 marks]

a.ii.

One correct substitution seen (M1)

$6.4 m$ A1

[2 marks]

b.i.

$7.2 m$ A1

[1 mark]

b.ii.

$A = \frac{1}{2} \times 2 ((2.4 + 0) + 2 (6.4 + 7.2))$ (A1)(M1)

Note: Award A1 for $h = 2$ seen. Award M1 for correct substitution into the trapezoidal rule (the zero can be omitted in working).

$= 29.6 m^{2}$ A1

[3 marks]

$A = \int_{2}^{8} - 0.1 x^{3} + 0.8 x^{2} d x$ OR $A = \int_{2}^{8} y d x$ A1A1

Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if $d x$ is omitted.

[2 marks]

d.i.

$A = 32.4 m^{2}$ A2

Note: As per the marking instructions, FT from their integral in part (d)(i). Award at most A1FTA0 if their area is $> 48$ , this is outside the constraints of the question (a $6 \times 8$ rectangle).

[2 marks]

d.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

A particle moves such that its displacement, $x$ metres, from a point $O$ at time $t$ seconds is given by the differential equation

$\frac{d^{2} x}{d t^{2}} + 5 \frac{d x}{d t} + 6 x = 0$

The equation for the motion of the particle is amended to

$\frac{d^{2} x}{d t^{2}} + 5 \frac{d x}{d t} + 6 x = 3 t + 4$ .

When $t = 0$ the particle is stationary at $O$ .

Use the substitution $y = \frac{d x}{d t}$ to show that this equation can be written as

$(\begin{matrix} \frac{d x}{d t} \\ \frac{d y}{d t} \end{matrix}) = (\begin{matrix} 0 & 1 \\ - 6 & - 5 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})$ .

[1]

a.i.

Find the eigenvalues for the matrix $(\begin{matrix} 0 & 1 \\ - 6 & - 5 \end{matrix})$ .

[3]

a.ii.

Hence state the long-term velocity of the particle.

[1]

a.iii.

Use the substitution $y = \frac{d x}{d t}$ to write the differential equation as a system of coupled, first order differential equations.

[2]

b.i.

Use Euler’s method with a step length of $0.1$ to find the displacement of the particle when $t = 1$ .

[5]

b.ii.

Find the long-term velocity of the particle.

[1]

b.iii.

Markscheme

$y = \frac{d x}{d t} \Rightarrow \frac{d y}{d t} + 5 \frac{d x}{d t} + 6 x = 0$ OR $\frac{d y}{d t} + 5 y + 6 x = 0$ M1

Note: Award M1 for substituting $\frac{d y}{d t}$ for $\frac{d^{2} x}{d t^{2}}$ .

$(\begin{matrix} \frac{d x}{d t} \\ \frac{d y}{d t} \end{matrix}) = (\begin{matrix} 0 & 1 \\ - 6 & - 5 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})$ AG

[1 mark]

a.i.

$det (\begin{matrix} - λ & 1 \\ - 6 & - 5 - λ \end{matrix}) = 0$ (M1)

Note: Award M1 for an attempt to find eigenvalues. Any indication that $det (M - λ I) = 0$ has been used is sufficient for the (M1).

$- λ (- 5 - λ) + 6 = 0$ OR $λ^{2} + 5 λ + 6 = 0$ (A1)

$λ = - 2, - 3$ A1

[3 marks]

a.ii.

(on a phase portrait the particle approaches $(0, 0)$ as $t$ increases so long term velocity ( $y$ ) is)

$0$ A1

Note: Only award A1 for $0$ if both eigenvalues in part (a)(ii) are negative. If at least one is positive accept an answer of ‘no limit’ or ‘infinity’, or in the case of one positive and one negative also accept ‘no limit or $0$ (depending on initial conditions)’.

[1 mark]

a.iii.

$y = \frac{d x}{d t}$

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

$\frac{d y}{d t} + 5 y + 6 x = 3 t + 4$ A1

[2 marks]

b.i.

recognition that $h = 0.1$ in any recurrence formula (M1)

$(t_{n + 1} = t_{n} + 0.1)$

$x_{n + 1} = x_{n} + 0.1 y_{n}$ (A1)

$y_{n + 1} = y_{n} + 0.1 (3 t_{n} + 4 - 5 y_{n} - 6 x_{n})$ (A1)

(when $t = 1$ ,) $x = 0.64402 \dots \approx 0.644 m$ A2

[5 marks]

b.ii.

recognizing that $y$ is the velocity

$0.5 {m s}^{- 1}$ A1

[2 marks]

b.iii.

Examiners report

It was clear that second order differential equations had not been covered by many schools. Fortunately, many were able to successfully answer part (ii) as this was independent of the other two parts. For part (iii) it was expected that candidates would know that two negative eigenvalues mean the system tends to the origin and so the long-term velocity is 0. Some candidates tried to solve the system. It should be noted that when the command term is ‘state’ then no further working out is expected to be seen.

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

Forming a coupled system from a second order differential equation and solving it using Euler’s method is a technique included in the course guide. Candidates who had learned this technique were successful in this question.

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

The curve $y = f\left( x \right)$ is shown in the graph, for $0 \leqslant x \leqslant 10$ .

The curve $y = f\left( x \right)$ passes through the following points.

It is required to find the area bounded by the curve, the $x$ -axis, the $y$ -axis and the line $x = 10$ .

One possible model for the curve $y = f\left( x \right)$ is a cubic function.

Use the trapezoidal rule to find an estimate for the area.

[3]

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

[3]

b.i.

Write down the coefficient of determination.

[1]

b.ii.

Write down an expression for the area enclosed by the cubic regression curve, the $x$ -axis, the $y$ -axis and the line $x = 10$ .

[1]

c.i.

Find the value of this area.

[2]

c.ii.

Markscheme

Area = $\frac{2}{2}\left( {2 + 2\left( {4.5 + 4.2 + 3.3 + 4.5} \right) + 8} \right)$ M1A1

Area = 43 A1

[3 marks]

$y = 0.0389{x^3} - 0.534{x^2} + 2.06x + 2.06$ M1A2

[3 marks]

b.i.

${R^2} = 0.991$ A1

[1 mark]

b.ii.

Area = $\int\limits_0^{10} {y\,dx}$ A1

[1 mark]

c.i.

42.5 A2

[2 marks]

c.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

At an archery tournament, a particular competition sees a ball launched into the air while an archer attempts to hit it with an arrow.

The path of the ball is modelled by the equation

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 5 \\ 0 \end{matrix}) + t (\begin{matrix} u_{x} \\ u_{y} - 5 t \end{matrix})$

where $x$ is the horizontal displacement from the archer and $y$ is the vertical displacement from the ground, both measured in metres, and $t$ is the time, in seconds, since the ball was launched.

$u_{x}$ is the horizontal component of the initial velocity
$u_{y}$ is the vertical component of the initial velocity.

In this question both the ball and the arrow are modelled as single points. The ball is launched with an initial velocity such that $u_{x} = 8$ and $u_{y} = 10$ .

An archer releases an arrow from the point $(0, 2)$ . The arrow is modelled as travelling in a straight line, in the same plane as the ball, with speed $60 {m s}^{- 1}$ and an angle of elevation of $10 °$ .

Find the initial speed of the ball.

[2]

a.i.

Find the angle of elevation of the ball as it is launched.

[2]

a.ii.

Find the maximum height reached by the ball.

[3]

Assuming that the ground is horizontal and the ball is not hit by the arrow, find the $x$ coordinate of the point where the ball lands.

[3]

For the path of the ball, find an expression for $y$ in terms of $x$ .

[3]

Determine the two positions where the path of the arrow intersects the path of the ball.

[4]

Determine the time when the arrow should be released to hit the ball before the ball reaches its maximum height.

[4]

Markscheme

$\sqrt{10^{2} + 8^{2}}$ (M1)

$= 12.8 (12.8062 \dots, \sqrt{164}) (m s^{- 1})$ A1

[2 marks]

a.i.

$\tan^{- 1} (\frac{10}{8})$ (M1)

$= 0.896$ OR $51.3$ ( $0.896055 \dots$ OR $51.3401 \dots °$ ) A1

Note: Accept $0.897$ or $51.4$ from use of $arcsin (\frac{10}{12.8})$ .

[2 marks]

a.ii.

$y = t (10 - 5 t)$ (M1)

Note: The M1 might be implied by a correct graph or use of the correct equation.

METHOD 1 – graphical Method

sketch graph (M1)

Note: The M1 might be implied by correct graph or correct maximum (eg $t = 1$ ).

max occurs when $y = 5 m$ A1

METHOD 2 – calculus

differentiating and equating to zero (M1)

$\frac{d y}{d t} = 10 - 10 t = 0$

$t = 1$

$y (= 1 (10 - 5)) = 5 m$ A1

METHOD 3 – symmetry

line of symmetry is $t = 1$ (M1)

$y (= 1 (10 - 5)) = 5 m$ A1

[3 marks]

attempt to solve $t (10 - 5 t) = 0$ (M1)

$t = 2$ (or $t = 0$ ) (A1)

$x (= 5 + 8 \times 2) = 21 m$ A1

Note: Do not award the final A1 if $x = 5$ is also seen.

[3 marks]

METHOD 1

$t = \frac{x - 5}{8}$ M1A1

$y = (\frac{x - 5}{8}) (10 - 5 \times \frac{x - 5}{8})$ A1

METHOD 2

$y = k (x - 5) (x - 21)$ A1

when $x = 13, y = 5$ so $k = \frac{5}{(13 - 5) (13 - 21)} = - \frac{5}{64}$ M1A1

$(y = - \frac{5}{64} (x - 5) (x - 21))$

METHOD 3

if $y = a x^{2} + b x + c$

$0 = 25 a + 5 b + c$
$5 = 169 a + 13 b + c$
$0 = 441 a + 21 b + c$ M1A1

solving simultaneously, $a = - \frac{5}{64}, b = \frac{130}{64}, c = - \frac{525}{64}$ A1

( $y = - \frac{5}{64} x^{2} + \frac{130}{64} x - \frac{525}{64}$ )

METHOD 4

use quadratic regression on $(5, 0), (13, 5), (21, 0)$ M1A1

$y = - \frac{5}{64} x^{2} + \frac{130}{64} x - \frac{525}{64}$ A1

Note: Question asks for expression; condone omission of " $y =$ ".

[3 marks]

trajectory of arrow is $y = x \tan 10 + 2$ (A1)

intersecting $y = x \tan 10 + 2$ and their answer to (d) (M1)

$(8.66, 3.53) ((8.65705 \dots, 3.52647 \dots))$ A1

$(15.1, 4.66) ((15.0859 \dots, 4.66006 \dots))$ A1

[4 marks]

when $x_{target} = 8.65705 \dots, t_{target} = \frac{8.65705 \dots - 5}{8} = 0.457132 \dots s$ (A1)

attempt to find the distance from point of release to intersection (M1)

$\sqrt{8.65705 \dots^{2} + {(3.52647 \dots - 2)}^{2}} (= 8.79060 \dots m)$

time for arrow to get there is $\frac{8.79060 \dots}{60} = 0.146510 \dots s$ (A1)

so the arrow should be released when

$t = 0.311 (s) (0.310622 \dots (s))$ A1

[4 marks]

Examiners report

This question was found to be the most difficult on the paper. There were a good number of good solutions to parts (a) and part (b), frequently with answers just written down with no working. Part (c) caused some difficulties with confusing variables. The most significant difficulties started with part (d) and became greater to the end of the question. Few candidates were able to work through the final two parts.

a.i.

[N/A]

a.ii.

[N/A]

A biologist introduces $100$ rabbits to an island and records the size of their population $(x)$ over a period of time. The population growth of the rabbits can be approximately modelled by the following differential equation, where $t$ is time measured in years.

$\frac{d x}{d t} = 2 x$

A population of $100$ foxes is introduced to the island when the population of rabbits has reached $1000$ . The subsequent population growth of rabbits and foxes, where $y$ is the population of foxes at time $t$ , can be approximately modelled by the coupled equations:

$\frac{d x}{d t} = x (2 - 0.01 y)$

$\frac{d y}{d t} = y (0.0002 x - 0.8)$

Use Euler’s method with a step size of $0.25$ , to find

The graph of the population sizes, according to this model, for the first $4$ years after the foxes were introduced is shown below.

Describe the changes in the populations of rabbits and foxes for these $4$ years at

Find the population of rabbits $1$ year after they were introduced.

[5]

(i) the population of rabbits 1 year after the foxes were introduced.

(ii) the population of foxes 1 year after the foxes were introduced.

[6]

point $A$ .

[1]

c.i.

point $B$ .

[2]

c.ii.

Find the non-zero equilibrium point for the populations of rabbits and foxes.

[3]

Markscheme

$\int \frac{1}{x} d x = \int 2 d t$ (M1)

$\ln x = 2 t + c$

$x = A e^{2 t}$ (A1)

$x (0) = 100 \Rightarrow A = 100$ (M1)

$x = 100 e^{2 t}$ (A1)

$x (1) = 739$ A1

Note: Accept $738$ for the final A1.

[5 marks]

$t_{n + 1} = t_{n} + 0.25$ (A1)

Note: This may be inferred from a correct $t$ column, where this is seen.

$x_{n + 1} = x_{n} + 0.25 x_{n} (2 - 0.01 y_{n})$ (A1)

$y_{n + 1} = y_{n} + 0.25 y_{n} (0.0002 x_{n} - 0.8)$ (A1)

(A1)

Note: Award A1 for whole line correct when $t = 0.5$ or $t = 0.75$ . The $t$ column may be omitted and implied by the correct $x$ and $y$ values. The formulas are implied by the correct $x$ and $y$ columns.

(i) $2840$ ( $2836$ OR $2837$ ) A1

(ii) $58$ OR $59$ A1

[6 marks]

both populations are increasing A1

[1 mark]

c.i.

rabbits are decreasing and foxes are increasing A1A1

[2 marks]

c.ii.

setting at least one $DE$ to zero (M1)

$x = 4000, y = 200$ A1A1

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Jorge is carefully observing the rise in sales of a new app he has created.

The number of sales in the first four months is shown in the table below.

Jorge believes that the increase is exponential and proposes to model the number of sales $N$ in month $t$ with the equation

$N = A e^{r t}, A, r \in ℝ$

Jorge plans to adapt Euler’s method to find an approximate value for $r$ .

With a step length of one month the solution to the differential equation can be approximated using Euler’s method where

$N (n + 1) \approx N (n) + 1 \times N' (n), n \in ℕ$

Jorge decides to take the mean of these values as the approximation of $r$ for his model. He also decides the graph of the model should pass through the point $(2, 52)$ .

The sum of the square residuals for these points for the least squares regression model is approximately $6.555$ .

Show that Jorge’s model satisfies the differential equation

$\frac{d N}{d t} = r N$

[2]

Show that $r \approx \frac{N (n + 1) - N (n)}{N (n)}$

[3]

Hence find three approximations for the value of $r$ .

[3]

Find the equation for Jorge’s model.

[3]

Find the sum of the square residuals for Jorge’s model using the values $t = 1, 2, 3, 4$ .

[2]

Comment how well Jorge’s model fits the data.

[1]

f.i.

Give two possible sources of error in the construction of his model.

[2]

f.ii.

Markscheme

$\frac{d N}{d t} = r A e^{r t}$ (M1)A1

Note: M1 is for an attempt to find $\frac{d N}{d t}$

$= r N$ AG

Note: Accept solution of the differential equation by separating variables

[2 marks]

$N (n + 1) \approx N (n) + 1 \times N' (n) \Rightarrow N' (n) \approx N (n + 1) - N (n)$ M1

$\Rightarrow r N (n) \approx N (n + 1) - N (n)$ M1A1

$\Rightarrow r \approx \frac{N (n + 1) - N (n)}{N (n)}$ AG

Note: Do not penalize the use of the $=$ sign.

[3 marks]

Correct method (M1)

$r \approx \frac{52 - 40}{40} = 0.3$

$r \approx \frac{70 - 52}{52} = 0.346$

$r \approx \frac{98 - 70}{70} = 0.4$ A2

Note: A1 for a single error A0 for two or more errors.

[3 marks]

$r = 0.349 (0.34871 \dots)$ or $\frac{68}{195}$ A1

$52 = A e^{0.34871 \dots \times 2}$ (M1)

$A = 25.8887 \dots$

$N = 25.9 e^{0.349 t}$ A1

[3 marks]

${(36.6904 \dots - 40)}^{2} + 0 + {(73.6951 \dots - 70)}^{2} + {(104.4435 \dots - 98)}^{2}$ (M1)

$= 66.1 (66.126 \dots)$ A1

[2 marks]

The sum of the square residuals is approximately $10$ times as large as the minimum possible, so Jorge’s model is unlikely to fit the data exactly R1

[1 mark]

f.i.

For example

Selecting a single point for the curve to pass through

Approximating the gradient of the curve by the gradient of a chord R1R1

[2 marks]

f.ii.

Examiners report

[N/A]

f.i.

[N/A]

f.ii.

A student investigating the relationship between chemical reactions and temperature finds the Arrhenius equation on the internet.

$k = A e^{- \frac{c}{T}}$

This equation links a variable $k$ with the temperature $T$ , where $A$ and $c$ are positive constants and $T > 0$ .

The Arrhenius equation predicts that the graph of $\ln k$ against $\frac{1}{T}$ is a straight line.

Write down

The following data are found for a particular reaction, where $T$ is measured in Kelvin and $k$ is measured in ${cm}^{3} {mol}^{- 1} s^{- 1}$ :

Find an estimate of

Show that $\frac{d k}{d T}$ is always positive.

[3]

Given that $\lim_{T \to \infty} k = A$ and $\lim_{T \to 0} k = 0$ , sketch the graph of $k$ against $T$ .

[3]

(i) the gradient of this line in terms of $c$ ;

(ii) the $y$ -intercept of this line in terms of $A$ .

[4]

Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$ .

[2]

$c$ .

It is not required to state units for this value.

[1]

e.i.

$A$ .

It is not required to state units for this value.

[2]

e.ii.

Markscheme

attempt to use chain rule, including the differentiation of $\frac{1}{T}$ (M1)

$\frac{d k}{d T} = A \times \frac{c}{T^{2}} \times e^{- \frac{c}{T}}$ A1

this is the product of positive quantities so must be positive R1

Note: The R1 may be awarded for correct argument from their derivative. R1 is not possible if their derivative is not always positive.

[3 marks]

A1A1A1

Note: Award A1 for an increasing graph, entirely in first quadrant, becoming concave down for larger values of $T$ , A1 for tending towards the origin and A1 for asymptote labelled at $k = A$ .

[3 marks]

taking $\ln$ of both sides OR substituting $y = \ln x$ and $x = \frac{1}{T}$ (M1)

$\ln k = \ln A - \frac{c}{T}$ OR $y = - c x + \ln A$ (A1)

(i) so gradient is $- c$ A1

(ii) $y$ -intercept is $\ln A$ A1

Note: The implied (M1) and (A1) can only be awarded if both correct answers are seen. Award zero if only one value is correct and no working is seen.

[4 marks]

an attempt to convert data to $\frac{1}{T}$ and $\ln k$ (M1)

e.g. at least one correct row in the following table

line is $\ln k = - 13400 \times \frac{1}{T} + 15.0 (= - 13383.1 \dots \times \frac{1}{T} + 15.0107 \dots)$ A1

[2 marks]

$c = 13400 (13383.1 \dots)$ A1

[1 mark]

e.i.

attempt to rearrange or solve graphically $\ln A = 15.0107 \dots$ (M1)

$A = 3 300 000 (3 304 258 \dots)$ A1

Note: Accept an $A$ value of $3269017$ … from use of $3 sf$ value.

[2 marks]

e.ii.

Examiners report

This question caused significant difficulties for many candidates and many did not even attempt the question. Very few candidates were able to differentiate the expression in part (a) resulting in difficulties for part (b). Responses to parts (c) to (e) illustrated a lack of understanding of linearizing a set of data. Those candidates that were able to do part (d) frequently lost a mark as their answer was given in x and y.

[N/A]

e.i.

[N/A]

e.ii.

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

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

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

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

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

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

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

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

[2]

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

[2]

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

[3]

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

[3]

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

[2]

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

[2]

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

[6]

Markscheme

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

3, −3 A1A1

[2 marks]

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

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

Note: Can be done in either order.

[2 marks]

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

[3 marks]

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

Note: Award M1A1 for either interval.

= 0.00730 A1

[3 marks]

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

= 2.87 A1

[2 marks]

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

the curve begins with the positive part of the cycle R1

[2 marks]

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

$a = 3.00$ A1

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

$d = 1.76$ A1

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

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

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

$c = 0.00100$ A1

[6 marks]

Examiners report

[N/A]

Consider the system of paired differential equations

$\dot x = 3x + 2y$

$\dot y = 2x + 3y$ .

This represents the populations of two species of symbiotic toadstools in a large wood.

Time $t$ is measured in decades.

Use the eigenvalue method to find the general solution to this system of equations.

[10]

Given the initial conditions that when $t = 0$ , $x = 150$ , $y = 50$ , find the particular solution.

[3]

b.i.

Hence find the solution when $t = 1$ .

[1]

b.ii.

As $t \to \infty$ , find an asymptote to the trajectory of the particular solution found in (b)(i) and state if this trajectory will be moving towards or away from the origin.

[4]

Markscheme

The characteristic equation is given by

$\left| {\begin{array}{*{20}{c}} {3 - \lambda }&2 \\ 2&{3 - \lambda } \end{array}} \right| = 0 \Rightarrow {\lambda ^2} - 6\lambda + 5 = 0 \Rightarrow \lambda = 1{\text{ or 5}}$ M1A1A1A1

$\lambda = 1{\text{ }}\left( {\begin{array}{*{20}{c}} 2&2 \\ 2&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right){\text{ gives an eigenvector of form }}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ M1A1

$\lambda = 5{\text{ }}\left( {\begin{array}{*{20}{c}} { - 2}&2 \\ 2&{ - 2} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right){\text{ gives an eigenvector of form }}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ M1A1

General solution is $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = A{e^t}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) + B{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ A1A1

[10 marks]

Require $A + B = 150,{\text{ }} - A + B = 50 \Rightarrow A = 50,{\text{ B = 100}}$ M1A1

Particular solution is $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = 50{e^t}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) + 100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ A1

[3 marks]

b.i.

$t = 1 \Rightarrow \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {15000} \\ {14700} \end{array}} \right){\text{ }}\left( {3sf} \right)$ A1

[1 mark]

b.ii.

The dominant term is $100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ so as $t \to \infty$ , $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) \simeq 100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ M1A1

Giving the asymptote as $y = x$ A1

The trajectory is moving away from the origin. A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Consider the curve $y = \sqrt{x}$ .

The shape of a piece of metal can be modelled by the region bounded by the functions $f$ , $g$ , the $x$ -axis and the line segment $[AB]$ , as shown in the following diagram. The units on the $x$ and $y$ axes are measured in metres.

The piecewise function $f$ is defined by

$f (x) = {\begin{matrix} \sqrt{x} & 0 \leq x \leq 0.16 \\ 1.25 x + 0.2 & 0.16 < x \leq 0.5 \end{matrix}$

The graph of $g$ is obtained from the graph of $f$ by:

a stretch scale factor of $\frac{1}{2}$ in the $x$ direction,
followed by a stretch scale factor $\frac{1}{2}$ in the $y$ direction,
followed by a translation of $0.2$ units to the right.

Point $A$ lies on the graph of $f$ and has coordinates $(0.5, 0.825)$ . Point $B$ is the image of $A$ under the given transformations and has coordinates $(p, q)$ .

The piecewise function $g$ is given by

$g (x) = {\begin{matrix} h (x) & 0.2 \leq x \leq a \\ 1.25 x + b & a < x \leq p \end{matrix}$

The area enclosed by $y = g (x)$ , the $x$ -axis and the line $x = p$ is $0.0627292 m^{2}$ correct to six significant figures.

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

[2]

a.i.

Hence show that the equation of the tangent to the curve at the point $(0.16, 0.4)$ is $y = 1.25 x + 0.2$ .

[2]

a.ii.

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

[2]

Find an expression for $h (x)$ .

[2]

c.i.

Find the value of $a$ .

[1]

c.ii.

Find the value of $b$ .

[2]

c.iii.

Find the area enclosed by $y = f (x)$ , the $x$ -axis and the line $x = 0.5$ .

[3]

d.i.

Find the area of the shaded region on the diagram.

[4]

d.ii.

Markscheme

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

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

[2 marks]

a.i.

gradient at $x = 0.16$ is $\frac{1}{2} \times \frac{1}{\sqrt{0.16}}$ M1

$= 1.25$

EITHER

$y - 0.4 = 1.25 (x - 0.16)$ M1

$0.4 = 1.25 (0.16) + b$ M1

Note: Do not allow working backwards from the given answer.

THEN

hence $y = 1.25 x + 0.2$ AG

[2 marks]

a.ii.

$p = 0.45, q = 0.4125$ (or $0.413$ ) (accept " $(0.45, 0.4125)$ ") A1A1

[2 marks]

$(h (x) =) \frac{1}{2} \sqrt{2 (x - 0.2)}$ A2

Note: Award A1 if only two correct transformations are seen.

[2 marks]

c.i.

$(a =) 0.28$ A1

[1 mark]

c.ii.

EITHER

Correct substitution of their part (b) (or $(0.28, 0.2)$ ) into the given expression (M1)

$\frac{1}{2} (1.25 \times 2 (x - 0.2) + 0.2)$ (M1)

Note: Award M1 for transforming the equivalent expression for $f$ correctly.

THEN

$(b =) - 0.15$ A1

[2 marks]

c.iii.

recognizing need to add two integrals (M1)

$\int_{0}^{0.16} \sqrt{x} d x + \int_{0.16}^{0.5} (1.25 x + 0.2) d x$ (A1)

Note: The second integral could be replaced by the formula for the area of a trapezoid $\frac{1}{2} \times 0.34 (0.4 + 0.825)$ .

$0.251 m^{2} (0.250916 \dots)$ A1

[3 marks]

d.i.

EITHER

area of a trapezoid $\frac{1}{2} \times 0.05 (0.4125 + 0.825) = 0.0309375$ (M1)(A1)

$\int_{0.45}^{0.5} (8.25 x - 3.3) d x = 0.0309375$ (M1)(A1)

Note: If the rounded answer of $0.413$ from part (b) is used, the integral is $\int_{0.45}^{0.5} (8.24 x - 3.295) d x = 0.03095$ which would be awarded (M1)(A1).

THEN

shaded area $= 0.250916 \dots - 0.0627292 - 0.0309375$ (M1)

Note: Award (M1) for the subtraction of both $0.0627292 \dots$ and their area for the trapezoid from their answer to (a)(i).

$= 0.157 m^{2} (0.15725)$ A1

[4 marks]

d.ii.

Examiners report

The differentiation using the power rule was well done. In part (ii) some candidates felt it was sufficient to refer to the equation being the same as the one generated by their calculator. Generally, for ‘show that’ questions an algebraic derivation is expected.

a.i.

[N/A]

a.ii.

The candidates were successful at applying transformations to points but very few were able to apply these transformations to derive the correct function h. In most cases it was due to not appreciating the effect the horizontal transformations have on x.

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Part (i) was frequently done well using the inbuilt functionality of the GDC. Part (ii) was less structured, and candidates needed to create a clear diagram so they could easily see which areas needed to be subtracted. Most of those who were successful used the formula for the trapezoid for the area they needed to find, though others were also successful through finding the equation of the line AB.

d.i.

[N/A]

d.ii.

A sector of a circle, centre $O$ and radius $4.5 m$ , is shown in the following diagram.

A square field with side $8 m$ has a goat tied to a post in the centre by a rope such that the goat can reach all parts of the field up to $4.5 m$ from the post.

^{[Source: mynamepong, n.d. Goat [image online] Available at: https://thenounproject.com/term/goat/1761571/}
^{This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)}
^{https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 22 April 2010] Source adapted.]}

Let $V$ be the volume of grass eaten by the goat, in cubic metres, and $t$ be the length of time, in hours, that the goat has been in the field.

The goat eats grass at the rate of $\frac{d V}{d t} = 0.3 t e^{- t}$ .

Find the angle $AÔB$ .

[3]

a.i.

Find the area of the shaded segment.

[5]

a.ii.

Find the area of the field that can be reached by the goat.

[4]

Find the value of $t$ at which the goat is eating grass at the greatest rate.

[2]

The goat is tied in the field for $8$ hours.

Find the total volume of grass eaten by the goat during this time.

[3]

Markscheme

$(\frac{1}{2} AÔB =) arccos (\frac{4}{4.5}) = 27.266 \dots$ (M1)(A1)

$AÔB = 54.532 \dots \approx 54.5 °$ ( $0.951764 \dots \approx 0.952$ radians) A1

Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.

[3 marks]

a.i.

finding area of triangle

EITHER

area of triangle $= \frac{1}{2} \times 4 . 5^{2} \times \sin (54.532 \dots)$ (M1)

Note: Award M1 for correct substitution into formula.

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

$AB = 2 \times \sqrt{4 . 5^{2} - 4^{2}} = 4.1231 \dots$ (M1)

area triangle $= \frac{4.1231 \dots \times 4}{2}$

$= 8.24621 \dots \approx 8.25 m^{2}$ (A1)

finding area of sector

EITHER

area of sector $= \frac{54.532 \dots}{360} \times π \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

area of sector $= \frac{1}{2} \times 0.9517641 \dots \times 4 . 5^{2}$ (M1)

$= 9.63661 \dots \approx 9.64 m^{2}$ (A1)

THEN

area of segment $= 9.63661 \dots - 8.24621 \dots$

$= 1.39 m^{2} (1.39040 \dots)$ A1

[5 marks]

a.ii.

METHOD 1

$π \times 4 . 5^{2} (63.6172 \dots)$ (A1)

$4 \times 1.39040 . . . (5.56160)$ (A1)

subtraction of four segments from area of circle (M1)

$= 58.1 m^{2} (58.055 \dots)$ A1

METHOD 2

angle of sector $= 90 - 54.532 \dots (\frac{π}{2} - 0.951764 \dots)$ (A1)

area of sector $= \frac{90 - 54.532 \dots}{360} \times π \times 4 . 5^{2} (= 6.26771 \dots)$ (A1)

area is made up of four triangles and four sectors (M1)

total area $= (4 \times 8.2462 \dots) + (4 \times 6.26771 \dots)$

$= 58.1 m^{2} (58.055 \dots)$ A1

[4 marks]

sketch of $\frac{d V}{d t}$ OR $\frac{d V}{d t} = 0.110363 \dots$ OR attempt to find where $\frac{d^{2} V}{d t^{2}} = 0$ (M1)

$t = 1$ hour A1

[2 marks]

recognizing $V = \int \frac{d V}{d t} d t$ (M1)

$\int_{0}^{8} 0.3 t e^{- 1} d t$ (A1)

volume eaten is $0.299 \dots m^{3} (0.299094 \dots)$ A1

[3 marks]

Examiners report

Generally, this question was answered well but provided a good example of final marks being lost due to premature rounding. Some candidates gave a correct three significant figure intermediate answer of 27.3˚ for the angle in the right-angles triangle and then doubled it to get 54.6˚ as a final answer. This did not receive the final answer mark as the correct answer is 54.5˚ to three significant figures. Premature rounding needs to be avoided in all questions.

a.i.

[N/A]

a.ii.

Unfortunately, many candidates failed to see the connection to part (a). Indeed, the most common answer was to assume the goat could eat all the grass in a circle of radius 4.5m.

Most candidates completed this question successfully by graphing the function. A few tried to differentiate the function again and, in some cases, also managed to obtain the correct answer.

This was a question that was pleasingly answered correctly by many candidates who recognized that integration was needed to find the answer. As in part (c) a few tried to do the integration ‘by hand’, and were largely unsuccessful.

Consider the function $f(x) = \frac{{\sqrt x }}{{\sin x}},{\text{ }}0 < x < \pi$ .

Consider the region bounded by the curve $y = f(x)$ , the $x$ -axis and the lines $x = \frac{\pi }{6},{\text{ }}x = \frac{\pi }{3}$ .

Show that the $x$ -coordinate of the minimum point on the curve $y = f(x)$ satisfies the equation $\tan x = 2x$ .

[5]

a.i.

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

[2]

a.ii.

Sketch the graph of $y = f(x)$ showing clearly the minimum point and any asymptotic behaviour.

[3]

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y = - x$ .

[4]

This region is now rotated through $2\pi$ radians about the $x$ -axis. Find the volume of revolution.

[3]

Markscheme

attempt to use quotient rule or product rule M1

$f’(x) = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}{\text{ }}\left( { = \frac{1}{{2\sqrt x \sin x}} - \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}} \right)$ A1A1

Note: Award A1 for $\frac{1}{{2\sqrt x \sin x}}$ or equivalent and A1 for $- \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}$ or equivalent.

setting $f’(x) = 0$ M1

$\frac{{\sin x}}{{2\sqrt x }} - \sqrt x \cos x = 0$

$\frac{{\sin x}}{{2\sqrt x }} = \sqrt x \cos x$ or equivalent A1

$\tan x = 2x$ AG

[5 marks]

a.i.

$x = 1.17$

$0 < x \leqslant 1.17$ A1A1

Note: Award A1 for $0 < x$ and A1 for $x \leqslant 1.17$ . Accept $x < 1.17$ .

[2 marks]

a.ii.

N17/5/MATHL/HP2/ENG/TZ0/10.b/M

concave up curve over correct domain with one minimum point above the $x$ -axis. A1

approaches $x = 0$ asymptotically A1

approaches $x = \pi$ asymptotically A1

Note: For the final A1 an asymptote must be seen, and $\pi$ must be seen on the $x$ -axis or in an equation.

[3 marks]

$f’(x){\text{ }}\left( { = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}} \right) = 1$ (A1)

attempt to solve for $x$ (M1)

$x = 1.96$ A1

$y = f(1.96 \ldots )$

$= 1.51$ A1

[4 marks]

$V = \pi \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{x{\text{d}}x}}{{{{\sin }^2}x}}}$ (M1)(A1)

Note: M1 is for an integral of the correct squared function (with or without limits and/or $\pi$ ).

$= 2.68{\text{ }}( = 0.852\pi )$ A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

An environmental scientist is asked by a river authority to model the effect of a leak from a power plant on the mercury levels in a local river. The variable $x$ measures the concentration of mercury in micrograms per litre.

The situation is modelled using the second order differential equation

$\frac{d^{2} x}{d t^{2}} + 3 \frac{d x}{d t} + 2 x = 0$

where $t \geq 0$ is the time measured in days since the leak started. It is known that when $t = 0, x = 0$ and $\frac{d x}{d t} = 1$ .

If the mercury levels are greater than $0.1$ micrograms per litre, fishing in the river is considered unsafe and is stopped.

The river authority decides to stop people from fishing in the river for $10 %$ longer than the time found from the model.

Show that the system of coupled first order equations:

$\frac{d x}{d t} = y$

$\frac{d y}{d t} = - 2 x - 3 y$

can be written as the given second order differential equation.

[2]

Find the eigenvalues of the system of coupled first order equations given in part (a).

[3]

Hence find the exact solution of the second order differential equation.

[5]

Sketch the graph of $x$ against $t$ , labelling the maximum point of the graph with its coordinates.

[2]

Use the model to calculate the total amount of time when fishing should be stopped.

[3]

Write down one reason, with reference to the context, to support this decision.

[1]

Markscheme

differentiating first equation. M1

$\frac{d^{2} x}{d t^{2}} = \frac{d y}{d t}$

substituting in for $\frac{d y}{d t}$ M1

$= - 2 x - 3 y = - 2 x - 3 \frac{d x}{d t}$

therefore $\frac{d^{2} x}{d t^{2}} + 3 \frac{d x}{d t} + 2 x = 0$ AG

Note: The AG line must be seen to award the final M1 mark.

[2 marks]

the relevant matrix is $(\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix})$ (M1)

Note: $(\begin{matrix} - 3 & - 2 \\ 1 & 0 \end{matrix})$ is also possible.

(this has characteristic equation) $- λ (- 3 - λ) + 2 = 0$ (A1)

$λ = - 1, - 2$ A1

[3 marks]

EITHER

the general solution is $x = A e^{- t} + B e^{- 2 t}$ M1

Note: Must have constants, but condone sign error for the M1.

so $\frac{d x}{d t} = - A e^{- t} - 2 B e^{- 2 t}$ M1A1

attempt to find eigenvectors (M1)

respective eigenvectors are $(\begin{matrix} 1 \\ - 1 \end{matrix})$ and $(\begin{matrix} 1 \\ - 2 \end{matrix})$ (or any multiple)

$(\begin{matrix} x \\ y \end{matrix}) = A e^{- t} (\begin{matrix} 1 \\ - 1 \end{matrix}) + B e^{- 2 t} (\begin{matrix} 1 \\ - 2 \end{matrix})$ (M1)A1

THEN

the initial conditions become:

$0 = A + B$

$1 = - A - 2 B$ M1

this is solved by $A = 1, B = - 1$

so the solution is $x = e^{- t} - e^{- 2 t}$ A1

[5 marks]

A1A1

Note: Award A1 for correct shape (needs to go through origin, have asymptote at $y = 0$ and a single maximum; condone $x < 0$ ). Award A1 for correct coordinates of maximum.

[2 marks]

intersecting graph with $y = 0.1$ (M1)

so the time fishing is stopped between $2.1830 \dots$ and $0.11957 \dots$ (A1)

$= 2.06 (343 \dots)$ days A1

[3 marks]

Any reasonable answer. For example:

There are greater downsides to allowing fishing when the levels may be dangerous than preventing fishing when the levels are safe.

The concentration of mercury may not be uniform across the river due to natural variation / randomness.

The situation at the power plant might get worse.

Mercury levels are low in water but still may be high in fish. R1

Note: Award R1 for a reasonable answer that refers to this specific context (and not a generic response that could apply to any model).

[1 mark]

Examiners report

Many candidates did not get this far, but the attempts at the question that were seen were generally good. The greater difficulties were seen in parts (e) and (f), but this could be a problem with time running out.

[N/A]

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

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

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

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

[3]

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

[4]

Markscheme

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

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

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

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

[3 marks]

METHOD 1

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

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

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

METHOD 2

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

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

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

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

[4 marks]

Examiners report

[N/A]

Charlotte decides to model the shape of a cupcake to calculate its volume.

From rotating a photograph of her cupcake she estimates that its cross-section passes through the points $(0, 3.5), (4, 6), (6.5, 4), (7, 3)$ and $(7.5, 0)$ , where all units are in centimetres. The cross-section is symmetrical in the $x$ -axis, as shown below:

She models the section from $(0, 3.5)$ to $(4, 6)$ as a straight line.

Charlotte models the section of the cupcake that passes through the points $(4, 6), (6.5, 4), (7, 3)$ and $(7.5, 0)$ with a quadratic curve.

Charlotte thinks that a quadratic with a maximum point at $(4, 6)$ and that passes through the point $(7.5, 0)$ would be a better fit.

Believing this to be a better model for her cupcake, Charlotte finds the volume of revolution about the $x$ -axis to estimate the volume of the cupcake.

Find the equation of the line passing through these two points.

[2]

Find the equation of the least squares regression quadratic curve for these four points.

[2]

b.i.

By considering the gradient of this curve when $x = 4$ , explain why it may not be a good model.

[1]

b.ii.

Find the equation of the new model.

[4]

Write down an expression for her estimate of the volume as a sum of two integrals.

[4]

d.i.

Find the value of Charlotte’s estimate.

[1]

d.ii.

Markscheme

$y = \frac{5}{8} x + \frac{7}{2} (y = 0.625 x + 3.5)$ A1A1

Note: Award A1 for $0.625 x$ , A1 for $3.5$ .
Award a maximum of A0A1 if not part of an equation.

[2 marks]

$y = - 0.975 x^{2} + 9.56 x - 16.7$ (M1)A1

$(y = - 0.974630 x^{2} + 9.55919 x - 16.6569 \dots)$

[2 marks]

b.i.

gradient of curve is positive at $x = 4$ R1

Note: Accept a sensible rationale that refers to the gradient.

[1 mark]

b.ii.

METHOD 1

let $y = a x^{2} + b x + c$

differentiating or using $x = \frac{- b}{2 a}$ (M1)

$8 a + b = 0$

substituting in the coordinates
$7 . 5^{2} a + 7.5 b + c = 0$ (A1)
$4^{2} a + 4 b + c = 6$ (A1)

solve to get
$y = - \frac{24}{49} x^{2} + \frac{192}{49} x - \frac{90}{49}$ OR $y = - 0.490 x^{2} + 3.92 x - 1.84$ A1

Note: Use of quadratic regression with points using the symmetry of the graph is a valid method.

METHOD 2

$y = a {(x - 4)}^{2} + 6$ (M1)

$0 = a {(7.5 - 4)}^{2} + 6$ (M1)

$a = - \frac{24}{49}$ (A1)

$y = - \frac{24}{49} {(x - 4)}^{2} + 6$ OR $y = - 0.490 {(x - 4)}^{2} + 6$ A1

[4 marks]

$π \int_{0}^{4} {(\frac{5}{8} x + 3.5)}^{2} d x + π \int_{4}^{7.5} {(- \frac{24}{49} {(x - 4)}^{2} + 6)}^{2} d x$ (M1)(M1) (M1)A1

Note: Award (M1)(M1)(M1)A0 if $π$ is omitted but response is otherwise correct. Award (M1) for an integral that indicates volume, (M1) for their part (a) within their volume integral, (M1) for their part (b)(i) within their volume integral, A1 for their correct two integrals with all correct limits.

[4 marks]

d.i.

$501 {cm}^{3} (501.189 \dots)$ A1

[1 mark]

d.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

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

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

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

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

[2]

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

[3]

Explain why $f$ is an even function.

[1]

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

[1]

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

[4]

Find $g'(x)$ .

[3]

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

[2]

g.i.

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

[2]

g.ii.

Markscheme

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

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

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

[2 marks]

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

shape A1

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

$x$ -intercepts A1

[3 marks]

EITHER

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

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

[1 mark]

EITHER

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

horizontal line cuts twice R1

Note: Accept any equivalent correct statement.

[1 mark]

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

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

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

[4 marks]

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

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

[3 marks]

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

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

[2 marks]

g.i.

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

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

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

[2 marks]

g.ii.

Examiners report

[N/A]

g.i.

[N/A]

g.ii.

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

Let $u = \tan x$ .

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

[2]

a.i.

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

[4]

a.ii.

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

[2]

a.iii.

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

[2]

b.i.

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

[3]

b.ii.

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

[2]

b.iii.

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

[3]

Markscheme

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

[2 marks]

a.i.

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

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

[4 marks]

a.ii.

$x = 0.0736$ A1

$x = 1.13$ A1

[2 marks]

a.iii.

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

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

[2 marks]

b.i.

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

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

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

[3 marks]

b.ii.

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

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

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

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

[2 marks]

b.iii.

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

$x = \arctan (1)$ A1

$x = \arctan (3)$ A1

Note: Only accept answers given the required form.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

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

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

[2]

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

[2]

b.i.

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

[1]

b.ii.

Markscheme

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

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

t₁ = 0.441(s) A1

[2 marks]

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

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

[2 marks]

b.i.

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

[1 mark]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

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

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

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

[5]

Find the coordinates of P and Q.

[4]

b.i.

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

[3]

b.ii.

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

[7]

Markscheme

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

attempt at implicit differentiation M1

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

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

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

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

[5 marks]

EITHER

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

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

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

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

THEN

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

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

[4 marks]

b.i.

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

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

m₁m₂= 1 AG

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

[3 marks]

b.ii.

equate derivative to −1 M1

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

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

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

(0.486,0.486) A1

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

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

[7 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

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

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

[7]

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

[5]

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

[3]

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

[7]

Markscheme

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

differentiating implicitly: M1

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

Note: Award A1 for each side.

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

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

$y = 0$ not possible (as 0 ≠ 5) R1

hence exactly two points AG

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

[7 marks]

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

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

gradient of normal is 2 M1

1 = 4 + c (M1)

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

[5 marks]

substituting (M1)

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

$x = 0.724$ A1

[3 marks]

recognition of two volumes (M1)

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

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

volume 2

EITHER

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

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

THEN

total volume = 19.9 A1

[7 marks]

Examiners report

[N/A]

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

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

Markscheme

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

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

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

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

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

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

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

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

⇒ $d = - 4$ A1

$x = 1$ , $y = 0$

⇒ $0 = 4 - {\text{ln}}\,2 + c - 4$

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

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

[7 marks]

Examiners report

[N/A]

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

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

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

Find his velocity when $t = 15$ .

[2]

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

[2]

Determine the value of $h$ .

[5]

Markscheme

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

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

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

[2 marks]

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

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

[2 marks]

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

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

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

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

[5 marks]

Examiners report

[N/A]

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

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

[4]

Calculate the area of $A$ .

[2]

Markscheme

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

METHOD 1

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

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

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

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

METHOD 2

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

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

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

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

METHOD 3

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

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

[4 marks]

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

[2 marks]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = {\left( {x - 1} \right)^2}$ , $x$ ≥ 1 and the function $g$ is defined by $g\left( x \right) = {x^2} + 1$ , $x$ ≥ 0.

The region $R$ is bounded by the curves $y = f\left( x \right)$ , $y = g\left( x \right)$ and the lines $y = 0$ , $x = 0$ and $y = 9$ as shown on the following diagram.

The shape of a clay vase can be modelled by rotating the region $R$ through 360˚ about the $y$ -axis.

Find the volume of clay used to make the vase.

Markscheme

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

volume $= \pi {\int_0^9 {\left( {{y^{\frac{1}{2}}} + 1} \right)} ^2}{\text{d}}y - \pi \int_1^9 {\left( {y - 1} \right)} {\text{d}}y$ (M1)(M1)(M1)(A1)(A1)

Note: Award (M1) for use of formula for rotating about $y$ -axis, (M1) for finding at least one inverse, (M1) for subtracting volumes, (A1)(A1)for each correct expression, including limits.

$= 268.6 \ldots - 100.5 \ldots \left( {85.5\pi - 32\pi } \right)$

$= 168\left( { = 53.5\pi } \right)$ A2

[7 marks]

Examiners report

[N/A]

The curve $C$ is defined by equation $xy - \ln y = 1,{\text{ }}y > 0$ .

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

[4]

Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$

[3]

Markscheme

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

$y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ M1A1A1

Note: Award A1 for the first two terms, A1 for the third term and the 0.

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

Note: Accept $\frac{{ - {y^2}}}{{\ln y}}$ .

Note: Accept $\frac{{ - y}}{{x - \frac{1}{y}}}$ .

[4 marks]

${m_T} = \frac{{{{\text{e}}^2}}}{{1 - {\text{e}} \times \frac{2}{{\text{e}}}}}$ (M1)

${m_T} = - {{\text{e}}^2}$ (A1)

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

$- {{\text{e}}^2}x - y + 3{\text{e}} = 0$ or equivalent A1

Note: Accept $y = - 7.39x + 8.15$ .

[3 marks]

Examiners report

[N/A]

An object is placed into the top of a long vertical tube, filled with a thick viscous fluid, at time $t = 0$ seconds.

Initially it is thought that the resistance of the fluid would be proportional to the velocity of the object. The following model was proposed, where the object’s displacement, $x$ , from the top of the tube, measured in metres, is given by the differential equation

$\frac{{{{\text{d}}^2}x}}{{{\text{d}}{t^2}}} = 9.81 - 0.9\left( {\frac{{{\text{d}}x}}{{{\text{d}}t}}} \right)$ .

The maximum velocity approached by the object as it falls is known as the terminal velocity.

An experiment is performed in which the object is placed in the fluid on a number of occasions and its terminal velocity recorded. It is found that the terminal velocity was consistently smaller than that predicted by the model used. It was suggested that the resistance to motion is actually proportional to the velocity squared and so the following model was set up.

$\frac{{{{\text{d}}^2}x}}{{{\text{d}}{t^2}}} = 9.81 - 0.9{\left( {\frac{{{\text{d}}x}}{{{\text{d}}t}}} \right)^2}$

At terminal velocity the acceleration of an object is equal to zero.

By substituting $v = \frac{{{\text{d}}x}}{{{\text{d}}t}}$ into the equation, find an expression for the velocity of the particle at time $t$ . Give your answer in the form $v = f(t)$ .

[7]

From your solution to part (a), or otherwise, find the terminal velocity of the object predicted by this model.

[2]

Write down the differential equation as a system of first order differential equations.

[2]

Use Euler’s method, with a step length of 0.2, to find the displacement and velocity of the object when $t = 0.6$ .

[4]

By repeated application of Euler’s method, find an approximation for the terminal velocity, to five significant figures.

[1]

Use the differential equation to find the terminal velocity for the object.

[2]

Use your answers to parts (d), (e) and (f) to comment on the accuracy of the Euler approximation to this model.

[2]

Markscheme

$\frac{{{\text{d}}v}}{{{\text{d}}t}} = 9.81 - 0.9v$ M1

$\int {\frac{1}{{9.81 - 0.9v}}} {\text{d}}v = \int 1 \,{\text{d}}t$ M1

$- \frac{1}{{0.9}}{\text{ln}}\left( {9.81 - 0.9v} \right) = t + c$ A1

$9.81 - 0.9v = A{{\text{e}}^{ - 0.9t}}$ A1

$v = \frac{{9.81 - A{{\text{e}}^{ - 0.9t}}}}{{0.9}}$ A1

when $t = 0$ , $v = 0$ hence $A = 9.81$ A1

$v = \frac{{9.81\left( {1 - {{\text{e}}^{ - 0.9t}}} \right)}}{{0.9}}$

$v = 10.9\left( {1 - {{\text{e}}^{ - 0.9t}}} \right)$ A1

[7 marks]

either let $t$ tend to infinity, or $\frac{{{\text{d}}v}}{{{\text{d}}t}} = 0$ (M1)

$v = 10.9$ A1

[2 marks]

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = y$ M1

$\frac{{{\text{dy}}}}{{{\text{d}}t}} = 9.81 - 0.9{y^2}$ A1

[2 marks]

${x_{n + 1}} = {x_n} + 0.2{y_n}$ , ${y_{n + 1}} = {y_n} + 0.2\left( {9.81 - 0.9{{\left( {{y_n}} \right)}^2}} \right)$ (M1)(A1)

$x = 1.04$ , $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 3.31$ (M1)A1

[4 marks]

3.3015 A1

[1 mark]

$0 = 9.81 - 0.9{\left( v \right)^2}$ M1

$\Rightarrow v = \sqrt {\frac{{9.81}}{{0.9}}} = 3.301511 \ldots \,\,\left( { = 3.30} \right)$ A1

[2 marks]

the model found the terminal velocity very accurately, so good approximation R1

intermediate values had object exceeding terminal velocity so not good approximation R1

[2 marks]

Examiners report

[N/A]

An earth satellite moves in a path that can be described by the curve $72.5{x^2} + 71.5{y^2} = 1$ where $x = x(t)$ and $y = y(t)$ are in thousands of kilometres and $t$ is time in seconds.

Given that $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 7.75 \times {10^{ - 5}}$ when $x = 3.2 \times {10^{ - 3}}$ , find the possible values of $\frac{{{\text{d}}y}}{{{\text{d}}t}}$ .

Give your answers in standard form.

Markscheme

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

METHOD 1

substituting for $x$ and attempting to solve for $y$ (or vice versa) (M1)

$y = ( \pm )0.11821 \ldots$ (A1)

EITHER

$145x + 143y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{145x}}{{143y}}} \right)$ M1A1

$145x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 143y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$ M1A1

THEN

attempting to find $\frac{{{\text{d}}x}}{{{\text{d}}t}}{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}t}} = - \frac{{145(3.2 \times {{10}^{ - 3}})}}{{143\left( {( \pm )0.11821 \ldots } \right)}} \times (7.75 \times {{10}^{ - 5}})} \right)$ (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ - 6}}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

METHOD 2

$y = ( \pm )\sqrt {\frac{{1 - 72.5{x^2}}}{{71.5}}}$ M1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = ( \pm )0.0274 \ldots$ (M1)(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = ( \pm )0.0274 \ldots \times 7.75 \times {10^{ - 5}}$ (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ - 6}}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

[6 marks]

Examiners report

[N/A]

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

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

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

[4]

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

[4]

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

[4]

c.i.

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

[4]

c.ii.

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

[3]

Markscheme

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

METHOD 1

$f'\left( x \right) = \frac{{\frac{{2\left( {x - 3} \right)}}{x} - \left( {2\,{\text{ln}}\,x + 1} \right)}}{{{{\left( {x - 3} \right)}^2}}}\left( { = \frac{{2\left( {x - 3} \right) - x\left( {2\,{\text{ln}}\,x + 1} \right)}}{{x{{\left( {x - 3} \right)}^2}}}} \right)$ (M1)A1A1A1

Note: Award M1 for attempt at quotient rule, A1A1 for numerator and A1 for denominator.

METHOD 2

$f\left( x \right) = \left( {2\,{\text{ln}}\,x + 1} \right){\left( {x - 3} \right)^{ - 1}}$ (A1)

$f'\left( x \right) = \left( {\frac{2}{x}} \right){\left( {x - 3} \right)^{ - 1}} - \left( {2\,{\text{ln}}\,x + 1} \right){\left( {x - 3} \right)^{ - 2}}\left( { = \frac{{2\left( {x - 3} \right) - x\left( {2\,{\text{ln}}\,x + 1} \right)}}{{x{{\left( {x - 3} \right)}^2}}}} \right)$ (M1)A1A1

Note: Award M1 for attempt at product rule, A1 for first term, A1 for second term.

[4 marks]

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

$x = 0.899$ A1

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

(0.899, −0.375)

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

[4 marks]

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

and is concave down for $x$ > 1 A1

$x$ -intercept at 0.607 A1

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

[4 marks]

c.i.

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

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

$y$ -intercept at 0.607 A1

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

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

[4 marks]

c.ii.

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

0 < $x$ < 0.372 A1

Note: Do not award FT marks.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

A particle moves along a horizontal line such that at time $t$ seconds, $t$ ≥ 0, its acceleration $a$ is given by $a$ = 2 $t$ − 1. When $t$ = 6 , its displacement $s$ from a fixed origin O is 18.25 m. When $t$ = 15, its displacement from O is 922.75 m. Find an expression for $s$ in terms of $t$ .

Markscheme

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

attempt to integrate $a$ to find $v$ M1

$v = \int {a\,{\text{d}}t = \int {\left( {2t - 1} \right)} } \,{\text{d}}t$

$= {t^2} - t + c$ A1

$s = \int {v\,{\text{d}}t = \int {\left( {{t^2} - t + c} \right)} } \,{\text{d}}t$

$= \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + ct + d$ A1

attempt at substitution of given values (M1)

at $t = 6{\text{,}}\,\,\,18.25 = 72 - 18 + 6c + d$

at $t = 15{\text{,}}\,\,\,922.75 = 1125 - 112.5 + 15c + d$

solve simultaneously: (M1)

$c = - 6{\text{,}}\,\,d = 0.25$ A1

$\Rightarrow s = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + - 6t + \frac{1}{4}$

[6 marks]

Examiners report

[N/A]

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

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

Markscheme

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

METHOD 1

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

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

$x = 0.667$

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

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

METHOD 1

Attempt to differentiate (M1)

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

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

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

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

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

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

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

[4 marks]

Examiners report

[N/A]