Edexcel A-Level Mathematics A2 Unit P4 Pure Mathematics 4 Questions

Practise P4 techniques across proof, parametric curves, functions, differentiation, integration and vectors in multi-step problems.

Covered syllabus points

P4.1 - Proof P4.2 - Algebra and functions P4.3 - Coordinate geometry in the (x, y) plane P4.4 - Binomial expansion P4.5 - Differentiation P4.6 - Integration P4.7 - Vectors

[Maximum number: 4]

Prove by contradiction that for all positive numbers k

k+\frac{9}{k} \geqslant 6

Assume that there exists a positive number k such that $k+\frac9k<6$ .
$k+\frac9k<6\Rightarrow k^2+9<6k\Rightarrow k^2-6k+9<0$ $(k-3)^2<0$
But numbers squared are $\ge 0$ , hence $k+\frac9k\ge 6$ .
A1*

[Maximum number: 16]

Figure 2

Figure 2 shows a sketch of the curve defined by the parametric equations

x=t^{2}+2 t \quad y=\frac{2}{t(3-t)} \quad a \leqslant t \leqslant b

where a and b are constants.
The ends of the curve lie on the line with equation y=1

(a)

Find the value of a and the value of b

The region R, shown shaded in Figure 2, is bounded by the curve and the line with equation y=1

[ 2 ]

$y=\frac{2}{t(3-t)}=1 \Rightarrow 2=3 t-t^{2} \Rightarrow t=\ldots$ $\left(t^{2}-3 t+2=(t-1)(t-2)=0 \Rightarrow\right) a=1, b=2$
(2)

(b)

Show that the area of region R is given by

M-k \int_{a}^{b} \frac{t+1}{t(3-t)} \mathrm{d} t

where M and k are constants to be found.

[ 5 ]

Area under curve $=\int_{t=11}^{t=" 2 "} y \mathrm{~d} x=\int_{1}^{2} y \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$ $t=1 \Rightarrow x=3, \quad t=2 \Rightarrow x=8$
So area of $R=1 \times(8-3)-\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$ $=5-4 \int_{1}^{2} \frac{t+1}{t(3-t)} \mathrm{d} t$
(5)

(c)

Write $\frac{t+1}{t(3-t)}$ in partial fractions.

[ 3 ]

(ii)

$\frac{t+1}{t(3-t)}=\frac{A}{t}+\frac{B}{3-t}$ $\Rightarrow t+1=A(3-t)+B t$

E.g. $t=0 \Rightarrow 1=3 A \Rightarrow A=\frac{1}{3} ; t=3 \Rightarrow 4=3 B \Rightarrow B=\frac{4}{3}$ $\frac{t+1}{t(3-t)}=\frac{1}{3 t}+\frac{4}{3(3-t)}$

(d)

Use algebraic integration to find the exact area of R, giving your answer in simplest form.

[ 6 ]

$\int\frac{t+1}{t(3-t)}\,dt =\int\left(\frac1{3t}+\frac4{3(3-t)}\right)\,dt =\frac13\ln t-\frac43\ln(3-t)$ $\begin{aligned} \text{Area} &=5-4\left[\frac13\ln t-\frac43\ln(3-t)\right]_1^2\\ &=5-4\left(\frac13\ln2+\frac43\ln2\right) \end{aligned}$ $=5-\frac{20}{3}\ln2$

[Maximum number: 18]

Figure 3

Figure 3 shows a sketch of the curve C with parametric equations

x=6 t-3 \sin 2 t \quad y=2 \cos t \quad 0 \leqslant t \leqslant \frac{\pi}{2}

The curve meets the y-axis at 2 and the x-axis at k, where k is a constant.

(a)

State the value of k.

[ 1 ]

$3\pi$

(b)

Use parametric differentiation to show that

\frac{\mathrm{d} y}{\mathrm{~d} x}=\lambda \operatorname{cosec} t

where $\lambda$ is a constant to be found.

The point P with parameter $t=\frac{\pi}{4}$ lies on C.
The tangent to C at the point P cuts the y-axis at the point N.

[ 4 ]

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 \sin t}{6-6 \cos 2 t}$

M1 A1

$=\frac{-2 \sin t}{6-6\left(1-2 \sin ^{2} t\right)}=\frac{-2 \sin t}{12 \sin ^{2} t}=-\frac{1}{6} \operatorname{cosec} t$

M1 A1*

(4)

(c)

Find the exact y coordinate of N, giving your answer in simplest form.

The region bounded by the curve, the x-axis and the y-axis is rotated through $2 \pi$ radians about the x-axis to form a solid of revolution.

[ 3 ]

$P=\left(\frac{3\pi}{2}-3,\sqrt2\right)$
For example,
$y-\sqrt2=-\frac{\sqrt2}{6}\left(0-\left(\frac{3\pi}{2}-3\right)\right)$
or
$\sqrt2=-\frac{\sqrt2}{6}\left(\frac{3\pi}{2}-3\right)+c$ $y=\frac{\pi\sqrt2}{4}+\frac{\sqrt2}{2}$

(d)

Show that the volume of this solid is given by

\int_{0}^{\alpha} \beta(1-\cos 4 t) \mathrm{d} t

where $\alpha$ and $\beta$ are constants to be found.

[ 4 ]

Attempts $y^{2} \frac{\mathrm{~d} x}{\mathrm{~d} t}=4 \cos ^{2} t(6-6 \cos 2 t)$ $=(2 \cos 2 t+2)(6-6 \cos 2 t)=6\left(2-2 \cos ^{2} 2 t\right)=6(2-(1+\cos 4 t))$
Volume $=\int_{0}^{\frac{\pi}{2}} \pi y^{2} \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{0}^{\frac{\pi}{2}} 6 \pi(1-\cos 4 t) \mathrm{d} t$

(e)

Hence, using algebraic integration, find the exact volume of this solid.

[ 6 ]

$\left[6 \pi t-\frac{6 \pi \sin 4 t}{4}\right]_{0}^{\frac{\pi}{2}},=3 \pi^{2}$

M1, A1

(6)

(14 marks)

[Maximum number: 8]

\mathrm{f}(x)=(8-3 x)^{\frac{4}{3}} \quad 0<x<\frac{8}{3}

(a)

Show that the binomial expansion of f(x) in ascending powers of x up to and including the term in $x^{3}$ is

A-8 x+\frac{x^{2}}{2}+B x^{3}+\ldots

where A and B are constants to be found.

[ 4 ]

$(8-3 x)^{\frac{4}{3}}=16\left(1-\frac{3}{8} x\right)^{\frac{4}{3}}$ but condone $16(1-k x)^{\frac{4}{3}}, k \neq 3$
"Correct" term 3 or term 4 in
$(1-k x)^{\frac{4}{3}}=1 \pm \frac{4}{3} k x+\frac{\frac{4}{3} \times \frac{1}{3}}{2}( \pm k x)^{2}+\frac{\frac{4}{3} \times \frac{1}{3} \times\left(-\frac{2}{3}\right)}{6}( \pm k x)^{3}+\ldots$
Two terms correct of $16-8 x+\frac{x^{2}}{2}+\frac{x^{3}}{24}+\ldots$ following M1, M1 and $k= \pm \frac{3}{8}$ $=16-8 x+\frac{x^{2}}{2}+\frac{x^{3}}{24}+\ldots$
(4)

(b)

Use proof by contradiction to prove that the curve with equation

y=8+8 x-\frac{15}{2} x^{2}

does not intersect the curve with equation

y=A-8 x+\frac{x^{2}}{2}+B x^{3} \quad 0<x<\frac{8}{3}

where A and B are the constants found in part (a).
(Solutions relying on calculator technology are not acceptable.)

[ 4 ]

Assume the curves meet:
$8x-\frac{15x^2}{2}+8=16-8x+\frac{x^2}{2}+\frac{x^3}{24}$ $0=8-16x+8x^2+\frac{x^3}{24} =8(x-1)^2+\frac{x^3}{24}$
For $0<x<\frac83$ ,
$8(x-1)^2\ge0,\qquad \frac{x^3}{24}>0$
so
$8(x-1)^2+\frac{x^3}{24}>0$
This contradicts the equation $0=8(x-1)^2+\frac{x^3}{24}$ , so the curves do not meet.
(4)

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 4

Question (a)

Question (b)