Pearson Edexcel IAL Mathematics Unit P3: Pure Mathematics 3 Question Bank

Practise P3 skills across functions, trigonometry, logarithms, differentiation, integration and numerical methods using algebraic and graphical reasoning.

Covered syllabus points

P3.1 - Algebra and functions P3.2 - Trigonometry P3.3 - Exponentials and logarithms P3.4 - Differentiation P3.5 - Integration P3.6 - Numerical methods

[Maximum number: 9]

The function f is defined by

f(x)=\frac{5x}{x^2+7x+12}+\frac{5x}{x+4},\qquad x>0

(a)

Show that $f(x)=\frac{5x}{x+3}$ .

[ 3 ]

\begin{aligned}\frac{5x}{x^2+7x+12}+\frac{5x}{x+4}&=\frac{5x+5x(x+3)}{(x+3)(x+4)}\\&=\frac{5x^2+20x}{(x+3)(x+4)}\\&=\frac{5x(x+4)}{(x+3)(x+4)}=\frac{5x}{x+3}.\end{aligned}

M1: common denominator.
A1: correct unsimplified fraction.
A1*: reaches the given result.

(b)

Find $f^{-1}$ .

[ 3 ]

\begin{aligned}y&=\frac{5x}{x+3}\\xy+3y&=5x\\5x-xy&=3y\\x&=\frac{3y}{5-y}.\end{aligned}

f^{-1}(x)=\frac{3x}{5-x}

Domain: 0<x<5.

M1, A1, A1

(c)

Find, in simplest form, f'(x).

[ 3 ]

f(x)=\frac{5x}{x+3}

f'(x)=\frac{5(x+3)-5x}{(x+3)^2}=\frac{15}{(x+3)^2}

M1: quotient/product rule.
A1: $\frac{15}{(x+3)^2}$ .

[Maximum number: 7]

A curve C has equation y=f(x), where

f(x)=\arcsin\left(\frac{x}{2}\right),\qquad -2\leqslant x\leqslant2,\qquad -\frac{\pi}{2}\leqslant y\leqslant\frac{\pi}{2}

(a)

Sketch C.

[ 1 ]

(b)

Given $x=2\sin y$ , show that

\frac{dy}{dx}=\frac{1}{\sqrt{A-x^2}}

where A is a constant to be found.

[ 3 ]

x=2\sin y\Rightarrow \frac{dx}{dy}=2\cos y

\frac{dy}{dx}=\frac{1}{2\cos y}

Since $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-\frac{x^2}{4}}$ ,

\frac{dy}{dx}=\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{\sqrt{4-x^2}}

Thus A=4.

M1 A1 A1cso

(c)

The point P lies on C and has y coordinate $\frac{\pi}{4}$ .

Find the equation of the tangent to C at P. Write your answer in the form y=mx+c, where m and c are constants to be found.

[ 3 ]

At P, $y=\frac{\pi}{4}$ , so

x=2\sin\frac{\pi}{4}=\sqrt2

\left.\frac{dy}{dx}\right|_{x=\sqrt2}=\frac{1}{\sqrt{4-2}}=\frac{\sqrt2}{2}

y-\frac{\pi}{4}=\frac{\sqrt2}{2}(x-\sqrt2)

y=\frac{\sqrt2}{2}x-1+\frac{\pi}{4}

M1 dM1 A1

[Maximum number: 13]

A scientist is studying a population of fish in a lake. The number of fish, N, in the population, t years after the start of the study, is modelled by the equation

N=\frac{600 \mathrm{e}^{0.3 t}}{2+\mathrm{e}^{0.3 t}} \quad t \geqslant 0

Use the equation of the model to answer parts (a), (b), (c), (d) and (e).

(a)

Find the number of fish in the lake at the start of the study.

[ 1 ]

(b)

Find the upper limit to the number of fish in the lake.

[ 1 ]

(c)

Find the time, after the start of the study, when there are predicted to be 500 fish in the lake. Give your answer in years and months to the nearest month.

[ 4 ]

$500=\frac{600e^{0.3t}}{2+e^{0.3t}}\Rightarrow 100e^{0.3t}=1000\Rightarrow e^{0.3t}=10$
M1 A1
$t=\frac{\ln 10}{0.3}=7\text{ years }8\text{ months}$
dM1 A1

(d)

Show that

$\frac{dN}{dt}=\frac{Ae^{0.3t}}{(2+e^{0.3t})^2}$

where A is a constant to be found.

[ 3 ]

$N=\frac{600e^{0.3t}}{2+e^{0.3t}}$ $\frac{dN}{dt}=\frac{(2+e^{0.3t})(180e^{0.3t})-600e^{0.3t}(0.3e^{0.3t})}{(2+e^{0.3t})^2}$
M1 A1
$\frac{dN}{dt}=\frac{360e^{0.3t}}{(2+e^{0.3t})^2}$
A1

(e)

Given that when t=T, $\frac{dN}{dt}=8$ ,
find the value of T to one decimal place.

(Solutions relying entirely on calculator technology are not acceptable.)

[ 4 ]

$8=\frac{360e^{0.3t}}{(2+e^{0.3t})^2}$ $8(e^{0.3t})^2-328(e^{0.3t})+32=0$ $e^{0.3t}=\frac{41+\sqrt{1665}}{2}$ $t=\frac{\ln\left(\frac{41+\sqrt{1665}}{2}\right)}{0.3}$ $T\approx12.4$

M1 dM1 ddM1 A1

Marking guidance:
M1 for setting dN/dt=8 and obtaining a quadratic in $e^{0.3t}$ .
dM1 for a correct attempt to solve the quadratic in $e^{0.3t}$ .
ddM1 for correct log work for $e^{0.3t}=k$ , k>0.
A1 for awrt 12.4.

[Maximum number: 7]

A curve C has equation

y=x^{\sin x} \quad x>0 \quad y>0

(a)

Find, by firstly taking natural logarithms, an expression for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ in terms of x and y.

[ 5 ]

$y=x^{\sin x}\Rightarrow \ln y=\sin x\ln x$

B1

$\frac1y\frac{dy}{dx}=\frac{\sin x}{x}+\ln x\cos x$

M1 M1 A1

$\frac{dy}{dx}=y\left(\frac{\sin x}{x}+\ln x\cos x\right)$

A1

(b)

Hence show that the x coordinates of the stationary points of C are solutions of the equation

\tan x+x \ln x=0

[ 2 ]

$\frac{dy}{dx}=0\Rightarrow y\left(\frac{\sin x}{x}+\ln x\cos x\right)=0$ $\frac{\sin x}{x}+\ln x\cos x=0$

M1

$\frac{\sin x}{\cos x}+x\ln x=0 \Rightarrow \tan x+x\ln x=0$

A1*

All question bank results loaded

Covered syllabus points

Unit P3: Pure Mathematics 3 question 1

Question (a)

Question (b)

Question (c)

Unit P3: Pure Mathematics 3 question 2

Question (a)

Question (b)

Question (c)

Unit P3: Pure Mathematics 3 question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Unit P3: Pure Mathematics 3 question 4

Question (a)

Question (b)