CAIE A-Level Further Math A2 2 Further Pure Mathematics 2 Questions

Practise Further Pure Mathematics 2 through hyperbolic functions, matrices, calculus, complex numbers and differential equations, using mark schemes to check exact methods.

Covered syllabus points

2.1 Hyperbolic functions 2.2 Matrices 2.3 Differentiation 2.4 Integration 2.5 Complex numbers 2.6 Differential equations

[Maximum number: 15]

(a)

Sketch the graph of $y=\operatorname{coth} x$ for x>0 and state the equations of the asymptotes.

[ 2 ]

Figure for Answer to Question (a) — CAIE A-Level Further Math A2

Correct shape and position, not too truncated.

$x=0, \quad y=1$
States equations of asymptotes.

2

(b)

Starting from the definitions of coth and cosech in terms of exponentials, prove that

\operatorname{coth}^{2} x-\operatorname{cosech}^{2} x=1 .

The curve C has equation $y=\ln \operatorname{coth}\left(\frac{1}{2} x\right)$ for x>0.

[ 3 ]

$\operatorname{coth} x=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{x}-\mathrm{e}^{-x}} \quad \operatorname{cosech} x=\frac{2}{\mathrm{e}^{x}-\mathrm{e}^{-x}}$ $\left(\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{x}-\mathrm{e}^{-x}}\right)^{2}-\frac{4}{\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}=\frac{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}-2}{\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}=1$
Writes over common denominator, AG.

3

(c)

Show that $\frac{\mathrm{d} y}{\mathrm{~d} x}=-\operatorname{cosech} x$ .

[ 3 ]

$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\operatorname{sech}^{2}\left(\frac{1}{2} x\right)}{2 \tanh \left(\frac{1}{2} x\right)}=-\frac{1}{2 \sinh \left(\frac{1}{2} x\right) \cosh \left(\frac{1}{2} x\right)} \text { Or } \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\operatorname{cosech}^{2}\left(\frac{1}{2} x\right)}{2 \operatorname{coth}\left(\frac{1}{2} x\right)}=-\frac{1}{2 \sinh \left(\frac{1}{2} x\right) \cosh \left(\frac{1}{2} x\right)} \end{aligned}$
Uses chain rule.

$=-\frac{1}{\sinh (x)}=-\operatorname{cosech} x$
AG

3

(d)

It is given that the arc length of C from x=a to x=2 a is $\ln 4$ , where a is a positive constant.

Show that $\cosh a=2$ and find, in logarithmic form, the exact value of a.

[ 7 ]

$\int_{a}^{2 a} \sqrt{1+\operatorname{cosech}^{2} x} \mathrm{~d} x$
Forms correct integral.

$\int_{a}^{2 a} \sqrt{\operatorname{coth}^{2} x} \mathrm{~d} x=\int_{a}^{2 a} \operatorname{coth} x \mathrm{~d} x$
Uses $\operatorname{coth}^{2} x-\operatorname{cosech}^{2} x=1$ .

$=[\ln \sinh x]_{a}^{2 a}=\ln \sinh 2 a-\ln \sinh a$
Integrates and substitutes limits.

$=\ln \frac{\sinh 2 a}{\sinh a}=\ln (2 \cosh a)$
Combines logarithms and uses double angle formula.

$\ln (2 \cosh a)=\ln 4 \Rightarrow \cosh a=2$
AG

$a=\ln \left(2+\sqrt{2^{2}-1}\right)=\ln (2+\sqrt{ } 3)$
Must reject $\ln (2-\sqrt{3})$ .

7

[Maximum number: 11]

The integral $I_{n}$ , where n is an integer, is defined by $I_{n}=\int_{0}^{\frac{4}{3}}\left(1+x^{2}\right)^{\frac{1}{2} n} \mathrm{~d} x$ .

(a)

Find the exact value of $I_{-1}$ giving your answer in the form $\ln a$ , where a is an integer to be determined.

[ 2 ]

$I_{-1}=\int_{0}^{\frac{4}{3}}\left(1+x^{2}\right)^{-\frac{1}{2}} \mathrm{~d} x=\left[\sinh ^{-1} x\right]_{0}^{\frac{4}{3}}=\ln 3$
Recognises integral.

2

(b)

By considering $\frac{\mathrm{d}}{\mathrm{d} x}\left(x\left(1+x^{2}\right)^{\frac{1}{2} n}\right)$ , or otherwise, show that

(n+1) I_{n}=n I_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^{n} .

[ 5 ]

$\frac{\mathrm{d}}{\mathrm{d} x}\left(x\left(1+x^{2}\right)^{\frac{1}{2} n}\right)=n x^{2}\left(1+x^{2}\right)^{\frac{1}{2} n-1}+\left(1+x^{2}\right)^{\frac{1}{2} n}$
Uses the product rule to differentiate.

$n\left(1+x^{2}-1\right)\left(1+x^{2}\right)^{\frac{1}{2} n-1}+\left(1+x^{2}\right)^{\frac{1}{2} n}$

M1*

Uses $x^{2}=1+x^{2}-1$ .

$\left[x\left(1+x^{2}\right)^{\frac{1}{2} n}\right]_{0}^{\frac{4}{3}}=n I_{n}-n I_{n-2}+I_{n}$
Integrates both sides using the limits given. Requires
previous method mark.

$\frac{4}{3}\left(\frac{5}{3}\right)^{n}=(n+1) I_{n}-n I_{n-2} \Rightarrow(n+1) I_{n}=n I_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^{n}$
Substitutes limits and rearranges. AG.

5

(c)

A curve has equation $y=x^{2}$ , for $0 \leqslant x \leqslant \frac{2}{3}$ . The arc length of the curve is denoted by s. Use the substitution u=2 x to show that $s=\frac{1}{2} I_{1}$ and find the exact value of s.

[ 4 ]

$s=\int_{0}^{\frac{2}{3}} \sqrt{1+4 x^{2}} \mathrm{~d} x$
Forms correct integral with correct limits.

$\frac{1}{2} \int_{0}^{\frac{4}{3}} \sqrt{1+u^{2}} \mathrm{~d} u=\frac{1}{2} I_{1}$
AG.

$2 I_{1}=I_{-1}+\frac{4}{3}\left(\frac{5}{3}\right)$
Applies reduction formula with n=1.

$s=\frac{1}{4} \ln 3+\frac{5}{9}$
4

[Maximum number: 11]

(a)

Show that an appropriate integrating factor for

\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+y \sqrt{x^{2}+1}=x^{2}-x \sqrt{x^{2}+1}

is $x+\sqrt{x^{2}+1}$ .

[ 4 ]

$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{y}{\sqrt{ }\left(x^{2}+1\right)}=\frac{x}{x^{2}+1}\left(x-\sqrt{ }\left(x^{2}+1\right)\right)$ $\mathrm{e}^{\int \frac{1}{\sqrt{\left(x^{2}+1\right)}} \mathrm{d} x}=\mathrm{e}^{\sinh ^{-1} x}$ $=x+\sqrt{ }\left(x^{2}+1\right)$
7a 4

(b)

Hence find the solution of the differential equation

\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+y \sqrt{x^{2}+1}=x^{2}-x \sqrt{x^{2}+1}

for which $y=\ln 2$ when x=0. Give your answer in the form y=f(x).

[ 7 ]

$\frac{\mathrm{d}}{\mathrm{d} x}\left(y\left(x+\sqrt{ }\left(x^{2}+1\right)\right)\right)=-\frac{x}{x^{2}+1}$ $y\left(x+\sqrt{ }\left(x^{2}+1\right)\right)=-\int \frac{x}{x^{2}+1} \mathrm{~d} x=-\frac{1}{2} \ln \left(x^{2}+1\right)+C$ $\ln 2=C$ $y=\frac{\ln 2-\frac{1}{2} \ln \left(x^{2}+1\right)}{x+\sqrt{ }\left(x^{2}+1\right)}=\left(x-\sqrt{ }\left(x^{2}+1\right)\right) \ln \left(\frac{1}{2} \sqrt{ }\left(x^{2}+1\right)\right)$
7b 7

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)