CAIE Further Math A2: Further Pure Mathematics 2

[Maximum number: 10]

Figure for Question 1 — CAIE A-Level Further Math A2

The diagram shows part of the curve $y=x \operatorname{sech}^{2} x$ and its maximum point M.

(a)

Show that, at M,

2 x \tanh x-1=0

and verify that this equation has a root between 0.7 and 0.8 .

[ 4 ]

$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x \operatorname{sech}^{2} x \tanh x+\operatorname{sech}^{2} x {\left[=1-\tanh ^{2} x-2 x \tanh x+2 x \tanh ^{3} x\right]} \end{aligned}$
Differentiating using product rule
and chain rule for M1.

$\operatorname{sech}^{2} x \neq 0 \Rightarrow 2 x \tanh x-1=0$
AG.

$\begin{aligned} 2(0.7) \tanh (0.7)-1=-0.154<0 2(0.8) \tanh (0.8)-1=0.062>0 \end{aligned}$
Shows sign change. Must write
down values correct to at least 1dp
for B 1 .

4

(b)

By considering a suitable set of rectangles, use the diagram to show that

\sum_{r=2}^{n} r \operatorname{sech}^{2} r<n \tanh n+\ln \operatorname{sech} n-\tanh 1-\ln \operatorname{sech} 1 .

[ 6 ]

$\sum_{r=2}^{n} r \operatorname{sech}^{2} r<\int_{1}^{n} x \operatorname{sech}^{2} x \mathrm{~d} x$
Compares sum with integral.
Consistent limits for M1.

$\int_{1}^{n} x \operatorname{sech}^{2} x \mathrm{~d} x=[x \tanh x]_{1}^{n}-\int_{1}^{n} \tanh x \mathrm{~d} x$
Integrates by parts.

$=[x \tanh x+\ln \operatorname{sech} x]_{1}^{n}$ $=n \tanh n+\ln \operatorname{sech} n-(\tanh 1+\ln \operatorname{sech} 1)$
AG. Must have gained all previous
marks.

6

[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: 9]

Find the solution of the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+3 y=\sin x

for which y=1 when x=0. Give your answer in the form y=f(x).

$\mathrm{e}^{\int 3 \mathrm{~d} x}=\mathrm{e}^{3 x}$
Finds integrating factor.

$\frac{\mathrm{d}}{\mathrm{d} x}\left(y \mathrm{e}^{3 x}\right)=\mathrm{e}^{3 x} \sin x$
Correct form on LHS and attempt
to integrate RHS.

EITHER
$\begin{aligned} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=-\mathrm{e}^{3 x} \cos x+3 \int \mathrm{e}^{3 x} \cos x \mathrm{~d} x \text { OR } \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=\frac{1}{3} \mathrm{e}^{3 x} \sin x-\frac{1}{3} \int \mathrm{e}^{3 x} \cos x \mathrm{~d} x \end{aligned}$
Integrates by parts once or uses
$\sin x=\frac{e^{\mathrm{i} x}-\mathrm{e}^{-\mathrm{i} x}}{2 \mathrm{i}}.$

EITHER
$\begin{aligned} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=-\mathrm{e}^{3 x} \cos x+3\left(\mathrm{e}^{3 x} \sin x-3 \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x\right) \text { OR } \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=\frac{1}{3} \mathrm{e}^{3 x} \sin x-\frac{1}{3}\left(\frac{1}{3} \mathrm{e}^{3 x} \cos x+\frac{1}{3} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x\right) \end{aligned}$
Integrates by parts again or substitutes $\frac{\mathrm{e}^{\mathrm{i} x}-\mathrm{e}^{-\mathrm{i} x}}{2 \mathrm{i}}=\sin x$ and
$\frac{\mathrm{e}^{\mathrm{i} x}+\mathrm{e}^{-\mathrm{i} x}}{2}=\cos x.$ $y \mathrm{e}^{3 x}=\frac{1}{10} \mathrm{e}^{3 x}(3 \sin x-\cos x)+C$
Must not see i.

$1=-\frac{1}{10}+C$
Finds C. Substitutes into their
expression (must be integrated).

$y=\frac{3}{10} \sin x-\frac{1}{10} \cos x+\frac{11}{10} e^{-3 x}$
Divides through by coefficient of y.

9

[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

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

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question 4

Question (a)

Question (b)

Question (c)

Question (d)