A Level Further Mathematics 9231 Paper 22, Oct/Nov 2023

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9231/22 · Oct/Nov 2023

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Find the Maclaurin’s series for $\ln (x + 2) + \ln (x^{2} + 5)$ up to and including the term in $x^{2}$ .

[5]

It is given that $x = 1 + \frac{1}{t} a n d y = t e^{t} .$

Show that $\frac{d y}{d x} = - e^{t} (t^{3} + t^{2})$ .

[3]

Find $\frac{d^{2} y}{d x^{2}}$ in terms of $t$ .

[4]

Use de Moivre’s theorem to show that $\cos 5 θ = 16 \cos^{5} θ - 20 \cos^{3} θ + 5 \cos θ .$

[4]

Hence obtain the roots of the equation $32 x^{5} - 40 x^{3} + 10 x - \sqrt{2} = 0$ in the form $\cos (q π)$ , where $q$ is a rational number.

[4]

Find the solution of the differential equation $\frac{d y}{d x} + 3 y = \sin x$ for which $y = 1$ when $x = 0$ . Give your answer in the form $y = f (x)$ .

[9]

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

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]

By considering a suitable set of rectangles, use the diagram to show that $\sum_{r = 2}^{n} r {sech}^{2} r < n \tanh n + \ln sech n - \tanh 1 - \ln sech 1.$

[6]

The matrix $P$ is given by $P = (\begin{array}{ccccccc} 1 & - 1 & 10 & 2 & 10 & 0 & - 1 \end{array}) .$

State the eigenvalues of $P$ .

[1]

Use the characteristic equation of $P$ to find $P^{- 1}$ .

[4]

The $3 \times 3$ matrix $A$ has distinct non-zero eigenvalues $a, \frac{1}{2}, 2$ with corresponding eigenvectors $(\begin{array}{c} 100 \end{array}), (\begin{array}{c} - 120 \end{array}), (\begin{array}{c} 11 - 1 \end{array}),$ respectively. Find $A^{- 1}$ in terms of $a$ .

[5]

Starting from the definitions of $\cosh$ and $\sinh$ in terms of exponentials, prove that $2 \sinh^{2} A = \cosh 2 A - 1.$

[3]

A curve has equation $y = x^{2}$ , for $0 ⩽ x ⩽ \frac{2}{3}$ . The area of the surface generated when the curve is rotated through $2 π$ radians about the $x$ -axis is denoted by $S$ . Use the substitution $x = \frac{1}{2} \sinh u$ to show that $S = \frac{1}{32} π (\frac{820}{81} - \ln 3)$ .

[9]

It is given that $v = y^{4}$ and $y^{3} \frac{d^{2} y}{d x^{2}} + 3 y^{2} {(\frac{d y}{d x})}^{2} + y^{3} \frac{d y}{d x} + y^{4} = e^{- 2 x} .$

Show that $\frac{d^{2} v}{d x^{2}} + \frac{d v}{d x} + 4 v = 4 e^{- 2 x} .$

[4]

Find $y$ in terms of $x$ , given that, when $x = 0$ , $y = 1$ and $\frac{d y}{d x} = - \frac{3}{8}$ .

[10]

Mark scheme

9231/22 · Oct/Nov 2023

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1 · 5 marks

M1
Changes $\ln (x + 2)$ or $\ln (x^{2} + 5)$ so that the formula given in the list of formulae (MF19) can be applied.
A1
$\ln (x + 2) = \ln 2 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$
A1
$\ln (x^{2} + 5) = \ln (5 (1 + \frac{1}{5} x^{2})) = \ln 5 + \frac{1}{5} x^{2} + \dots$
M1
Sums power series
A1
$\ln (x + 2) + \ln (x^{2} + 5) = \ln 10 + \frac{1}{2} x + \frac{3}{40} x^{2}$

2a · 3 marks

B1
$\frac{d y}{d t} = e^{t} (t + 1)$ , $\frac{d x}{d t} = - t^{- 2}$
M1
Uses chain rule, AG.
A1
$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = - t^{2} e^{t} (t + 1) = - e^{t} (t^{3} + t^{2})$

2b · 4 marks

M1
Applies product rule.
A1
$\frac{d}{d t} (\frac{d y}{d x}) = - e^{t} (3 t^{2} + 2 t) - e^{t} (t^{3} + t^{2}) = - e^{t} (t^{3} + 4 t^{2} + 2 t)$
M1
Uses chain rule, multiplies by their $\frac{d t}{d x}$ for M1.
A1
$\frac{d^{2} y}{d x^{2}} = t^{3} e^{t} (t^{2} + 4 t + 2)$

3a · 4 marks

M1
Expands and takes real part. Accept RHS to LHS using $2 \cos θ = z + \frac{1}{z}$ .
A1
$\cos 5 θ = Re (\cos θ + i \sin θ)^{5} = \cos^{5} θ - 10 \sin^{2} θ \cos^{3} θ + 5 \sin^{4} θ \cos θ$
M1
Applies $\sin^{2} θ = 1 - \cos^{2} θ$ .
A1
AG

3b · 4 marks

M1
Applies identify given in (b).
M1
Solves $\cos 5 θ = \frac{1}{2} \sqrt{2}$
A1
Gives one correct solution. Accept $q = \frac{1}{20}$ .
A1
Gives other solutions. OE. A0 for repeated roots.

4 · 9 marks

M1
Finds integrating factor.
A1
$e^{\int 3 d x} = e^{3 x}$
M1
Correct form on LHS and attempt to integrate RHS.
M1
Integrates by parts once or uses $\sin x = \frac{e^{i x} - e^{- i x}}{2 i}$ .
A1
$\int e^{3 x} \sin x d x = - e^{3 x} \cos x + 3 \int e^{3 x} \cos x d x$ OR $\int e^{3 x} \sin x d x = \frac{1}{3} e^{3 x} \sin x - \frac{1}{3} \int e^{3 x} \cos x d x$
M1
Integrates by parts again or substitutes $\sin x = \frac{e^{i x} - e^{- i x}}{2 i}$ and $\cos x = \frac{e^{i x} + e^{- i x}}{2}$ .
A1
Must not see i. $y e^{3 x} = \frac{1}{10} e^{3 x} (3 \sin x - \cos x) + C$
M1
Finds C. Substitutes into their expression (must be integrated).
A1
Divides through by coefficient of y. $y = \frac{3}{10} \sin x - \frac{1}{10} \cos x + \frac{11}{10} e^{- 3 x}$

5a · 4 marks

M1
Differentiating using product rule and chain rule for M1.
A1
$\frac{d y}{d x} = - 2 x {sech}^{2} x \tanh x + {sech}^{2} x$
A1
AG. ${sech}^{2} x \neq 0 ⟹ 2 x \tanh x - 1 = 0$
B1
Shows sign change. Must write down values correct to at least 1dp for B1.

5b · 6 marks

M1
Compares sum with integral. Consistent limits for M1.
A1
$\sum_{r = 2}^{n} r {sech}^{2} r < \int_{1}^{n} x {sech}^{2} x d x$
M1
Integrates by parts.
A1
$[x \tanh x]_{1}^{n} - \int_{1}^{n} \tanh x d x$
A1
$= [x \tanh x + \ln sech x]_{1}^{n}$
A1
AG. Must have gained all previous marks. $= n \tanh n + \ln sech n - (\tanh 1 + \ln sech 1)$

6a · 1 mark

B1
1, 2, $- 1$

6b · 4 marks

B1
States that P satisfies its characteristic equation.
M1
Multiplies through by $P^{- 1}$ .
M1
Evaluates $P^{2}$ and forms expression for $P^{- 1}$ .
A1
$P^{- 1} = (\begin{array}{ccc} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & - 1 \end{array})$

6c · 5 marks

B1
$D = (\begin{array}{ccc} \frac{1}{a} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2} \end{array})$
M1
Applies $A^{- 1} = P D P^{- 1}$ .
M1
Multiplies two adjacent matrices.
A1
$= (\begin{array}{ccc} \frac{1}{a} & - 2 & \frac{1}{2} \\ 0 & 4 & \frac{1}{2} \\ 0 & 0 & - \frac{1}{2} \end{array}) (\begin{array}{ccc} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & - 1 \end{array})$
A1
$(\begin{array}{ccc} \frac{1}{a} & \frac{1 - 2 a}{2 a} & \frac{3 - 3 a}{2 a} \\ 0 & 2 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2} \end{array})$

7a · 3 marks

B1
$\cosh 2 A = \frac{1}{2} (e^{2 A} + e^{- 2 A}) \sinh A = \frac{1}{2} (e^{A} - e^{- A})$
M1
Expands, AG. A0 for mixing variables e.g. $\sinh A = \frac{1}{2} (e^{x} - e^{- x})$ .
A1
$2 \sinh^{2} A = \frac{1}{2} (e^{A} - e^{- A})^{2} = \frac{1}{2} (e^{2 A} - 2 + e^{- 2 A}) = \cosh 2 A - 1$

7b · 9 marks

M1
Correct formula with correct limits. Correct limits for M1. (Limits may be recovered.)
A1
$S = \int_{0}^{\frac{2}{3}} 2 π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x = 2 π \int_{0}^{\frac{2}{3}} x^{2} \sqrt{1 + 4 x^{2}} d x$
M1
Applies given substitution to their expression with correct limits. (Limits may be recovered.)
A1
Must be simplified. $= \frac{1}{4} π \int_{0}^{\sinh^{- 1} \frac{4}{3}} \sinh^{2} u \cosh^{2} u d u$
M1
Applies $\sinh 2 u = 2 \sinh u \cosh u$ . May use double angle formulae for cosh and sinh instead.
M1
Applies $\sinh^{2} A = \frac{1}{2} (\cosh 2 A - 1)$ . S must have the form $a \int \sinh^{2} u \cosh^{2} u d u$ .
A1
$= \frac{1}{32} π {[\frac{1}{4} \sinh 4 u - u]}_{0}^{\sinh^{- 1} \frac{4}{3}}$
B1
$\sinh^{- 1} \frac{4}{3} = \ln 3$
A1
AG. $S = \frac{1}{32} π (\frac{1}{8} (e^{\ln 81} - e^{- \ln 81}) - \ln 3) = \frac{1}{32} π (\frac{1}{8} (81 - \frac{1}{81}) - \ln 3) = \frac{1}{32} π (\frac{820}{81} - \ln 3)$

8a · 4 marks

B1
$\frac{d v}{d x} = 4 y^{3} \frac{d y}{d x}$
B1
$\frac{d^{2} v}{d x^{2}} = 4 y^{3} \frac{d^{2} y}{d x^{2}} + 12 y^{2} {(\frac{d y}{d x})}^{2}$
M1
Uses substitution to find v-x equation, AG.
A1
AG. $= 4 (y^{3} \frac{d^{2} y}{d x^{2}} + 3 y^{2} {(\frac{d y}{d x})}^{2} + y^{3} \frac{d y}{d x} + y^{4}) = 4 e^{- 2 x}$

8b · 10 marks

M1
Auxiliary equation.
A1
Complimentary function. Allow ‘v =’ missing.
B1
Particular integral and its derivatives.
M1
Substitutes and equates coefficients.
A1
$k = \frac{2}{3}$
A1
Substitutes for y and find v in terms of x. Must not see i.
B1
$v^{'} = 4 y^{3} \frac{d y}{d x} = e^{- \frac{1}{2} x} (- \frac{\sqrt{15}}{2} A \sin \frac{\sqrt{15}}{2} x + \frac{\sqrt{15}}{2} B \cos \frac{\sqrt{15}}{2} x) - \frac{1}{2} e^{- \frac{1}{2} x} (A \cos \frac{\sqrt{15}}{2} x + B \sin \frac{\sqrt{15}}{2} x) - \frac{4}{3} e^{- 2 x}$
M1
Substitutes initial conditions and forms simultaneous equations. Must have used the product rule when differentiating their v for M1.
A1
$A = \frac{1}{3}, B = 0$
A1
$y = {(\frac{1}{3} e^{- \frac{1}{2} x} \cos \frac{\sqrt{15}}{2} x + \frac{2}{3} e^{- 2 x})}^{\frac{1}{4}}$

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