A Level Further Mathematics 9231 Paper 21, Oct/Nov 2024

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9231/21 · Oct/Nov 2024

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Find the set of values of $k$ for which the system of equations $\begin{aligned} x + 5 y + 6 z & = 1, \\ k x + 2 y + 2 z & = 2, \\ - 3 x + 4 y + 8 z & = 3, \end{aligned}$ has a unique solution and interpret this situation geometrically.

[4]

It is given that $x = 1 + \frac{1}{t} and y = \cos^{- 1} t for 0 < t < 1.$

Show that $\frac{d y}{d x} = \frac{t^{2}}{\sqrt{1 - t^{2}}}$ .

[2]

Show that $\frac{d^{2} y}{d x^{2}} = - t^{a} (1 - t^{2})^{b} (2 - t^{2})$ , where $a$ and $b$ are constants to be determined.

[4]

A curve has equation $y = e^{x}$ for $\ln \frac{4}{3} ⩽ x ⩽ \ln \frac{12}{5}$ . The area of the surface generated when the curve is rotated through $2 π$ radians about the $x$ -axis is denoted by $A$ .

Use the substitution $u = e^{x}$ to show that $A = 2 π \int_{\frac{4}{3}}^{\frac{12}{5}} \sqrt{1 + u^{2}} d u .$

[2]

Use the substitution $u = \sinh v$ to show that $A = π (\frac{904}{225} + \ln \frac{5}{3}) .$

[6]

The matrix $A$ is given by $A = (\begin{array}{ccc} - 11 & 1 & 8 \\ 0 & - 2 & 0 \\ - 16 & 1 & 13 \end{array}) .$

Show that $(\begin{array}{c} 1 \\ 1 \\ 1 \end{array})$ is an eigenvector of $A$ and state the corresponding eigenvalue.

[2]

Show that the characteristic equation of $A$ is $λ^{3} - 19 λ - 30 = 0$ and hence find the other eigenvalues of $A$ .

[3]

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

[4]

Find the particular solution of the differential equation $6 \frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + x = t^{2} + t + 1,$ given that, when $t = 0$ , $x = 12$ and $\frac{d x}{d t} = - 6$ .

[10]

The diagram shows the curve with equation $y = {(\frac{1}{2})}^{x}$ for $0 ⩽ x ⩽ 1$ , together with a set of $N$ rectangles each of width $\frac{1}{N}$ .

By considering the sum of the areas of these rectangles, show that $\int_{0}^{1} {(\frac{1}{2})}^{x} d x > L_{N}$ , where $L_{N} = \frac{1}{2 N (2^{\frac{1}{N}} - 1)} .$

[4]

Use a similar method to find, in terms of $N$ , an upper bound $U_{N}$ for $\int_{0}^{1} {(\frac{1}{2})}^{x} d x$ .

[4]

Find the least value of $N$ such that $U_{N} - L_{N} ⩽ 10^{- 3}$ .

[2]

Given that $\int_{0}^{1} {(\frac{1}{2})}^{x} d x = \frac{1}{2 \ln 2}$ , use the value of $N$ found in part (c) to find upper and lower bounds for $\ln 2$ .

[4]

Show that an appropriate integrating factor for $\sqrt{x^{2} + 16} \frac{d y}{d x} + y = x \sqrt{x^{2} + 16}$ is $\frac{1}{4} x + \frac{1}{4} \sqrt{x^{2} + 16}$ .

[4]

Hence find the solution of the differential equation $\sqrt{x^{2} + 16} \frac{d y}{d x} + y = x \sqrt{x^{2} + 16}$ for which $y = 6$ when $x = 3$ .

[6]

By considering the binomial expansion of ${(z + \frac{1}{z})}^{7}$ , where $z = \cos θ + i \sin θ$ , use de Moivre's theorem to show that $\cos^{7} θ = a \cos 7 θ + b \cos 5 θ + c \cos 3 θ + d \cos θ,$ where $a, b, c$ and $d$ are constants to be determined.

[5]

Let $I_{n} = \int_{0}^{\frac{1}{4} π} \cos^{n} θ d θ$ . Show that $n I_{n} = 2^{- \frac{1}{2} n} + (n - 1) I_{n - 2} .$

[4]

Using the results given in parts (a) and (b), find the exact value of $I_{9}$ .

[5]

Mark scheme

9231/21 · Oct/Nov 2024

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1 · 4 marks

M1
$∣ \begin{array}{ccc} 1 & 5 & 6 \\ k & 2 & 2 \\ - 3 & 4 & 8 \end{array} ∣ = ∣ \begin{array}{cc} 2 & 2 \\ 4 & 8 \end{array} ∣ - 5 ∣ \begin{array}{cc} k & 2 \\ - 3 & 8 \end{array} ∣ + 6 ∣ \begin{array}{cc} k & 2 \\ - 3 & 4 \end{array} ∣ = 8 - 5 (8 k + 6) + 6 (4 k + 6)$
A1
$∣ \begin{array}{ccc} 1 & 5 & 6 \\ k & 2 & 2 \\ - 3 & 4 & 8 \end{array} ∣ = ∣ \begin{array}{cc} 2 & 2 \\ 4 & 8 \end{array} ∣ - 5 ∣ \begin{array}{cc} k & 2 \\ - 3 & 8 \end{array} ∣ + 6 ∣ \begin{array}{cc} k & 2 \\ - 3 & 4 \end{array} ∣ = 8 - 5 (8 k + 6) + 6 (4 k + 6)$
A1
$k \neq \frac{7}{8}$
B1
Three planes intersect at a unique point.

2a · 2 marks

M1
$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{1}{\sqrt{1 - t^{2}}} \times (- t^{2}) = - \frac{t^{2}}{\sqrt{1 - t^{2}}}$
A1
$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{1}{\sqrt{1 - t^{2}}} \times (- t^{2}) = - \frac{t^{2}}{\sqrt{1 - t^{2}}}$

2b · 4 marks

M1
$\frac{d}{d t} (\frac{d y}{d x}) = \frac{2 t \sqrt{1 - t^{2}} + t^{3} (1 - t^{2})^{- \frac{1}{2}}}{1 - t^{2}}$
A1
$\frac{d}{d t} (\frac{d y}{d x}) = \frac{2 t \sqrt{1 - t^{2}} + t^{3} (1 - t^{2})^{- \frac{1}{2}}}{1 - t^{2}}$
M1
$\frac{d^{2} y}{d x^{2}} = \frac{2 t \sqrt{1 - t^{2}} + t^{3} (1 - t^{2})^{- \frac{1}{2}}}{1 - t^{2}} (- t^{2}) = - 2 t^{3} (1 - t^{2})^{- \frac{1}{2}} - t^{5} (1 - t^{2})^{- \frac{3}{2}} = - t^{3} (1 - t^{2})^{- \frac{3}{2}} (2 (1 - t^{2}) + t^{2}) = - t^{3} (1 - t^{2})^{- \frac{3}{2}} (2 - t^{2})$
A1
$\frac{d^{2} y}{d x^{2}} = \frac{2 t \sqrt{1 - t^{2}} + t^{3} (1 - t^{2})^{- \frac{1}{2}}}{1 - t^{2}} (- t^{2}) = - 2 t^{3} (1 - t^{2})^{- \frac{1}{2}} - t^{5} (1 - t^{2})^{- \frac{3}{2}} = - t^{3} (1 - t^{2})^{- \frac{3}{2}} (2 (1 - t^{2}) + t^{2}) = - t^{3} (1 - t^{2})^{- \frac{3}{2}} (2 - t^{2})$

3a · 2 marks

M1
$A = 2 π \int_{\ln \frac{4}{3}}^{\ln \frac{12}{5}} e^{x} \sqrt{1 + e^{2 x}} d x = 2 π \int_{\frac{4}{3}}^{\frac{12}{5}} u \sqrt{1 + u^{2}} (\frac{1}{u}) d u$
A1
$A = 2 π \int_{\ln \frac{4}{3}}^{\ln \frac{12}{5}} e^{x} \sqrt{1 + e^{2 x}} d x = 2 π \int_{\frac{4}{3}}^{\frac{12}{5}} u \sqrt{1 + u^{2}} (\frac{1}{u}) d u$

3b · 6 marks

B1
$u = \sinh v \Rightarrow d u = \cosh v d v$
M1
$A = 2 π \int_{\ln 3}^{\ln 5} \sqrt{1 + \sinh^{2} v} \cosh v d v A = 2 π \int_{\ln 3}^{\ln 5} \cosh^{2} v d v$
A1
$A = 2 π \int_{\ln 3}^{\ln 5} \sqrt{1 + \sinh^{2} v} \cosh v d v A = 2 π \int_{\ln 3}^{\ln 5} \cosh^{2} v d v$
M1
$π \int_{\ln 3}^{\ln 5} (\cosh 2 v + 1) d v$
A1
$= π {[\frac{1}{2} \sinh 2 v + v]}_{\ln 3}^{\ln 5}$
A1
$π (\frac{312}{50} + \ln 5 - \frac{40}{18} - \ln 3) = π (\frac{904}{225} + \ln \frac{5}{3})$

4a · 2 marks

M1
$(\begin{array}{ccc} - 11 & 1 & 8 \\ 0 & - 2 & 0 \\ - 16 & 1 & 13 \end{array}) (\begin{array}{c} 1 \\ 1 \\ 1 \end{array}) = (\begin{array}{c} - 2 \\ - 2 \\ - 2 \end{array})$
A1
$λ = - 2$

4b · 3 marks

M1
$∣ \begin{array}{ccc} - 11 - λ & 1 & 8 \\ 0 & - 2 - λ & 0 \\ - 16 & 1 & 13 - λ \end{array} ∣ = 0$
A1
$(- 11 - λ) (- 2 - λ) (13 - λ) + 128 (- 2 - λ) = 0$ $(- 2 - λ) (λ^{2} - 2 λ - 15) = 0$ $λ^{3} - 19 λ - 30 = 0$
B1
$λ = 5, - 3$

4c · 4 marks

B1
$A^{3} - 19 A - 30 I = 0$
M1
$30 A^{- 1} = A^{2} - 19 I$
M1
$A^{2} = (\begin{array}{ccc} - 7 & - 5 & 16 \\ 0 & 4 & 0 \\ - 32 & - 5 & 41 \end{array})$
A1
$A^{- 1} = \frac{1}{30} (\begin{array}{ccc} - 26 & - 5 & 16 \\ 0 & - 15 & 0 \\ - 32 & - 5 & 22 \end{array})$

5 · 10 marks

M1
$6 m^{2} - 5 m + 1 = 0 \Rightarrow m = \frac{1}{2}, \frac{1}{3}$
A1
$x = A e^{\frac{1}{2} t} + B e^{\frac{1}{3} t}$
B1
$x = p t^{2} + q t + r \Rightarrow x^{'} = 2 p t + q \Rightarrow x^{''} = 2 p$
M1
$p = 1 - 10 p + q = 1 12 p - 5 q + r = 1$
A1
$p = 1 q = 11 r = 44$
A1
$x = A e^{\frac{1}{2} t} + B e^{\frac{1}{3} t} + t^{2} + 11 t + 44$
M1
$x^{'} = \frac{1}{2} A e^{\frac{1}{2} t} + \frac{1}{3} B e^{\frac{1}{3} t} + 2 t + 11$
M1
$A + B + 44 = 12 \frac{1}{2} A + \frac{1}{3} B + 11 = - 6 \Rightarrow A = - 38, B = 6$
A1
$A + B + 44 = 12 \frac{1}{2} A + \frac{1}{3} B + 11 = - 6 \Rightarrow A = - 38, B = 6$
A1
$x = - 38 e^{\frac{1}{2} t} + 6 e^{\frac{1}{3} t} + t^{2} + 11 t + 44$

6a · 4 marks

M1
$[\int_{0}^{1} {(\frac{1}{2})}^{x} d x >] (\frac{1}{N}) {(\frac{1}{2})}^{\frac{1}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{2}{N}} + \dots + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 1}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N}{N}}$
A1
$[\int_{0}^{1} {(\frac{1}{2})}^{x} d x >] (\frac{1}{N}) {(\frac{1}{2})}^{\frac{1}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{2}{N}} + \dots + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 1}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N}{N}}$
M1
$= \frac{1}{N} \sum_{n = 1}^{N} {(\frac{1}{2})}^{\frac{n}{N}} = \frac{{(\frac{1}{2})}^{\frac{1}{N}} (\frac{1}{2} - 1)}{N ({(\frac{1}{2})}^{\frac{1}{N}} - 1)} = \frac{(\frac{1}{2}) {(\frac{1}{2})}^{\frac{1}{N}}}{N - N {(\frac{1}{2})}^{\frac{1}{N}}} = \frac{1}{2 N (2^{\frac{1}{N}} - 1)}$
A1
$= \frac{1}{N} \sum_{n = 1}^{N} {(\frac{1}{2})}^{\frac{n}{N}} = \frac{{(\frac{1}{2})}^{\frac{1}{N}} (\frac{1}{2} - 1)}{N ({(\frac{1}{2})}^{\frac{1}{N}} - 1)} = \frac{(\frac{1}{2}) {(\frac{1}{2})}^{\frac{1}{N}}}{N - N {(\frac{1}{2})}^{\frac{1}{N}}} = \frac{1}{2 N (2^{\frac{1}{N}} - 1)}$

6b · 4 marks

M1
$\int_{0}^{1} {(\frac{1}{2})}^{x} d x < (\frac{1}{N}) + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{1}{N}} + \dots + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 2}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 1}{N}}$
A1
$\int_{0}^{1} {(\frac{1}{2})}^{x} d x < (\frac{1}{N}) + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{1}{N}} + \dots + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 2}{N}} + (\frac{1}{N}) {(\frac{1}{2})}^{\frac{N - 1}{N}}$
M1
$\frac{1}{N} \sum_{n = 0}^{N - 1} {(\frac{1}{2})}^{\frac{n}{N}} = \frac{\frac{1}{2} - 1}{N ({(\frac{1}{2})}^{\frac{1}{N}} - 1)} = \frac{1}{2 N (1 - {(\frac{1}{2})}^{\frac{1}{N}})} = \frac{2^{\frac{1}{N}}}{2 N (2^{\frac{1}{N}} - 1)}$
A1
$\frac{1}{N} \sum_{n = 0}^{N - 1} {(\frac{1}{2})}^{\frac{n}{N}} = \frac{\frac{1}{2} - 1}{N ({(\frac{1}{2})}^{\frac{1}{N}} - 1)} = \frac{1}{2 N (1 - {(\frac{1}{2})}^{\frac{1}{N}})} = \frac{2^{\frac{1}{N}}}{2 N (2^{\frac{1}{N}} - 1)}$

6c · 2 marks

M1
$\frac{2^{\frac{1}{N}}}{2 N (2^{\frac{1}{N}} - 1)} - \frac{1}{2 N (2^{\frac{1}{N}} - 1)} = \frac{1}{2 N} \leq 10^{- 3} \Rightarrow 2 N \geq 10^{3}$
A1
Least value of $N$ is 500

6d · 4 marks

M1
$L_{N} < \frac{1}{2 \ln 2} < U_{N} \Rightarrow \frac{1}{2 U_{N}} < \ln 2 < \frac{1}{2 L_{N}}$
A1
$L_{N} < \frac{1}{2 \ln 2} < U_{N} \Rightarrow \frac{1}{2 U_{N}} < \ln 2 < \frac{1}{2 L_{N}}$
A1
$0.693 = \frac{1}{2 U_{N}} < \ln 2 < \frac{1}{2 L_{N}} = 0.694$
A1
$0.693 = \frac{1}{2 U_{N}} < \ln 2 < \frac{1}{2 L_{N}} = 0.694$

7a · 4 marks

B1
$\frac{d y}{d x} + \frac{y}{\sqrt{x^{2} + 16}} = x$
M1
$e^{\int \frac{1}{\sqrt{x^{2} + 16}} d x} = e^{\sinh^{- 1} (\frac{x}{4})}$
A1
$e^{\int \frac{1}{\sqrt{x^{2} + 16}} d x} = e^{\sinh^{- 1} (\frac{x}{4})}$
A1
$= \frac{1}{4} x + \frac{1}{4} \sqrt{x^{2} + 16}$

7b · 6 marks

M1
$\frac{d}{d x} (y (x + \sqrt{x^{2} + 16})) = x^{2} + x \sqrt{x^{2} + 16}$
A1
$\frac{d}{d x} (y (x + \sqrt{x^{2} + 16})) = x^{2} + x \sqrt{x^{2} + 16}$
M1
$y (x + \sqrt{x^{2} + 16}) = \frac{1}{3} x^{3} + \frac{1}{3} {(x^{2} + 16)}^{\frac{3}{2}} + C$
A1
$y (x + \sqrt{x^{2} + 16}) = \frac{1}{3} x^{3} + \frac{1}{3} {(x^{2} + 16)}^{\frac{3}{2}} + C$
M1
$6 (3 + \sqrt{25}) = \frac{27}{3} + \frac{25}{3} \sqrt{25} + C$
A1
$y (x + \sqrt{x^{2} + 16}) = \frac{1}{3} x^{3} + \frac{1}{3} {(x^{2} + 16)}^{\frac{3}{2}} - \frac{8}{3}$

8a · 5 marks

B1
$z + z^{- 1} = 2 \cos θ$
M1
${(z + z^{- 1})}^{7} = (z^{7} + z^{- 7}) + 7 (z^{5} + z^{- 5}) + 21 (z^{3} + z^{- 3}) + 35 (z + z^{- 1})$
A1
${(z + z^{- 1})}^{7} = (z^{7} + z^{- 7}) + 7 (z^{5} + z^{- 5}) + 21 (z^{3} + z^{- 3}) + 35 (z + z^{- 1})$
M1
$2^{7} \cos^{7} θ = 2 \cos 7 θ + 7 (2 \cos 5 θ) + 21 (2 \cos 3 θ) + 35 (2 \cos θ)$
A1
$\cos^{7} θ = \frac{1}{64} \cos 7 θ + \frac{7}{64} \cos 5 θ + \frac{21}{64} \cos 3 θ + \frac{35}{64} \cos θ$

8b · 4 marks

M1
$I_{n} = \int_{0}^{\frac{1}{4} π} \cos^{n - 1} θ \cos θ d θ = {[\cos^{n - 1} θ \sin θ]}_{0}^{\frac{1}{4} π} + (n - 1) \int_{0}^{\frac{1}{4} π} \cos^{n - 2} θ \sin^{2} θ d θ$
A1
$I_{n} = \int_{0}^{\frac{1}{4} π} \cos^{n - 1} θ \cos θ d θ = {[\cos^{n - 1} θ \sin θ]}_{0}^{\frac{1}{4} π} + (n - 1) \int_{0}^{\frac{1}{4} π} \cos^{n - 2} θ \sin^{2} θ d θ$
M1
$I_{n} = {[\cos^{n - 1} θ \sin θ]}_{0}^{\frac{1}{4} π} + (n - 1) \int_{0}^{\frac{1}{4} π} \cos^{n - 2} θ (1 - \cos^{2} θ) d θ$
A1
$I_{n} = 2^{- \frac{n}{2}} + (n - 1) I_{n - 2} - (n - 1) I_{n} \Rightarrow n I_{n} = 2^{- \frac{n}{2}} + (n - 1) I_{n - 2}$

8c · 5 marks

M1
$9 I_{9} = 2^{- \frac{9}{2}} + 8 I_{7} = \frac{1}{16} (2^{- \frac{1}{2}}) + 8 I_{7}$
A1
$9 I_{9} = 2^{- \frac{9}{2}} + 8 I_{7} = \frac{1}{16} (2^{- \frac{1}{2}}) + 8 I_{7}$
M1
$I_{7} = \int_{0}^{\frac{1}{4} π} \frac{1}{64} \cos 7 θ + \frac{7}{64} \cos 5 θ + \frac{21}{64} \cos 3 θ + \frac{35}{64} \cos θ d θ$
A1
$I_{7} = \frac{1}{64} {[\frac{1}{7} \sin 7 θ + \frac{7}{5} \sin 5 θ + \frac{21}{3} \sin 3 θ + 35 \sin θ]}_{0}^{\frac{1}{4} π}$
A1
$9 I_{9} = \frac{1}{16} (2^{- \frac{1}{2}}) + \frac{1}{8} (\frac{1}{7} (- 2^{- \frac{1}{2}}) + \frac{7}{5} (- 2^{- \frac{1}{2}}) + \frac{21}{3} (2^{- \frac{1}{2}}) + 35 (2^{- \frac{1}{2}}))$ $I_{9} = \frac{2867}{5040} (2^{- \frac{1}{2}}) = \frac{2867}{10080} \sqrt{2}$

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