A Level Further Mathematics 9231 Paper 13, May/June 2023

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9231/13 · May/June 2023

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Prove by mathematical induction that, for all positive integers $n$ , $5^{3 n} + 32^{n} - 33$ is divisible by 31. [6]

[6]

Use standard results from the list of formulae (MF19) to show that $\sum_{r = 1}^{n} (6 r^{2} + 6 r - 5) = a n^{3} + b n^{2} + c n,$ where $a$ , $b$ and $c$ are integers to be determined. [2]

[2]

Use the method of differences to find $\sum_{r = 1}^{n} \frac{6 r^{2} + 6 r - 5}{r^{2} + r}$ in terms of $n$ . [4]

[4]

Find also $\sum_{r = n + 1}^{2 n} \frac{6 r^{2} + 6 r - 5}{r^{2} + r}$ in terms of $n$ . [2]

[2]

The equation $x^{4} - x^{2} + 2 x + 5 = 0$ has roots $α, β, γ, δ$ .

Find a quartic equation whose roots are $α^{2}, β^{2}, γ^{2}, δ^{2}$ and state the value of $α^{2} + β^{2} + γ^{2} + δ^{2}$ . [4]

[4]

Find the value of $\frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}} + \frac{1}{δ^{2}}$ . [3]

[3]

Find the value of $α^{4} + β^{4} + γ^{4} + δ^{4}$ . [2]

[2]

The matrix $M$ is given by $M = (\begin{array}{cc} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & \cos 2 θ \end{array}) (\begin{array}{cc} 1 & k \\ 0 & 1 \end{array})$ , where $0 < θ < π$ and $k$ is a non-zero constant. The matrix $M$ represents a sequence of two geometrical transformations, one of which is a shear.

Describe fully the other transformation and state the order in which the transformations are applied. [3]

[3]

Write $M^{- 1}$ as the product of two matrices, neither of which is $I$ . [2]

[2]

Find, in terms of $k$ , the value of $\tan θ$ for which $M - I$ is singular. [5]

[5]

Given that $k = 2 \sqrt{3}$ and $θ = \frac{1}{3} π$ , show that the invariant points of the transformation represented by $M$ lie on the line $3 y + \sqrt{3} x = 0$ . [4]

[4]

Show that the curve with Cartesian equation $x^{2} - y^{2} = a,$ where $a$ is a positive constant, has polar equation $r^{2} = a \sec 2 θ$ . [3]

[3]

The curve $C$ has polar equation $r^{2} = a \sec 2 θ$ , where $a$ is a positive constant, for $0 ⩽ θ < \frac{1}{4} π$ . Sketch $C$ and state the minimum distance of $C$ from the pole. [3]

[3]

Find, in terms of $a$ , the exact value of the area of the region enclosed by $C$ , the initial line, and the half-line $θ = \frac{1}{12} π$ . [You may use any result from the list of formulae (MF19) without proof.] [4]

[4]

The points $A, B, C$ have position vectors $i + j, - i + 2 j + 4 k, - 2 i + j + 3 k,$ respectively, relative to the origin $O$ .

Find the equation of the plane $A B C$ , giving your answer in the form $a x + b y + c z = d$ . [5]

[5]

Find the perpendicular distance from $O$ to the plane $A B C$ . [2]

[2]

Find a vector equation of the common perpendicular to the lines $O C$ and $A B$ . [8]

[8]

The curve $C$ has equation $y = \frac{x^{2} + 2 x + 1}{x - 3}$ .

Find the equations of the asymptotes of $C$ . [3]

[3]

Find the coordinates of the turning points on $C$ . [3]

[3]

Sketch $C$ . [3]

[3]

Sketch the curves with equations $y = ∣ \frac{x^{2} + 2 x + 1}{x - 3} ∣$ and $y^{2} = \frac{x^{2} + 2 x + 1}{x - 3}$ on a single diagram, clearly identifying each curve. [4]

[4]

Mark scheme

9231/13 · May/June 2023

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1 · 6 marks

B1
$5^{3} + 32 - 33 = 124$ is divisible by 31.
B1
Assume that $5^{3 k} + 32^{k} - 33$ is divisible by 31 for some positive integer $k$ .
M1
$5^{3 k + 3} + 32^{k + 1} - 33 = (124 + 1) 5^{3 k} + (31 + 1) 32^{k} - 33$
A1
$5^{3 k + 3} + 32^{k + 1} - 33 = (124 + 1) 5^{3 k} + (31 + 1) 32^{k} - 33$
A1
is divisible by 31 because $124 \times 5^{3 k} + 31 \times 32^{k}$ is divisible by 31.
A1
Hence, by induction, true for every positive integer $n$ .

2a · 2 marks

M1
$6 (\frac{1}{6} n (n + 1) (2 n + 1)) + 6 (\frac{1}{2} n (n + 1)) [- 5 n]$
A1
$2 n^{3} + 6 n^{2} - n$

2b · 4 marks

B1
$\frac{6 r^{2} + 6 r - 5}{r^{2} + r} = 6 - \frac{5}{r (r + 1)}$
B1
$\frac{1}{r (r + 1)} = \frac{1}{r} - \frac{1}{r + 1}$
M1
$\sum_{r = 1}^{n} \frac{1}{r (r + 1)} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{n} - \frac{1}{n + 1}$
A1
$\sum_{r = 1}^{n} \frac{6 r^{2} + 6 r - 5}{r^{2} + r} = 6 n - 5 + \frac{5}{n + 1}$

2c · 2 marks

M1
$\sum_{r = 1}^{2 n} \frac{6 r^{2} + 6 r - 5}{r^{2} + r} - \sum_{r = 1}^{n} \frac{6 r^{2} + 6 r - 5}{r^{2} + r} = 12 n - 5 + \frac{5}{2 n + 1} - 6 n + 5 - \frac{5}{n + 1}$
A1
$12 n - 5 + \frac{5}{2 n + 1} - 6 n + 5 - \frac{5}{n + 1} = 6 n + \frac{5}{2 n + 1} - \frac{5}{n + 1}$

3a · 4 marks

B1
$y = x^{2}$ so, $x = y^{\frac{1}{2}}$
M1
$y^{2} - y + 2 y^{\frac{1}{2}} + 5 = 0$ so, $(2 y^{\frac{1}{2}})^{2} = (- y^{2} + y - 5)^{2}$
A1
$y^{4} - 2 y^{3} + 11 y^{2} - 14 y + 25 = 0$
B1
$α^{2} + β^{2} + γ^{2} + δ^{2} = 2$

3b · 3 marks

B1
$α^{2} β^{2} γ^{2} δ^{2} = 25$
M1
$\frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}} + \frac{1}{δ^{2}} = \frac{α^{2} β^{2} δ^{2} + α^{2} β^{2} γ^{2} + β^{2} γ^{2} δ^{2} + α^{2} γ^{2} δ^{2}}{α^{2} β^{2} γ^{2} δ^{2}}$
A1
$\frac{14}{25}$

3c · 2 marks

M1
$α^{4} + β^{4} + γ^{4} + δ^{4} = (α^{2} + β^{2} + γ^{2} + δ^{2})^{2} - 2 (α^{2} β^{2} + α^{2} γ^{2} + α^{2} δ^{2} + β^{2} γ^{2} + β^{2} δ^{2} + γ^{2} δ^{2}) = 2^{2} - 2 (11)$
A1
$- 18$

4a · 3 marks

B1
Rotation [anticlockwise]
B1
about the origin through angle $2 θ$ .
B1
Shear [in the $x$ -direction] followed by a rotation [anticlockwise about the origin through angle $2 θ$ ].

4b · 2 marks

B1
${(\begin{array}{cc} 1 & k \\ 0 & 1 \end{array})}^{- 1} = (\begin{array}{cc} 1 & - k \\ 0 & 1 \end{array})$ , ${(\begin{array}{cc} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & \cos 2 θ \end{array})}^{- 1} = (\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array})$
B1
$M^{- 1} = (\begin{array}{cc} 1 & - k \\ 0 & 1 \end{array}) (\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array})$

4c · 5 marks

B1
$M - I = (\begin{array}{cc} \cos 2 θ - 1 & k \cos 2 θ - \sin 2 θ \\ \sin 2 θ & k \sin 2 θ + \cos 2 θ - 1 \end{array})$
M1
$(\cos 2 θ - 1) (k \sin 2 θ + \cos 2 θ - 1) - k \sin 2 θ \cos 2 θ + \sin^{2} 2 θ [= 0]$
A1
$2 - 2 \cos 2 θ - k \sin 2 θ = 0$
M1
$4 \sin^{2} θ = 2 k \sin θ \cos θ$
A1
$\tan θ = \frac{1}{2} k$

4d · 4 marks

B1
$M = (\begin{array}{cc} - \frac{1}{2} & - \frac{3}{2} \sqrt{3} \\ \frac{1}{2} \sqrt{3} & \frac{5}{2} \end{array})$ .
B1
$(\begin{array}{cc} - \frac{1}{2} & - \frac{3}{2} \sqrt{3} \\ \frac{1}{2} \sqrt{3} & \frac{5}{2} \end{array}) (\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} - \frac{1}{2} x - \frac{3}{2} \sqrt{3} y \\ \frac{1}{2} \sqrt{3} x + \frac{5}{2} y \end{array})$
M1
$- \frac{1}{2} x - \frac{3}{2} \sqrt{3} y = x [\Rightarrow - \frac{3}{2} x - \frac{3}{2} \sqrt{3} y = 0 \Rightarrow x + \sqrt{3} y = 0]$ and $\frac{1}{2} \sqrt{3} x + \frac{5}{2} y = y [\Rightarrow \frac{1}{2} \sqrt{3} x + \frac{3}{2} y = 0 \Rightarrow \sqrt{3} x + 3 y = 0]$
A1
$\sqrt{3} x + 3 y = 0$

5a · 3 marks

B1
$r^{2} (\cos^{2} θ - \sin^{2} θ) = a$
M1
$r^{2} \cos 2 θ = a$
A1
$r^{2} = a \sec 2 θ$

5b · 3 marks

B1
Initial line drawn. Correct domain and position, r strictly increasing.
1 mark
Also sloping to right, concave on opposite side to pole, correct gradient when $θ = 0$ and $θ \to π / 4$ .
B1
$\sqrt{a}$

5c · 4 marks

M1
$\frac{1}{2} a \int_{0}^{\frac{1}{12} π} \sec 2 θ d θ$
M1
$\frac{1}{4} a {[\ln \tan (θ + \frac{1}{4} π)]}_{0}^{\frac{1}{12} π}$ or $\frac{1}{4} a {[\ln (\tan 2 θ + \sec 2 θ)]}_{0}^{\frac{1}{12} π}$
A1
$\frac{1}{4} a {[\ln \tan (θ + \frac{1}{4} π)]}_{0}^{\frac{1}{12} π}$ or $\frac{1}{4} a {[\ln (\tan 2 θ + \sec 2 θ)]}_{0}^{\frac{1}{12} π}$
A1
$\frac{1}{4} a \ln \sqrt{3} = \frac{1}{8} a \ln 3$

6a · 5 marks

B1
$\vec{A B} = - 2 i + j + 4 k \vec{A C} = - 3 i + 3 k \vec{B C} = - i - j - k$
M1
$∣ \begin{array}{ccc} i & j & k \\ - 2 & 1 & 4 \\ - 3 & 0 & 3 \end{array} ∣ = (\begin{array}{c} 3 \\ - 6 \\ 3 \end{array}) \sim (\begin{array}{c} 1 \\ - 2 \\ 1 \end{array})$
A1
$∣ \begin{array}{ccc} i & j & k \\ - 2 & 1 & 4 \\ - 3 & 0 & 3 \end{array} ∣ = (\begin{array}{c} 3 \\ - 6 \\ 3 \end{array}) \sim (\begin{array}{c} 1 \\ - 2 \\ 1 \end{array})$
M1
$1 (1) - 2 (1) + 1 (0) = - 1 \Rightarrow x - 2 y + z = - 1$
A1
$1 (1) - 2 (1) + 1 (0) = - 1 \Rightarrow x - 2 y + z = - 1$

6b · 2 marks

M1
$\frac{1}{\sqrt{1^{2} + 2^{2} + 1^{2}}} = \frac{1}{\sqrt{6}} = 0.408$
A1
$\frac{1}{\sqrt{1^{2} + 2^{2} + 1^{2}}} = \frac{1}{\sqrt{6}} = 0.408$

6c · 8 marks

M1
$\vec{O P} = (\begin{array}{c} - 2 λ \\ λ \\ 3 λ \end{array}), \vec{O Q} = (\begin{array}{c} 1 - 2 μ \\ 1 + μ \\ 4 μ \end{array}) \Rightarrow \vec{P Q} = (\begin{array}{c} 1 - 2 μ + 2 λ \\ 1 + μ - λ \\ 4 μ - 3 λ \end{array})$
A1
$\vec{O P} = (\begin{array}{c} - 2 λ \\ λ \\ 3 λ \end{array}), \vec{O Q} = (\begin{array}{c} 1 - 2 μ \\ 1 + μ \\ 4 μ \end{array}) \Rightarrow \vec{P Q} = (\begin{array}{c} 1 - 2 μ + 2 λ \\ 1 + μ - λ \\ 4 μ - 3 λ \end{array})$
M1
$(\begin{array}{c} 1 - 2 μ + 2 λ \\ 1 + μ - λ \\ 4 μ - 3 λ \end{array}) \cdot (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array}) = 0$
1 mark
$17 μ - 14 λ = 1$
1 mark
$(\begin{array}{c} 1 - 2 μ + 2 λ \\ 1 + μ - λ \\ 4 μ - 3 λ \end{array}) \cdot (\begin{array}{c} - 2 \\ 1 \\ 4 \end{array}) = 0 \Rightarrow 21 μ - 17 λ = 1$
1 mark
$λ = - \frac{4}{5} \Rightarrow \vec{O P} = - \frac{4}{5} (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array})$
A1
$λ = - \frac{4}{5} \Rightarrow \vec{O P} = - \frac{4}{5} (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array})$
A1
$r = - \frac{4}{5} (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array}) + k (\begin{array}{c} 1 \\ 2 \\ 0 \end{array})$

7a · 3 marks

B1
$x = 3$
M1
$y = \frac{(x - 3) (x + 5) + 16}{x - 3} = x + 5 + \frac{16}{x - 3}$
A1
$y = x + 5$

7b · 3 marks

M1
$\frac{d y}{d x} = 1 - \frac{16}{(x - 3)^{2}} = 0 \Rightarrow (x - 3)^{2} = 16$
A1
$x = - 1, 7$
A1
$(- 1, 0), (7, 16)$

7c · 3 marks

B1
Axes and labelled asymptotes.
B1
Upper branch correct.
B1
Lower branch correct and no additional branches.

7d · 4 marks

B1
Clear labels, axes and their vertical asymptote.
B1
$y = ∣ \frac{x^{2} + 2 x + 1}{x - 3} ∣$ correct, FT from their sketch in (c).
B1
Upper branch of $y^{2} = \frac{x^{2} + 2 x + 1}{x - 3}$ (positive square root).
B1
Lower branch of $y^{2} = \frac{x^{2} + 2 x + 1}{x - 3}$ (negative square root). FT from previous mark.

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