A Level Further Mathematics 9231 Paper 11, May/June 2024

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9231/11 · May/June 2024

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The cubic equation $2 x^{3} + x^{2} - p x - 5 = 0$ , where $p$ is a positive constant, has roots $α, β, γ$ .

State, in terms of $p$ , the value of $α β + β γ + γ α$ .

[1]

Find the value of $α^{2} β γ + α β^{2} γ + α β γ^{2}$ .

[2]

Deduce a cubic equation whose roots are $α β, β γ, α γ$ .

[1]

Given that $α^{2} + β^{2} + γ^{2} = \frac{1}{3}$ , find the value of $p$ .

[2]

Prove by mathematical induction that $6^{4 n} + 38^{n} - 2$ is divisible by 74 for all positive integers $n$ .

[6]

Use standard results from the list of formulae (MF19) to show that $\sum_{r = 1}^{N} r (r + 1) (3 r + 4) = \frac{1}{12} N (N + 1) (N + 2) (9 N + 19) .$

[3]

Express $\frac{3 r + 4}{r (r + 1)}$ in partial fractions and hence use the method of differences to find $\sum_{r = 1}^{N} \frac{3 r + 4}{r (r + 1)} {(\frac{1}{4})}^{r + 1}$ in terms of $N$ .

[4]

Deduce the value of $\sum_{r = 1}^{\infty} \frac{3 r + 4}{r (r + 1)} {(\frac{1}{4})}^{r + 1}$ .

[1]

The matrix $M$ is given by $M = (\begin{array}{cc} \frac{1}{2} & - \frac{1}{2} \sqrt{3} \\ \frac{1}{2} \sqrt{3} & \frac{1}{2} \end{array}) (\begin{array}{cc} 14 & 0 \\ 0 & 1 \end{array})$ .

The matrix $M$ represents a sequence of two geometrical transformations in the $x$ - $y$ plane. Give full details of each transformation, and make clear the order in which they are applied.

[4]

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

[2]

Find the equations of the invariant lines, through the origin, of the transformation represented by $M$ .

[5]

The triangle $A B C$ in the $x$ - $y$ plane is transformed by $M$ onto triangle $D E F$ . Given that the area of triangle $D E F$ is $28 {cm}^{2}$ , find the area of triangle $A B C$ .

[2]

The points $A, B, C$ have position vectors $2 i + 2 j + 4 k, 2 i + 4 j - k, - 3 i - 3 j + 4 k,$ 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]

The point $D$ has position vector $2 i + j + 3 k$ . Find the perpendicular distance from $D$ to the plane $A B C$ .

[2]

Find the shortest distance between the lines $A B$ and $C D$ .

[5]

The curve $C$ has equation $y = \frac{x^{2} + a x + 1}{x + 2}$ , where $a > \frac{5}{2}$ .

Find the equations of the asymptotes of $C$ .

[3]

Show that $C$ has no stationary points.

[4]

Sketch $C$ , stating the coordinates of the point of intersection with the $y$ -axis and labelling the asymptotes.

[3]

d(i)

Sketch the curve with equation $y = ∣ \frac{x^{2} + a x + 1}{x + 2} ∣$ .

[2]

d(ii)

On your sketch in part (i), draw the line $y = a$ .

[1]

d(iii)

It is given that $∣ \frac{x^{2} + a x + 1}{x + 2} ∣ < a$ for $- 5 - \sqrt{14} < x < - 3$ and $- 5 + \sqrt{14} < x < 3$ . Find the value of $a$ .

[2]

The curve $C$ has polar equation $r^{2} = (π - θ) \tan^{- 1} (π - θ)$ , for $0 ⩽ θ ⩽ π$ .

Sketch $C$ and state the polar coordinates of the point of $C$ furthest from the pole.

[3]

Using the substitution $u = π - θ$ , or otherwise, find the area of the region enclosed by $C$ and the initial line.

[7]

Show that, at the point of $C$ furthest from the initial line, $2 (π - θ) \tan^{- 1} (π - θ) \cot θ - \frac{π - θ}{1 + (π - θ)^{2}} - \tan^{- 1} (π - θ) = 0$ and verify that this equation has a root for $θ$ between 1.2 and 1.3.

[5]

Mark scheme

9231/11 · May/June 2024

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1a · 1 mark

B1
$- \frac{1}{2} p$

1b · 2 marks

M1
$α β^{2} γ + α^{2} β γ + α β γ^{2} = α β γ (β + γ + α) = \frac{5}{2} (- \frac{1}{2})$
A1
$- \frac{5}{4}$

1c · 1 mark

B1
$4 z^{3} + 2 p z^{2} - 5 z - 25 = 0$

1d · 2 marks

M1
$\frac{1}{3} = \frac{1}{4} + p$
A1
$p = \frac{1}{12}$

2 · 6 marks

B1
$6^{4} + 38 - 2 = 1332$ is divisible by 74.
B1
Assume that $6^{4 k} + 38^{k} - 2$ is divisible by 74 for some positive integer $k$ .
M1
Then $6^{4 k + 4} + 38^{k + 1} - 2 = (1295 + 1) \times 6^{4 k} + (37 + 1) \times 38^{k} - 2$
A1
Separates $6^{4 k} + 38^{k} - 2$ or considers difference.
A1
Is divisible by 74 because $1295 \times 6^{4 k} + 37 \times 38^{k}$ is divisible by 74.
A1
Hence, by induction, $6^{4 k} + 38^{k} - 2$ is divisible by 74, is true for every positive integer $n$ .

3a · 3 marks

B1
$\sum_{r = 1}^{N} r (r + 1) (3 r + 4) = 3 \sum_{r = 1}^{N} r^{3} + 7 \sum_{r = 1}^{N} r^{2} + 4 \sum_{r = 1}^{N} r$
M1
$\frac{3}{4} N^{2} (N + 1)^{2} + \frac{7}{6} N (N + 1) (2 N + 1) + 2 N (N + 1)$
A1
$\frac{1}{12} N (9 N (N^{2} + 2 N + 1) + 14 (2 N^{2} + 3 N + 1) + 24 N + 24) \frac{1}{12} N (9 N^{3} + 46 N^{2} + 75 N + 38) = \frac{1}{12} N (N + 1) (N + 2) (9 N + 19)$

3b · 4 marks

M1
Finds partial fractions: method
A1
$\frac{3 r + 4}{r (r + 1)} = \frac{4}{r} - \frac{1}{r + 1}$
M1
$\sum_{r = 1}^{N} \frac{3 r + 4}{r (r + 1)} {(\frac{1}{4})}^{r + 1} = (\frac{1}{4^{2}}) (\frac{4}{1} - \frac{1}{2}) + (\frac{1}{4^{3}}) (\frac{4}{2} - \frac{1}{3}) + \dots + (\frac{1}{4^{N + 1}}) (\frac{4}{N} - \frac{1}{N + 1})$
A1
$\frac{1}{4} - \frac{1}{4^{N + 1} (N + 1)}$

3c · 1 mark

B1
$\frac{1}{4}$

4a · 4 marks

B1
[One way] stretch, rotation
B1
Stretch followed by rotation
B1
Stretch parallel to the $x$ -axis, scale factor 14
B1
Rotation, $\frac{1}{3} π$ [anticlockwise] about the origin.

4b · 2 marks

B1
${(\begin{array}{cc} \frac{1}{2} & - \frac{1}{2} \sqrt{3} \\ \frac{1}{2} \sqrt{3} & \frac{1}{2} \end{array})}^{- 1} = (\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \sqrt{3} \\ - \frac{1}{2} \sqrt{3} & \frac{1}{2} \end{array})$ , ${(\begin{array}{cc} 14 & 0 \\ 0 & 1 \end{array})}^{- 1} = \frac{1}{14} (\begin{array}{cc} 1 & 0 \\ 0 & 14 \end{array})$
B1
$M^{- 1} = (\begin{array}{cc} \frac{1}{14} & 0 \\ 0 & 1 \end{array}) (\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \sqrt{3} \\ - \frac{1}{2} \sqrt{3} & \frac{1}{2} \end{array})$

4c · 5 marks

B1
$M = (\begin{array}{cc} 7 & - \frac{1}{2} \sqrt{3} \\ 7 \sqrt{3} & \frac{1}{2} \end{array})$
B1
$(\begin{array}{cc} 7 & - \frac{1}{2} \sqrt{3} \\ 7 \sqrt{3} & \frac{1}{2} \end{array}) (\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} 7 x - \frac{1}{2} \sqrt{3} y \\ 7 \sqrt{3} x + \frac{1}{2} y \end{array})$
M1
$7 \sqrt{3} x + \frac{1}{2} m x = m (7 x - \frac{1}{2} \sqrt{3} m x)$
A1
$7 \sqrt{3} + \frac{1}{2} m = 7 m - \frac{1}{2} \sqrt{3} m^{2} \Rightarrow \frac{1}{2} \sqrt{3} m^{2} - \frac{13}{2} m + 7 \sqrt{3} = 0$
A1
$y = 2 \sqrt{3} x$ and $y = \frac{7}{3} \sqrt{3} x$

4d · 2 marks

M1
$28 = 14 \times ∣ A B C ∣$
A1
$2 {cm}^{2}$

5a · 5 marks

B1
$\vec{A B} = 2 j - 5 k \vec{A C} = - 5 i - 5 j \vec{B C} = - 5 i - 7 j + 5 k$
M1
Finds normal to the plane ABC: method using cross product
A1
$∣ \begin{array}{ccc} i & j & k \\ 0 & 2 & - 5 \\ - 5 & - 5 & 0 \end{array} ∣ = (\begin{array}{c} - 25 \\ 25 \\ 10 \end{array}) \sim (\begin{array}{c} - 5 \\ 5 \\ 2 \end{array})$
M1
Substitutes point
A1
$- 5 (2) + 5 (2) + 2 (4) = 8 \Rightarrow - 5 x + 5 y + 2 z = 8$

5b · 2 marks

M1
$\frac{8 - (2 i + j + 3 k) \cdot (- 5 i + 5 j + 2 k)}{\sqrt{5^{2} + 5^{2} + 2^{2}}}$
A1
$\frac{7}{\sqrt{54}} = 0.953$

5c · 5 marks

B1
$\vec{C D} = (\begin{array}{c} 2 \\ 1 \\ 3 \end{array}) - (\begin{array}{c} - 3 \\ - 3 \\ 4 \end{array}) = (\begin{array}{c} 5 \\ 4 \\ - 1 \end{array})$
M1
Finds cross product for common perpendicular: method
A1
$∣ \begin{array}{ccc} i & j & k \\ 0 & 2 & - 5 \\ 5 & 4 & - 1 \end{array} ∣ = (\begin{array}{c} 18 \\ - 25 \\ - 10 \end{array})$
M1
$\frac{1}{\sqrt{1049}} (\begin{array}{c} - 5 \\ - 5 \\ 0 \end{array}) \cdot (\begin{array}{c} 18 \\ - 25 \\ - 10 \end{array})$
A1
$\frac{35}{\sqrt{1049}} = 1.08$

6a · 3 marks

B1
$x = - 2$
M1
$y = \frac{(x + 2) (x + a - 2) - 2 a + 5}{x + 2} = x + a - 2 + \frac{5 - 2 a}{x + 2}$
A1
$y = x + a - 2$

6b · 4 marks

M1
$\frac{d y}{d x} = \frac{(x + 2) (2 x + a) - (x^{2} + a x + 1)}{(x + 2)^{2}}$
A1
$x^{2} + 4 x + 2 a - 1 = 0 (or \frac{d y}{d x} = 1 + \frac{2 a - 5}{(x + 2)^{2}})$
M1
Consideration of discriminant or sign of $y^{'}$
A1
$16 - 4 (2 a - 1) = 20 - 8 a < 0 (or y^{'} > 0) \Rightarrow No stationary points$

6c · 3 marks

B1
Axes and asymptotes labelled.
B1
Branches correct. (Asymptotes may cross above, on or below the $x$ -axis.)
B1
$(0, \frac{1}{2})$

6d(i) · 2 marks

B1
FT from their attempt in part 6(c).
B1
Everything correct (approach to vertical asymptote and cusps correct).

6d(ii) · 1 mark

B1
Line $y = a$ drawn on sketch.

6d(iii) · 2 marks

M1
$\frac{x^{2} + a x + 1}{x + 2} = a or \frac{x^{2} + a x + 1}{x + 2} = - a x^{2} + 1 - 2 a = 0 or x^{2} + 2 a x + 1 + 2 a = 0 - a - \sqrt{a^{2} - 2 a - 1} < x < - \sqrt{2 a - 1}, - a + \sqrt{a^{2} - 2 a - 1} < x < \sqrt{2 a - 1}$
A1
$a = 5$

7a · 3 marks

B1
Correct shape.
B1
Section $\frac{1}{2} π ⩽ θ ⩽ π$ correct.
B1
$(\sqrt{π \tan^{- 1} π}, 0) = (1.99, 0)$

7b · 7 marks

M1
$A = \frac{1}{2} \int_{0}^{π} (π - θ) \tan^{- 1} (π - θ) d θ or A = - \frac{1}{2} \int_{π}^{0} u \tan^{- 1} (u) d u$
M1
Integrates by parts: method
A1
$\frac{1}{2} {[\frac{1}{2} (u)^{2} \tan^{- 1} (u)]}_{0}^{π} - \frac{1}{4} \int_{0}^{π} \frac{u^{2}}{1 + u^{2}} d u$
M1
$\int_{0}^{π} \frac{u^{2}}{1 + u^{2}} d u = \int_{0}^{π} \frac{1 + u^{2} - 1}{1 + u^{2}} d u = \int_{0}^{π} 1 - \frac{1}{1 + u^{2}} d u$
A1
$[u - \tan^{- 1} u]_{0}^{π}$
A1
$A = \frac{1}{2} {[\frac{1}{2} (u)^{2} \tan^{- 1} (u)]}_{0}^{π} - \frac{1}{4} [u - \tan^{- 1} (u)]_{0}^{π}$
A1
$A = \frac{1}{4} π^{2} \tan^{- 1} π - \frac{1}{4} (π - \tan^{- 1} π) = \frac{1}{4} (π^{2} + 1) \tan^{- 1} π - \frac{1}{4} π = 2.65$

7c · 5 marks

B1
$y = \sqrt{(π - θ) \tan^{- 1} (π - θ)} \sin θ$
M1
Finds derivative: method
A1
$\frac{d y}{d θ} = \sqrt{(π - θ) \tan^{- 1} (π - θ)} \cos θ - \frac{(π - θ) (1 + (π - θ)^{2})^{- 1} + \tan^{- 1} (π - θ)}{2 \sqrt{(π - θ) \tan^{- 1} (π - θ)}} \sin θ$
A1
$[θ \neq 0], π \Rightarrow 2 (π - θ) \tan^{- 1} (π - θ) \cot θ - \frac{π - θ}{1 + (π - θ)^{2}} - \tan^{- 1} (π - θ) = 0$
B1
$2 (π - 1.2) \tan^{- 1} (π - 1.2) \cot 1.2 - \frac{π - 1.2}{1 + (π - 1.2)^{2}} - \tan^{- 1} (π - 1.2) = 0.151$ and $2 (π - 1.3) \tan^{- 1} (π - 1.3) \cot 1.3 - \frac{π - 1.3}{1 + (π - 1.3)^{2}} - \tan^{- 1} (π - 1.3) = - 0.395$

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