A Level Further Mathematics 9231 Paper 12, Oct/Nov 2023

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9231/12 · Oct/Nov 2023

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Use standard results from the list of formulae (MF19) to find $\sum_{r = 1}^{n} (3 r^{2} + 3 r + 1)$ in terms of $n$ , simplifying your answer.

[3]

Show that $\frac{1}{r^{3}} - \frac{1}{(r + 1)^{3}} = \frac{3 r^{2} + 3 r + 1}{r^{3} (r + 1)^{3}}$ and hence use the method of differences to find $\sum_{r = 1}^{n} \frac{3 r^{2} + 3 r + 1}{r^{3} (r + 1)^{3}}$ .

[5]

Deduce the value of $\sum_{r = 1}^{\infty} \frac{3 r^{2} + 3 r + 1}{r^{3} (r + 1)^{3}}$ .

[1]

Prove by mathematical induction that, for all positive integers $n$ , $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = (x^{2} + 2 n x + n (n - 1)) e^{x} .$

[6]

The matrix $M$ is given by $M = (\begin{array}{cc} k & 0 \\ 0 & 1 \end{array}) (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ , where $k$ is a constant and $k \neq 0$ and $k \neq 1$ .

The matrix $M$ represents a sequence of two geometrical transformations. State the type of each transformation, and make clear the order in which they are applied.

[2]

The unit square in the $x – y$ plane is transformed by $M$ onto parallelogram $O P Q R$ . Find, in terms of $k$ , the area of parallelogram $O P Q R$ and the matrix which transforms $O P Q R$ onto the unit square.

[3]

Show that the line through the origin with gradient $\frac{1}{k - 1}$ is invariant under the transformation in the $x – y$ plane represented by $M$ .

[3]

The cubic equation $27 x^{3} + 18 x^{2} + 6 x - 1 = 0$ has roots $α$ , $β$ , $γ$ .

Show that a cubic equation with roots $3 α + 1$ , $3 β + 1$ , $3 γ + 1$ is $y^{3} - y^{2} + y - 2 = 0.$

[3]

The sum $(3 α + 1)^{n} + (3 β + 1)^{n} + (3 γ + 1)^{n}$ is denoted by $S_{n}$ . Find the values of $S_{2}$ and $S_{3}$ .

[4]

Find the values of $S_{- 1}$ and $S_{- 2}$ .

[3]

The plane $Π_{1}$ has equation $r = i - j - 2 k + λ (i - 2 j - 3 k) + μ (3 i - k)$ .

Find an equation for $Π_{1}$ in the form $a x + b y + c z = d$ .

[4]

The line $l$ , which does not lie in $Π_{1}$ , has equation $r = - 3 i + k + t (i + j + k)$ . Show that $l$ is parallel to $Π_{1}$ .

[2]

Find the distance between $l$ and $Π_{1}$ .

[3]

The plane $Π_{2}$ has equation $3 x + 3 y + 2 z = 1$ . Find a vector equation of the line of intersection of $Π_{1}$ and $Π_{2}$ .

[4]

The curve $C$ has polar equation $r = e^{- θ} - e^{- \frac{1}{2} π}$ , where $0 ⩽ θ ⩽ \frac{1}{2} π$ .

Sketch $C$ and state, in exact form, the greatest distance of a point on $C$ from the pole.

[3]

Find the exact value of the area of the region bounded by $C$ and the initial line.

[5]

Show that, at the point on $C$ furthest from the initial line, $1 - e^{θ - \frac{1}{2} π} - \tan θ = 0$ and verify that this equation has a root between 0.56 and 0.57.

[5]

The curve $C$ has equation $y = f (x)$ , where $f (x) = \frac{x^{2}}{x + 1}$ .

Find the equations of the asymptotes of $C$ .

[3]

Find the coordinates of any stationary points on $C$ .

[2]

Sketch $C$ .

[3]

Find the coordinates of any stationary points on the curve with equation $y = \frac{1}{f (x)}$ .

[2]

Sketch the curve with equation $y = \frac{1}{f (x)}$ and find, in exact form, the set of values for which $\frac{1}{f (x)} > f (x) .$

[6]

Mark scheme

9231/12 · Oct/Nov 2023

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1a · 3 marks

M1
$\frac{1}{2} n (n + 1) (2 n + 1) + \frac{3}{2} n (n + 1) + n$
A1
$\frac{1}{2} n (n + 1) (2 n + 1) + \frac{3}{2} n (n + 1) + n$
A1
$n^{3} + 3 n^{2} + 3 n$

1b · 5 marks

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

1c · 1 mark

1 mark
1

2 · 6 marks

M1
$\frac{d}{d x} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} = (x^{2} + 2 x) e^{x}$ so true when $n = 1$ .
A1
$\frac{d}{d x} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} = (x^{2} + 2 x) e^{x}$ so true when $n = 1$ .
B1
Assume that $\frac{d^{k}}{d x^{k}} (x^{2} e^{x}) = (x^{2} + 2 k x + k (k - 1)) e^{x}$ [for some value of $k$ ].
M1
$\frac{d^{k + 1}}{d x^{k + 1}} (x^{2} e^{x}) = (x^{2} + 2 k x + k (k - 1)) e^{x} + e^{x} (2 x + 2 k)$
A1
$(x^{2} + 2 (k + 1) x + k (k + 1)) e^{x}$
A1
So true when $n = k + 1$ . By induction, true for all positive integers $n$ .

3a · 2 marks

B2
Shear followed by a stretch.

3b · 3 marks

B1
$∣ O P Q R ∣ = ∣ \det M ∣ = ∣ k ∣$
M1
$M^{- 1} = \frac{1}{k} (\begin{array}{cc} 1 & 0 \\ - 1 & k \end{array})$
A1
$M^{- 1} = \frac{1}{k} (\begin{array}{cc} 1 & 0 \\ - 1 & k \end{array})$

3c · 3 marks

B1
$(\begin{array}{cc} k & 0 \\ 1 & 1 \end{array}) (\begin{array}{c} x \\ \frac{1}{k - 1} x \end{array})$
M1
$(\begin{array}{cc} k & 0 \\ 1 & 1 \end{array}) (\begin{array}{c} x \\ \frac{1}{k - 1} x \end{array}) = (\begin{array}{c} k x \\ x + \frac{1}{k - 1} x \end{array}) = (\begin{array}{c} k x \\ \frac{k}{k - 1} x \end{array})$
A1
$k (\begin{array}{c} x \\ \frac{1}{k - 1} x \end{array})$

4a · 3 marks

B1
$y = 3 x + 1 \Rightarrow x = \frac{1}{3} (y - 1)$ $\Rightarrow 27 {(\frac{y - 1}{3})}^{3} + 18 {(\frac{y - 1}{3})}^{2} + 6 (\frac{y - 1}{3}) - 1 = 0$
M1
$\Rightarrow (y - 1)^{3} + 2 (y - 1)^{2} + 2 (y - 1) - 1 = 0$ $\Rightarrow y^{3} - 3 y^{2} + 3 y - 1 + 2 y^{2} - 4 y + 2 + 2 y - 2 - 1 = 0$
A1
$y^{3} - y^{2} + y - 2 = 0$

4b · 4 marks

M1
$S_{2} = 1^{2} - 2 (1) = - 1$
A1
$S_{2} = 1^{2} - 2 (1) = - 1$
M1
$S_{3} = (3 α + 1)^{3} + (3 β + 1)^{3} + (3 γ + 1)^{3} = - 1 - (1) + 6$
A1
4

4c · 3 marks

B1
$S_{- 1} = \frac{(3 α + 1) (3 β + 1) + (3 β + 1) (3 γ + 1) + (3 γ + 1) (3 α + 1)}{(3 α + 1) (3 β + 1) (3 γ + 1)} = \frac{1}{2}$
M1
$2 S_{- 2} = S_{1} - 3 + S_{- 1} = 1 - 3 + \frac{1}{2}$
A1
$S_{- 2} = - \frac{3}{4}$

5a · 4 marks

M1
$∣ \begin{array}{ccc} i & j & k \\ 1 & - 2 & - 3 \\ 3 & 0 & - 1 \end{array} ∣ = (\begin{array}{c} 2 \\ - 8 \\ 6 \end{array}) \sim (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array})$
A1
$∣ \begin{array}{ccc} i & j & k \\ 1 & - 2 & - 3 \\ 3 & 0 & - 1 \end{array} ∣ = (\begin{array}{c} 2 \\ - 8 \\ 6 \end{array}) \sim (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array})$
M1
$1 (1) - 4 (- 1) + 3 (- 2) = - 1$
A1
$x - 4 y + 3 z = - 1$

5b · 2 marks

M1
$(\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) \cdot (\begin{array}{c} 1 \\ 1 \\ 1 \end{array}) = 1 - 4 + 3 = 0$
A1
$(\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) \cdot (\begin{array}{c} 1 \\ 1 \\ 1 \end{array}) = 1 - 4 + 3 = 0$

5c · 3 marks

M1
$\frac{1}{\sqrt{1^{2} + 4^{2} + 3^{2}}} ∣ (\begin{array}{c} - 4 \\ 1 \\ 3 \end{array}) \cdot (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) ∣$ or $\frac{1}{\sqrt{1^{2} + 4^{2} + 3^{2}}} ∣ ((\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) \cdot (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) + 1) ∣$
A1
$\frac{1}{\sqrt{1^{2} + 4^{2} + 3^{2}}} ∣ (\begin{array}{c} - 4 \\ 1 \\ 3 \end{array}) \cdot (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) ∣$ or $\frac{1}{\sqrt{1^{2} + 4^{2} + 3^{2}}} ∣ ((\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) \cdot (\begin{array}{c} 1 \\ - 4 \\ 3 \end{array}) + 1) ∣$
A1
$\frac{1}{\sqrt{26}} (= 0.196)$

5d · 4 marks

B1
States point common to both planes e.g. $(\begin{array}{c} - \frac{1}{15} \\ \frac{4}{15} \\ 0 \end{array})$ .
M1
$∣ \begin{array}{ccc} i & j & k \\ 1 & - 4 & 3 \\ 3 & 3 & 2 \end{array} ∣ = (\begin{array}{c} - 17 \\ 7 \\ 15 \end{array})$
A1
$∣ \begin{array}{ccc} i & j & k \\ 1 & - 4 & 3 \\ 3 & 3 & 2 \end{array} ∣ = (\begin{array}{c} - 17 \\ 7 \\ 15 \end{array})$
A1
$r = (\begin{array}{c} \frac{5}{7} \\ 0 \\ - \frac{4}{7} \end{array}) + λ (\begin{array}{c} - 17 \\ 7 \\ 15 \end{array})$

6a · 3 marks

B1
Initial line drawn. Correct shape, $r$ strictly decreasing.
B1
Correct shape at extremities.
B1
$1 - e^{- \frac{1}{2} π}$

6b · 5 marks

M1
$\frac{1}{2} \int_{0}^{\frac{1}{2} π} (e^{- θ} - e^{- \frac{1}{2} π})^{2} d θ$
A1
$\frac{1}{2} \int_{0}^{\frac{1}{2} π} e^{- 2 θ} - 2 e^{- θ - \frac{1}{2} π} + e^{- π} d θ$
M1
$\frac{1}{2} {[- \frac{1}{2} e^{- 2 θ} + 2 e^{- θ - \frac{1}{2} π} + e^{- π} θ]}_{0}^{\frac{1}{2} π}$
A1
$\frac{1}{2} {[- \frac{1}{2} e^{- 2 θ} + 2 e^{- θ - \frac{1}{2} π} + e^{- π} θ]}_{0}^{\frac{1}{2} π}$
A1
$\frac{1}{2} (- \frac{1}{2} e^{- π} + 2 e^{- π} + \frac{1}{2} π e^{- π} + \frac{1}{2} - 2 e^{- \frac{1}{2} π}) = \frac{3}{4} e^{- π} + \frac{1}{4} π e^{- π} - e^{- \frac{1}{2} π} + \frac{1}{4}$

6c · 5 marks

B1
$y = (e^{- θ} - e^{- \frac{1}{2} π}) \sin θ$
M1
$\frac{d y}{d θ} = (e^{- θ} - e^{- \frac{1}{2} π}) \cos θ + \sin θ (- e^{- θ}) = 0$
A1
$\frac{d y}{d θ} = (e^{- θ} - e^{- \frac{1}{2} π}) \cos θ + \sin θ (- e^{- θ}) = 0$
A1
$[θ \neq \frac{1}{2} π \Rightarrow] 1 + (\frac{- e^{- θ}}{e^{- θ} - e^{- \frac{1}{2} π}}) \tan θ = 0 \Rightarrow 1 - e^{θ - \frac{1}{2} π} - \tan θ = 0$
B1
$1 - e^{0.56 - \frac{1}{2} π} - \tan 0.56 = 0.00912$ and $1 - e^{0.57 - \frac{1}{2} π} - \tan 0.57 = - 0.00856$

7a · 3 marks

B1
$x = - 1$
M1
$y = \frac{(x + 1) (x - 1) + 1}{x + 1}$
A1
$y = x - 1$

7b · 2 marks

M1
$\frac{d y}{d x} = \frac{x^{2} + 2 x}{(x + 1)^{2}} = 0$
A1
$(0, 0), (- 2, - 4)$

7c · 3 marks

B1
Axes and asymptotes.
B1
Left branch correct.
B1
Right branch correct.

7d · 2 marks

B1
$(- 2, - \frac{1}{4})$
B1
$(- 2, - \frac{1}{4})$

7e · 6 marks

B1
Left branch correct.
B1
Right branch correct.
1 mark
$\frac{x^{2}}{x + 1} = 1$
1 mark
$\frac{x^{2}}{x + 1} = - 1$ $x^{2} - x - 1 = 0$
A1
$x = \frac{1}{2} - \frac{1}{2} \sqrt{5}$ or $x = \frac{1}{2} + \frac{1}{2} \sqrt{5}$
B1
$x < - 1, \frac{1}{2} - \frac{1}{2} \sqrt{5} < x < \frac{1}{2} + \frac{1}{2} \sqrt{5}, x \neq 0$

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