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A-Level Further Maths – Core Pure

Complex numbers, matrices, proof by induction, hyperbolic functions, further calculus.

11 questions  ·  75 marks  ·  answers included  ·  A-Level Further Maths (Core Pure 1 & 2)
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This worksheet covers:

  • Core Pure (Edexcel CP1/CP2, AQA and OCR equivalents)
  • Complex numbers and de Moivre
  • Matrices
  • Roots of polynomials
  • Proof by induction
  • Hyperbolic functions
  • Maclaurin series
  • Polar coordinates
  • Further integration
  • Volumes of revolution

A calculator may be used throughout. Allow about 90 minutes.

Questions

Section A – Core skills

Give exact answers unless otherwise stated.
1

Given that $1 + 2\text{i}$ is a root of $z^3 – 5z^2 + 11z – 15 = 0$, find the other two roots.

Show answer
$1 – 2\text{i}$ is also a root; $(z – 1 – 2\text{i})(z – 1 + 2\text{i}) = z^2 – 2z + 5$, and $z^3 – 5z^2 + 11z – 15 = (z^2 – 2z + 5)(z – 3)$. Roots: $1 – 2\text{i}$ and $3$
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2

Let $A = \begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}$.(a) Find $A^{-1}$.(b) The matrix $A$ represents a transformation $T$. Describe the effect of $T$ on the area of a shape.(c) Find the eigenvalues and corresponding eigenvectors of $M = \begin{pmatrix}4 & 1\\2 & 3\end{pmatrix}$.

Show answer
(a) $A^{-1} = \dfrac{1}{5}\begin{pmatrix}4 & -1\\-3 & 2\end{pmatrix}$(b) $\det A = 5$, so areas are multiplied by 5.(c) $\lambda^2 – 7\lambda + 10 = 0$: $\lambda = 5$, eigenvector $\begin{pmatrix}1\\1\end{pmatrix}$; $\lambda = 2$, eigenvector $\begin{pmatrix}1\\-2\end{pmatrix}$
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3

The roots of $z^3 – 2z^2 + 3z – 1 = 0$ are $\alpha$, $\beta$ and $\gamma$.(a) Write down $\alpha + \beta + \gamma$, $\alpha\beta + \beta\gamma + \gamma\alpha$ and $\alpha\beta\gamma$.(b) Find $\alpha^2 + \beta^2 + \gamma^2$.(c) Find $\alpha^3 + \beta^3 + \gamma^3$.

Show answer
(a) $2,\ 3,\ 1$(b) $2^2 – 2(3) = -2$(c) Each root satisfies $\alpha^3 = 2\alpha^2 – 3\alpha + 1$, so sum $= 2(-2) – 3(2) + 3 = -7$
[6]
4

Prove by induction that $7^n + 5$ is divisible by 6 for all positive integers $n$.

Show answer
$n = 1$: $12 = 6 \times 2$. Assume $7^k + 5 = 6m$. Then $7^{k+1} + 5 = 7(7^k + 5) – 30 = 42m – 30 = 6(7m – 5)$, divisible by 6. True for $n = 1$ and $k \Rightarrow k+1$, so true for all $n \in \mathbb{Z}^+$.
[5]

Section B – Exam-style

5

Let $z = -2 + 2\sqrt{3}\,\text{i}$.(a) Write $z$ in the form $r\text{e}^{\text{i}\theta}$, $-\pi < \theta \leq \pi$.(b) Hence find $z^4$ in the form $a + b\text{i}$.(c) Find the four roots of $w^4 = -16$, giving your answers in the form $a + b\text{i}$.

Show answer
(a) $z = 4\text{e}^{2\pi\text{i}/3}$(b) $z^4 = 256\text{e}^{8\pi\text{i}/3} = 256\text{e}^{2\pi\text{i}/3} = -128 + 128\sqrt{3}\,\text{i}$(c) $-16 = 16\text{e}^{\text{i}\pi}$, $w = 2\text{e}^{\text{i}(\pi/4 + k\pi/2)}$: $w = \pm\sqrt{2} \pm \sqrt{2}\,\text{i}$ (all four sign combinations)
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6

(a) Solve $3\cosh x – 2\sinh x = 3$, giving answers in exact logarithmic form.(b) Find $\displaystyle\int \frac{1}{\sqrt{x^2 + 4x + 13}}\,\text{d}x$.

Show answer
(a) $\tfrac{1}{2}\text{e}^x + \tfrac{5}{2}\text{e}^{-x} = 3 \Rightarrow \text{e}^{2x} – 6\text{e}^x + 5 = 0 \Rightarrow \text{e}^x = 1$ or $5$: $x = 0$ or $x = \ln 5$(b) $\int \dfrac{1}{\sqrt{(x+2)^2 + 9}}\,\text{d}x = \text{arsinh}\left(\dfrac{x + 2}{3}\right) + c$
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7

(a) Use standard results to show that $\displaystyle\sum_{r=1}^{n} (2r – 1)^2 = \frac{n(2n-1)(2n+1)}{3}$.(b) Use the method of differences to find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$ and hence state the sum to infinity.

Show answer
(a) $\sum (4r^2 – 4r + 1) = \tfrac{4n(n+1)(2n+1)}{6} – 2n(n+1) + n = \tfrac{n}{3}[2(n+1)(2n+1) – 6(n+1) + 3] = \tfrac{n}{3}(4n^2 – 1) = \tfrac{n(2n-1)(2n+1)}{3}$(b) $\dfrac{1}{r(r+1)} = \dfrac{1}{r} – \dfrac{1}{r+1}$; sum $= 1 – \dfrac{1}{n+1} = \dfrac{n}{n+1}$; sum to infinity $= 1$
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8

Let $B = \begin{pmatrix}1 & 2 & 0\\0 & 1 & 3\\2 & 0 & 1\end{pmatrix}$.(a) Show that $\det B = 13$.(b) Find $B^{-1}$.(c) Hence solve the system $x + 2y = 3$, $y + 3z = 4$, $2x + z = 5$.

Show answer
(a) $\det B = 1(1 – 0) – 2(0 – 6) + 0 = 13$(b) $B^{-1} = \dfrac{1}{13}\begin{pmatrix}1 & -2 & 6\\6 & 1 & -3\\-2 & 4 & 1\end{pmatrix}$(c) $\begin{pmatrix}x\\y\\z\end{pmatrix} = \dfrac{1}{13}\begin{pmatrix}3 – 8 + 30\\18 + 4 – 15\\-6 + 16 + 5\end{pmatrix} = \begin{pmatrix}25/13\\7/13\\15/13\end{pmatrix}$
[7]

Section C – Extended

9

(a) Find the Maclaurin series of $\ln(1 + 2x)$ up to and including the term in $x^3$, and state the range of $x$ for which it is valid.(b) Hence find an approximation for $\ln 1.2$ to 4 decimal places and compare it with the calculator value.

Show answer
(a) $2x – 2x^2 + \tfrac{8}{3}x^3$, valid for $-\tfrac{1}{2} < x \leq \tfrac{1}{2}$(b) $x = 0.1$: $0.2 – 0.02 + 0.002667 = 0.1827$; calculator $\ln 1.2 = 0.1823$
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10

The curve $C$ has polar equation $r = 2 + \cos\theta$, $0 \leq \theta < 2\pi$.(a) Find the area enclosed by $C$.(b) Find the exact values of $r$ at the points where the tangent to $C$ is parallel to the initial line, other than where $\theta = \pi$. [Hint: consider $y = r\sin\theta$.]

Show answer
(a) $\tfrac{1}{2}\int_0^{2\pi}(4 + 4\cos\theta + \cos^2\theta)\,\text{d}\theta = \tfrac{1}{2}(8\pi + \pi) = \tfrac{9\pi}{2}$(b) $y = 2\sin\theta + \sin\theta\cos\theta$; $\dfrac{\text{d}y}{\text{d}\theta} = 2\cos\theta + \cos 2\theta = 2\cos^2\theta + 2\cos\theta – 1 = 0 \Rightarrow \cos\theta = \dfrac{-1 + \sqrt{3}}{2}$, so $r = 2 + \dfrac{\sqrt{3} – 1}{2} = \dfrac{3 + \sqrt{3}}{2}$
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11

(a) The region bounded by $y = \text{e}^{-x}$, the $x$-axis and the lines $x = 0$ and $x = \ln 2$ is rotated through $2\pi$ radians about the $x$-axis. Find the exact volume of the solid formed.(b) Find the mean value of $\cosh x$ on the interval $[0, \ln 3]$, giving your answer in exact form and to 3 significant figures.(c) Evaluate $\displaystyle\int_0^1 \frac{1}{x^2 + 2x + 5}\,\text{d}x$ exactly.

Show answer
(a) $\pi\int_0^{\ln 2}\text{e}^{-2x}\,\text{d}x = \dfrac{\pi}{2}\left(1 – \tfrac{1}{4}\right) = \dfrac{3\pi}{8}$(b) $\dfrac{\sinh(\ln 3)}{\ln 3} = \dfrac{4/3}{\ln 3} = \dfrac{4}{3\ln 3} = 1.21$(c) $\left[\tfrac{1}{2}\arctan\dfrac{x+1}{2}\right]_0^1 = \dfrac{\pi}{8} – \dfrac{1}{2}\arctan\dfrac{1}{2}$
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