Exam Drill · Further Maths A*

The equation $x^{3} - 5 x^{2} + 3 x + 9 = 0$ has roots $α$ , $β$ , $γ$ .

Show that $α^{2} + β^{2} + γ^{2} = 19$ . [2]
Find the cubic equation, with integer coefficients, whose roots are $α + 2$ , $β + 2$ , $γ + 2$ . [4]

$\sum α = 5$ , $\sum α β = 3$ B1
$α^{2} + β^{2} + γ^{2} = {(\sum α)}^{2} - 2 \sum α β = 25 - 6 = 19$ M1 A1 → 2
The identity itself is the method mark. Quoting it and substituting wrongly still earns the M1.
Substitute $y = x + 2$ , so $x = y - 2$ M1
$(y - 2)^{3} - 5 (y - 2)^{2} + 3 (y - 2) + 9 = 0$ A1
$y^{3} - 6 y^{2} + 12 y - 8 - 5 y^{2} + 20 y - 20 + 3 y - 6 + 9 = 0$ A1
$y^{3} - 11 y^{2} + 35 y - 25 = 0$ A1 → 4
Check without expanding: new $\sum$ of roots $= 5 + 6 = 11$ ✓, new product $= - 9 + 2 (3) + 4 (5) + 8 = 25$ ✓.

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

Write down the equation of the vertical asymptote, and show that $C$ has an oblique asymptote $y = x + 1$ . [3]
Show that no point of $C$ has a $y$ -coordinate strictly between $2 - 2 \sqrt{3}$ and $2 + 2 \sqrt{3}$ . [5]

$x = 1$ B1
Division: $\frac{x^{2} + 2}{x - 1} = x + 1 + \frac{3}{x - 1}$ M1
$\frac{3}{x - 1} \to 0$ as $x \to \pm \infty$ , so $y = x + 1$ is an asymptote A1 → 3
$y (x - 1) = x^{2} + 2 \Rightarrow x^{2} - y x + (y + 2) = 0$ M1 A1
Real $x$ requires $Δ ⩾ 0$ : $y^{2} - 4 (y + 2) ⩾ 0$ M1
$y^{2} - 4 y - 8 ⩾ 0 \Rightarrow y = \frac{4 \pm \sqrt{48}}{2} = 2 \pm 2 \sqrt{3}$ A1
Hence $y ⩽ 2 - 2 \sqrt{3}$ or $y ⩾ 2 + 2 \sqrt{3}$ A1 → 5
Trap: writing $Δ > 0$ loses the endpoints, which are attained at $x = 1 \pm \sqrt{3}$ .

Express $\frac{1}{r (r + 2)}$ in partial fractions. [2]
Hence show that $\sum_{r = 1}^{n} \frac{1}{r (r + 2)} = \frac{3}{4} - \frac{2 n + 3}{2 (n + 1) (n + 2)}$ . [5]
Deduce the sum to infinity. [1]

$\frac{1}{r (r + 2)} = \frac{1}{2} (\frac{1}{r} - \frac{1}{r + 2})$ M1 A1 → 2
$\frac{1}{2} \sum_{r = 1}^{n} (\frac{1}{r} - \frac{1}{r + 2})$ , listing enough terms to show the cancellation M1
$= \frac{1}{2} (1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2})$ A1 A1
$= \frac{3}{4} - \frac{1}{2} \cdot \frac{(n + 2) + (n + 1)}{(n + 1) (n + 2)}$ M1
$= \frac{3}{4} - \frac{2 n + 3}{2 (n + 1) (n + 2)}$ A1 → 5
Two terms survive at each end because the gap is 2. Losing one of them is the classic slip.
$\frac{3}{4}$ B1 → 1

$𝐌 = (\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array})$ .

Find $\det 𝐌$ and $𝐌^{- 1}$ . [3]
Find the equations of the two lines through the origin that are invariant under $𝐌$ . [4]
The unit square is transformed by $𝐌$ . State the area of its image. [2]

$\det 𝐌 = 4 - 1 = 3$ B1
$𝐌^{- 1} = \frac{1}{3} (\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array})$ M1 A1 → 3
Take $y = m x$ ; the image of $(x, m x)$ is $(2 x + m x, x + 2 m x)$ M1
Invariance needs $x + 2 m x = m (2 x + m x)$ M1
$1 + 2 m = 2 m + m^{2} \Rightarrow m^{2} = 1$ A1
$y = x$ and $y = - x$ A1 → 4
An invariant line maps onto itself; the points on it need not stay put.
Area $= | \det 𝐌 | \times 1 = 3$ M1 A1 → 2

The curve $C$ has polar equation $r = 1 + \cos θ$ , for $0 ⩽ θ < 2 π$ .

Sketch $C$ . [2]
Find the exact area of the region enclosed by $C$ . [5]
Find the greatest distance of a point of $C$ from the initial line, giving an exact answer. [3]

Cardioid: $r = 2$ at $θ = 0$ , $r = 0$ at $θ = π$ , symmetric in the initial line B1 B1 → 2
$A = \frac{1}{2} \int_{0}^{2 π} (1 + \cos θ)^{2} d θ$ M1
$= \frac{1}{2} \int_{0}^{2 π} (1 + 2 \cos θ + \cos^{2} θ) d θ$ A1
$\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ M1
$= \frac{1}{2} (2 π + 0 + π)$ A1
$A = \frac{3 π}{2}$ A1 → 5
Distance from the initial line $= r \sin θ = (1 + \cos θ) \sin θ$ M1
Derivative $\cos θ + \cos 2 θ = 2 \cos^{2} θ + \cos θ - 1 = (2 \cos θ - 1) (\cos θ + 1) = 0 \Rightarrow \cos θ = \frac{1}{2}$ A1
$θ = \frac{π}{3}$ : distance $= \frac{3}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}$ A1 → 3
Maximising $r$ is not the same as maximising the distance from the initial line — $r$ is largest at $θ = 0$ , where that distance is zero.

The lines $l_{1}$ and $l_{2}$ have vector equations $𝐫 = 𝐢 + 2 𝐣 + 3 𝐤 + t (2 𝐢 - 𝐣 + 2 𝐤)$ and $𝐫 = 2 𝐢 - 𝐣 + 𝐤 + s (𝐢 + 3 𝐣 - 𝐤)$ .

Show that $l_{1}$ and $l_{2}$ are skew. [4]
Find the exact shortest distance between them. [4]
Find the acute angle between them, correct to the nearest $0 . 1^{\circ}$ . [3]

Directions $(2, - 1,2)$ and $(1,3, - 1)$ are not multiples of one another, so not parallel B1
Equating components gives $1 + 2 t = 2 + s$ , $2 - t = - 1 + 3 s$ , $3 + 2 t = 1 - s$ M1
First two give $t = \frac{4}{7}$ ... any correct pair solved A1
Values fail the third equation, so the lines do not meet — not parallel and not intersecting, hence skew A1 → 4
$𝐧 = (2, - 1,2) \times (1,3, - 1) = (- 5,4,7)$ M1 A1
$𝐚_{2} - 𝐚_{1} = (1, - 3, - 2)$ ; distance $= \frac{| (1, - 3, - 2) \cdot (- 5,4,7) |}{| (- 5,4,7) |}$ M1
$= \frac{| - 5 - 12 - 14 |}{\sqrt{90}} = \frac{31}{3 \sqrt{10}}$ A1 → 4
$\approx 3.27$ . Forgetting the modulus and reporting a negative distance is a needless A-mark loss.
$\cos ϕ = \frac{| (2, - 1,2) \cdot (1,3, - 1) |}{3 \sqrt{11}}$ M1
$= \frac{| 2 - 3 - 2 |}{3 \sqrt{11}} = \frac{1}{\sqrt{11}}$ A1
$ϕ = 72 . 5^{\circ}$ A1 → 3
Without the modulus you get $107 . 5^{\circ}$ ; the question asked for the acute angle.

Prove by induction that $\sum_{r = 1}^{n} r 2^{r} = (n - 1) 2^{n + 1} + 2$ for all integers $n ⩾ 1$ . [7]
The matrix $𝐀 = (\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array})$ . Prove by induction that $𝐀^{n} = (\begin{array}{cc} 1 & 2^{n} - 1 \\ 0 & 2^{n} \end{array})$ for all integers $n ⩾ 1$ . [5]

Base: $n = 1$ : LHS $= 2$ , RHS $= 0 \cdot 4 + 2 = 2$ ✓ B1
Assume true for $n = k$ : $\sum_{r = 1}^{k} r 2^{r} = (k - 1) 2^{k + 1} + 2$ M1
$\sum_{r = 1}^{k + 1} r 2^{r} = (k - 1) 2^{k + 1} + 2 + (k + 1) 2^{k + 1}$ M1
$= 2^{k + 1} [(k - 1) + (k + 1)] + 2 = 2 k \cdot 2^{k + 1} + 2$ A1
$= k \cdot 2^{k + 2} + 2$ A1
which is the result with $n = k + 1$ A1
Conclusion: true for $n = 1$ , and truth for $k$ implies truth for $k + 1$ , so true for all integers $n ⩾ 1$ A1 → 7
The final mark is for the sentence, and it is the one most often thrown away. It must mention the base case, the implication, and “for all $n ⩾ 1$ ”.
Base: $n = 1$ gives $(\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array})$ ✓ B1
Assume $𝐀^{k} = (\begin{array}{cc} 1 & 2^{k} - 1 \\ 0 & 2^{k} \end{array})$ ; then $𝐀^{k + 1} = 𝐀^{k} 𝐀$ M1
$= (\begin{array}{cc} 1 & 1 + 2 (2^{k} - 1) \\ 0 & 2^{k + 1} \end{array})$ A1
$= (\begin{array}{cc} 1 & 2^{k + 1} - 1 \\ 0 & 2^{k + 1} \end{array})$ A1
Conclusion stated properly A1 → 5

The equation $x^{4} - 2 x^{3} + 5 x - 1 = 0$ has roots $α$ , $β$ , $γ$ , $δ$ .

Find $\sum α^{2}$ . [3]
Find $\sum \frac{1}{α}$ . [3]
Find the quartic equation, with integer coefficients, whose roots are $\frac{1}{α}$ , $\frac{1}{β}$ , $\frac{1}{γ}$ , $\frac{1}{δ}$ . [5]

$\sum α = 2$ , $\sum α β = 0$ , $\sum α β γ = - 5$ , $α β γ δ = - 1$ B1
$\sum α^{2} = {(\sum α)}^{2} - 2 \sum α β$ M1
$= 4 - 0 = 4$ A1 → 3
The missing $x^{2}$ term means $\sum α β = 0$ , not that the term is absent from the identity.
$\sum \frac{1}{α} = \frac{\sum α β γ}{α β γ δ}$ M1 A1
$= \frac{- 5}{- 1} = 5$ A1 → 3
Put $y = \frac{1}{x}$ , so $x = \frac{1}{y}$ M1
$\frac{1}{y^{4}} - \frac{2}{y^{3}} + \frac{5}{y} - 1 = 0$ A1
Multiply by $- y^{4}$ : M1
$y^{4} - 5 y^{3} + 2 y - 1 = 0$ A1 A1 → 5
Check: the new sum of roots is $5$ , agreeing with part (b). Reversing the coefficient list is the same operation, but only when you keep the signs straight.