# Question 1

## (a) [3 marks]
M1: Multiply both sides by $(5-x)(1+6x)$
M1: Compare coefficients or substitute values for $x$
A1: $A = 2$, $B = 3$

## (b) [6 marks]
M1: Integrate $\frac{2}{5-x}$ to get $-2\ln(5-x)$
M1: Integrate $\frac{3}{1+6x}$ to get $\frac{1}{2}\ln(1+6x)$
M1: Apply limits $0$ to $4$
M1: Evaluate at $x=4$: $-2\ln(1) + \frac{1}{2}\ln(25)$
M1: Evaluate at $x=0$: $-2\ln(5) + \frac{1}{2}\ln(1)$
A1: $k = -\frac{3}{2}$ (or equivalent rational form)

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# Question 2

## (a) [3 marks]
M1: Use $R = \sqrt{4+25} = \sqrt{29}$
M1: Find $a$ where $\tan(a) = \frac{5}{2}$
A1: $R = \sqrt{29}$, $a = 1.19$ radians (to 3 s.f.)

## (b)(i) [2 marks]
M1: Maximum occurs when $\cos(x+a) = 1$, so $x + a = 0$ (or $2\pi$, etc.)
A1: $x = 2\pi - 1.19 = 5.09$ (to 3 s.f.)

## (b)(ii) [3 marks]
M1: Solve $\sqrt{29}\cos(x+a) = -1$
M1: $\cos(x+a) = -\frac{1}{\sqrt{29}}$
M1: Find both values of $x+a$ in range and subtract $a$
A1: $x = 1.77$ and $x = 4.51$ (to 3 s.f.)

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# Question 3

## (a) [2 marks]
M1: Apply Remainder Theorem: $f(-\frac{1}{2}) = -2$
A1: $d = 3$

## (b)(i) [3 marks]
M1: Perform polynomial division by $(2x+1)$
M1: Factor the quotient as $(4x^2 - 8x + 3)$ or equivalent
A1: $g(x) = (2x+1)(2x-1)(2x-3)$

## (b)(ii) [4 marks]
M1: Factorize numerator: $4x^2 - 1 = (2x-1)(2x+1)$
M1: Cancel to get $h(x) = \frac{2x-1}{2x-3}$
M1: Differentiate: $h'(x) = \frac{2(2x-3) - (2x-1)(2)}{(2x-3)^2} = \frac{-4}{(2x-3)^2}$
A1: Since $h'(x) < 0$ for $x > 2$, $h$ is decreasing

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# Question 4

## (a) [2 marks]
M1: Apply binomial expansion to $(1+5x)^{1/5}$
A1: $1 + x - \frac{10x^2}{2}$ (or $1 + x - 5x^2$)

## (b)(i) [3 marks]
M1: Rewrite as $8(1 + \frac{3x}{8})^{-2/3}$
M1: Apply binomial expansion with $n = -\frac{2}{3}$
A1: $1 - \frac{x}{4} - \frac{5x^2}{128}$ (or equivalent)

## (b)(ii) [2 marks]
M1: Set $3x = \frac{1}{27}$ to get $x = \frac{1}{81}$, substitute into expansion
A1: $0.3158$ (to 4 d.p.)

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# Question 5

## (a) [3 marks]
M1: Find $\frac{dx}{dt} = -2\sin(2t)$ and $\frac{dy}{dt} = \cos(t)$
M1: At $t = \frac{\pi}{6}$: $\frac{dx}{dt} = -\sqrt{3}$, $\frac{dy}{dt} = \frac{\sqrt{3}}{2}$
A1: Gradient $= -\frac{1}{2}$

## (b) [3 marks]
M1: Gradient of normal is $2$
M1: Find coordinates of P: $(\frac{3}{4}, \frac{1}{2})$
A1: $y = 2x - 1$ (or $2x - y - 1 = 0$)

## (c) [5 marks]
M1: Substitute $y = 2x - 1$ into $y = \sin(q)$ and $x = \cos(2q)$
M1: Use $\sin(q) = 2\cos(2q) - 1$
M1: Substitute $\cos(2q) = 1 - 2\sin^2(q)$ to form quadratic
M1: Rearrange to $4\sin^2(q) + \sin(q) - 3 = 0$
M1: Solve to get $\sin(q) = -1$ (giving $x = -1$) or $\sin(q) = \frac{3}{4}$
A1: $x = -1$ or $x = -\frac{7}{16}$

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# Question 6

## (a) [4 marks]
M1: Find direction vector of AB: $(2, -4, -6)$
M1: Find direction vector of $l$: $(3, 1, -2)$
M1: Use dot product formula: $\cos(\theta) = \frac{|6-4+12|}{\sqrt{56}\sqrt{14}}$
A1: $\theta = 22.2°$ (or $0.387$ radians, to 3 s.f.)

## (b) [4 marks]
M1: Let $C = (3+3\lambda, 2+\lambda, 10-2\lambda)$
M1: Find $\vec{BC} = (-2+3\lambda, 4+\lambda, 6-2\lambda)$
M1: Condition $\vec{BC} \cdot \vec{AB} = 0$: $-4+6\lambda - 4 - \lambda + (-12) + 4\lambda = 0$
M1: Solve: $9\lambda = 20$, $\lambda = \frac{20}{9}$
A1: $C = (\frac{87}{9}, \frac{38}{9}, \frac{50}{9})$ or $(\frac{29}{3}, \frac{38}{9}, \frac{50}{9})$

## (c) [4 marks]
M1: Find $\vec{AC} = (\frac{60}{9}, \frac{20}{9}, -\frac{40}{9})$; $|AC| = \frac{20\sqrt{13}}{9}$
M1: $\vec{BD}$ parallel to $\vec{AC}$ means $D = B + k\vec{AC}$ for some $k$
M1: Condition $\vec{CD} \cdot \vec{BD} = 0$ and $|DE| = \frac{1}{2}|AC|$
M1: E lies on line through B and D with $