Advanced level 2026 North west regional mock pure mathematics 2 guide

Question 1

(i) $3x^2 – 4x + 2 = 0$; roots: $\alpha$ and $\beta$

(a) Sum of roots $= -\frac{b}{a} \Rightarrow \alpha + \beta = \frac{4}{3}$

Product of roots $= \frac{c}{a} \Rightarrow \alpha\beta = \frac{2}{3}$

(b) New roots: $\frac{1}{\alpha^3}$ and $\frac{1}{\beta^3}$

$SOR = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{\alpha^3\beta^3} = \frac{-8/27}{(2/3)^3} = -1$

$POR = \frac{1}{\alpha^3} \times \frac{1}{\beta^3} = \frac{1}{\alpha^3\beta^3} = 1 \div \left(\frac{2}{3}\right)^3 = \frac{27}{8}$

Eqn: $x^2 – (SOR)x + POR = 0 \Rightarrow x^2 – (-1)x + \frac{27}{8} = 0$

$\Rightarrow$ New equation $8x^2 + 8x + 27 = 0$

(ii) $\left| \frac{x+3}{x-5} \right| \Rightarrow \frac{x+3}{x-5} \Rightarrow \frac{x+3}{x-5} > 0; CV = -3, 5$

(Image: Number line showing critical values at -3 and 5. The region $x < -3$ is positive, $-3 < x < 5$ is negative, and $x > 5$ is positive.)

Range of values: $(x < -3) \cup (x > 5)$

Question 2

(i) $y = x^2e^{2x}$

(a) At stationary points, $\frac{dy}{dx} = 0$

$\frac{dy}{dx} = x^2(2e^{2x}) + (2x)e^{2x} \Rightarrow \frac{dy}{dx} = 2xe^{2x}(x + 1)$

$\frac{dy}{dx} = 0 \Rightarrow \begin{cases} 2xe^{2x} = 0 \Rightarrow x = 0 \\ x + 1 = 0 \Rightarrow x = -1 \end{cases}$

The nature of the stationary points are determined using the second derivative, $\frac{d^2y}{dx^2}$

$\frac{dy}{dx} = 2e^{2x}(x^2 + x) \Rightarrow \frac{d^2y}{dx^2} = 2e^{2x}(2x + 1) + 4e^{2x}(x^2 + x)$

$\frac{d^2y}{dx^2} = 2e^{2x}(2x + 1 + 2x^2 + 2x) = 2e^{2x}(2x^2 + 4x + 1)$

$\frac{d^2y}{dx^2} \Big|_{x=0} = 2e^0(0 + 0 + 1) = 2 > 0 \Rightarrow x = 0$ is a minimum stationary point.

$\frac{d^2y}{dx^2} \Big|_{x=-1} = 2e^{-1}(2 – 4 + 1) = -0.7 < 0 \Rightarrow x = -1$ is a max stationary point.

Coordinates of stationary points:

At $x = -1, y = (-1)^2e^{2(-1)} = e^{-2} \Rightarrow (-1, e^{-2})$

(b) At point of inflexion, $\frac{d^2y}{dx^2} = 0 \Rightarrow 2e^{2x}(2x^2 + 4x + 1) = 0$

$\Rightarrow 2e^{2x} = 0$ or $2x^2 + 4x + 1 = 0$

(ii) $(a+b)^n = \sum_{r=0}^{n} {}^n C_r \cdot a^{n-r} \cdot b^r; n \in \mathbb{N}$

$\Rightarrow \left(2x – \frac{1}{x^2}\right)^9 = \sum_{r=0}^{9} {}^9 C_r \cdot (2x)^{9-r} \cdot \left(-\frac{1}{x^2}\right)^r$

Term independent of $x$, $x^0, k \Rightarrow {}^9 C_r \cdot (2x)^{9-r} \cdot \left(-\frac{1}{x^2}\right)^r = kx^0$

$\Rightarrow 9 – r – 2r = 0 \Rightarrow 9 = 3r \Rightarrow r = 3$

$\Rightarrow k = ({}^9 C_3)(2)^6(-1)^3 = (84)(64)(-1) \Rightarrow k = -5376$

Question 3

a) $\sin 2\theta + \cos 2\theta + 1 = 0 \equiv R\cos(2\theta – x) + 1 = 0$

$R\cos(2\theta – x) = R(\cos 2\theta \cos x + \sin 2\theta \sin x)$; hence

$\sin 2\theta + \cos 2\theta = R\cos(2\theta – x) \Rightarrow \begin{cases} R\cos x = 1 \dots (1) \\ R\sin x = 1 \dots (2) \end{cases}$

$R^2 = 2 \Rightarrow R = \sqrt{2}$; $\tan x = 1 \Rightarrow x = 45^\circ$

$\Rightarrow \sin 2\theta + \cos 2\theta = \sqrt{2}\cos(2\theta – \frac{\pi}{4})$

$\sqrt{2}\cos(2\theta – \frac{\pi}{4}) + 1 = 0 \Rightarrow \cos(2\theta – \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$

$2\theta – \frac{\pi}{4} = 2n\pi \pm \frac{3\pi}{4}; n \in \mathbb{N}$

Solution to the equation are: $\left\{ \frac{3\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{4} \right\}$

b) $\frac{\cos A}{\sec A – \tan A} = \frac{\cos A}{\frac{1}{\cos A} – \frac{\sin A}{\cos A}} = \frac{\cos A}{\frac{1 – \sin A}{\cos A}} = \frac{\cos^2 A}{1 – \sin A} = \frac{1 – \sin^2 A}{1 – \sin A} = 1 + \sin A$

c) Following the steps in a), $5\cos x – 12\sin x = 13\cos(x + 67.38^\circ)$