Advanced level 2026 technical North West regional mock mathematics 2 guide

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Mathematics Exam Marking Scheme

Section 1: Polynomials and Logarithms

1a) $\frac{P(x)}{x+1} = 2x^2 – 7x + 6$ M1M1

$\rightarrow P(x) = (x+1)(2x^2 – 7x + 6)$ M1

$= (x+1)(2x-3)(x-2)$ A0.5A0.5

b) $P(x) = 0 \rightarrow x = -1; x = \frac{3}{2}; x = 2$ M1A0.5A0.5

c) i. $2(\log x)^3 – 5(\log x)^2 – (\log x) + 6 = 0$

Comparing with $P(x) = 0$, we have:

$\log x = -1 \rightarrow x = e^{-1}$ M0.5

$\log x = \frac{3}{2} \rightarrow x = e^{\frac{3}{2}}$ A0.5A0.5A0.5

$\log x = 2 \rightarrow x = e^2$

ii. $2(e^x)^2 – 5e^x + \frac{6}{e^x} – 1 = 0$

$\rightarrow 2(e^x)^3 – 5(e^x)^2 – e^x + 6 = 0$

Comparing with $P(x) = 0$, we have:

$e^x = -1$ (impossible) M0.5

$e^x = \frac{3}{2} \rightarrow x = \ln \frac{3}{2}$ A0.5A0.5A0.5

$e^x = 2 \rightarrow x = \ln 2$

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Section 2: Calculus and Logarithmic Equations

2a) $\int \frac{3x}{x^2+1} dx = \frac{3}{2} \int \frac{2x}{x^2+1} dx = \frac{3}{2} \ln |x^2+1| + k$ M1M1A1

b) i. $3\log_3 x + \log_x 3 – 10 = 0$

$\rightarrow 3\log_3 x + \frac{\log_3 3}{\log_3 x} – 10 = 0$ M1

$\rightarrow 3(\log_3 x)^2 – 10\log_3 x + 1 = 0$ M1

$\rightarrow x = 3^{\frac{5+\sqrt{22}}{3}}; x = 3^{\frac{5-\sqrt{22}}{3}}$ A0.5A0.5

ii. $(\frac{16}{25})^{x^2-12} = (\frac{125}{64})^3$

$\rightarrow (\frac{4}{5})^{2x^2-24} = (\frac{4}{5})^{-9} \rightarrow 2x^2 – 24 = -9$

$\therefore x = \pm \sqrt{\frac{15}{2}}$ M1M1M1A0.5A0.5

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Section 3: Probability (Dresses in Boxes)

3a) i. No. of dresses in Bessem’s box:

$25 + y + 15 – y = 20$ M0.5A0.5

ii. No. of dresses in Ambe’s box:

$25 – y + y + 5 = 10$ M0.5A0.5

b) i. $P(\text{Blue from Bessem}) = \frac{15-y}{20} = \frac{10}{20} = \frac{1}{2}$ M1A1

ii. $P(\text{Blue from Ambe}) = \frac{y+5}{10} = \frac{10}{10} = 1$ M1A1

c) i. $P(\text{Black from Bessem}) = \frac{5+y}{20} = \frac{12}{20} = \frac{3}{5}$ M1A1

ii. $P(\text{Black from Ambe}) = \frac{5-y}{10} = \frac{2}{10} = \frac{1}{5}$ M1A1

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Section 4: Geometry (Circles)

4a) $x^2 + y^2 + 2ax + 2by + c = 0$

$\begin{cases} -8b + c = -16 \\ 4a + 8b + c = -20 \\ 4b + c = -4 \end{cases} \rightarrow a = -5; b = 1; c = -8$

$\therefore (\Omega): x^2 + y^2 – 10x + 2y – 8 = 0$ M1M1M1A1A1A1

b) Center: $(5, -1)$ M1A1

Radius: $r = \sqrt{34}$ units M1A1

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Section 5: Integration and Functions

5a) $f(x) = \frac{1}{2} – \frac{6}{(x+1)(x-3)} \equiv \frac{1}{2} + \frac{a}{x+1} – \frac{b}{x-3}$

$f(x) = \frac{1}{2} + \frac{a}{x+1} + \frac{b}{x-3}$

$\therefore a = 3; b = -3$ M1M1M1A0.5A0.5

b) $f'(x) = -\frac{3}{(x+1)^2} + \frac{3}{(x-3)^2}$

$= \frac{48(x-1)}{(2x^2-4x-6)^2}$ M1M1A1

c) $\int_0^2 f(x) dx = \left[ \frac{1}{2}x + 3\ln|x+1| – 3\ln|x-3| \right]_0^2$

$= 7.6$ M1M1A1

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Section 6: Systems of Equations

6a) $\begin{cases} 2\ln x – 3\ln y + 4 = 0 \\ \ln(xy) – 3 = 0 \end{cases}$

$\rightarrow \begin{cases} 2\ln x – 3\ln y = -4 \\ \ln x + \ln y = 3 \end{cases}$

$S = \{(e, e^2)\}$ M1M1A1A1

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