2023 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

We can solve the inequality by squaring both sides:
\((4x - 3)^2 \le (x + 6)^2\)
\(16x^2 - 24x + 9 \le x^2 + 12x + 36\)
\(15x^2 - 36x - 27 \le 0\)

Divide the entire inequality by 3:
\(5x^2 - 12x - 9 \le 0\)

Factorise the quadratic expression:
\((5x + 3)(x - 3) \le 0\)

Thus, the critical values are \(x = -\frac{3}{5}\) and \(x = 3\).
Since the inequality is \(\le 0\), the solution lies between the critical values:
\(-\frac{3}{5} \le x \le 3\).

M1 for squaring both sides or setting up two linear equations/inequalities: \(4x - 3 = \pm(x + 6)\)
M1 for obtaining a three-term quadratic: \(15x^2 - 36x - 27 \le 0\) (or equivalent)
M1 for attempting to solve their quadratic to find critical values
A1 for critical values \(-\frac{3}{5}\) (or \(-0.6\)) and \(3\)
A1 for the correct final inequality: \(-\frac{3}{5} \le x \le 3\) (or \(-0.6 \le x \le 3\))

Using the laws of logarithms:
\(\log_3(x-2)^2 - \log_3(x+4) = 1\)
\(\log_3\left(\frac{(x-2)^2}{x+4}\right) = 1\)

Convert the logarithmic equation to exponential form:
\(\frac{(x-2)^2}{x+4} = 3^1\)
\((x-2)^2 = 3(x+4)\)
\(x^2 - 4x + 4 = 3x + 12\)

Rearrange into a standard quadratic form:
\(x^2 - 7x - 8 = 0\)

Factorise the quadratic equation:
\((x-8)(x+1) = 0\)
This gives \(x = 8\) or \(x = -1\).

Since the logarithm is only defined for positive arguments, we require \(x-2 > 0\) and \(x+4 > 0\), which means \(x > 2\).
Therefore, the only valid solution is \(x = 8\).

B1 for using the power law: \(\log_3(x-2)^2\)
M1 for using the subtraction law to write as a single logarithm: \(\log_3\left(\frac{(x-2)^2}{x+4}\right)\)
M1 for removing the logarithm to form a quadratic equation: \((x-2)^2 = 3(x+4)\)
A1 for getting the correct quadratic equation: \(x^2 - 7x - 8 = 0\)
A1 for solving and selecting the correct solution \(x=8\) only and rejecting \(x=-1\)

We use the identity \(\cos^2\theta = 1 - \sin^2\theta\):
\(3(1 - \sin^2\theta) - 5\sin\theta - 1 = 0\)
\(3 - 3\sin^2\theta - 5\sin\theta - 1 = 0\)
\(3\sin^2\theta + 5\sin\theta - 2 = 0\)

Factorise the quadratic in \(\sin\theta\):
\((3\sin\theta - 1)(\sin\theta + 2) = 0\)

This gives:
\(\sin\theta = \frac{1}{3}\) or \(\sin\theta = -2\)

Since \(-1 \le \sin\theta \le 1\), the equation \(\sin\theta = -2\) has no solution.

For \(\sin\theta = \frac{1}{3}\) in the interval \(0 \le \theta \le 2\pi\):
Reference angle: \(\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 0.3398\) radians.
In Quadrant I: \(\theta \approx 0.340\) (to 3 s.f.)
In Quadrant II: \(\theta = \pi - 0.3398 \approx 2.80\) (to 3 s.f.)

Therefore, the solutions are \(\theta = 0.340\) and \(\theta = 2.80\).

M1 for using the identity \(\cos^2\theta = 1 - \sin^2\theta\) to obtain an equation in terms of \(\sin\theta\) only
A1 for the correct quadratic equation: \(3\sin^2\theta + 5\sin\theta - 2 = 0\)
M1 for solving their quadratic equation to find \(\sin\theta = \frac{1}{3}\) (and showing \(\sin\theta = -2\) is rejected)
A1 for finding one correct angle: \(\theta = 0.340\) (accept 0.34)
A1 for finding the second correct angle: \(\theta = 2.80\) (accept 2.8), and no extra solutions in range

Using the Factor Theorem, since \(x - 2\) is a factor of \(p(x)\), we have \(p(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)
\(16 + 4a + 2b - 6 = 0\)
\(4a + 2b = -10\)
\(2a + b = -5\) --- (1)

Using the Remainder Theorem, since dividing \(p(x)\) by \(x + 1\) gives a remainder of \(-15\), we have \(p(-1) = -15\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15\)
\(-2 + a - b - 6 = -15\)
\(a - b - 8 = -15\)
\(a - b = -7\) --- (2)

Now we solve equations (1) and (2) simultaneously. Adding the two equations:
\((2a + b) + (a - b) = -5 + (-7)\)
\(3a = -12\)
\(a = -4\)

Substitute \(a = -4\) into equation (2):
\(-4 - b = -7\)
\(b = 3\)

Therefore, \(a = -4\) and \(b = 3\).

M1 for applying the Factor Theorem: \(p(2) = 0\) to get \(4a + 2b = -10\) (or equivalent)
M1 for applying the Remainder Theorem: \(p(-1) = -15\) to get \(a - b = -7\) (or equivalent)
M1 for attempting to solve the two simultaneous equations in \(a\) and \(b\)
A1 for finding \(a = -4\)
A1 for finding \(b = 3\)

First, find the y-coordinate of the point where \(x = 1\):
\(y = \frac{4(1) - 1}{1^2 + 2} = \frac{3}{3} = 1\)
So the point of contact is \((1, 1)\).

Next, differentiate \(y\) with respect to \(x\) using the quotient rule:
Let \(u = 4x - 1 \implies \frac{du}{dx} = 4\)
Let \(v = x^2 + 2 \implies \frac{dv}{dx} = 2x\)
\(\frac{dy}{dx} = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2}\)
\(\frac{dy}{dx} = \frac{4(x^2 + 2) - (4x - 1)(2x)}{(x^2 + 2)^2}\)
\(\frac{dy}{dx} = \frac{4x^2 + 8 - 8x^2 + 2x}{(x^2 + 2)^2} = \frac{-4x^2 + 2x + 8}{(x^2 + 2)^2}\)

Evaluate \(\frac{dy}{dx}\) at \(x = 1\) to find the gradient of the tangent, \(m\):
\(m = \frac{-4(1)^2 + 2(1) + 8}{(1^2 + 2)^2} = \frac{6}{9} = \frac{2}{3}\)

The equation of the tangent at \((1, 1)\) with gradient \(\frac{2}{3}\) is:
\(y - 1 = \frac{2}{3}(x - 1)\)
\(y = \frac{2}{3}x - \frac{2}{3} + 1\)
\(y = \frac{2}{3}x + \frac{1}{3}\)

B1 for finding the correct y-coordinate \(y = 1\)
M1 for attempting to use the quotient rule to differentiate (with at least one term in the numerator correct)
A1 for the correct derivative: \(\frac{dy}{dx} = \frac{-4x^2 + 2x + 8}{(x^2 + 2)^2}\) (or equivalent unsimplified expression)
M1 for substituting \(x = 1\) into their derivative to find the gradient of the tangent
A1 for the correct equation of the tangent: \(y = \frac{2}{3}x + \frac{1}{3}\) (or equivalent fraction forms)

(a) Given \( y = e^{2x} - 8e^x + 5x \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = 2e^{2x} - 8e^x + 5 \)
For stationary points, set \( \frac{dy}{dx} = 0 \):
\( 2(e^x)^2 - 8e^x + 5 = 0 \)
Let \( u = e^x \):
\( 2u^2 - 8u + 5 = 0 \)
Using the quadratic formula:
\( u = \frac{8 \pm \sqrt{(-8)^2 - 4(2)(5)}}{4} = \frac{8 \pm \sqrt{24}}{4} = 2 \pm \frac{\sqrt{6}}{2} \)
Since both values are positive, we have:
\( x = \ln\left(2 + \frac{\sqrt{6}}{2}\right) \) and \( x = \ln\left(2 - \frac{\sqrt{6}}{2}\right) \).

(b) Differentiating a second time:
\( \frac{d^2y}{dx^2} = 4e^{2x} - 8e^x = 4e^x(e^x - 2) \)
For \( x = \ln\left(2 + \frac{\sqrt{6}}{2}\right) \), we have \( e^x = 2 + \frac{\sqrt{6}}{2} \approx 3.22 > 2 \).
Thus, \( \frac{d^2y}{dx^2} = 4\left(2 + \frac{\sqrt{6}}{2}\right)\left(\frac{\sqrt{6}}{2}\right) > 0 \), so this point is a minimum.
For \( x = \ln\left(2 - \frac{\sqrt{6}}{2}\right) \), we have \( e^x = 2 - \frac{\sqrt{6}}{2} \approx 0.78 < 2 \).
Thus, \( \frac{d^2y}{dx^2} = 4\left(2 - \frac{\sqrt{6}}{2}\right)\left(-\frac{\sqrt{6}}{2}\right) < 0 \), so this point is a maximum.

(c) Using the chain rule:
\( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \)
At \( x = 0 \), \( \frac{dy}{dx} = 2e^0 - 8e^0 + 5 = -1 \).
Given \( \frac{dx}{dt} = 0.2 \):
\( \frac{dy}{dt} = -1 \times 0.2 = -0.2 \) units per second.

(a)
M1: For a correct attempt to differentiate \( y \).
A1: For \( \frac{dy}{dx} = 2e^{2x} - 8e^x + 5 \).
M1: For setting \( \frac{dy}{dx} = 0 \) and attempting to solve the quadratic equation in \( e^x \).
A1: For finding \( e^x = 2 \pm \frac{\sqrt{6}}{2} \) (or decimal equivalent \( 3.22 \) and \( 0.78 \)).
A1: For both exact answers: \( x = \ln\left(2 \pm \frac{\sqrt{6}}{2}\right) \).

(b)
M1: For attempting to find \( \frac{d^2y}{dx^2} \).
A1: For \( \frac{d^2y}{dx^2} = 4e^{2x} - 8e^x \).
A1: For correctly substituting the values and determining that \( x = \ln\left(2 + \frac{\sqrt{6}}{2}\right) \) is a minimum and \( x = \ln\left(2 - \frac{\sqrt{6}}{2}\right) \) is a maximum.

(c)
M1: For using the chain rule \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \).
A1: For finding \( \frac{dy}{dx} = -1 \) at \( x = 0 \).
A1: For \( -0.2 \).

(a) The 3rd, 7th, and 15th terms of the arithmetic progression are:
\( T_3 = a + 2d \)
\( T_7 = a + 6d \)
\( T_{15} = a + 14d \)
Since these are consecutive terms of a geometric progression:
\( \frac{a+6d}{a+2d} = \frac{a+14d}{a+6d} \)
\( (a+6d)^2 = (a+2d)(a+14d) \)
\( a^2 + 12ad + 36d^2 = a^2 + 16ad + 28d^2 \)
\( 8d^2 = 4ad \)
Since \( d \neq 0 \) for a non-constant AP, we divide by \( 4d \):
\( 2d = a \)
Substituting \( a = 2d \) back into the terms:
\( T_3 = 2d + 2d = 4d \)
\( T_7 = 2d + 6d = 8d \)
\( T_{15} = 2d + 14d = 16d \)
So the common ratio \( r \) of the GP is:
\( r = \frac{8d}{4d} = 2 \).

(b) The sum of the first 10 terms of the AP is:
\( S_{10} = \frac{10}{2}(2a + 9d) = 325 \)
\( 5(2a + 9d) = 325 \Rightarrow 2a + 9d = 65 \)
Since \( a = 2d \):
\( 2(2d) + 9d = 65 \Rightarrow 13d = 65 \Rightarrow d = 5 \)
And \( a = 2(5) = 10 \).

(c) The first term of the GP is \( T_3 = 4d = 4(5) = 20 \).
The common ratio is \( r = 2 \).
The sum of the first 8 terms of the GP is:
\( S_8 = \frac{20(2^8 - 1)}{2 - 1} = 20(256 - 1) = 20 \times 255 = 5100 \).

(a)
M1: For writing the three terms of the AP in terms of \( a \) and \( d \).
M1: For setting up the geometric progression ratio equation: \( (a+6d)^2 = (a+2d)(a+14d) \).
A1: For simplifying the equation to obtain \( a = 2d \).
A1: For substituting back to show that \( r = 2 \).

(b)
M1: For using the AP sum formula: \( \frac{10}{2}(2a+9d) = 325 \).
A1: For the equation \( 2a + 9d = 65 \) or equivalent.
M1: For substituting \( a = 2d \) into their sum equation and attempting to solve for \( d \).
A1: For \( d = 5 \) and \( a = 10 \).

(c)
M1: For finding the first term of the GP: \( a_{GP} = 20 \).
M1: For applying the GP sum formula with \( n = 8 \) and \( r = 2 \).
A1: For \( 5100 \).

(a) Expressing the LHS with a common denominator:
LHS = \( \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)} \)
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
LHS = \( \frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2\operatorname{cosec} \theta \) (Proved).

(b) Using the identity in (a) with \( \theta = 2x \):
\( 2\operatorname{cosec} 2x = 4 \Rightarrow \operatorname{cosec} 2x = 2 \Rightarrow \sin 2x = \frac{1}{2} \)
Since \( 0 \le x \le \pi \), we have \( 0 \le 2x \le 2\pi \).
\( 2x = \frac{\pi}{6} \) or \( 2x = \frac{5\pi}{6} \)
\( x = \frac{\pi}{12} \) or \( x = \frac{5\pi}{12} \).

(c) Factoring the equation:
\( \cos y (3\cos y - 4\sin y) = 0 \)
Either \( \cos y = 0 \Rightarrow y = 90^\circ \)
Or \( 3\cos y - 4\sin y = 0 \Rightarrow \tan y = \frac{3}{4} \Rightarrow y \approx 36.9^\circ \).

(a)
M1: For combining the fractions with a common denominator.
A1: For correctly expanding the numerator to \( \sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta \).
M1: For using \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify the numerator to \( 2(1 + \cos \theta) \).
A1: For simplifying and obtaining \( 2\operatorname{cosec} \theta \) clearly.

(b)
M1: For using the identity to rewrite the equation as \( 2\operatorname{cosec} 2x = 4 \).
A1: For obtaining \( \sin 2x = \frac{1}{2} \).
M1: For solving \( 2x = \frac{\pi}{6}, \frac{5\pi}{6} \) (or finding one correct value of \( x \)).
A1: For both exact answers: \( x = \frac{\pi}{12}, \frac{5\pi}{12} \).

(c)
M1: For factoring the expression to obtain \( \cos y(3\cos y - 4\sin y) = 0 \).
A1: For \( y = 90^\circ \).
A1: For \( y = 36.9^\circ \) (accept \( 36.9 \) to \( 37 \)).

(a) From the first equation:
\( \log_2(3x - y) = 3 \Rightarrow 3x - y = 2^3 = 8 \Rightarrow y = 3x - 8 \) --- (1)
From the second equation:
\( 4^{x-1} \times 2^{y+1} = 8^{x-y} \)
Expressing all bases as 2:
\( 2^{2(x-1)} \times 2^{y+1} = 2^{3(x-y)} \)
\( 2x - 2 + y + 1 = 3x - 3y \)
\( 2x + y - 1 = 3x - 3y \Rightarrow x - 4y = -1 \) --- (2)
Substitute (1) into (2):
\( x - 4(3x - 8) = -1 \)
\( x - 12x + 32 = -1 \)
\( -11x = -33 \Rightarrow x = 3 \)
Substitute \( x = 3 \) back into (1):
\( y = 3(3) - 8 = 1 \).
Check domain: \( 3x - y = 9 - 1 = 8 > 0 \), which is valid.

(b) Using the laws of logarithms:
\( \log_5((w + 3)(2w - 1)) = \log_5(3^2) \)
\( (w + 3)(2w - 1) = 9 \)
\( 2w^2 + 5w - 3 = 9 \)
\( 2w^2 + 5w - 12 = 0 \)
\( (2w - 3)(w + 4) = 0 \)
This gives \( w = 1.5 \) or \( w = -4 \).
We must check the validity of the roots in the original logarithmic terms:
If \( w = 1.5 \), both \( w+3 = 4.5 > 0 \) and \( 2w-1 = 2 > 0 \) are valid.
If \( w = -4 \), \( w+3 = -1 < 0 \), which is undefined.
Therefore, the only solution is \( w = 1.5 \).

(a)
M1: For removing logarithms in the first equation to obtain \( 3x - y = 8 \).
M1: For expressing the powers in the second equation with base 2.
A1: For the linear equation \( 2x + y - 1 = 3x - 3y \) or equivalent.
M1: For a correct method to solve the simultaneous equations.
A1: For \( x = 3 \).
A1: For \( y = 1 \).

(b)
M1: For applying the product law to LHS: \( \log_5((w + 3)(2w - 1)) \).
M1: For applying the power law to RHS: \( \log_5 9 \).
A1: For the quadratic equation \( 2w^2 + 5w - 12 = 0 \).
M1: For attempting to solve their quadratic equation to obtain two roots.
A1: For rejecting \( w = -4 \) and identifying \( w = 1.5 \) as the only valid solution.

(a) (i) Total number of ways to choose 5 people from 12 with no restrictions is:
\( \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \).

(ii) Choose exactly 2 children from the 3 children, and the remaining 3 committee members from the remaining 9 adults (5 men + 4 women):
\( \binom{3}{2} \times \binom{9}{3} = 3 \times 84 = 252 \).

(iii) 'At least 3 women' means either exactly 3 women or exactly 4 women:
Case 1: Exactly 3 women and 2 non-women (men or children):
\( \binom{4}{3} \times \binom{8}{2} = 4 \times 28 = 112 \).
Case 2: Exactly 4 women and 1 non-woman:
\( \binom{4}{4} \times \binom{8}{1} = 1 \times 8 = 8 \).
Total ways = \( 112 + 8 = 120 \).

(b) (i) Odd digits available: 1, 3, 5, 7 (4 options).
Even digits available: 2, 4, 6, 8 (4 options).
Number of choices for the first digit (odd) = 4.
Number of choices for the last digit (even) = 4.
The remaining 4 positions can be filled by any of the remaining 6 digits:
\( P(6, 4) = 6 \times 5 \times 4 \times 3 = 360 \).
Total codes = \( 4 \times 4 \times 360 = 5760 \).

(ii) Treat the digits 2 and 3 as a single block. They can be arranged within the block in \( 2! = 2 \) ways ((2,3) or (3,2)).
Choose 4 other digits from the remaining 6 digits:
\( \binom{6}{4} = 15 \).
We now have 5 items to arrange (the block + 4 other digits):
\( 5! = 120 \).
Total codes = \( 2 \times 15 \times 120 = 3600 \).

(a)
(i)
B1: For \( 792 \).

(ii)
M1: For \( \binom{3}{2} \times \binom{9}{3} \) (concept of selecting from children and adults separately).
A1: For \( 252 \).

(iii)
M1: For considering both cases (3 women and 4 women).
M1: For evaluating at least one case correctly: \( \binom{4}{3} \times \binom{8}{2} = 112 \) or \( \binom{4}{4} \times \binom{8}{1} = 8 \).
A1: For \( 120 \).

(b)
(i)
M1: For identifying the choices for first (4) and last (4) positions.
M1: For \( P(6, 4) \) (or equivalent permutation calculation for the middle positions).
A1: For \( 5760 \).

(ii)
M1: For using a block method or position method (e.g., 5 positions for the block, 2 ways within the block, and arranging the remaining digits).
A1: For \( 3600 \).

Using laws of logarithms, we can write the equation as:
\(\log_3(x - 2)^2 - \log_3(3x - 12) = 1\)
\(\log_3\left(\frac{(x - 2)^2}{3x - 12}\right) = 1\)

Converting to exponential form:
\(\frac{(x - 2)^2}{3x - 12} = 3^1\)
\((x - 2)^2 = 3(3x - 12)\)
\(x^2 - 4x + 4 = 9x - 36\)
\(x^2 - 13x + 40 = 0\)

Factoring the quadratic equation:
\((x - 5)(x - 8) = 0\)

Thus, \(x = 5\) or \(x = 8\).

Checking the validity:
For \(x = 5\), \(x - 2 = 3 > 0\) and \(3x - 12 = 3 > 0\) (valid).
For \(x = 8\), \(x - 2 = 6 > 0\) and \(3x - 12 = 12 > 0\) (valid).

Both solutions are valid.

M1: For applying the power law to obtain \(\log_3(x - 2)^2\)
M1: For applying the subtraction law to obtain \(\log_3\left(\frac{(x - 2)^2}{3x - 12}\right)\)
M1: For converting to index form correctly to obtain \((x - 2)^2 = 3(3x - 12)\)
A1: For obtaining the correct quadratic equation \(x^2 - 13x + 40 = 0\)
A1: For both correct solutions \(x = 5\) and \(x = 8\)

To find the stationary points, we first find the first derivative \(\frac{\text{d}y}{\text{d}x}\) using the product rule:
\(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x}(x^2 - 3)\text{e}^{-x} + (x^2 - 3)\frac{\text{d}}{\text{d}x}(\text{e}^{-x})\)
\(\frac{\text{d}y}{\text{d}x} = 2x\text{e}^{-x} - (x^2 - 3)\text{e}^{-x}\)
\(\frac{\text{d}y}{\text{d}x} = (3 + 2x - x^2)\text{e}^{-x}\)

Set \(\frac{\text{d}y}{\text{d}x} = 0\):
\((3 + 2x - x^2)\text{e}^{-x} = 0\)
Since \(\text{e}^{-x} \neq 0\), we have:
\(x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0\)
So, the \(x\)-coordinates of the stationary points are \(x = 3\) and \(x = -1\).

Now we find the corresponding \(y\)-coordinates:
When \(x = -1\): \(y = ((-1)^2 - 3)\text{e}^{-(-1)} = -2\text{e}\).
When \(x = 3\): \(y = (3^2 - 3)\text{e}^{-3} = 6\text{e}^{-3}\).

To find the nature of each point, we find the second derivative \(\frac{\text{d}^2y}{\text{d}x^2}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left[(3 + 2x - x^2)\text{e}^{-x}\right] = (2 - 2x)\text{e}^{-x} - (3 + 2x - x^2)\text{e}^{-x} = (x^2 - 4x - 1)\text{e}^{-x}\).

Evaluate at \(x = -1\):
\(\frac{\text{d}^2y}{\text{d}x^2} = ((-1)^2 - 4(-1) - 1)\text{e}^{-(-1)} = 4\text{e} > 0\), so \((-1, -2\text{e})\) is a minimum point.

Evaluate at \(x = 3\):
\(\frac{\text{d}^2y}{\text{d}x^2} = (3^2 - 4(3) - 1)\text{e}^{-3} = -4\text{e}^{-3} < 0\), so \((3, 6\text{e}^{-3})\) is a maximum point.

M1: For attempt to differentiate using the product rule
A1: For obtaining correct first derivative \(\frac{\text{d}y}{\text{d}x} = (3 + 2x - x^2)\text{e}^{-x}\)
A1: For finding correct exact coordinates: \((-1, -2\text{e})\) and \((3, 6\text{e}^{-3})\)
M1: For finding the second derivative \(\frac{\text{d}^2y}{\text{d}x^2} = (x^2 - 4x - 1)\text{e}^{-x}\) and evaluating it at one of their \(x\) values
A1: For correctly identifying \((-1, -2\text{e})\) as a minimum and \((3, 6\text{e}^{-3})\) as a maximum with valid evidence

First, find the expansions of each bracket up to the term in \(x^2\):
\((1 + 3x)^5 = 1 + \binom{5}{1}(3x) + \binom{5}{2}(3x)^2 + \dots\)
\((1 + 3x)^5 = 1 + 5(3x) + 10(9x^2) + \dots = 1 + 15x + 90x^2 + …\)

\((2 - x)^4 = 2^4 + \binom{4}{1}(2^3)(-x) + \binom{4}{2}(2^2)(-x)^2 + \dots\)
\((2 - x)^4 = 16 + 4(8)(-x) + 6(4)(x^2) + \dots = 16 - 32x + 24x^2 + …\)

Now, find the terms in the product that result in \(x^2\):
\(\text{Coefficient of } x^2 = (1 \times 24) + (15 \times -32) + (90 \times 16)\)
\(\text{Coefficient of } x^2 = 24 - 480 + 1440 = 984\).

B1: For obtaining correct first three terms of \((1 + 3x)^5\) as \(1 + 15x + 90x^2\) (or equivalent)
B1: For obtaining correct first three terms of \((2 - x)^4\) as \(16 - 32x + 24x^2\) (or equivalent)
M1: For a correct expression of the sum of the three products to find the coefficient of \(x^2\)
A1: For correct calculation of individual products (\(24\), \(-480\), \(1440\))
A1: For the final answer \(984\)

From the linear equation, express \(x\) in terms of \(y\):
\(x = 5 - y\)

Substitute \(x = 5 - y\) into the quadratic equation:
\(2(5 - y)^2 - (5 - y)y + y^2 = 14\)

Expand and simplify:
\(2(25 - 10y + y^2) - 5y + y^2 + y^2 = 14\)
\(50 - 20y + 2y^2 - 5y + y^2 + y^2 = 14\)
\(4y^2 - 25y + 36 = 0\)

Factor the quadratic equation:
\((y - 4)(4y - 9) = 0\)

This gives \(y = 4\) or \(y = \frac{9}{4}\).

Find the corresponding values of \(x\):
When \(y = 4\), \(x = 5 - 4 = 1\).
When \(y = \frac{9}{4}\), \(x = 5 - \frac{9}{4} = \frac{11}{4}\).

So the solutions are \(x = 1, y = 4\) and \(x = \frac{11}{4}, y = \frac{9}{4}\).

M1: For expressing \(x\) in terms of \(y\) (or vice versa) correctly
M1: For substituting their expression into the quadratic equation
A1: For obtaining a correct simplified 3-term quadratic equation, e.g., \(4y^2 - 25y + 36 = 0\)
M1: For solving their 3-term quadratic equation to find two values of \(y\) (or \(x\))
A1: For both correct pairs of solutions: \((1, 4)\) and \(\left(\frac{11}{4}, \frac{9}{4}\right)\)

Using the identity \(\cos^2\theta = 1 - \sin^2\theta\), we can rewrite the equation as:
\(3(1 - \sin^2\theta) + 5\sin\theta - 1 = 0\)
\(3 - 3\sin^2\theta + 5\sin\theta - 1 = 0\)
\(3\sin^2\theta - 5\sin\theta - 2 = 0\)

Factor the quadratic in terms of \(\sin\theta\):
\((3\sin\theta + 1)(\sin\theta - 2) = 0\)

This gives two possible cases:

Case 1: \(\sin\theta = -\frac{1}{3}\)
Since \(\sin\theta\) is negative, \(\theta\) is in the third or fourth quadrant.
The basic angle is \(\theta_{\text{ref}} = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ\).
In Quadrant III: \(\theta = 180^\circ + 19.47^\circ \approx 199.5^\circ\).
In Quadrant IV: \(\theta = 360^\circ - 19.47^\circ \approx 340.5^\circ\).

Case 2: \(\sin\theta = 2\)
This case has no solutions since the range of sine is \([-1, 1]\).

Thus, the solutions are \(\theta = 199.5^\circ\) and \(\theta = 340.5^\circ\).

M1: For using the identity \(\cos^2\theta = 1 - \sin^2\theta\) to form a quadratic equation in \(\sin\theta\)
A1: For obtaining correct quadratic equation \(3\sin^2\theta - 5\sin\theta - 2 = 0\)
M1: For solving their quadratic equation to obtain \(\sin\theta = -\frac{1}{3}\) (and identifying \(\sin\theta = 2\) has no solution)
A1: For one correct angle, either \(199.5^\circ\) or \(340.5^\circ\)
A1: For the other correct angle and no extra solutions in the range

(i) Using the quotient rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8(2x+1) - 8x(2)}{(2x+1)^2} = \frac{16x + 8 - 16x}{(2x+1)^2} = \frac{8}{(2x+1)^2}\). (ii) When \(x = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8}{(2(1)+1)^2} = \frac{8}{9}\). Using small changes: \(\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x} \delta x = \frac{8}{9}h\). (iii) Using the chain rule: \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\). When \(x = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{8}{9} \times 0.09 = 0.08\) units per second. (iv) The gradient of the line \(y = -2x + 5\) is \(-2\). Since the normal is parallel to this line, the gradient of the normal is \(-2\), which means the gradient of the tangent is \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\). Solving \(\frac{8}{(2x+1)^2} = \frac{1}{2}\) gives \((2x+1)^2 = 16\), so \(2x+1 = 4\) or \(2x+1 = -4\). This yields \(x = 1.5\) or \(x = -2.5\). Substituting these back into the curve equation gives the points \((1.5, 3)\) and \((-2.5, 5)\).

(i) M1 for applying the quotient rule, A1 for correct simplified derivative. (ii) M1 for substituting \(x = 1\) into their derivative, A1 for \(\frac{8}{9}h\). (iii) M1 for utilizing the chain rule formula, A1 for 0.08. (iv) M1 for finding tangent gradient is 0.5, M1 for setting their derivative equal to 0.5 and solving for \(x\), A1 for correct \(x\) values, A1 for correct y coordinates.

(i) Taking natural logarithms on both sides: \(\ln y = \ln(p \cdot q^x) = \ln p + \ln(q^x) = \ln p + x \ln q\). (ii) The equation represents a straight line \(Y = mX + C\) where \(Y = \ln y\), \(X = x\), gradient \(m = \ln q\) and intercept \(C = \ln p\). The gradient is \(m = \frac{9.5 - 3.5}{6 - 2} = \frac{6}{4} = 1.5\). So, \(\ln q = 1.5\). Using the point \((2, 3.5)\): \(3.5 = \ln p + 2(1.5) \implies \ln p = 0.5\). (iii) Since \(\ln p = 0.5 \implies p = \mathrm{e}^{0.5}\) and \(\ln q = 1.5 \implies q = \mathrm{e}^{1.5}\). (iv) When \(x = 4\), \(\ln y = \ln p + 4\ln q = 0.5 + 4(1.5) = 6.5\). Thus, \(y = \mathrm{e}^{6.5} \approx 665\).

(i) B1 for logarithmic expansion, B1 for final correct form. (ii) M1 for finding the gradient of the line, A1 for \(\ln q = 1.5\), M1 for substituting a point to find intercept, A1 for \(\ln p = 0.5\). (iii) B1 for \(p = \mathrm{e}^{0.5}\), B1 for \(q = \mathrm{e}^{1.5}\). (iv) M1 for substitution of \(x = 4\) into the linear equation, A1 for \(y = \mathrm{e}^{6.5}\) (accept 665).

(i) Since \(OP : PA = 2 : 1\), \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Since \(OQ : QB = 1 : 3\), \(\vec{OQ} = \frac{1}{4}\mathbf{b}\). Then, \(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\). Also, \(\vec{BP} = \vec{BO} + \vec{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (ii) First expression: \(\vec{OX} = \vec{OA} + \vec{AX} = \mathbf{a} + \lambda(-\mathbf{a} + \frac{1}{4}\mathbf{b}) = (1-\lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\). Second expression: \(\vec{OX} = \vec{OB} + \vec{BX} = \mathbf{b} + \mu(\frac{2}{3}\mathbf{a} - \mathbf{b}) = \frac{2}{3}\mu\mathbf{a} + (1-\mu)\mathbf{b}\). (iii) Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): For \(\mathbf{a}\): \(1-\lambda = \frac{2}{3}\mu\). For \(\mathbf{b}\): \(\frac{1}{4}\lambda = 1-\mu \implies \lambda = 4 - 4\mu\). Substitute \(\lambda\) into the first equation: \(1 - (4 - 4\mu) = \frac{2}{3}\mu \implies -3 + 4\mu = \frac{2}{3}\mu \implies \frac{10}{3}\mu = 3 \implies \mu = 0.9\). Then, \(\lambda = 4 - 4(0.9) = 0.4\).

(i) B1 for \(\vec{AQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\), B2 for \(\vec{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (ii) M1 for using vector addition for \(\vec{OX}\), A1 for \((1-\lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\), A1 for \(\frac{2}{3}\mu\mathbf{a} + (1-\mu)\mathbf{b}\). (iii) M1 for equating the coefficients to form simultaneous equations, M1 for solving the simultaneous equations, A1 for \(\mu = 0.9\), A1 for \(\lambda = 0.4\).

(i) Let \(\phi = 2\theta - \frac{\pi}{6}\). Since \(0 \le \theta \le \pi\), we have \(-\frac{\pi}{6} \le \phi \le \frac{11\pi}{6}\). Now, \(\tan \phi = \sqrt{3} \implies \phi = \frac{\pi}{3}, \frac{4\pi}{3}\). If \(2\theta - \frac{\pi}{6} = \frac{\pi}{3} \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}\). If \(2\theta - \frac{\pi}{6} = \frac{4\pi}{3} \implies 2\theta = \frac{3\pi}{2} \implies \theta = \frac{3\pi}{4}\). (ii) \(LHS = \frac{(1 + \sin x)^2 + \cos^2 x}{\cos x(1 + \sin x)} = \frac{1 + 2\sin x + \sin^2 x + \cos^2 x}{\cos x(1 + \sin x)} = \frac{2 + 2\sin x}{\cos x(1 + \sin x)} = \frac{2(1 + \sin x)}{\cos x(1 + \sin x)} = \frac{2}{\cos x} = 2 \sec x\) (proved). (iii) Using the identity, the equation simplifies to \(2 \sec x = 4 \implies \sec x = 2 \implies \cos x = \frac{1}{2}\). In the interval \(0^\circ \le x \le 360^\circ\), \(x = 60^\circ\) or \(x = 300^\circ\).

(i) M1 for finding the primary angles for \(\tan \phi = \sqrt{3}\), M1 for setting up the equations for \(\theta\), A1 for \(\theta = \frac{\pi}{4}\), A1 for \(\theta = \frac{3\pi}{4}\). (ii) M1 for combining fractions with a common denominator, M1 for applying the identity \(\sin^2 x + \cos^2 x = 1\), A1 for simplifying and factoring out 2, A1 for final correct proof. (iii) M1 for simplifying the equation to \(\cos x = 0.5\), A1 for \(x = 60^\circ\), A1 for \(x = 300^\circ\).

(i) The first, third and seventh terms of the AP are: \(T_1 = a\), \(T_3 = a + 2d\), \(T_7 = a + 6d\). The first, second and third terms of the GP are: \(G_1 = a\), \(G_2 = ar\), \(G_3 = ar^2\). Since \(T_1 = G_1\), \(T_3 = G_2\), \(T_7 = G_3\): \(a + 2d = ar \implies 2d = a(r - 1)\) (Eq 1). \(a + 6d = ar^2 \implies 6d = a(r^2 - 1)\) (Eq 2). Since \(6d = 3(2d)\), we can substitute Eq 1 into Eq 2: \(3a(r-1) = a(r^2-1)\). Since \(d \ne 0\), then \(a \ne 0\) and \(r \ne 1\). Dividing by \(a(r-1)\) yields: \(3 = r + 1 \implies r = 2\). (ii) From Eq 1, since \(r = 2\): \(2d = a(2-1) \implies d = \frac{a}{2}\). (iii) The sum of the first 10 terms of the AP: \(S_{10} = \frac{10}{2}(2a + 9d) = 130 \implies 5(2a + 9d) = 130 \implies 2a + 9d = 26\). Substituting \(d = \frac{a}{2}\): \(2a + 9(\frac{a}{2}) = 26 \implies 6.5a = 26 \implies a = 4\). Thus, \(d = \frac{4}{2} = 2\).

(i) M1 for expressing AP and GP terms correctly, M1 for forming the simultaneous equations, A1 for substituting to eliminate \(d\), A1 for correct proof of \(r = 2\) (must reject \(r=1\)). (ii) M1 for substituting \(r=2\) back into one of the equations, A1 for \(d = \frac{a}{2}\). (iii) M1 for using the AP sum formula and setting it to 130, M1 for substituting \(d\) in terms of \(a\), A1 for \(a = 4\), A1 for \(d = 2\).