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

(a) Equating the line and the curve:
\(x^2 + (k - 1)x + 1 = 2kx - 3\)
\(x^2 + (k - 1 - 2k)x + 4 = 0\)
\(x^2 - (k + 1)x + 4 = 0\)

For no intersection, the discriminant \(\Delta < 0\):
\(\Delta = [-(k + 1)]^2 - 4(1)(4) < 0\)
\((k + 1)^2 - 16 < 0\)
\((k + 1)^2 < 16\)
\(-4 < k + 1 < 4\)
\(-5 < k < 3\)

(b) Completing the square for the quadratic curve:
\(y = 3(x^2 - 4x) + m\)
\(y = 3[(x - 2)^2 - 4] + m\)
\(y = 3(x - 2)^2 - 12 + m\)

The minimum value of the curve is \(-12 + m\).
Setting the minimum value to 2:
\(-12 + m = 2 \implies m = 14\)

(a)
- **M1**: Sets \(x^2 + (k - 1)x + 1 = 2kx - 3\) and rearranges to a 3-term quadratic in \(x\).
- **A1**: Correct quadratic equation: \(x^2 - (k + 1)x + 4 = 0\) (or equivalent).
- **M1**: Uses discriminant \(b^2 - 4ac < 0\).
- **A1**: Correct final range: \(-5 < k < 3\).

(b)
- **M1**: Attempt to complete the square or use calculus (differentiate and set to 0) to find the vertex.
- **A1**: Correct \(x\)-coordinate of the vertex, \(x = 2\).
- **M1**: Substutes \(x = 2\) into the equation and sets equal to 2 (or equates completed square constant term to 2).
- **A1**: Correct value of \(m = 14\).

(a) Since \(x - 2\) is a factor, \(P(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(16 + 4a + 2b - 12 = 0\)
\(4a + 2b = -4 \implies 2a + b = -2\) --- (Equation 1)

Since \(P(x)\) divided by \(x + 1\) has a remainder of \(-15\), \(P(-1) = -15\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 12 = -15\)
\(-2 + a - b - 12 = -15\)
\(a - b - 14 = -15 \implies a - b = -1\) --- (Equation 2)

Adding Equation 1 and Equation 2:
\((2a + b) + (a - b) = -2 + (-1)\)
\(3a = -3 \implies a = -1\)

Substituting \(a = -1\) into Equation 2:
\(-1 - b = -1 \implies b = 0\)

(b) Using \(a = -1\) and \(b = 0\), the polynomial is \(P(x) = 2x^3 - x^2 - 12\).
To find the remainder when divided by \(2x - 1\), we evaluate \(P\left(\frac{1}{2}\right)\):
\(P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - 12\)
\(P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - \frac{1}{4} - 12\)
\(P\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{4} - 12 = -12\)

(a)
- **M1**: Uses the factor theorem \(P(2) = 0\) to form a linear equation in \(a\) and \(b\).
- **A1**: Obtains \(2a + b = -2\) (or equivalent).
- **M1**: Uses the remainder theorem \(P(-1) = -15\) to form a second linear equation in \(a\) and \(b\).
- **A1**: Obtains \(a - b = -1\) (or equivalent).
- **A1**: Solves the simultaneous equations to find \(a = -1\) and \(b = 0\).

(b)
- **M1**: Identifies that the remainder is obtained by evaluating \(P\left(\frac{1}{2}\right)\).
- **M1**: Substitutes \(x = \frac{1}{2}\) into their \(P(x)\).
- **A1**: Correctly evaluates to find the remainder is \(-12\).

(a) The perimeter \(P\) of the sector is given by:
\(P = 2r + r\theta\)

Given \(P = 20\):
\(2r + r\theta = 20 \implies r\theta = 20 - 2r \implies \theta = \frac{20 - 2r}{r}\)

The area \(A\) of the sector is:
\(A = \frac{1}{2}r^2\theta\)

Substitute \(\theta = \frac{20 - 2r}{r}\) into the area formula:
\(A = \frac{1}{2}r^2 \left(\frac{20 - 2r}{r}\right) = \frac{1}{2}r(20 - 2r) = 10r - r^2\) (Shown)

(b) To find the maximum area, differentiate \(A\) with respect to \(r\):
\(\frac{dA}{dr} = 10 - 2r\)

Set \(\frac{dA}{dr} = 0\) for stationary values:
\(10 - 2r = 0 \implies r = 5\)

Find the second derivative to justify the maximum:
\(\frac{d^2A}{dr^2} = -2\)

Since \(\frac{d^2A}{dr^2} < 0\), the area is a maximum when \(r = 5\).
The maximum area is:
\(A = 10(5) - (5)^2 = 50 - 25 = 25\text{ cm}^2\)

(c) Using \(r = 5\) in the perimeter equation:
\(2(5) + 5\theta = 20 \implies 10 + 5\theta = 20 \implies 5\theta = 10 \implies \theta = 2\text{ radians}\)

(a)
- **B1**: Writes down the correct formula for the perimeter: \(2r + r\theta = 20\).
- **M1**: Expresses \(\theta\) in terms of \(r\) or \(r\theta\) as \(20 - 2r\), and substitutes into the area formula \(A = \frac{1}{2}r^2\theta\).
- **A1**: Simplifies to obtain the given expression \(A = 10r - r^2\) clearly.

(b)
- **M1**: Differentiates \(A\) with respect to \(r\) and sets the derivative equal to 0.
- **A1**: Finds \(r = 5\) and evaluates \(A = 25\).
- **B1**: Uses second derivative \(\frac{d^2A}{dr^2} = -2 < 0\) (or equivalent first derivative test) to justify that it is a maximum.

(c)
- **M1**: Substitutes their value of \(r = 5\) back into the formula for perimeter or area to find \(\theta\).
- **A1**: Obtains \(\theta = 2\).

(a) Combining the fractions on the left-hand side (LHS):
\(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)
\(\text{LHS} = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(\text{LHS} = \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(\text{LHS} = \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)
\(\text{LHS} = \frac{2}{\sin \theta} = 2\csc \theta\) (Shown)

(b) Let \(\theta = 2x\). The equation becomes:
\(2\csc 2x = 4 \implies \csc 2x = 2\)
\(\sin 2x = \frac{1}{2}\)

Since \(0 < x < \pi\), the range for \(2x\) is \(0 < 2x < 2\pi\).

Solving \(\sin 2x = \frac{1}{2}\) in this range:
\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\)

Dividing by 2:
\(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\)

(a)
- **M1**: Puts both terms over a common denominator.
- **A1**: Obtains \(\frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\).
- **M1**: Uses \(\sin^2 \theta + \cos^2 \theta = 1\) and factorises the numerator to \(2(1 + \cos \theta)\).
- **A1**: Cancels the common factor \(1 + \cos \theta\) to obtain \(2\csc \theta\) convincingly.

(b)
- **M1**: Applies the identity from part (a) with \(\theta = 2x\) to get \(2\csc 2x = 4\).
- **A1**: Reduces to \(\sin 2x = \frac{1}{2}\).
- **M1**: Finds at least one correct value for \(2x\) (e.g. \(\frac{\pi}{6}\) or \(\frac{5\pi}{6}\)).
- **A1**: Obtains both correct answers: \(x = \frac{\pi}{12}\) and \(x = \frac{5\pi}{12}\) (and no others in the range).

(a) A committee of 5 with at least 3 women can have 3, 4, or 5 women.

Case 1: Exactly 3 women and 2 men
Number of ways = \(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\)

Case 2: Exactly 4 women and 1 man
Number of ways = \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)

Case 3: Exactly 5 women and 0 men
Number of ways = \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)

Total number of ways = \(120 + 60 + 6 = 186\).

(b) (i) Treat the 4 men as a single block. There are 6 women and 1 block of men, giving 7 items to arrange.
These 7 items can be arranged in \(7!\) ways.
The 4 men within their block can be arranged in \(4!\) ways.
Total number of ways = \(7! \times 4! = 5040 \times 24 = 120\,960\).

(ii) First arrange the 6 women in a line: \(6! = 720\) ways.
This creates 7 spaces (including the ends) where the men can stand ( _ W _ W _ W _ W _ W _ W _ ).
We choose 4 of these 7 spaces for the 4 men and arrange them:
Number of ways = \(P^7_4 = 7 \times 6 \times 5 \times 4 = 840\) ways.
Total number of ways = \(6! \times P^7_4 = 720 \times 840 = 604\,800\).

(a)
- **M1**: Identifies the three cases (3W 2M, 4W 1M, 5W 0M).
- **M1**: Calculates any two of these cases correctly using combinations.
- **A1**: Calculates all three cases correctly (120, 60, 6).
- **A1**: Adds the cases to find the final answer of 186.

(b)(i)
- **M1**: Treats the men as a single block to get \(7! \times 4!\) (or equivalent).
- **A1**: Correct final answer: 120,960.

(b)(ii)
- **M1**: Arranges women (\(6!\)) and considers the 7 gaps for the men (uses \(P^7_4\) or \(\binom{7}{4} \times 4!\)).
- **A1**: Correct final answer: 604,800.

(a) Differentiating once to get \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \int (6x - 4) dx = 3x^2 - 4x + C_1\)

Since \((2, 5)\) is a stationary point, \(\frac{dy}{dx} = 0\) when \(x = 2\):
\(3(2)^2 - 4(2) + C_1 = 0 \implies 12 - 8 + C_1 = 0 \implies C_1 = -4\)

So, \(\frac{dy}{dx} = 3x^2 - 4x - 4\).

Now, integrate \(\frac{dy}{dx}\) to find \(y\):
\(y = \int (3x^2 - 4x - 4) dx = x^3 - 2x^2 - 4x + C_2\)

Since the curve passes through \((2, 5)\), when \(x = 2\), \(y = 5\):
\((2)^3 - 2(2)^2 - 4(2) + C_2 = 5\)
\(8 - 8 - 8 + C_2 = 5 \implies -8 + C_2 = 5 \implies C_2 = 13\)

Therefore, the equation of the curve is:
\(y = x^3 - 2x^2 - 4x + 13\)

(b) To determine the nature of the stationary point at \(x = 2\), evaluate \(\frac{d^2y}{dx^2}\) at \(x = 2\):
\(\frac{d^2y}{dx^2} = 6(2) - 4 = 8\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point \((2, 5)\) is a local minimum.

(c) To find other stationary points, set \(\frac{dy}{dx} = 0\):
\(3x^2 - 4x - 4 = 0\)
\((3x + 2)(x - 2) = 0\)

Thus, the other stationary point is at \(x = -\frac{2}{3}\).

(a)
- **M1**: Integrates \(6x - 4\) once.
- **A1**: Obtains \(\frac{dy}{dx} = 3x^2 - 4x + C_1\).
- **M1**: Uses \(\frac{dy}{dx} = 0\) when \(x = 2\) to find \(C_1 = -4\).
- **M1**: Integrates their \(\frac{dy}{dx}\) and uses the point \((2, 5)\) to find the constant of integration \(C_2\).
- **A1**: Correct final equation: \(y = x^3 - 2x^2 - 4x + 13\).

(b)
- **M1**: Substitutes \(x = 2\) into \(\frac{d^2y}{dx^2} = 6x - 4\).
- **A1**: Obtains value \(8 > 0\) and correctly states it is a minimum.

(c)
- **B1**: Solves \(3x^2 - 4x - 4 = 0\) to obtain the other root \(x = -\frac{2}{3}\) (or \(-0.667\)).

(a) Using logarithmic properties:
\(\log_3 \left(\frac{2x + 1}{x - 2}\right) = 2\)

Converting to exponential form:
\(\frac{2x + 1}{x - 2} = 3^2 = 9\)
\(2x + 1 = 9(x - 2)\)
\(2x + 1 = 9x - 18\)
\(7x = 19 \implies x = \frac{19}{7}\)

Checking constraints: \(x > 2\). Since \(\frac{19}{7} \approx 2.71 > 2\), this is a valid solution.

(b) Rewrite the equation:
\(5 \cdot (5^y)^2 - 7(5^y) + 2 = 0\)

Let \(u = 5^y\):
\(5u^2 - 7u + 2 = 0\)
\((5u - 2)(u - 1) = 0\)

This gives \(u = \frac{2}{5} = 0.4\) or \(u = 1\).

If \(5^y = 1 \implies y = 0\).
If \(5^y = 0.4 \implies y = \frac{\log(0.4)}{\log(5)} \approx -0.57\).

(a)
- **M1**: Combines logs to form \(\log_3 \left(\frac{2x + 1}{x - 2}\right) = 2\).
- **M1**: Converts to exponential form: \(\frac{2x + 1}{x - 2} = 9\).
- **M1**: Solves the linear equation for \(x\).
- **A1**: Correct final answer \(x = \frac{19}{7}\) (or exact equivalent, e.g., \(2.71\)).

(b)
- **M1**: Uses substitution \(u = 5^y\) to write as a quadratic: \(5u^2 - 7u + 2 = 0\).
- **A1**: Solves quadratic to find \(u = 0.4\) and \(u = 1\).
- **M1**: Solves exponential equations \(5^y = 0.4\) and \(5^y = 1\) using logarithms where necessary.
- **A1**: Correctly obtains both \(y = 0\) and \(y = -0.57\) (to 2 decimal places).

(a) Since the terms are in a geometric progression, the common ratio \(r\) is constant:
\(r = \frac{k}{k + 2} = \frac{2k - 3}{k}\)

Cross-multiplying:
\(k^2 = (2k - 3)(k + 2)\)
\(k^2 = 2k^2 + 4k - 3k - 6\)
\(k^2 = 2k^2 + k - 6\)
\(0 = k^2 + k - 6 - k^2 \implies k^2 - k - 6 = 0\) (Shown)

Factoring to find \(k\):
\((k - 3)(k + 2) = 0\)
\(k = 3\) or \(k = -2\)

Since \(k\) is a positive constant, \(k = 3\).

(b) Substituting \(k = 3\) into the terms:
- First term \(a = k + 2 = 3 + 2 = 5\)
- Second term \(ar = k = 3\)

Common ratio \(r = \frac{3}{5} = 0.6\).

Since \(|r| < 1\), we can find the sum to infinity \(S_\infty\):
\(S_\infty = \frac{a}{1 - r} = \frac{5}{1 - 0.6} = \frac{5}{0.4} = 12.5\)

(a)
- **M1**: Sets up equal ratios: \(\frac{k}{k + 2} = \frac{2k - 3}{k}\).
- **M1**: Multiplies out and expands brackets to form a quadratic.
- **A1**: Arrives at \(k^2 - k - 6 = 0\) clearly.
- **A1**: Solves and selects the positive value \(k = 3\) (rejecting \(k = -2\)).

(b)
- **M1**: Finds first term \(a = 5\).
- **M1**: Finds common ratio \(r = 0.6\).
- **M1**: Uses sum to infinity formula \(S_\infty = \frac{a}{1 - r}\).
- **A1**: Correctly evaluates to find \(S_\infty = 12.5\) (or \(\frac{25}{2}\)).

(a) Using the product rule with \(u = 2x + 1\) and \(v = (4x - 3)^{1/2}\):
\(\frac{\text{d}u}{\text{d}x} = 2\)
\(\frac{\text{d}v}{\text{d}x} = \frac{1}{2}(4x - 3)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x - 3}}\)

Then:
\(\frac{\text{d}y}{\text{d}x} = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}\)
\(\frac{\text{d}y}{\text{d}x} = (2x + 1)\frac{2}{\sqrt{4x - 3}} + 2\sqrt{4x - 3}\)

Combining over a common denominator:
\(\frac{\text{d}y}{\text{d}x} = \frac{2(2x + 1) + 2(4x - 3)}{\sqrt{4x - 3}}\)

\(\frac{\text{d}y}{\text{d}x} = \frac{4x + 2 + 8x - 6}{\sqrt{4x - 3}}\)

\(\frac{\text{d}y}{\text{d}x} = \frac{12x - 4}{\sqrt{4x - 3}}\)

Thus, \(A = 12\) and \(B = -4\).

(b) When \(x = 1\), \(\frac{\text{d}y}{\text{d}x} = \frac{12(1) - 4}{\sqrt{4(1) - 3}} = \frac{8}{1} = 8\).

Using small changes: \(\delta y \approx \frac{\text{d}y}{\text{d}x} \delta x\).

Since \(\delta x = p\), the approximate change in \(y\) is \(\delta y \approx 8p\).

Part (a)
M1: For attempt to use product rule or quotient rule with correct structure.
B1: For correct differentiation of \((4x - 3)^{1/2}\) to get \(2(4x - 3)^{-1/2}\) or equivalent.
A1: For correct unsimplified derivative.
M1: For combining terms over a common denominator.
A1: For correct values A = 12 and B = -4.

Part (b)
M1: For substituting x = 1 into their derivative from part (a).
A1: For finding the gradient at x = 1 is 8 (must follow from correct derivative or correct FT).
A1: For correct approximate change 8p.

(a) For a geometric progression, the common ratio \(r\) must be constant:
\(\frac{x}{x + 3} = \frac{x - 2}{x}\)

Cross-multiplying gives:
\(x^2 = (x - 2)(x + 3)\)

\(x^2 = x^2 + x - 6\)

\(x = 6\)

(b) Substituting \(x = 6\), the first three terms are \(9\), \(6\), and \(4\).

First term, \(a = 9\).
Common ratio, \(r = \frac{6}{9} = \frac{2}{3}\).

The sum to infinity, \(S_\infty = \frac{a}{1 - r}\):
\(S_\infty = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27\).

(c) The sum of the first 6 terms, \(S_6 = \frac{a(1 - r^6)}{1 - r}\):
\(S_6 = \frac{9(1 - (2/3)^6)}{1 - 2/3} = 27\left(1 - \frac{64}{729}\right) = 27 - \frac{64}{27} = \frac{665}{27}\).

The difference between the sum to infinity and the sum of the first 6 terms is:
\(S_\infty - S_6 = 27 - \frac{665}{27} = \frac{729 - 665}{27} = \frac{64}{27}\).

Part (a)
M1: For setting up the ratio equation \(\frac{x}{x + 3} = \frac{x - 2}{x}\) or equivalent.
M1: For expanding and solving the linear equation in x.
A1: For x = 6.

Part (b)
M1: For identifying a = 9 and finding r = 2/3.
A1: For correct sum to infinity, 27.

Part (c)
M1: For correct formula used for \(S_6\) with their a and r.
A1: For finding \(S_6 = \frac{665}{27}\).
A1: For correct exact difference \(\frac{64}{27}\).

(a)
Case 1: \(3x - 5 = 2x + 1 \implies x = 6\).
Checking: \(|3(6) - 5| = 13\) and \(2(6) + 1 = 13\). This is a valid solution.
Case 2: \(3x - 5 = -(2x + 1) \implies 5x = 4 \implies x = 0.8\).
Checking: \(|3(0.8) - 5| = 2.6\) and \(2(0.8) + 1 = 2.6\). This is also a valid solution.
So, the solutions are \(x = 0.8\) and \(x = 6\).

(b)
- The graph of \(y = |3x - 5|\) is a V-shape with its vertex at \(\left(\frac{5}{3}, 0\right)\) and crossing the y-axis at \((0, 5)\).
- The graph of \(y = 2x + 1\) is a straight line with gradient 2, crossing the y-axis at \((0, 1)\) and the x-axis at \((-0.5, 0)\).
- The two graphs intersect in the first quadrant at the points where \(x = 0.8\) and \(x = 6\).

(a)
M1: Attempting to solve both cases, e.g., \(3x - 5 = 2x + 1\) and \(3x - 5 = -(2x + 1)\).
A1: Finding \(x = 6\).
M1: Attempting to solve \(5x = 4\).
A1: Finding \(x = 0.8\) (or \(\frac{4}{5}\)).

(b)
B1: For a correct V-shape for \(y = |3x - 5|\) with vertex on the x-axis at \((1.67, 0)\) and y-intercept at \((0, 5)\).
B1: For a straight line with positive gradient, y-intercept at \((0, 1)\) and x-intercept at \((-0.5, 0)\).
B1: For indicating both intersection points clearly in the first quadrant.

(a) Let \(u = 3^x\). The equation becomes \(3u^2 - 10u + 3 = 0\).
Factoring gives \((3u - 1)(u - 3) = 0\), which yields \(u = \frac{1}{3}\) or \(u = 3\).
Thus:
- \(3^x = \frac{1}{3} \implies x = -1\)
- \(3^x = 3 \implies x = 1\).

(b) Using the change of base rule: \(\log_4 y = \frac{\log_2 y}{\log_2 4} = \frac{1}{2}\log_2 y = \log_2 \sqrt{y}\).
The equation can be written as:
\(\log_2 \sqrt{y} + \log_2 (y - 3) = 1\)
\(\log_2 (\sqrt{y}(y - 3)) = 1\)
\(\sqrt{y}(y - 3) = 2\)
Squaring both sides:
\(y(y - 3)^2 = 4 \implies y(y^2 - 6y + 9) = 4\)
\(y^3 - 6y^2 + 9y - 4 = 0\)
Factoring the cubic gives \((y - 1)^2(y - 4) = 0\).
- If \(y = 1\), the term \(\log_2(y-3) = \log_2(-2)\) is undefined. Thus, \(y = 1\) is rejected.
- Therefore, the only valid solution is \(y = 4\).

(a)
M1: For using the substitution \(u = 3^x\) to obtain the quadratic equation \(3u^2 - 10u + 3 = 0\).
A1: For solving to get \(u = \frac{1}{3}\) and \(u = 3\).
M1: For solving \(3^x = \frac{1}{3}\) or \(3^x = 3\).
A1: For \(x = -1\) and \(x = 1\).

(b)
M1: For expressing \(\log_4 y\) as \(\frac{1}{2}\log_2 y\) or equivalent change of base.
M1: For setting up and attempting to solve the equation \(y(y-3)^2 = 4\).
A1: For obtaining \(y = 4\) and explicitly rejecting \(y = 1\).

(a) Let \(X = x^2\) and \(Y = \ln y\). The straight line passes through \((2, 5)\) and \((6, 17)\).
The gradient \(m\) is given by:
\(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\).
Using the point-slope form with \((2, 5)\):
\(Y - 5 = 3(X - 2)\)
\(Y = 3X - 1\).
Therefore, \(\ln y = 3x^2 - 1\).

(b) Taking exponentials on both sides of the equation from part (a):
\(y = e^{3x^2 - 1}\)
\(y = e^{-1} \cdot e^{3x^2}\)
\(y = e^{-1} \left(e^3\right)^{x^2}\).
This is in the form \(y = A b^{x^2}\), where:
\(A = e^{-1} \approx 0.368\)
\(b = e^3 \approx 20.1\).

(a)
M1: For calculating the gradient \(m = \frac{17 - 5}{6 - 2} = 3\).
M1: For substituting their gradient and one point into a straight-line formula, e.g., \(\ln y - 5 = 3(x^2 - 2)\).
A1: For obtaining the correct equation: \(\ln y = 3x^2 - 1\).

(b)
M1: For raising both sides to the base \(e\), i.e., \(y = e^{3x^2 - 1}\).
M1: For correctly using exponent laws to write \(y = e^{-1} \cdot (e^3)^{x^2}\).
A1: For finding the exact or approximate value of \(A = e^{-1}\) (or awrt 0.368).
A1: For finding the exact or approximate value of \(b = e^3\) (or awrt 20.1).

(a) Using vector addition:
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}\mathbf{b} - \mathbf{a}\).
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(b) Expressing \(\overrightarrow{OR}\) in two ways:
Way 1:
\(\overrightarrow{OR} = \overrightarrow{OA} + \overrightarrow{AR} = \mathbf{a} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda \left(\frac{1}{2}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \frac{1}{2}\lambda \mathbf{b}\).

Way 2:
\(\overrightarrow{OR} = \overrightarrow{OB} + \overrightarrow{BR} = \mathbf{b} + \mu \overrightarrow{BP} = \mathbf{b} + \mu \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\).

By equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
1) \(1 - \lambda = \frac{2}{3}\mu\)
2) \(\frac{1}{2}\lambda = 1 - \mu \implies \lambda = 2 - 2\mu\)

Substituting (2) into (1):
\(1 - (2 - 2\mu) = \frac{2}{3}\mu\)
\(-1 + 2\mu = \frac{2}{3}\mu\)
\(\frac{4}{3}\mu = 1 \implies \mu = \frac{3}{4}\).

Now substitute \(\mu = \frac{3}{4}\) back into (2):
\(\lambda = 2 - 2\left(\frac{3}{4}\right) = \frac{1}{2}\).

(a)
B1: For \(\overrightarrow{AQ} = \frac{1}{2}\mathbf{b} - \mathbf{a}\) or equivalent.
B1: For \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\) or equivalent.

(b)
M1: Expressing \(\overrightarrow{OR}\) in terms of \(\lambda\) as \((1 - \lambda)\mathbf{a} + \frac{1}{2}\lambda \mathbf{b}\).
M1: Expressing \(\overrightarrow{OR}\) in terms of \(\mu\) as \(\frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\).
M1: Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations.
M1: For a complete method to solve the simultaneous equations.
A1: For \(\lambda = \frac{1}{2}\).
A1: For \(\mu = \frac{3}{4}\).

(a) Let \(u = x^2\) and \(v = (2x + 1)^{1/2}\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\).
Using the chain rule, \(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(2x + 1)^{-1/2} \times 2 = (2x + 1)^{-1/2}\).
Applying the quotient rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x + 1)^{1/2}(2x) - x^2(2x + 1)^{-1/2}}{2x + 1}\)
Multiply the numerator and denominator by \((2x + 1)^{1/2}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x(2x + 1) - x^2}{(2x + 1)^{3/2}}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x^2 + 2x - x^2}{(2x + 1)^{3/2}} = \frac{3x^2 + 2x}{(2x + 1)^{3/2}} = \frac{x(3x + 2)}{(2x + 1)^{3/2}}\).

(b) For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{x(3x + 2)}{(2x + 1)^{3/2}} = 0 \implies x(3x + 2) = 0\).
This yields \(x = 0\) or \(x = -\frac{2}{3}\).
Since \(x > -0.5\), the value \(x = -\frac{2}{3}\) is outside the domain and is rejected.
Substituting \(x = 0\) into the curve's equation:
\(y = \frac{0^2}{\sqrt{2(0) + 1}} = 0\).
Thus, the stationary point is \((0, 0)\).

(a)
M1: For differentiating the numerator to get \(2x\).
M1: For differentiating the denominator using the chain rule to get \((2x + 1)^{-1/2}\).
M1: For correctly substituting their derivatives into the quotient rule formula.
M1: For algebraic simplification, including multiplying the numerator and denominator by \((2x + 1)^{1/2}\).
A1: For fully correct completion to the given expression.

(b)
M1: Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find critical values.
A1: For selecting \(x = 0\) and correctly rejecting \(x = -2/3\) based on the domain.
A1: For the coordinates \((0, 0)\).

(a) Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\).
- Third term: \(a + 2d = 8\) (Equation 1)
- Ninth term: \(a + 8d = 26\) (Equation 2)

Subtract Equation 1 from Equation 2:
\(6d = 18 \implies d = 3\).

Substitute \(d = 3\) into Equation 1:
\(a + 2(3) = 8 \implies a = 2\).

The sum of the first 20 terms is:
\(S_{20} = \frac{20}{2} [2a + 19d] = 10 [2(2) + 19(3)] = 10 [4 + 57] = 10 \times 61 = 610\).

(b) In a geometric progression, the ratio between consecutive terms is constant:
\(\frac{x}{x + 4} = \frac{x - 2}{x}\)

Cross-multiplying:
\(x^2 = (x + 4)(x - 2)\)
\(x^2 = x^2 + 2x - 8\)
\(2x = 8 \implies x = 4\).

Substituting \(x = 4\), the first three terms are:
- First term, \(a = 4 + 4 = 8\)
- Second term, \(x = 4\)
- Third term, \(x - 2 = 2\)

The common ratio is \(r = \frac{4}{8} = \frac{1}{2}\).
Since \(|r| < 1\), the sum to infinity exists and is:
\(S_{\infty} = \frac{a}{1 - r} = \frac{8}{1 - 1/2} = 16\).

(a)
M1: For setting up the simultaneous equations \(a + 2d = 8\) and \(a + 8d = 26\).
A1: For finding \(d = 3\) and \(a = 2\).
M1: For using the correct arithmetic progression sum formula.
A1: For 610.

(b)
M1: For setting up the ratio equation \(\frac{x}{x+4} = \frac{x-2}{x}\).
A1: For solving to find \(x = 4\).
M1: For finding the first term \(a = 8\) and common ratio \(r = 0.5\) and attempting to apply the sum to infinity formula.
A1: For \(S_{\infty} = 16\).

(a) Combining the fractions on the Left Hand Side (LHS) over a common denominator:
\(\text{LHS} = \frac{(1 - \sin \theta)^2 + \cos^2 \theta}{\cos \theta(1 - \sin \theta)}\)
\(\text{LHS} = \frac{1 - 2\sin \theta + \sin^2 \theta + \cos^2 \theta}{\cos \theta(1 - \sin \theta)}\)
Using \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{1 - 2\sin \theta + 1}{\cos \theta(1 - \sin \theta)} = \frac{2 - 2\sin \theta}{\cos \theta(1 - \sin \theta)}\)
Factoring out 2 from the numerator:
\(\text{LHS} = \frac{2(1 - \sin \theta)}{\cos \theta(1 - \sin \theta)}\)
Cancelling the common factor \((1 - \sin \theta)\):
\(\text{LHS} = \frac{2}{\cos \theta} = 2\sec \theta = \text{RHS}\).

(b) Using the identity proven in part (a), the equation can be written as:
\(2\sec 2x = 4 \implies \sec 2x = 2 \implies \cos 2x = \frac{1}{2}\).

Since \(0^\circ \le x \le 180^\circ\), the range for \(2x\) is \(0^\circ \le 2x \le 360^\circ\).
Solving \(\cos 2x = \frac{1}{2}\) in this range:
\(2x = 60^\circ \implies x = 30^\circ\)
\(2x = 300^\circ \implies x = 150^\circ\).
Both solutions lie within the specified interval.

(a)
M1: Putting the left-hand side over a common denominator.
M1: Expanding \((1 - \sin \theta)^2\) to get \(1 - 2\sin \theta + \sin^2 \theta\).
M1: Substituting \(\sin^2 \theta + \cos^2 \theta = 1\) and factoring the numerator.
A1: Correct completion to show \(2\sec \theta\).

(b)
M1: Using the identity to write \(\cos 2x = \frac{1}{2}\) or equivalent.
M1: Finding at least one correct value for \(2x\) (e.g. \(60^\circ\) or \(300^\circ\)).
A1: For \(x = 30^\circ\).
A1: For \(x = 150^\circ\) (with no extra angles in the range).

(a) The total number of people is \(6 + 8 = 14\).
The number of ways to choose 5 people from 14 is given by:
\(\binom{14}{5} = \frac{14!}{5! \times 9!} = 2002\).

(b) To have more women than men on a committee of 5, the possibilities are:
Case 1: 3 women and 2 men
Number of ways = \(\binom{8}{3} \times \binom{6}{2} = 56 \times 15 = 840\).
Case 2: 4 women and 1 man
Number of ways = \(\binom{8}{4} \times \binom{6}{1} = 70 \times 6 = 420\).
Case 3: 5 women and 0 men
Number of ways = \(\binom{8}{5} \times \binom{6}{0} = 56 \times 1 = 56\).

Total number of ways = \(840 + 420 + 56 = 1316\).

(a)
M1: For writing \(\binom{14}{5}\) or \({}^{14}\text{C}_5\).
A1: For 2002.

(b)
B1: For identifying the three correct cases (3W, 2M), (4W, 1M), and (5W, 0M).
M1: For calculating the number of ways for Case 1 (840).
M1: For calculating the number of ways for Case 2 (420).
M1: For calculating the number of ways for Case 3 (56).
A1: For the correct sum of 1316.

**Part (a):**
Using the laws of logarithms:
\(\log_2((x - 3)(x + 1)) = 5\)
\((x - 3)(x + 1) = 2^5\)
\(x^2 - 2x - 3 = 32\)
\(x^2 - 2x - 35 = 0\)
\((x - 7)(x + 5) = 0\)

This gives \(x = 7\) or \(x = -5\).
Since the logarithmic terms require \(x - 3 > 0\) and \(x + 1 > 0\) (meaning \(x > 3\)), the solution \(x = -5\) is invalid.
Therefore, the only valid solution is \(x = 7\).

**Part (b):**
Taking logarithms on both sides (base \(e\) or base 10):
\(\ln(5^{2y - 1}) = \ln(7^y)\)
\((2y - 1)\ln(5) = y\ln(7)\)
\(2y\ln(5) - \ln(5) = y\ln(7)\)
\(y(2\ln(5) - \ln(7)) = \ln(5)\)
\(y = \frac{\ln(5)}{2\ln(5) - \ln(7)}\)

Using a calculator:
\(y \approx \frac{1.6094}{3.2189 - 1.9459} = \frac{1.6094}{1.2730} \approx 1.26\)

**Part (a):**
* **M1** for combining logarithms into a single log: \(\log_2((x - 3)(x + 1))\).
* **M1** for removing logarithms: \((x - 3)(x + 1) = 2^5\) or 32.
* **A1** for expanding and forming the quadratic equation: \(x^2 - 2x - 35 = 0\).
* **A1** for obtaining \(x = 7\) and explicitly rejecting \(x = -5\).

**Part (b):**
* **M1** for taking logarithms on both sides and applying power rule: \((2y-1)\log(5) = y\log(7)\).
* **M1** for rearranging to make \(y\) the subject: \(y = \frac{\log(5)}{2\log(5) - \log(7)}\).
* **A1** for finding \(y \approx 1.26\) (accept awrt 1.26).

**Part (a):**
The perimeter \(P\) of a sector is given by:
\(P = 2r + r\theta\)
Given that \(P = 40\):
\(2r + r\theta = 40 \implies r\theta = 40 - 2r \implies \theta = \frac{40 - 2r}{r}\)

The area \(A\) of a sector is given by:
\(A = \frac{1}{2}r^2\theta\)

Substituting \(\theta\) into the area formula:
\(A = \frac{1}{2}r^2\left(\frac{40 - 2r}{r}\right)\)
\(A = \frac{1}{2}r(40 - 2r)\)
\(A = 20r - r^2\) (as required).

**Part (b):**
Given that \(A = 75\):
\(20r - r^2 = 75\)
\(r^2 - 20r + 75 = 0\)
\((r - 5)(r - 15) = 0\)

This gives the possible values of \(r\):
\(r = 5\) or \(r = 15\)

Using \(\theta = \frac{40 - 2r}{r}\) to find the corresponding values of \(\theta\):
* If \(r = 5\):
\(\theta = \frac{40 - 2(5)}{5} = \frac{30}{5} = 6\) radians.
* If \(r = 15\):
\(\theta = \frac{40 - 2(15)}{15} = \frac{10}{15} = \frac{2}{3}\) radians (or \(0.667\)).

**Part (a):**
* **B1** for writing a correct perimeter equation: \(2r + r\theta = 40\).
* **M1** for substituting their expression for \(\theta\) (or \(r\theta\)) into \(A = \frac{1}{2}r^2\theta\).
* **A1** for obtaining \(A = 20r - r^2\) clearly with no errors.

**Part (b):**
* **M1** for setting \(20r - r^2 = 75\) and attempting to solve the quadratic equation.
* **A1** for finding both correct values of \(r\): \(r = 5\) and \(r = 15\).
* **M1** for using a correct expression to find at least one value of \(\theta\).
* **A1** for both correct corresponding pairs: \(r = 5, \theta = 6\) and \(r = 15, \theta = \frac{2}{3}\) (or \(0.667\)).

**Part (a):**
First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left(\frac{2 + 6}{2}, \frac{3 + 11}{2}\right) = (4, 7)\)

Next, find the gradient \(m\) of the line segment \(AB\):
\(m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2\)

The gradient of the perpendicular bisector is the negative reciprocal of \(m\):
\(m_{\perp} = -\frac{1}{2}\)

The equation of the perpendicular bisector passing through the midpoint \(M(4, 7)\) is:
\(y - 7 = -\frac{1}{2}(x - 4)\)
\(y - 7 = -\frac{1}{2}x + 2\)
\(y = -\frac{1}{2}x + 9\) (or \(x + 2y = 18\))

**Part (b):**
To find where the perpendicular bisector meets the coordinate axes:
* On the \(x\)-axis (point \(C\), where \(y = 0\)):
\(0 = -\frac{1}{2}x + 9 \implies x = 18\), so \(C\) has coordinates \((18, 0)\).
* On the \(y\)-axis (point \(D\), where \(x = 0\)):
\(y = 9\), so \(D\) has coordinates \((0, 9)\).

The area of the right-angled triangle \(OCD\) is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 18 \times 9 = 81\) square units.

**Part (a):**
* **B1** for finding the correct midpoint \((4, 7)\).
* **B1** for finding the correct gradient of the perpendicular bisector \(m_{\perp} = -\frac{1}{2}\) (from gradient of \(AB = 2\)).
* **M1** for attempting to find the equation of a line passing through their midpoint with their perpendicular gradient.
* **A1** for a correct equation of the line in any equivalent form, e.g., \(y = -\frac{1}{2}x + 9\) or \(x + 2y = 18\).

**Part (b):**
* **M1** for setting \(x=0\) and \(y=0\) to find the coordinates of \(C\) and \(D\).
* **A1** for obtaining correct coordinates \((18, 0)\) and \((0, 9)\) (or lengths of 18 and 9).
* **A1** for calculating the correct area of the triangle as 81.