2023 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

Let the first term be \(a\) and the common difference be \(d\).

Using the formula for the \(n\)-th term of an arithmetic progression, \(u_n = a + (n-1)d\):
\(u_3 = a + 2d = 8\) (Equation 1)

Using the formula for the sum of the first \(n\) terms, \(S_n = \frac{n}{2}(2a + (n-1)d)\):
\(S_9 = \frac{9}{2}(2a + 8d) = 108\)
\(9(a + 4d) = 108\)
\(a + 4d = 12\) (Equation 2)

Subtract Equation 1 from Equation 2:
\((a + 4d) - (a + 2d) = 12 - 8\)
\(2d = 4\)
\(d = 2\)

Substitute \(d = 2\) back into Equation 1:
\(a + 2(2) = 8\)
\(a + 4 = 8\)
\(a = 4\)

M1: State a correct equation for either \(u_3\) or \(S_9\) in terms of \(a\) and \(d\).
A1: Obtain both correct simplified equations: \(a + 2d = 8\) and \(a + 4d = 12\) (or equivalent).
M1: Attempt to solve the two simultaneous equations for \(a\) and \(d\).
A0.5: Obtain both correct values: \(a = 4\) and \(d = 2\).

First, rewrite the integrand with fractional and negative exponents:
\(\int_{1}^{4} \left( 3x^{1/2} - 2x^{-2} \right) \text{d}x\)

Integrate term by term:
\(\int \left( 3x^{1/2} - 2x^{-2} \right) \text{d}x = \left[ \frac{3x^{3/2}}{3/2} - \frac{2x^{-1}}{-1} \right]_{1}^{4}\)
\(= \left[ 2x^{3/2} + \frac{2}{x} \right]_{1}^{4}\)

Now substitute the limits 4 and 1 into the integrated expression:
At \(x = 4\):
\(2(4)^{3/2} + \frac{2}{4} = 2(8) + 0.5 = 16.5\)

At \(x = 1\):
\(2(1)^{3/2} + \frac{2}{1} = 2 + 2 = 4\)

Subtract the lower limit value from the upper limit value:
\(16.5 - 4 = 12.5\) (or \(\frac{25}{2}\))

M1: Integrate at least one term correctly (increasing power of \(x\) by 1).
A1: Obtain the fully correct integrated expression: \(2x^{3/2} + \frac{2}{x}\) (or equivalent).
M1: Substitute limits 4 and 1 into their integrated expression and subtract.
A0.5: Obtain the correct final answer \(12.5\) (or \(\frac{25}{2}\)).

To find the points of intersection, equate the line and the curve:
\(x^2 - 4x + 1 = kx - 3\)

Rearrange to form a quadratic equation equal to zero:
\(x^2 - (4+k)x + 4 = 0\)

For the line to not intersect the curve, this quadratic equation must have no real roots. Therefore, the discriminant must be negative (\(b^2 - 4ac < 0\)):
\(a = 1\), \(b = -(4+k)\), \(c = 4\)

\(b^2 - 4ac = (-(4+k))^2 - 4(1)(4) < 0\)
\((k+4)^2 - 16 < 0\)
\((k+4)^2 < 16\)

Taking the square root on both sides:
\(-4 < k+4 < 4\)

Subtracting 4 from all parts:
\(-8 < k < 0\)

M1: Eliminate \(y\) to form a quadratic equation in \(x\).
A1: Obtain the correct quadratic equation in standard form: \(x^2 - (4+k)x + 4 = 0\) (or equivalent).
M1: Use the discriminant condition \(b^2 - 4ac < 0\).
A0.5: Obtain the correct range \(-8 < k < 0\) (or equivalent interval notation).

The perimeter \(P\) of a sector is given by:
\(P = 2r + r\theta = 30\) (Equation 1)

The area \(A\) of a sector is given by:
\(A = \frac{1}{2}r^2\theta = 50\) (Equation 2)

From Equation 1, express \(\theta\) in terms of \(r\):
\(r\theta = 30 - 2r \implies \theta = \frac{30-2r}{r}\)

Substitute this expression for \(\theta\) into Equation 2:
\(\frac{1}{2}r^2 \left( \frac{30-2r}{r} \right) = 50\)
\(\frac{1}{2}r(30 - 2r) = 50\)
\(15r - r^2 = 50\)
\(r^2 - 15r + 50 = 0\)

Factorise the quadratic equation:
\((r - 5)(r - 10) = 0\)

Thus, the possible values of \(r\) are \(r = 5\) and \(r = 10\). Both lead to positive values of \(\theta\) (\(\theta = 4\) and \(\theta = 1\) respectively), so both are valid.

M1: State the correct formulas for perimeter and area: \(2r + r\theta = 30\) and \(\frac{1}{2}r^2\theta = 50\).
A1: Eliminate \(\theta\) to obtain a correct quadratic equation in terms of \(r\) only (e.g., \(r^2 - 15r + 50 = 0\)).
M1: Solve the quadratic equation by factorisation or formula.
A0.5: Obtain both correct values of \(r\): \(r = 5\) and \(r = 10\).

First, use the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\) to rewrite the equation in terms of \(\cos x\):
\(3(1 - \cos^2 x) - 5 \cos x - 1 = 0\)
\(3 - 3 \cos^2 x - 5 \cos x - 1 = 0\)
\(3 \cos^2 x + 5 \cos x - 2 = 0\)

Let \(y = \cos x\). The equation becomes:
\(3y^2 + 5y - 2 = 0\)

Factorise the quadratic expression:
\((3y - 1)(y + 2) = 0\)

This gives two possible values for \(\cos x\):
1) \(\cos x = \frac{1}{3}\)
2) \(\cos x = -2\) (This equation has no real solutions because \(-1 \le \cos x \le 1\)).

Now, solve \(\cos x = \frac{1}{3}\) within the range \(0^\circ \le x \le 360^\circ\):
\(x = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ\)

Using the symmetry of the cosine function, the second solution is:
\(x = 360^\circ - 70.53^\circ \approx 289.47^\circ\)

Rounding both answers to 1 decimal place gives:
\(x = 70.5^\circ\) and \(x = 289.5^\circ\)

M1: Use the identity \(1 - \cos^2 x\) to express the equation as a quadratic in \(\cos x\).
A1: Obtain the correct quadratic equation: \(3\cos^2 x + 5\cos x - 2 = 0\).
M1: Solve the quadratic equation to find \(\cos x = \frac{1}{3}\).
A0.5: Obtain both correct angles: \(70.5^\circ\) and \(289.5^\circ\) (rounded to 1 d.p.).

(a) The first three terms of the arithmetic progression (AP) are:
\(T_1 = a\)
\(T_2 = a + d\)
\(T_3 = a + 2d\)

The first three terms of the geometric progression (GP) are:
\(G_1 = a\)
\(G_2 = ar\)
\(G_3 = ar^2\)

We are given that:
\(T_1 = G_1 \implies a = a\)
\(T_3 = G_2 \implies a + 2d = ar \implies 2d = a(r - 1) \implies d = \frac{1}{2}a(r - 1)\) (as required).

We are also given that:
\(T_2 = G_3 \implies a + d = ar^2\)

Substitute the expression for \(d\) into this equation:
\(a + \frac{1}{2}a(r-1) = ar^2\)

Since \(a \neq 0\) (otherwise the GP would have a sum to infinity of 0, which contradicts part b), we can divide both sides by \(a\):
\(1 + \frac{1}{2}(r-1) = r^2\)

Multiply through by 2:
\(2 + r - 1 = 2r^2 \implies 2r^2 - r - 1 = 0\)

Factorize the quadratic:
\((2r + 1)(r - 1) = 0\)

Thus, the two possible values of \(r\) are:
\(r = -\frac{1}{2}\) and \(r = 1\).

(b) Since we are given that \(r \neq 1\), we must have \(r = -\frac{1}{2}\).

The sum to infinity of the GP is:
\(S_\infty = \frac{a}{1-r} = 12\)
\(\frac{a}{1 - (-1/2)} = 12 \implies \frac{a}{1.5} = 12 \implies a = 18\).

Now, substitute \(a = 18\) and \(r = -\frac{1}{2}\) to find \(d\):
\(d = \frac{1}{2}(18)\left(-\frac{1}{2} - 1\right) = 9\left(-\frac{3}{2}\right) = -13.5\).

The sum of the first 20 terms of the AP is given by:
\(S_{20} = \frac{20}{2} \left[ 2a + 19d \right]\)
\(S_{20} = 10 \left[ 2(18) + 19(-13.5) \right]\)
\(S_{20} = 10 \left[ 36 - 256.5 \right] = 10 \left[ -220.5 \right] = -2205\).

Part (a) [5.5 marks]:
- M1: For setting up the equations relating AP terms and GP terms, specifically \(a + 2d = ar\) and \(a + d = ar^2\).
- A1: For correctly deriving \(d = \frac{1}{2}a(r-1)\).
- M1: For substituting \(d\) into the second equation to get an equation in terms of \(a\) and \(r\).
- A1: For obtaining the simplified quadratic equation \(2r^2 - r - 1 = 0\) (or equivalent).
- A1.5: For solving the quadratic to find both values: \(r = -\frac{1}{2}\) and \(r = 1\) (award 1 mark for one correct value).

Part (b) [4 marks]:
- M1: For using \(S_\infty = 12\) with the correct value of \(r = -\frac{1}{2}\) to formulate an equation for \(a\).
- A1: For obtaining \(a = 18\).
- M1: For calculating \(d = -13.5\) and attempting to use the AP sum formula for \(S_{20}\).
- A1: For obtaining the correct final answer of \(-2205\).

(a) Since there is a stationary point at \(P(2, 5)\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 2\).
Substitute \(x = 2\) into the given expression:
\(\frac{k}{(2(2) - 3)^2} - 2 = 0\)
\(\frac{k}{(4 - 3)^2} = 2 \implies k = 2\) (as required).

(b) Substitute \(k = 2\) back into the derivative:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(2x - 3)^{-2} - 2\)

To find the equation of the curve, integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(x\):
\(y = \int \left( 2(2x - 3)^{-2} - 2 \right) \mathrm{d}x\)
\(y = \frac{2(2x - 3)^{-1}}{(-1) \times 2} - 2x + C\)
\(y = -\frac{1}{2x-3} - 2x + C\)

Since the curve passes through the point \(P(2, 5)\), substitute \(x = 2\) and \(y = 5\) to find \(C\):
\(5 = -\frac{1}{2(2)-3} - 2(2) + C\)
\(5 = -1 - 4 + C\)
\(5 = -5 + C \implies C = 10\)

Thus, the equation of the curve is:
\(y = 10 - 2x - \frac{1}{2x-3}\).

(c) To find the stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{2}{(2x-3)^2} - 2 = 0\)
\((2x-3)^2 = 1\)
\(2x-3 = 1 \implies x = 2\) (this corresponds to point \(P\))
\(2x-3 = -1 \implies 2x = 2 \implies x = 1\)

Substitute \(x = 1\) into the curve equation to find the \(y\)-coordinate:
\(y = 10 - 2(1) - \frac{1}{2(1)-3} = 10 - 2 - (-1) = 9\).

So the coordinates of the other stationary point are \((1, 9)\).

To find its nature, compute the second derivative of \(y\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left( 2(2x-3)^{-2} - 2 \right)\)
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4(2x-3)^{-3} \times 2 = -\frac{8}{(2x-3)^3}\)

Evaluate the second derivative at \(x = 1\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{8}{(2(1)-3)^3} = -\frac{8}{-1} = 8\)

Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\) at \(x = 1\), the stationary point \((1, 9)\) is a minimum point.

Part (a) [1.5 marks]:
- M1: For substituting \(x = 2\) and setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
- A0.5: For demonstrating that \(k = 2\) clearly.

Part (b) [3.5 marks]:
- M1: For attempting to integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) (power increased by 1 and divided by new power, involving chain rule dividing by 2).
- A1: For obtaining correct integration including constant \(C\): \(y = -\frac{1}{2x-3} - 2x + C\).
- M1: For substituting \((2, 5)\) to calculate \(C\).
- A0.5: For stating the correct equation of the curve.

Part (c) [4.5 marks]:
- M1: For setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\).
- A1: For identifying the other stationary point's \(x\)-coordinate as \(x = 1\).
- A1: For calculating the corresponding \(y\)-coordinate as \(y = 9\).
- M1: For finding the second derivative \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{8}{(2x-3)^3}\) and substituting \(x = 1\).
- A0.5: For concluding that the point is a minimum because \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8 > 0\).

(a) To prove the identity, start with the Left-Hand Side (LHS) and combine the fractions over a common denominator:
\(\text{LHS} = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}\)
\(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}\)
\(\text{LHS} = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}\)

Use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{(1) + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
\(\text{LHS} = \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
\(\text{LHS} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}\)

Cancel the common term \((1 + \cos \theta)\) from the numerator and denominator:
\(\text{LHS} = \frac{2}{\sin \theta}\) (which is the RHS).

(b) The equation to solve is:
\(\frac{\sin 2x}{1 + \cos 2x} + \frac{1 + \cos 2x}{\sin 2x} = \frac{4}{3}\tan 2x\)

Apply the identity proved in part (a), replacing \(\theta\) with \(2x\):
\(\frac{2}{\sin 2x} = \frac{4}{3}\tan 2x\)

Substitute \(\tan 2x = \frac{\sin 2x}{\cos 2x}\):
\(\frac{2}{\sin 2x} = \frac{4\sin 2x}{3\cos 2x}\)

Cross-multiply to eliminate the fractions:
\(6\cos 2x = 4\sin^2 2x\)
\(3\cos 2x = 2\sin^2 2x\)

Use \(\sin^2 2x = 1 - \cos^2 2x\) to express the entire equation in terms of \(\cos 2x\):
\(3\cos 2x = 2(1 - \cos^2 2x)\)
\(3\cos 2x = 2 - 2\cos^2 2x\)
\(2\cos^2 2x + 3\cos 2x - 2 = 0\)

Factorize the quadratic in terms of \(\cos 2x\):
\((2\cos 2x - 1)(\cos 2x + 2) = 0\)

This yields two possible equations:
1) \(\cos 2x = -2\) (No solution, as \(-1 \le \cos \theta \le 1\))
2) \(\cos 2x = \frac{1}{2}\)

Given the domain \(0^\circ < x < 180^\circ\), the domain for \(2x\) is \(0^\circ < 2x < 360^\circ\).
Solve \(\cos 2x = \frac{1}{2}\):
\(2x = 60^\circ\) or \(2x = 300^\circ\)

Thus:
\(x = 30^\circ\) or \(x = 150^\circ\).

Part (a) [3.5 marks]:
- M1: For combining fractions with a common denominator of \(\sin \theta(1 + \cos \theta)\).
- A1: For expanding the numerator correctly to \(\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta\).
- M1: For using \(\sin^2 \theta + \cos^2 \theta = 1\) and factoring the numerator to \(2(1 + \cos \theta)\).
- A0.5: For canceling the bracket and correctly completing the proof.

Part (b) [6 marks]:
- M1: For utilizing the identity from part (a) to rewrite the equation as \(\frac{2}{\sin 2x} = \frac{4}{3}\tan 2x\).
- M1: For substituting \(\tan 2x = \frac{\sin 2x}{\cos 2x}\) and cross-multiplying.
- M1: For using \(\sin^2 2x = 1 - \cos^2 2x\) to set up a quadratic equation in \(\cos 2x\).
- A1: For finding the correct quadratic equation: \(2\cos^2 2x + 3\cos 2x - 2 = 0\).
- M1: For solving the quadratic equation to get \(\cos 2x = 0.5\) (and showing that \(\cos 2x = -2\) yields no solution).
- A1: For the final answers \(x = 30^\circ\) and \(x = 150^\circ\) (award 0.5 marks for each correct value; deduct 0.5 marks for any extra values in the range).

(a) Since \(OC\) bisects the angle \(AOB\), the angle of the sector \(OAC\) is:
\(\angle AOC = \frac{\theta}{2}\) radians.

The line segment \(PC\) is tangent to the circle at \(C\). Since \(OC\) is a radius, the tangent is perpendicular to the radius at the point of contact, making triangle \(OPC\) right-angled at \(C\).

In right-angled triangle \(OPC\):
\(\tan\left(\frac{\theta}{2}\right) = \frac{PC}{OC} = \frac{PC}{r} \implies PC = r \tan\left(\frac{\theta}{2}\right)\).

The area of the right-angled triangle \(OPC\) is:
\(\text{Area}(\triangle OPC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OC \times PC\)
\(\text{Area}(\triangle OPC) = \frac{1}{2} \times r \times r \tan\left(\frac{\theta}{2}\right) = \frac{1}{2} r^2 \tan\left(\frac{\theta}{2}\right)\).

The area of the sector \(OAC\) with angle \(\frac{\theta}{2}\) is:
\(\text{Area}(\text{sector } OAC) = \frac{1}{2} r^2 \left( \frac{\theta}{2} \right) = \frac{1}{4} r^2 \theta\).

The area of the region \(R\) is obtained by subtracting the area of the sector \(OAC\) from the area of the triangle \(OPC\):
\(\text{Area}(R) = \text{Area}(\triangle OPC) - \text{Area}(\text{sector } OAC)\)
\(\text{Area}(R) = \frac{1}{2} r^2 \tan\left(\frac{\theta}{2}\right) - \frac{1}{4} r^2 \theta = \frac{1}{2} r^2 \left( \tan\left(\frac{\theta}{2}\right) - \frac{\theta}{2} \right)\) (as required).

(b) The boundary of the region \(R\) consists of three parts: the straight line segment \(AP\), the tangent segment \(PC\), and the arc \(AC\).
Given \(\theta = 1.2\) radians and \(r = 8\text{ cm}\):
- The angle \(\angle AOC = \frac{1.2}{2} = 0.6\) radians.
- Length of arc \(AC = r \times \frac{\theta}{2} = 8 \times 0.6 = 4.8\text{ cm}\).
- Length of tangent \(PC = r \tan\left(\frac{\theta}{2}\right) = 8 \tan(0.6) \approx 8(0.68414) \approx 5.4731\text{ cm}\).
- In the right-angled triangle \(OPC\):
\(\cos(0.6) = \frac{OC}{OP} = \frac{8}{OP} \implies OP = \frac{8}{\cos(0.6)} \approx \frac{8}{0.82534} \approx 9.6930\text{ cm}\).
- Length of segment \(AP = OP - OA = 9.6930 - 8 = 1.6930\text{ cm}\).

The perimeter of region \(R\) is the sum of these three lengths:
\(\text{Perimeter} = AP + PC + \text{arc } AC\)
\(\text{Perimeter} \approx 1.6930 + 5.4731 + 4.8 = 11.966\text{ cm}\).

To 3 significant figures, the perimeter of the region \(R\) is \(12.0\text{ cm}\).

Part (a) [5.5 marks]:
- M1: For identifying that angle \(\angle AOC = \frac{\theta}{2}\).
- A1: For deriving \(PC = r \tan\left(\frac{\theta}{2}\right)\) from right-angled triangle \(OPC\).
- A1: For showing the area of triangle \(OPC = \frac{1}{2} r^2 \tan\left(\frac{\theta}{2}\right)\).
- M1: For writing the correct formula for the area of sector \(OAC\) with angle \(\frac{\theta}{2}\).
- A1.5: For subtracting the sector area from the triangle area to achieve the final shown result clearly.

Part (b) [4 marks]:
- M1: For computing the length of the arc \(AC = 4.8\text{ cm}\).
- M1: For computing the length of the tangent \(PC \approx 5.47\text{ cm}\).
- M1: For finding \(OP\) using cosine ratio, and subtracting radius to get \(AP \approx 1.69\text{ cm}\).
- A1: For summing the three parts to obtain \(12.0\text{ cm}\) (must be rounded to 3 significant figures).

(a) To complete the square for \(\mathrm{f}(x) = 2x^2 - 12x + 13\):
First, factor out 2 from the \(x^2\) and \(x\) terms:
\(\mathrm{f}(x) = 2(x^2 - 6x) + 13\)

Now, complete the square inside the bracket:
\(x^2 - 6x = (x - 3)^2 - 9\)

Substitute this back into the expression:
\(\mathrm{f}(x) = 2[(x - 3)^2 - 9] + 13\)
\(\mathrm{f}(x) = 2(x - 3)^2 - 18 + 13\)
\(\mathrm{f}(x) = 2(x - 3)^2 - 5\).

Thus, \(a = 2\), \(b = 3\), and \(c = -5\).

(b) For a quadratic function to have an inverse, it must be a one-to-one function. The vertex of this parabola is at \((3, -5)\), which means the axis of symmetry is at \(x = 3\).
Since the domain is given as \(x \le k\), the function is strictly decreasing (and thus one-to-one) for all values of \(x\) up to the vertex.
Therefore, the largest value of \(k\) for which the function is one-to-one is \(k = 3\).

(c) For \(k = 3\), the domain is \(x \le 3\). To find the inverse function, set \(y = \mathrm{f}(x)\) and solve for \(x\):
\(y = 2(x-3)^2 - 5\)
\(y + 5 = 2(x-3)^2\)
\(\frac{y+5}{2} = (x-3)^2\)

Taking the square root of both sides:
\(x - 3 = \pm\sqrt{\frac{y+5}{2}}\)

Since the domain of \(\mathrm{f}\) is \(x \le 3\), we have \(x - 3 \le 0\). Therefore, we must choose the negative square root:
\(x - 3 = -\sqrt{\frac{y+5}{2}}\)
\(x = 3 - \sqrt{\frac{y+5}{2}}\)

Replacing \(y\) with \(x\), we find:
\(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).

To find the domain of \(\mathrm{f}^{-1}\), we find the range of \(\mathrm{f}\) on its domain \(x \le 3\).
At the vertex \(x = 3\), the minimum value of \(\mathrm{f}(x)\) is \(-5\).
As \(x \to -\infty\), \(\mathrm{f}(x) \to \infty\).
Thus, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -5\).

Consequently, the domain of \(\mathrm{f}^{-1}\) is \(x \ge -5\).

Part (a) [3 marks]:
- M1: For attempting to complete the square by factoring out 2.
- A1: For obtaining the correct squared bracket expression: \(2(x-3)^2\).
- A1: For the fully correct completed square form: \(2(x-3)^2 - 5\).

Part (b) [1 mark]:
- B1: For stating \(k = 3\) correctly.

Part (c) [5.5 marks]:
- M1: For setting \(y = 2(x-3)^2 - 5\) and attempting to isolate the squared term.
- A1: For obtaining \((x-3)^2 = \frac{y+5}{2}\).
- M1: For taking the square root and correctly selecting the negative root due to the domain condition \(x \le 3\).
- A1: For expressing the inverse function correctly as \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).
- M1: For stating that the domain of \(\mathrm{f}^{-1}\) corresponds to the range of \(\mathrm{f}\).
- A0.5: For stating the domain as \(x \ge -5\).

(a) First, write the equation of the curve in index form:
\(y = 6x^{-1} + 2x\)

Differentiate with respect to \(x\) to find the gradient function:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = -6x^{-2} + 2 = 2 - \frac{6}{x^2}\)

Find the gradient of the tangent at point \(A\) where \(x = 2\):
\(m_t = 2 - \frac{6}{2^2} = 2 - 1.5 = 0.5\)

The gradient of the normal, \(m_n\), is the negative reciprocal of the tangent gradient:
\(m_n = -\frac{1}{0.5} = -2\)

Find the equation of the normal passing through \(A(2, 7)\) with gradient \(-2\):
\(y - 7 = -2(x - 2)\)
\(y - 7 = -2x + 4\)
\(y + 2x = 11\)

This is in the required form \(ay + bx = c\), where \(a = 1\), \(b = 2\), and \(c = 11\).

(b) To find where the normal intersects the curve again, solve the equations \(y + 2x = 11\) and \(y = \frac{6}{x} + 2x\) simultaneously:
Substitute \(y = 11 - 2x\) into the curve's equation:
\(11 - 2x = \frac{6}{x} + 2x\)

Since \(x > 0\), multiply the entire equation by \(x\):
\(11x - 2x^2 = 6 + 2x^2\)

Rearrange into a quadratic equation equal to zero:
\(4x^2 - 11x + 6 = 0\)

Since \(x = 2\) is the x-coordinate of point \(A\), \((x - 2)\) must be a factor of the quadratic expression. Factorize the quadratic:
\((x - 2)(4x - 3) = 0\)

This gives two solutions:
\(x = 2\) (at point \(A\))
\(x = \frac{3}{4} = 0.75\) (at point \(B\))

To find the \(y\)-coordinate of \(B\), substitute \(x = 0.75\) into the normal equation:
\(y = 11 - 2(0.75) = 11 - 1.5 = 9.5\)

Thus, the coordinates of the other intersection point \(B\) are \((0.75, 9.5)\) or \(\left(\frac{3}{4}, \frac{19}{2}\right)\).

Part (a) [4.5 marks]:
- M1: For attempting to differentiate the curve equation, achieving at least one correct term of \(-6x^{-2}\) or \(2\).
- A1: For the completely correct derivative: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - \frac{6}{x^2}\).
- M1: For substituting \(x = 2\) to find the tangent gradient, and taking the negative reciprocal to find the normal gradient \(-2\).
- A1.5: For writing the equation of the normal in the form \(y + 2x = 11\) (or any integer multiple thereof).

Part (b) [5 marks]:
- M1: For setting up the simultaneous equation by equating the curve and the normal: \(11 - 2x = \frac{6}{x} + 2x\).
- M1: For multiplying by \(x\) and forming a 3-term quadratic equation.
- A1: For obtaining the correct quadratic equation: \(4x^2 - 11x + 6 = 0\).
- M1: For factorizing or solving the quadratic to find \(x = 0.75\).
- A1: For finding \(y = 9.5\) and stating the final coordinates of \(B\) as \((0.75, 9.5)\) or equivalent fraction form.

To solve the equation:

\[2\ln(x+1) - \ln(x-1) = \ln(x+5)\]

First, apply the power law of logarithms to the first term:

\[\ln(x+1)^2 - \ln(x-1) = \ln(x+5)\]

Next, apply the subtraction law of logarithms on the left-hand side:

\[\ln\left(\frac{(x+1)^2}{x-1}\right) = \ln(x+5)\]

Since the logarithms on both sides have the same base, we can equate the arguments:

\[\frac{(x+1)^2}{x-1} = x+5\]

Multiply both sides by \((x-1)\):

\[(x+1)^2 = (x-1)(x+5)\]

Expand both sides:

\[x^2 + 2x + 1 = x^2 + 4x - 5\]

Subtract \(x^2\) from both sides and rearrange to solve the linear equation:

\[2x + 1 = 4x - 5\]
\[2x = 6\]
\[x = 3\]

We must check if \(x = 3\) is valid in the original logarithmic terms:
- \(x+1 = 4 > 0\)
- \(x-1 = 2 > 0\)
- \(x+5 = 8 > 0\)

Since all terms are well-defined, \(x = 3\) is the unique valid solution.

M1: For applying the power law to obtain \(\ln(x+1)^2\).
M1: For combining the terms on the left-hand side using subtraction law to obtain \(\ln\left(\frac{(x+1)^2}{x-1}\right)\) (or equivalent rearrangement).
M1: For equating the arguments to obtain \((x+1)^2 = (x-1)(x+5)\) and expanding correctly.
A0.8: For obtaining \(x = 3\) and confirming it as the only valid solution.

We start with the trigonometric equation:

\[3\sec^2 \theta - 5\tan \theta - 5 = 0\]

Use the Pythagorean identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to rewrite the equation in terms of \(\tan \theta\):

\[3(1 + \tan^2 \theta) - 5\tan \theta - 5 = 0\]

Expand the brackets:

\[3 + 3\tan^2 \theta - 5\tan \theta - 5 = 0\]

Simplify to obtain a quadratic equation in terms of \(\tan \theta\):

\[3\tan^2 \theta - 5\tan \theta - 2 = 0\]

Factorise the quadratic equation:

\[(3\tan \theta + 1)(\tan \theta - 2) = 0\]

This gives two possible equations for \(\tan \theta\):

1) \(\tan \theta = 2\)
For \(0^\circ \le \theta \le 180^\circ\):
\[\theta = \tan^{-1}(2) \approx 63.43^\circ \Rightarrow \theta = 63.4^\circ\] (to 1 d.p.)

2) \(\tan \theta = -\frac{1}{3}\)
For \(0^\circ \le \theta \le 180^\circ\):
\[\theta = 180^\circ - \tan^{-1}\left(\frac{1}{3}\right) \approx 180^\circ - 18.43^\circ = 161.57^\circ \Rightarrow \theta = 161.6^\circ\] (to 1 d.p.)

Thus, the solutions in the given range are \(\theta = 63.4^\circ\) and \(\theta = 161.6^\circ\).

M1: For using \(\sec^2 \theta = 1 + \tan^2 \theta\) to form a quadratic equation in \(\tan \theta\).
M1: For successfully solving the quadratic equation to find \(\tan \theta = 2\) and \(\tan \theta = -\frac{1}{3}\).
A1: For obtaining \(\theta = 63.4^\circ\) (or correct to 1 decimal place).
A0.8: For obtaining \(\theta = 161.6^\circ\) (or correct to 1 decimal place), and no other solutions in range.

To find the coordinates of the stationary point, we first differentiate the curve using the product rule.

Let \(u = 3x - 1 \Rightarrow \frac{du}{dx} = 3\).
Let \(v = e^{-3x} \Rightarrow \frac{dv}{dx} = -3e^{-3x}\).

Applying the product rule \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\):

\[\frac{dy}{dx} = (3x - 1)(-3e^{-3x}) + 3e^{-3x}\]

Factorise out the common factor \(e^{-3x}\):

\[\frac{dy}{dx} = e^{-3x} [-3(3x - 1) + 3]\]
\[\frac{dy}{dx} = e^{-3x} [-9x + 3 + 3]\]
\[\frac{dy}{dx} = e^{-3x}(6 - 9x)\]

At a stationary point, \(\frac{dy}{dx} = 0\):

\[e^{-3x}(6 - 9x) = 0\]

Since \(e^{-3x} \neq 0\) for all real values of \(x\), we solve:

\[6 - 9x = 0 \Rightarrow 9x = 6 \Rightarrow x = \frac{2}{3}\]

To find the corresponding \(y\)-coordinate, substitute \(x = \frac{2}{3}\) back into the equation of the curve:

\[y = \left(3\left(\frac{2}{3}\right) - 1\right) e^{-3\left(\frac{2}{3}\right)}\]
\[y = (2 - 1) e^{-2} = e^{-2}\]

Thus, the exact coordinates of the stationary point are \(\left(\frac{2}{3}, e^{-2}\right)\) or \(\left(\frac{2}{3}, \frac{1}{e^2}\right)\).

M1: For applying the product rule correctly to differentiate \((3x-1)e^{-3x}\).
A1: For obtaining the correct simplified derivative \(\frac{dy}{dx} = e^{-3x}(6-9x)\).
M1: For setting their \(\frac{dy}{dx} = 0\) and solving for \(x\).
A0.8: For obtaining the correct exact coordinates \(\left(\frac{2}{3}, e^{-2}\right)\) or equivalent.

(a) Since \((x - 2)\) is a factor of \(P(x)\), we have:
\[P(2) = 0\]
\[2(2)^3 + a(2)^2 + b(2) - 6 = 0\]
\[16 + 4a + 2b - 6 = 0\]
\[4a + 2b = -10\]
\[2a + b = -5 \quad \text{--- (1)}\]

When \(P(x)\) is divided by \((2x + 1)\), the remainder is \(-10\), which means:
\[P\left(-\frac{1}{2}\right) = -10\]
\[2\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) - 6 = -10\]
\[2\left(-\frac{1}{8}\right) + a\left(\frac{1}{4}\right) - \frac{b}{2} - 6 = -10\]
\[-\frac{1}{4} + \frac{a}{4} - \frac{b}{2} - 6 = -10\]
Multiplying the entire equation by 4:
\[-1 + a - 2b - 24 = -40\]
\[a - 2b - 25 = -40\]
\[a - 2b = -15 \quad \text{--- (2)}\]

From (1), we can write \(b = -5 - 2a\). Substituting this into (2):
\[a - 2(-5 - 2a) = -15\]
\[a + 10 + 4a = -15\]
\[5a = -25 \implies a = -5\]

Substituting \(a = -5\) back into the expression for \(b\):
\[b = -5 - 2(-5) = 5\]

Thus, \(a = -5\) and \(b = 5\).

(b) Substituting \(a = -5\) and \(b = 5\) into \(P(x)\):
\[P(x) = 2x^3 - 5x^2 + 5x - 6\]

Since \((x - 2)\) is a factor, we can divide \(P(x)\) by \((x - 2)\) using polynomial division or by comparing coefficients:
\[2x^3 - 5x^2 + 5x - 6 = (x - 2)(2x^2 + kx + 3)\]
Comparing the coefficient of \(x^2\):
\[-4 + k = -5 \implies k = -1\]

So, the quadratic factor is \(2x^2 - x + 3\).
Thus, the fully factorised polynomial is:
\[P(x) = (x - 2)(2x^2 - x + 3)\]

To find the real roots of \(P(x) = 0\), we set each factor to zero:
1) \(x - 2 = 0 \implies x = 2\)
2) \(2x^2 - x + 3 = 0\)

We check the discriminant of the quadratic factor:
\[\Delta = (-1)^2 - 4(2)(3) = 1 - 24 = -23\]
Since the discriminant is negative (\(\Delta < 0\)), the quadratic equation \(2x^2 - x + 3 = 0\) has no real roots.

Therefore, the only exact real root of \(P(x) = 0\) is \(x = 2\).

(a)
- M1: Attempts to use the factor theorem by setting \(P(2) = 0\).
- M1: Attempts to use the remainder theorem by setting \(P(-0.5) = -10\).
- A1: Obtains two correct linear equations in terms of \(a\) and \(b\), e.g., \(2a + b = -5\) and \(a - 2b = -15\).
- M1: Solves the two linear equations simultaneously.
- A1: Obtains correct values \(a = -5\) and \(b = 5\).

(b)
- M1: Attempts division of \(P(x)\) by \(x-2\) or uses equating coefficients to find the quadratic factor.
- A1: Correctly identifies the quadratic factor as \(2x^2 - x + 3\).
- M1: Evaluates the discriminant or attempts to solve the quadratic factor to check for real roots.
- A1.6: Concludes correctly that \(x = 2\) is the only real root because the quadratic factor has no real solutions.

(a) We start with the equation:
\[y = A e^{kx}\]
Taking the natural logarithm of both sides:
\[\ln y = \ln(A e^{kx}) = \ln A + \ln(e^{kx}) = kx + \ln A\]
This equation is of the linear form \(Y = mX + C\), where \(Y = \ln y\), \(X = x\), \(m = k\), and \(C = \ln A\). Hence, the graph of \(\ln y\) against \(x\) is a straight line with:
- \(\text{Gradient} = k\)
- \(\text{Vertical intercept} = \ln A\)

(b) Using the given data points for \((x, \ln y)\), which are \((1.2, 3.84)\) and \((3.5, 7.06)\):
The gradient \(k\) of the line is:
\[k = \frac{7.06 - 3.84}{3.5 - 1.2} = \frac{3.22}{2.3} = 1.40\]

Using the linear equation \(\ln y = kx + \ln A\) and substituting \(x = 1.2\) and \(\ln y = 3.84\):
\[3.84 = 1.4(1.2) + \ln A\]
\[3.84 = 1.68 + \ln A\]
\[\ln A = 2.16\]

To find \(A\), we take the exponential of both sides:
\[A = e^{2.16} \approx 8.6711\]
To 3 significant figures, \(k = 1.40\) and \(A = 8.67\).

(c) When \(x = 2.0\), we use the relation \(\ln y = kx + \ln A\):
\[\ln y = 1.4(2.0) + 2.16 = 2.8 + 2.16 = 4.96\]
Thus,
\[y = e^{4.96} \approx 142.59\]
To 3 significant figures, \(y = 143\).

(a)
- M1: Applies natural logarithms correctly to both sides of the equation \(y = A e^{kx}\).
- A1: Obtains \(\ln y = kx + \ln A\) and states that this matches the equation of a straight line.
- A1: Correctly identifies the gradient as \(k\) and the vertical intercept as \(\ln A\).

(b)
- M1: Sets up the equation to find the gradient \(k\) from the given points.
- A1: Obtains \(k = 1.4\) (or \(1.40\)).
- M1: Uses their value of \(k\) and one data point to set up an equation for \(\ln A\).
- A1: Obtains \(\ln A = 2.16\).
- A1: Correctly calculates \(A = e^{2.16} \approx 8.67\).

(c)
- M1: Substitutes \(x = 2.0\) into either \(y = A e^{kx}\) or \(\ln y = kx + \ln A\).
- A0.6: Obtains \(y \approx 143\) (accept answers in the range 142 to 143 due to rounding of intermediate values).

(a) We begin with the equation:
\[3 \cos 2\theta + 7 \sin \theta - 5 = 0\]
Using the double-angle identity for cosine, \(\cos 2\theta = 1 - 2 \sin^2 \theta\), we substitute this into the equation:
\[3(1 - 2 \sin^2 \theta) + 7 \sin \theta - 5 = 0\]
Expanding the terms:
\[3 - 6 \sin^2 \theta + 7 \sin \theta - 5 = 0\]
Simplifying and collecting like terms:
\[-6 \sin^2 \theta + 7 \sin \theta - 2 = 0\]
Multiplying the entire equation by \(-1\) gives the desired form:
\[6 \sin^2 \theta - 7 \sin \theta + 2 = 0\]

(b) Let \(u = \sin \theta\). The quadratic equation becomes:
\[6u^2 - 7u + 2 = 0\]
We can solve this quadratic by factorising:
\[(3u - 2)(2u - 1) = 0\]
This gives two possible values for \(u\):
\[u = \frac{2}{3} \quad \text{or} \quad u = \frac{1}{2}\]

Case 1: \(\sin \theta = \frac{1}{2}\)
In the range \(0^\circ \le \theta \le 360^\circ\), the solutions are:
\[\theta = 30^\circ\]
\[\theta = 180^\circ - 30^\circ = 150^\circ\]

Case 2: \(\sin \theta = \frac{2}{3}\)
The principal value is:
\[\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ \approx 41.8^\circ\]
The second solution in the given range is:
\[\theta = 180^\circ - 41.81^\circ = 138.19^\circ \approx 138.2^\circ\]

All solutions in the interval \(0^\circ \le \theta \le 360^\circ\) are \(\theta = 30.0^\circ\), \(41.8^\circ\), \(138.2^\circ\), and \(150.0^\circ\).

(a)
- M1: Uses the identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\).
- M1: Expands the brackets correctly and simplifies the resulting equation.
- A1: Obtains the correct equation \(6 \sin^2 \theta - 7 \sin \theta + 2 = 0\) with no errors shown.

(b)
- M1: Attempts to solve the quadratic equation \(6u^2 - 7u + 2 = 0\) by factorisation, formula, or completing the square.
- A1: Correctly identifies \(\sin\theta = 1/2\) and \(\sin\theta = 2/3\).
- M1: Evaluates inverse sine to find the principal value for at least one of their solutions.
- A1: Correctly finds \(\theta = 30^\circ\) and \(\theta = 150^\circ\).
- A1: Correctly finds \(\theta = 41.8^\circ\) and \(\theta = 138.2^\circ\) (rounded to 1 decimal place).
- A1.6: Final answers must include all four correct solutions with no extra solutions in the given range.

(a) To find \(\frac{dy}{dx}\), we can use the quotient rule on \(y = \frac{u}{v}\), where:
\[u = x + 2 \implies \frac{du}{dx} = 1\]
\[v = e^{3x} \implies \frac{dv}{dx} = 3e^{3x}\]

Using the quotient rule formula, \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\[\frac{dy}{dx} = \frac{e^{3x}(1) - (x + 2)(3e^{3x})}{(e^{3x})^2}\]
Factoring out \(e^{3x}\) from the numerator:
\[\frac{dy}{dx} = \frac{e^{3x}[1 - 3(x + 2)]}{e^{6x}}\]
\[\frac{dy}{dx} = \frac{1 - 3x - 6}{e^{3x}}\]
\[\frac{dy}{dx} = \frac{-3x - 5}{e^{3x}}\]

Alternatively, using the product rule on \(y = (x + 2)e^{-3x}\):
\[\frac{dy}{dx} = (1)e^{-3x} + (x + 2)(-3e^{-3x}) = e^{-3x}(1 - 3x - 6) = (-3x - 5)e^{-3x}\]
This is equivalent to \(\frac{-3x - 5}{e^{3x}}\).

(b) To find the stationary point, we set \(\frac{dy}{dx} = 0\):
\[\frac{-3x - 5}{e^{3x}} = 0\]
Since \(e^{3x} \ne 0\) for all real values of \(x\), we solve:
\[-3x - 5 = 0 \implies 3x = -5 \implies x = -\frac{5}{3}\]

To find the corresponding exact \(y\)-coordinate, we substitute \(x = -\frac{5}{3}\) back into the equation of the curve:
\[y = \frac{-\frac{5}{3} + 2}{e^{3(-\frac{5}{3})}} = \frac{\frac{1}{3}}{e^{-5}} = \frac{1}{3} e^5\]

Thus, the exact coordinates of the stationary point are \(\left(-\frac{5}{3}, \frac{1}{3}e^5\right)\).

(c) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\). Representing \(\frac{dy}{dx}\) as \((-3x - 5)e^{-3x}\):
\[\frac{d^2y}{dx^2} = (-3)e^{-3x} + (-3x - 5)(-3e^{-3x})\]
\[\frac{d^2y}{dx^2} = e^{-3x}[-3 - 3(-3x - 5)] = e^{-3x}[-3 + 9x + 15] = (9x + 12)e^{-3x}\]

Now, evaluating this second derivative at \(x = -\frac{5}{3}\):
\[\left.\frac{d^2y}{dx^2}\right|_{x = -5/3} = \left(9\left(-\frac{5}{3}\right) + 12\right)e^{-3(-5/3)} = (-15 + 12)e^5 = -3e^5\]
Since \(e^5 > 0\), \(-3e^5 < 0\). Because the second derivative is negative, the stationary point is a maximum.

(a)
- M1: Applies the product rule or quotient rule to differentiate \(y\).
- A1: Obtains correct components of differentiation (e.g., \(1\) and \(3e^{3x}\)).
- M1: Correctly substitutes components into the quotient rule or product rule formula.
- A1: Simplifies the derivative correctly to show \(\frac{dy}{dx} = \frac{-3x - 5}{e^{3x}}\).

(b)
- M1: Sets their \(\frac{dy}{dx} = 0\) and solves for \(x\).
- A1: Obtains \(x = -\frac{5}{3}\) (or equivalent).
- M1: Substitutes their \(x\) value back into the original curve equation.
- A0.6: Obtains the exact coordinate \(y = \frac{1}{3}e^5\) (or \(\frac{e^5}{3}\)).

(c)
- M1: Finds the second derivative and evaluates it at their \(x\)-coordinate, or tests the sign of the gradient on either side of \(x = -\frac{5}{3}\).
- A1: Obtains negative value (e.g., \(-3e^5\)) and concludes it is a maximum (with clear reasoning).