2025 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

To have two distinct real roots, the discriminant of the quadratic equation must be strictly greater than zero. For the equation \(2x^2 + kx + (k+6) = 0\), we have \(a = 2\), \(b = k\), and \(c = k + 6\). Calculating the discriminant \(\Delta\): \(\Delta = b^2 - 4ac = k^2 - 4(2)(k + 6) = k^2 - 8k - 48\). Since we require \(\Delta > 0\): \(k^2 - 8k - 48 > 0\). Factorising the quadratic expression gives \((k - 12)(k + 4) > 0\). The critical values are \(k = 12\) and \(k = -4\). Since the inequality is strictly greater than zero, the required range of values is \(k < -4\) or \(k > 12\).

M1: Attempt to use the discriminant \(b^2 - 4ac\) and set the inequality \(b^2 - 4ac > 0\). A1: Obtain the correct quadratic inequality \(k^2 - 8k - 48 > 0\) or correct critical values \(-4\) and \(12\). A1: State the correct final range of values: \(k < -4\) or \(k > 12\) (or equivalent notation).

In a geometric progression, the ratio between consecutive terms is constant. Therefore: \(\frac{x}{x + 4} = \frac{x - 3}{x}\). Cross-multiplying gives: \(x^2 = (x + 4)(x - 3)\). Expanding the right-hand side yields: \(x^2 = x^2 + x - 12\). Subtracting \(x^2\) from both sides: \(0 = x - 12\), which simplifies to \(x = 12\).

M1: Set up a correct ratio equation using the definition of a geometric progression, e.g., \(\frac{x}{x+4} = \frac{x-3}{x}\). M1: Clear fractions and expand to obtain a linear equation (e.g., \(x^2 = x^2 + x - 12\)). A1: Obtain \(x = 12\).

First, find the mid-point, \(M\), of the line segment \(AB\): \(M = \left( \frac{1 + 5}{2}, \frac{-3 + 5}{2} \right) = (3, 1)\). Next, find the gradient of the line \(AB\), denoted by \(m_{AB}\): \(m_{AB} = \frac{5 - (-3)}{5 - 1} = \frac{8}{4} = 2\). Since the perpendicular bisector is perpendicular to \(AB\), its gradient \(m_{\perp}\) is the negative reciprocal: \(m_{\perp} = -\frac{1}{2}\). Now, using the point-slope form with the mid-point \((3, 1)\) and gradient \(-\frac{1}{2}\): \(y - 1 = -\frac{1}{2}(x - 3)\). Multiplying by 2 and simplifying: \(2y - 2 = -x + 3 \implies x + 2y = 5\) (or \(y = -\frac{1}{2}x + \frac{5}{2}\)).

M1: Correctly calculate the mid-point \((3, 1)\) and find the gradient of \(AB\) (which is \(2\)). M1: Attempt to find the perpendicular gradient \(-\frac{1}{2}\) and use it with the mid-point to form a line equation. A1: Obtain the correct equation of the line in any equivalent form, e.g., \(x + 2y = 5\) or \(y = -\frac{1}{2}x + \frac{5}{2}\).

The perimeter of the sector is given by: \(P = 2r + r\theta = 30 \implies r\theta = 30 - 2r\). The area of the sector is: \(A = \frac{1}{2}r^2\theta = 50\). This can be written as \(\frac{1}{2}r(r\theta) = 50\). Substituting the perimeter equation into the area equation gives: \(\frac{1}{2}r(30 - 2r) = 50 \implies 15r - r^2 = 50 \implies r^2 - 15r + 50 = 0\). Factorising this quadratic equation yields: \((r - 5)(r - 10) = 0\). Therefore, the possible values of \(r\) are \(r = 5\) and \(r = 10\).

M1: Set up two correct equations for the perimeter and area of the sector, e.g., \(2r + r\theta = 30\) and \(\frac{1}{2}r^2\theta = 50\). M1: Eliminate \(\theta\) to form a quadratic equation in \(r\), e.g., \(r^2 - 15r + 50 = 0\). A1: Obtain both correct values: \(r = 5\) and \(r = 10\).

Rewrite the integrand in index form: \(\int_{1}^{4} \left( 6x^{-1/2} - x \right) \text{d}x\). Integrating term by term: \(\left[ 12x^{1/2} - \frac{1}{2}x^2 \right]_{1}^{4}\). Evaluating at the upper limit \(x = 4\): \(12\sqrt{4} - \frac{1}{2}(4)^2 = 24 - 8 = 16\). Evaluating at the lower limit \(x = 1\): \(12\sqrt{1} - \frac{1}{2}(1)^2 = 12 - 0.5 = 11.5\). Subtracting the two values: \(16 - 11.5 = 4.5\) (or \(\frac{9}{2}\)).

M1: Integrate at least one term correctly to obtain either \(k x^{1/2}\) or \(k x^2\). A1: Obtain the correct integrated expression \(12\sqrt{x} - \frac{1}{2}x^2\). A1: Substitute limits correctly to obtain the final exact value \(4.5\) or \(\frac{9}{2}\).

Equating the line and the curve: \( x^2 - 4x + 7 = mx + 3 \), which simplifies to \( x^2 - (4+m)x + 4 = 0 \). For tangency, the discriminant must be zero: \( b^2 - 4ac = 0 \), so \( (-(4+m))^2 - 4(1)(4) = 0 \). This simplifies to \( (4+m)^2 = 16 \), giving \( 4+m = 4 \) or \( 4+m = -4 \). Hence, \( m = 0 \) or \( m = -8 \). When \( m = 0 \), the quadratic becomes \( x^2 - 4x + 4 = 0 \), giving \( x = 2 \). Substituting into the line equation gives \( y = 3 \), so the point of contact is \( (2, 3) \). When \( m = -8 \), the quadratic becomes \( x^2 + 4x + 4 = 0 \), giving \( x = -2 \). Substituting into the line equation gives \( y = -8(-2) + 3 = 19 \), so the point of contact is \( (-2, 19) \).

M1: For equating the line and the curve and forming a 3-term quadratic equation. M1: For applying \( b^2 - 4ac = 0 \) to their quadratic equation. A1: For finding the two correct values of \( m \) (\( m = 0 \) and \( m = -8 \)). A1: For finding the correct coordinates \( (2, 3) \) for \( m = 0 \). A1: For finding the correct coordinates \( (-2, 19) \) for \( m = -8 \).

(a) The 3rd, 5th, and 11th terms of the arithmetic progression are \( a + 2d \), \( a + 4d \), and \( a + 10d \) respectively. Since these form a geometric progression, the common ratio is constant, so \( \frac{a+4d}{a+2d} = \frac{a+10d}{a+4d} \). Multiplying out gives \( (a+4d)^2 = (a+2d)(a+10d) \), which expands to \( a^2 + 8ad + 16d^2 = a^2 + 12ad + 20d^2 \). Simplifying this gives \( 4d^2 + 4ad = 0 \), which factors as \( 4d(d+a) = 0 \). Given that \( d \neq 0 \), we must have \( d + a = 0 \), which proves \( d = -a \). (b) The common ratio \( r \) of the geometric progression is given by \( r = \frac{a+4d}{a+2d} \). Substituting \( d = -a \) into this expression gives \( r = \frac{a + 4(-a)}{a + 2(-a)} = \frac{-3a}{-a} = 3 \).

(a) M1: For writing the terms in terms of \( a \) and \( d \) and setting up the geometric progression condition \( (a+4d)^2 = (a+2d)(a+10d) \). M1: For expanding and simplifying to a quadratic equation in terms of \( a \) and \( d \). A1: For obtaining \( d = -a \) with clear, convincing working. (b) M1: For writing an expression for \( r \) in terms of \( a \) and \( d \) and substituting \( d = -a \). A1: For obtaining \( r = 3 \).

(a) We express \( f(x) \) in completed square form: \( f(x) = 2(x^2 - 6x) + 13 = 2((x-3)^2 - 9) + 13 = 2(x-3)^2 - 5 \). The vertex of the parabola is at \( (3, -5) \). For the function to be one-to-one (and hence have an inverse), the domain \( x \ge k \) must not include the vertex on both sides of the axis of symmetry \( x = 3 \). Thus, the least value of \( k \) is the x-coordinate of the vertex, so \( k = 3 \). (b) Setting \( y = 2(x-3)^2 - 5 \), we solve for \( x \): \( y + 5 = 2(x-3)^2 \implies \frac{y+5}{2} = (x-3)^2 \). Taking the square root, since \( x \ge 3 \), we select the positive square root: \( x - 3 = \sqrt{\frac{y+5}{2}} \implies x = 3 + \sqrt{\frac{y+5}{2}} \). Replacing \( y \) with \( x \), we get \( f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}} \). The domain of \( f^{-1} \) is the range of \( f \). Since the vertex of \( f(x) \) is at \( (3, -5) \) and the coefficient of \( x^2 \) is positive, the range of \( f \) for \( x \ge 3 \) is \( f(x) \ge -5 \). Hence, the domain of \( f^{-1} \) is \( x \ge -5 \).

(a) M1: For expressing \( f(x) \) in completed square form \( 2(x-3)^2 - 5 \). A1: For identifying \( k = 3 \). (b) M1: For attempting to make \( x \) the subject of \( y = 2(x-3)^2 - 5 \) (or using the quadratic formula). A1: For obtaining \( f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}} \) (or equivalent correct form, must be in terms of \( x \)). A1: For stating the correct domain \( x \ge -5 \).

To find the equation of the curve, we integrate the derivative: \( y = \int \frac{6}{\sqrt{3x+1}} \, dx = \int 6(3x+1)^{-1/2} \, dx \). Using the rule for integrating linear compositions, we get: \( y = \frac{6(3x+1)^{1/2}}{\frac{1}{2} \times 3} + C = \frac{6(3x+1)^{1/2}}{1.5} + C = 4\sqrt{3x+1} + C \). Since the curve passes through the point \( (8, 12) \), we substitute \( x = 8 \) and \( y = 12 \): \( 12 = 4\sqrt{3(8)+1} + C \implies 12 = 4\sqrt{25} + C \implies 12 = 20 + C \), which gives \( C = -8 \). Thus, the equation of the curve is \( y = 4\sqrt{3x+1} - 8 \).

M1: For integrating to obtain a term of the form \( k(3x+1)^{1/2} \). A1: For dividing by the coefficient of \( x \) (which is 3) or having \( 2 \times 6 \) in the numerator. A1: For the correct integrated expression \( 4(3x+1)^{1/2} \) (with or without \( +C \)). M1: For substituting \( x = 8 \) and \( y = 12 \) into their integrated expression to find \( C \). A1: For obtaining \( y = 4\sqrt{3x+1} - 8 \) (or equivalent).

(a) Completing the square: \(2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5\). So, \(a = 2\), \(b = 3\), \(c = -5\). (b) For \(f\) to have an inverse, the function must be one-to-one. The vertex of the quadratic curve is at \(x = 3\). Since the domain is restricted to \(x \le k\), the largest possible value of \(k\) is 3. (c) Let \(y = 2(x-3)^2 - 5\). Rearranging to make \(x\) the subject: \(y + 5 = 2(x-3)^2\) leads to \(\frac{y+5}{2} = (x-3)^2\). Since \(x \le 3\), we must take the negative square root: \(x - 3 = -\sqrt{\frac{y+5}{2}} \implies x = 3 - \sqrt{\frac{y+5}{2}}\). Hence, \(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\). The domain of \(f^{-1}\) is the range of \(f\). Since \(x \le 3\), the minimum value of \(f(x)\) is \(-5\), and \(f(x)\) increases as \(x\) decreases from 3. Thus, the range of \(f\) is \(f(x) \ge -5\), so the domain of \(f^{-1}\) is \(x \ge -5\).

(a) M1: Attempt to complete the square, reaching the form \(2(x-p)^2 + q\). A1: Correctly obtain \(2(x-3)^2\). A1: Correctly obtain \(-5\). (b) B1: State \(k = 3\) (or allow \(k \le 3\) if identified as the boundary). (c) M1: Rearrange equation of the form \(y = a(x-b)^2+c\) to express \((x-b)^2\) in terms of \(y\). A1: Correctly choose the negative square root due to domain restriction \(x \le 3\). A1: State correct inverse function \(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\) (must be in terms of \(x\)). B1: State correct domain \(x \ge -5\).

(a) Combine the terms on the LHS over a common denominator: \(LHS = \frac{(1 - \cos\theta)^2 + \sin^2\theta}{\sin\theta(1 - \cos\theta)}\). Expand the numerator: \(LHS = \frac{1 - 2\cos\theta + \cos^2\theta + \sin^2\theta}{\sin\theta(1 - \cos\theta)}\). Use the identity \(\sin^2\theta + \cos^2\theta = 1\): \(LHS = \frac{1 - 2\cos\theta + 1}{\sin\theta(1 - \cos\theta)} = \frac{2 - 2\cos\theta}{\sin\theta(1 - \cos\theta)}\). Factorise the numerator: \(LHS = \frac{2(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)}\). Cancel the common factor \((1 - \cos\theta)\): \(LHS = \frac{2}{\sin\theta}\). Hence, the identity is proved. (b) Using the identity from part (a) with \(\theta = 2x - 30^\circ\), the equation becomes: \(\frac{2}{\sin(2x - 30^\circ)} = 3 \implies \sin(2x - 30^\circ) = \frac{2}{3}\). Let \(\theta = 2x - 30^\circ\). Since \(0^\circ \le x \le 180^\circ\), the range for \(\theta\) is \(-30^\circ \le \theta \le 330^\circ\). Solve \(\sin\theta = \frac{2}{3}\): The basic angle is \(\alpha = \sin^{-1}(\frac{2}{3}) \approx 41.81^\circ\). Since sine is positive, \(\theta\) lies in the first or second quadrant: \(\theta = 41.81^\circ\) or \(\theta = 180^\circ - 41.81^\circ = 138.19^\circ\). For \(\theta = 41.81^\circ\): \(2x - 30^\circ = 41.81^\circ \implies 2x = 71.81^\circ \implies x \approx 35.9^\circ\) (to 1 decimal place). For \(\theta = 138.19^\circ\): \(2x - 30^\circ = 138.19^\circ \implies 2x = 168.19^\circ \implies x \approx 84.1^\circ\) (to 1 decimal place).

(a) M1: Combine fractions with a common denominator of \(\sin\theta(1 - \cos\theta)\). M1: Expand numerator correctly to obtain \(1 - 2\cos\theta + \cos^2\theta + \sin^2\theta\) and apply \(\sin^2\theta + \cos^2\theta = 1\). A1: Factorise numerator to \(2(1-\cos\theta)\) and cancel correctly to achieve the RHS. (b) M1: Recognise the connection to part (a) and write \(\sin(2x - 30^\circ) = \frac{2}{3}\). A1: Find the first value of the angle \(2x - 30^\circ \approx 41.8^\circ\) (or basic angle). M1: Find second angle in the correct range, i.e., \(180^\circ - 41.8^\circ = 138.2^\circ\). A1: Obtain \(x \approx 35.9^\circ\). A1: Obtain \(x \approx 84.1^\circ\). Deduct 1 mark for any extra incorrect solutions within the range.

To solve the inequality \( |2x - 3| < |x + 4| \), we can square both sides since both sides are non-negative:

\((2x - 3)^2 < (x + 4)^2\)

Expanding both sides:

\(4x^2 - 12x + 9 < x^2 + 8x + 16\)

Rearranging to form a quadratic inequality:

\(3x^2 - 20x - 7 < 0\)

Factorising the quadratic expression:

\((3x + 1)(x - 7) < 0\)

This yields the critical values \(x = -\frac{1}{3}\) and \(x = 7\). Since the inequality is less than zero, the solution is the region between these critical values:

\(-\frac{1}{3} < x < 7\)

M1: For squaring both sides and expanding to obtain a 3-term quadratic expression, or for solving two linear equations to find both critical values.
A1: For obtaining the correct critical values of \(-\frac{1}{3}\) and \(7\).
A1: For the correct final inequality range \(-\frac{1}{3} < x < 7\).

Let \(u = e^x\). The equation becomes a quadratic in terms of \(u\):

\(u^2 - 5u + 4 = 0\)

Factorising the quadratic equation:

\((u - 1)(u - 4) = 0\)

This gives:

\(u = 1\) or \(u = 4\)

Substituting back \(u = e^x\):

1) \(e^x = 1 \implies x = \ln 1 = 0\)

2) \(e^x = 4 \implies x = \ln 4\)

Thus, the exact solutions are \(x = 0\) and \(x = \ln 4\) (which can also be written as \(2\ln 2\)).

M1: For forming a quadratic equation in terms of \(e^x\) and solving to find two values for \(e^x\).
A1: For obtaining \(e^x = 1\) and \(e^x = 4\).
A1: For finding both correct exact solutions \(x = 0\) and \(x = \ln 4\) (or \(2\ln 2\)).

To find the stationary point, we differentiate \( y = \frac{\ln x}{x^2} \) using the quotient rule:

\(\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{(x^2)^2}\)

\(\frac{dy}{dx} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}\)

Setting \(\frac{dy}{dx} = 0\) to find stationary points:

\(1 - 2\ln x = 0 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}\)

Substitute \(x = e^{1/2}\) back into the equation for \(y\):

\(y = \frac{\ln(e^{1/2})}{(e^{1/2})^2} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}\)

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

M1: For applying the quotient rule (or product rule) correctly to find the derivative \(\frac{dy}{dx}\).
A1: For setting \(\frac{dy}{dx} = 0\) and solving to find \(x = e^{1/2}\) (or \(\sqrt{e}\)).
A1: For obtaining the correct exact y-coordinate \(\frac{1}{2e}\) and stating the final coordinates.

First, we find the antiderivative of the function:

\(\int \cos\left(3x - \frac{\pi}{6}\right) \, dx = \frac{1}{3} \sin\left(3x - \frac{\pi}{6}\right)\)

Now, we evaluate this antiderivative at the upper and lower limits of integration:

At the upper limit \(x = \frac{\pi}{6}\):

\(\frac{1}{3} \sin\left(3\left(\frac{\pi}{6}\right) - \frac{\pi}{6}\right) = \frac{1}{3} \sin\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \frac{1}{3} \sin\left(\frac{\pi}{3}\right) = \frac{1}{3} \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{6}\)

At the lower limit \(x = 0\):

\(\frac{1}{3} \sin\left(3(0) - \frac{\pi}{6}\right) = \frac{1}{3} \sin\left(-\frac{\pi}{6}\right) = \frac{1}{3} \left(-\frac{1}{2}\right) = -\frac{1}{6}\)

Subtracting the lower limit value from the upper limit value:

\(\int_{0}^{\frac{\pi}{6}} \cos\left(3x - \frac{\pi}{6}\right) \, dx = \frac{\sqrt{3}}{6} - \left(-\frac{1}{6}\right) = \frac{\sqrt{3} + 1}{6}\)

M1: For integrating to obtain an expression of the form \(k \sin(3x - \frac{\pi}{6})\) where \(k \neq 0\).
A1: For the correct integrated expression \( \frac{1}{3} \sin(3x - \frac{\pi}{6}) \).
A1: For substituting the limits correctly to obtain the exact value of \(\frac{\sqrt{3} + 1}{6}\).

1. Write the right-hand side as a single logarithm using the laws of logarithms:
\(2\ln x - \ln 3 = \ln(x^2) - \ln 3 = \ln\left(\frac{x^2}{3}\right)\).

2. Equate the arguments of the logarithms on both sides:
\(2x + 3 = \frac{x^2}{3}\).

3. Multiply by 3 and rearrange into standard quadratic form:
\(x^2 - 6x - 9 = 0\).

4. Solve the quadratic equation using the quadratic formula:
\(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2} = \frac{6 \pm \sqrt{72}}{2} = \frac{6 \pm 6\sqrt{2}}{2} = 3 \pm 3\sqrt{2}\).

5. Since \(\ln x\) is only defined for real numbers where \(x > 0\), we must check the validity of both roots:
- \(3 + 3\sqrt{2} \approx 7.24 > 0\) (valid)
- \(3 - 3\sqrt{2} \approx -1.24 < 0\) (invalid, must be rejected)

Thus, the only valid solution is \(x = 3 + 3\sqrt{2}\).

M1: Apply logarithm laws correctly to express \(2\ln x - \ln 3\) as a single logarithm \(\ln\left(\frac{x^2}{3}\right)\).
M1: Remove logarithms to obtain a quadratic equation, e.g., \(2x + 3 = \frac{x^2}{3}\).
M1: Attempt to solve their 3-term quadratic equation \(x^2 - 6x - 9 = 0\) using the quadratic formula or completing the square.
A1: Obtain the roots \(3 \pm 3\sqrt{2}\).
A1: Reject the negative root with a valid reason (e.g., noting that \(x > 0\) is required for \(\ln x\)) and state the final exact answer as \(3 + 3\sqrt{2}\).

1. To find the stationary point, we first differentiate \(y\) with respect to \(x\) using the quotient rule:
Let \(u = e^{2x}\) and \(v = x - 1\).
Then \(u' = 2e^{2x}\) and \(v' = 1\).

2. Apply the quotient rule formula \(\frac{dy}{dx} = \frac{u'v - uv'}{v^2}\):
\(\frac{dy}{dx} = \frac{2e^{2x}(x-1) - e^{2x}(1)}{(x-1)^2}\)

3. Factorise and simplify the numerator:
\(\frac{dy}{dx} = \frac{e^{2x}(2x - 2 - 1)}{(x-1)^2} = \frac{e^{2x}(2x - 3)}{(x-1)^2}\)

4. Set the derivative to 0 to find the stationary point:
\(\frac{e^{2x}(2x - 3)}{(x-1)^2} = 0\)
Since \(e^{2x} \neq 0\) for all real \(x\), we solve:
\(2x - 3 = 0 \implies x = \frac{3}{2}\) (or \(1.5\)). This value is valid as it lies in the domain \(x > 1\).

5. Substitute \(x = \frac{3}{2}\) back into the original equation to find the exact \(y\)-coordinate:
\(y = \frac{e^{2(3/2)}}{\frac{3}{2} - 1} = \frac{e^3}{\frac{1}{2}} = 2e^3\).

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

M1: Attempt differentiation using the quotient rule (or product rule) with correct general structure.
A1: Obtain correct derivative \(\frac{dy}{dx} = \frac{e^{2x}(2x-3)}{(x-1)^2}\) or equivalent.
M1: Equate their derivative to 0 and solve for \(x\).
M1: Substitute their \(x\)-value back into the original curve equation to find the corresponding \(y\)-value.
A1: State the exact coordinates as \(\left(\frac{3}{2}, 2e^3\right)\) or as \(x = \frac{3}{2}, y = 2e^3\).

To find the stationary point, we first differentiate the equation \(y = x^2 \ln(3x)\) using the product rule: \(\frac{dy}{dx} = 2x \ln(3x) + x^2 \cdot \frac{3}{3x} = 2x \ln(3x) + x\). Setting \(\frac{dy}{dx} = 0\) gives \(x(2\ln(3x) + 1) = 0\). Since \(x > 0\), we must have \(2\ln(3x) + 1 = 0\), which simplifies to \(\ln(3x) = -\frac{1}{2}\). Solving for \(x\) gives \(3x = e^{-1/2} = \frac{1}{\sqrt{e}}\) so \(x = \frac{1}{3\sqrt{e}}\). Substituting this back into the original curve equation: \(y = \left(\frac{1}{3\sqrt{e}}\right)^2 \ln\left(3 \cdot \frac{1}{3\sqrt{e}}\right) = \frac{1}{9e} \left(-\frac{1}{2}\right) = -\frac{1}{18e}\). To determine the nature of this point, we find the second derivative: \(\frac{d^2y}{dx^2} = 2\ln(3x) + 2x \cdot \frac{1}{x} + 1 = 2\ln(3x) + 3\). Evaluating this at the stationary point: \(\frac{d^2y}{dx^2} = 2\left(-\frac{1}{2}\right) + 3 = 2 > 0\). Since the second derivative is positive, this is a minimum point.

M1: Correct use of the product rule to find \(\frac{dy}{dx}\). A1: Correct expression \(2x \ln(3x) + x\). M1: Setting their derivative to 0 and solving for \(x\). A1: Correct exact x-coordinate \(x = \frac{1}{3\sqrt{e}}\) (or equivalent). A1: Correct exact y-coordinate \(y = -\frac{1}{18e}\). M1: Correctly determining the nature of the stationary point (e.g. via second derivative or first derivative sign change) to conclude it is a minimum.

(a) Using algebraic long division or equating coefficients: \(2x^2 + 5x - 1 = (ax + b)(2x + 1) + c = 2ax^2 + (a + 2b)x + (b + c)\). Comparing coefficients: \(2a = 2 \implies a = 1\), \(a + 2b = 5 \implies 1 + 2b = 5 \implies b = 2\), \(b + c = -1 \implies 2 + c = -1 \implies c = -3\). Thus, the expression is \(x + 2 - \frac{3}{2x+1}\). (b) Using the result from part (a): \(\int_{1}^{3} \left(x + 2 - \frac{3}{2x+1}\right) dx = \left[ \frac{1}{2}x^2 + 2x - \frac{3}{2}\ln|2x+1| \right]_{1}^{3}\). Substituting the upper limit \(3\): \(\left(\frac{1}{2}(3)^2 + 2(3) - \frac{3}{2}\ln|2(3)+1|\right) = \frac{21}{2} - \frac{3}{2}\ln 7\). Substituting the lower limit \(1\): \(\left(\frac{1}{2}(1)^2 + 2(1) - \frac{3}{2}\ln|2(1)+1|\right) = \frac{5}{2} - \frac{3}{2}\ln 3\). Subtracting the lower limit evaluation from the upper limit evaluation: \(\left(\frac{21}{2} - \frac{3}{2}\ln 7\right) - \left(\frac{5}{2} - \frac{3}{2}\ln 3\right) = 8 + \frac{3}{2}\ln 3 - \frac{3}{2}\ln 7\). Thus, \(p = 8\), \(q = 1.5\) (or \(\frac{3}{2}\)), and \(r = -1.5\) (or \(-\frac{3}{2}\)).

(a) M1: For attempting algebraic division or equating coefficients. A1: For obtaining any two of \(a = 1\), \(b = 2\), \(c = -3\) correctly. A1: For all three constants correct. (b) M1: For integrating the expression from (a) to obtain \(kx^2 + mx + n\ln(2x+1)\) with correct ln term. M1: For substituting the limits 3 and 1 correctly into their integrated expression. A1: For obtaining the correct final exact expression \(8 + 1.5\ln 3 - 1.5\ln 7\) (or equivalent with rational coefficients).