Edexcel IGCSE · Thinka-original Practice Paper

2025 Edexcel IGCSE Further Pure Mathematics Practice Paper with Answers

Thinka Jun 2025 Cambridge International A Level-Style Mock — Further Pure Mathematics

200 marks240 mins2025
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 Cambridge International A Level Further Pure Mathematics paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (4PM1/01)

Answer all eleven questions. Write your answers in the spaces provided. Without sufficient working, correct answers may be awarded no marks.
11 Question · 100 marks
Question 1 · Kinematics
7 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds (\(t \ge 0\)), the velocity, \(v\text{ m/s}\), of \(P\) is given by \(v = 6t^2 - 18t + 12\). (a) Find the values of \(t\) for which \(P\) is instantaneously at rest. (b) Find the acceleration of \(P\) when \(t = 3\). (c) Find the total distance travelled by \(P\) in the first 3 seconds of its motion.
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Worked solution

For (a): We set the velocity equal to zero to find when the particle is instantaneously at rest: \(6t^2 - 18t + 12 = 0\). Dividing by 6 gives \(t^2 - 3t + 2 = 0\). Factoring the quadratic, we get \((t - 1)(t - 2) = 0\). Thus, \(t = 1\) and \(t = 2\). For (b): The acceleration \(a\) is the derivative of the velocity \(v\) with respect to \(t\): \(a = \frac{dv}{dt} = 12t - 18\). Substituting \(t = 3\) into the expression: \(a = 12(3) - 18 = 36 - 18 = 18\text{ m/s}^2\). For (c): The particle changes its direction of motion at \(t = 1\) and \(t = 2\). To find the total distance travelled in the first 3 seconds, we must find the distance travelled in the intervals \([0, 1]\), \([1, 2]\), and \([2, 3]\). The displacement \(s(t)\) is given by \(s(t) = \int (6t^2 - 18t + 12) dt = 2t^3 - 9t^2 + 12t + C\). Assuming \(s(0) = 0\), we have \(C = 0\), so \(s(t) = 2t^3 - 9t^2 + 12t\). Evaluating \(s(t)\) at the critical times: \(s(0) = 0\), \(s(1) = 2(1)^3 - 9(1)^2 + 12(1) = 5\), \(s(2) = 2(2)^3 - 9(2)^2 + 12(2) = 16 - 36 + 24 = 4\), \(s(3) = 2(3)^3 - 9(3)^2 + 12(3) = 54 - 81 + 36 = 9\). The distance travelled in the interval \([0, 1]\) is \(|s(1) - s(0)| = |5 - 0| = 5\) m. The distance travelled in the interval \([1, 2]\) is \(|s(2) - s(1)| = |4 - 5| = 1\) m. The distance travelled in the interval \([2, 3]\) is \(|s(3) - s(2)| = |9 - 4| = 5\) m. The total distance travelled is \(5 + 1 + 5 = 11\) m.

Marking scheme

Part (a): M1 for setting \(v = 0\) and attempting to solve the quadratic equation (e.g. factoring to obtain two linear brackets). A1 for both \(t = 1\) and \(t = 2\). Part (b): M1 for differentiating \(v\) to find an expression for the acceleration \(a = 12t - 18\). A1 for substituting \(t = 3\) to get \(18\). Part (c): M1 for integrating \(v\) with respect to \(t\) to find the displacement function (at least two terms integrated correctly). M1 for calculating \(s(t)\) at the boundaries \(t = 0, 1, 2, 3\) and attempting to sum the absolute differences. A1 for a correct total distance of \(11\) m.
Question 2 · Structured
13 marks
The points \(A\) and \(B\) have coordinates \((2, 3)\) and \((8, 6)\) respectively. The line \(L_1\) passes through \(A\) and \(B\).

(a) Find an equation for \(L_1\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\), and \(c\) are integers. (3 marks)

The line \(L_2\) is perpendicular to \(L_1\) and passes through \(B\).

(b) Find an equation for \(L_2\). (3 marks)

The point \(C\) lies on \(L_2\) such that the length of \(BC\) is \(3\sqrt{5}\).

Given that the \(x\)-coordinate of \(C\) is less than the \(x\)-coordinate of \(B\),

(c) find the coordinates of \(C\). (4 marks)

(d) Find the area of triangle \(ABC\). (3 marks)
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Worked solution

(a) The gradient of the line \(L_1\) passing through \(A(2, 3)\) and \(B(8, 6)\) is:

\(m_1 = \dfrac{6 - 3}{8 - 2} = \dfrac{3}{6} = \dfrac{1}{2}\)

The equation of \(L_1\) is:

\(y - 3 = \dfrac{1}{2}(x - 2)\)

\(2(y - 3) = x - 2\)

\(2y - 6 = x - 2\)

\(x - 2y + 4 = 0\)

(b) Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) satisfies:

\(m_2 = -\dfrac{1}{m_1} = -2\)

Since \(L_2\) passes through \(B(8, 6)\), its equation is:

\(y - 6 = -2(x - 8)\)

\(y - 6 = -2x + 16\)

\(2x + y - 22 = 0\)

(c) Let the coordinates of \(C\) be \((x, y)\). Since \(C\) lies on \(L_2\), its coordinates satisfy \(y = 22 - 2x\).

The length of \(BC\) is \(3\sqrt{5}\), so:

\(BC^2 = (x - 8)^2 + (y - 6)^2 = (3\sqrt{5})^2 = 45\)

Substitute \(y = 22 - 2x\) into this equation:

\((x - 8)^2 + ((22 - 2x) - 6)^2 = 45\)

\((x - 8)^2 + (16 - 2x)^2 = 45\)

\((x - 8)^2 + 4(8 - x)^2 = 45\)

Since \((8 - x)^2 = (x - 8)^2\):

\(5(x - 8)^2 = 45\)

\((x - 8)^2 = 9\)

\(x - 8 = \pm 3\)

We are given that the \(x\)-coordinate of \(C\) is less than the \(x\)-coordinate of \(B\) (which is \(8\)), so \(x < 8\). Thus:

\(x - 8 = -3 \implies x = 5\)

Substitute \(x = 5\) back to find \(y\):

\(y = 22 - 2(5) = 12\)

So the coordinates of \(C\) are \((5, 12)\).

(d) Since \(L_1\) and \(L_2\) are perpendicular, triangle \(ABC\) has a right angle at \(B\).

The length of \(AB\) is:

\(AB = \sqrt{(8 - 2)^2 + (6 - 3)^2} = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}\)

Since \(BC = 3\sqrt{5}\):

\(\text{Area} = \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times 3\sqrt{5} \times 3\sqrt{5} = \dfrac{45}{2} = 22.5\)

Alternative Method:
Using the coordinate area formula with \(A(2, 3)\), \(B(8, 6)\), and \(C(5, 12)\):

\(\text{Area} = \dfrac{1}{2} |2(6 - 12) + 8(12 - 3) + 5(3 - 6)| = \dfrac{1}{2} |-12 + 72 - 15| = \dfrac{1}{2} |45| = 22.5\)

Marking scheme

Part (a):
- M1: For an attempt to find the gradient of \(L_1\).
- M1: For an attempt to find the equation of \(L_1\) using their gradient.
- A1: Correct equation in the form \(ax + by + c = 0\), e.g., \(x - 2y + 4 = 0\) (or any integer multiple).

Part (b):
- M1: For using the perpendicular gradient rule \(m_2 = -1/m_1\).
- M1: For an attempt to find the equation of \(L_2\) using their perpendicular gradient and point \(B\).
- A1: Correct equation in any form, e.g., \(y = -2x + 22\) or \(2x + y - 22 = 0\).

Part (c):
- M1: For setting up an expression for the distance \(BC^2\) substituting the equation of \(L_2\).
- M1: For setting their distance equation equal to \(45\) and attempting to solve the quadratic equation.
- A1: For obtaining \(x = 5\) (rejecting \(x = 11\) due to the condition \(x < 8\)).
- A1: For the correct coordinates of \(C(5, 12)\).

Part (d):
- M1: For finding the length of \(AB = 3\sqrt{5}\) (or \(\sqrt{45}\)).
- M1: For applying the area formula \(\frac{1}{2} \times AB \times BC\) or the coordinate area formula.
- A1: For the correct area of \(22.5\) (or \(45/2\)).
Question 3 · Trigonometric circles & sector areas
6 marks
A sector of a circle of radius \(r\) cm has an angle of \(\theta\) radians. A second sector of a circle has a radius of \((r+2)\) cm and an angle of \((\theta - 0.5)\) radians, where \(\theta > 0.5\).

The two sectors have the same arc length.

The area of the second sector is \(1.5\) times the area of the first sector.

Find the value of \(r\) and the value of \(\theta\).
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Worked solution

Using the formula for the arc length of a sector, \(s = r\theta\):
For the first sector: \(s_1 = r\theta\)
For the second sector: \(s_2 = (r+2)(\theta - 0.5)\)

Since the arc lengths are equal:
\(r\theta = (r+2)(\theta - 0.5)\)
\(r\theta = r\theta - 0.5r + 2\theta - 1\)
\(0.5r = 2\theta - 1 \implies r = 4\theta - 2\) (Equation 1)

Using the formula for the area of a sector, \(A = \frac{1}{2}r^2\theta\):
For the first sector: \(A_1 = \frac{1}{2}r^2\theta\)
For the second sector: \(A_2 = \frac{1}{2}(r+2)^2(\theta - 0.5)\)

Since the area of the second sector is \(1.5\) times the area of the first sector:
\(\frac{1}{2}(r+2)^2(\theta - 0.5) = 1.5 \times \frac{1}{2}r^2\theta\)
\((r+2)^2(\theta - 0.5) = 1.5r^2\theta\) (Equation 2)

Substitute \(r+2 = 4\theta\) and \(r = 4\theta - 2\) from Equation 1 into Equation 2:
\((4\theta)^2(\theta - 0.5) = 1.5(4\theta - 2)^2\theta\)
\(16\theta^2(\theta - 0.5) = 1.5(16\theta^2 - 16\theta + 4)\theta\)

Since \(\theta > 0.5\), we can divide both sides by \(\theta\):
\(16\theta(\theta - 0.5) = 1.5(16\theta^2 - 16\theta + 4)\)
\(16\theta^2 - 8\theta = 24\theta^2 - 24\theta + 6\)

Rearranging the terms to form a quadratic equation:
\(8\theta^2 - 16\theta + 6 = 0\)
\(4\theta^2 - 8\theta + 3 = 0\)

Factorising the quadratic equation:
\((2\theta - 3)(2\theta - 1) = 0\)

This gives \(\theta = 1.5\) or \(\theta = 0.5\).

Since \(\theta > 0.5\) (or because \(\theta = 0.5\) would make \(r = 0\), which is invalid for a circle), we choose:
\(\theta = 1.5\)

Substituting \(\theta = 1.5\) back into Equation 1:
\(r = 4(1.5) - 2 = 4\).

Marking scheme

M1: Equates the two arc lengths to set up a linear equation between \(r\) and \(\theta\): \(r\theta = (r+2)(\theta - 0.5)\).
A1: Correctly simplifies the equation to obtain \(r = 4\theta - 2\) (or equivalent).
M1: Sets up the area relation: \(\frac{1}{2}(r+2)^2(\theta - 0.5) = 1.5 \times \frac{1}{2}r^2\theta\).
M1: Substitutes the linear relation into the area equation to form a quadratic equation in either \(\theta\) or \(r\), e.g., \(4\theta^2 - 8\theta + 3 = 0\).
A1: Solves the quadratic equation to obtain \(\theta = 1.5\) and justifies the rejection of \(\theta = 0.5\).
A1: Obtains \(r = 4\) and \(\theta = 1.5\) (both values must be correct for this mark).
Question 4 · Inequalities & Linear programming
8 marks
(a) Find the set of values of \(x\) for which \(\frac{4x - 8}{x - 3} < x + 1\). (5) (b) Hence, find the set of values of \(y\) for which \(\frac{4y^2 - 8}{y^2 - 3} < y^2 + 1\). (3)
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Worked solution

Part (a) Rearrange the inequality to have 0 on one side: \(\frac{4x - 8}{x - 3} - (x + 1) < 0\). Combine into a single fraction: \(\frac{4x - 8 - (x + 1)(x - 3)}{x - 3} < 0\). Expanding the numerator gives \((x + 1)(x - 3) = x^2 - 2x - 3\), so the numerator becomes \(4x - 8 - (x^2 - 2x - 3) = -x^2 + 6x - 5\). The inequality is therefore \(\frac{-x^2 + 6x - 5}{x - 3} < 0\). Multiplying both sides by \(-1\) and reversing the inequality sign gives \(\frac{x^2 - 6x + 5}{x - 3} > 0\). Factorising the numerator gives \(\frac{(x - 1)(x - 5)}{x - 3} > 0\). The critical values are \(x = 1\), \(x = 3\), and \(x = 5\). Testing intervals: for \(x < 1\), the expression is negative; for \(1 < x < 3\), positive; for \(3 < x < 5\), negative; for \(x > 5\), positive. Thus, the solution is \(1 < x < 3\) or \(x > 5\). Part (b) Replacing \(x\) with \(y^2\) gives \(1 < y^2 < 3\) or \(y^2 > 5\). Solving \(1 < y^2 < 3\) yields \(1 < |y| < \sqrt{3}\), which gives \(1 < y < \sqrt{3}\) or \(-\sqrt{3} < y < -1\). Solving \(y^2 > 5\) yields \(|y| > \sqrt{5}\), which gives \(y > \sqrt{5}\) or \(y < -\sqrt{5}\). Combining these, the complete set of values is \(y < -\sqrt{5}\) or \(-\sqrt{3} < y < -1\) or \(1 < y < \sqrt{3}\) or \(y > \sqrt{5}\).

Marking scheme

Part (a) M1: For attempting to rearrange the inequality into a single fraction on one side, or multiplying by \((x-3)^2\). A1: For obtaining the correct simplified numerator or finding critical values \(1, 3, 5\). M1: For a valid method of finding the correct regions (e.g. testing values or sketching). A1: For identifying one correct interval (either \(1 < x < 3\) or \(x > 5\)). A1: For the fully correct set of values: \(1 < x < 3\) or \(x > 5\). Part (b) M1: For recognising that \(x = y^2\) and writing down \(1 < y^2 < 3\) or \(y^2 > 5\). A1: For correctly solving \(1 < y^2 < 3\) to get \(1 < y < \sqrt{3}\) and \(-\sqrt{3} < y < -1\). A1: For correctly solving \(y^2 > 5\) to get \(y > \sqrt{5}\) and \(y < -\sqrt{5}\).
Question 5 · structured
8 marks
The roots of the quadratic equation \(2x^2 - 5x + 4 = 0\) are \(\alpha\) and \(\beta\).

(a) Find the value of \(\alpha^2 + \beta^2\).

(b) Find a quadratic equation with integer coefficients which has roots \(\alpha^2 + \frac{1}{\beta}\) and \(\beta^2 + \frac{1}{\alpha}\).
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Worked solution

**(a)**
From the equation \(2x^2 - 5x + 4 = 0\), we have:
\(\alpha + \beta = \frac{5}{2}\)
\
\(\alpha\beta = \frac{4}{2} = 2\)

We use the identity:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)

Substitute the values:
\(\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2)\)
\(\alpha^2 + \beta^2 = \frac{25}{4} - 4 = \frac{9}{4}\) (or \(2.25\))

**(b)**
Let the roots of the new equation be \(\gamma = \alpha^2 + \frac{1}{\beta}\) and \(\delta = \beta^2 + \frac{1}{\alpha}\).

First, find the sum of the new roots:
\(\text{Sum} = \gamma + \delta = \alpha^2 + \beta^2 + \frac{1}{\alpha} + \frac{1}{\beta}\)
\(\text{Sum} = \left(\alpha^2 + \beta^2\right) + \frac{\alpha + \beta}{\alpha\beta}\)

Substitute the known values:
\(\text{Sum} = \frac{9}{4} + \frac{\frac{5}{2}}{2}\)
\(\text{Sum} = \frac{9}{4} + \frac{5}{4} = \frac{14}{4} = \frac{7}{2}\)

Next, find the product of the new roots:
\(\text{Product} = \gamma\delta = \left(\alpha^2 + \frac{1}{\beta}\right)\left(\beta^2 + \frac{1}{\alpha}\right)\)
\(\text{Product} = \alpha^2\beta^2 + \frac{\alpha^2}{\alpha} + \frac{\beta^2}{\beta} + \frac{1}{\alpha\beta}\)
\(\text{Product} = (\alpha\beta)^2 + (\alpha + \beta) + \frac{1}{\alpha\beta}\)

Substitute the known values:
\(\text{Product} = (2)^2 + \frac{5}{2} + \frac{1}{2}\)
\(\text{Product} = 4 + 3 = 7\)

Now, the new quadratic equation is given by:
\(x^2 - (\text{Sum})x + \text{Product} = 0\)
\(x^2 - \frac{7}{2}x + 7 = 0\)

Multiplying by 2 to obtain integer coefficients:
\(2x^2 - 7x + 14 = 0\)

Marking scheme

**Part (a)**
* **B1**: For correctly stating \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\) (or equivalent).
* **M1**: Use of the algebraic identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) with substitution of their sum and product.
* **A1**: Correct value of \(\frac{9}{4}\) (or \(2.25\)).

**Part (b)**
* **M1**: Expressing the sum of the new roots as \((\alpha^2 + \beta^2) + \frac{\alpha + \beta}{\alpha\beta}\) and substituting values.
* **A1**: Finding the correct sum \(\frac{7}{2}\).
* **M1**: Expanding the product of the new roots to obtain \((\alpha\beta)^2 + (\alpha + \beta) + \frac{1}{\alpha\beta}\) and substituting values.
* **A1**: Finding the correct product \(7\).
* **A1**: Forming the final equation \(2x^2 - 7x + 14 = 0\) (must include "\(= 0\)" and have integer coefficients, or any integer multiple thereof).
Question 6 · binomial_expansion_and_estimation
8 marks
The binomial expansion of \((4 - 9x)^{\frac{1}{2}}\), where \(|x| < \frac{4}{9}\), can be used to find an approximation for \(\sqrt{3.1}\).

(a) Find the binomial expansion of \((4 - 9x)^{\frac{1}{2}}\) in ascending powers of \(x\) up to and including the term in \(x^3\), simplifying each coefficient.

(b) By substituting \(x = 0.1\) into your expansion, find an approximation for \(\sqrt{3.1}\). Give your answer to 5 decimal places.
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Worked solution

**(a)**

First, factor out \(4^{\frac{1}{2}}\) from the expression:

\((4 - 9x)^{\frac{1}{2}} = 4^{\frac{1}{2}}\left(1 - \frac{9}{4}x\right)^{\frac{1}{2}} = 2\left(1 - \frac{9}{4}x\right)^{\frac{1}{2}}\)

Using the binomial expansion formula for \((1 + y)^n\) with \(y = -\frac{9}{4}x\) and \(n = \frac{1}{2}\):

\(\left(1 - \frac{9}{4}x\right)^{\frac{1}{2}} = 1 + \left(\frac{1}{2}\right)\left(-\frac{9}{4}x\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(-\frac{9}{4}x\right)^2 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}\left(-\frac{9}{4}x\right)^3 + \dots\)

Simplify each term inside the bracket:

\(= 1 - \frac{9}{8}x - \frac{1}{8}\left(\frac{81}{16}x^2\right) + \frac{1}{16}\left(-\frac{729}{64}x^3\right) + \dots\)

\(= 1 - \frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 - \dots\)

Now, multiply the entire expansion by 2:

\((4 - 9x)^{\frac{1}{2}} = 2\left(1 - \frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3\right)\)

\(= 2 - \frac{9}{4}x - \frac{81}{64}x^2 - \frac{729}{512}x^3\)

**(b)**

We want to approximate \(\sqrt{3.1}\).

Setting \(4 - 9x = 3.1\) gives:

\(9x = 0.9 \implies x = 0.1\)

Since \(|0.1| < \frac{4}{9}\), the expansion is valid.

Substitute \(x = 0.1\) into the expansion obtained in part (a):

\(\sqrt{3.1} \approx 2 - \frac{9}{4}(0.1) - \frac{81}{64}(0.1)^2 - \frac{729}{512}(0.1)^3\)

\(\sqrt{3.1} \approx 2 - 2.25(0.1) - 1.265625(0.01) - 1.423828125(0.001)\)

\(\sqrt{3.1} \approx 2 - 0.225 - 0.01265625 - 0.001423828125\)

\(\sqrt{3.1} \approx 1.760919921875\)

Rounding to 5 decimal places:

\(\sqrt{3.1} \approx 1.76092\)

Marking scheme

**Part (a)**
* **B1**: For taking out a factor of \(4^{\frac{1}{2}}\) (or \(2\)) to write the expression in the form \(2\left(1 - \frac{9}{4}x\right)^{\frac{1}{2}}\).
* **M1**: For an attempt at the binomial expansion of \((1 + kx)^{\frac{1}{2}}\), with at least two of the second, third, or fourth terms having correct binomial coefficient structure (i.e., correct application of \(\frac{n(n-1)}{2!}\) or \(\frac{n(n-1)(n-2)}{3!}\)).
* **A1**: For any two of the simplified variable terms: \(-\frac{9}{4}x\), \(-\frac{81}{64}x^2\) or \(-\frac{729}{512}x^3\) correct.
* **A1**: For the fully correct expansion: \(2 - \frac{9}{4}x - \frac{81}{64}x^2 - \frac{729}{512}x^3\).

**Part (b)**
* **B1**: For stating that \(x = 0.1\) is the value to substitute (can be implied by substitution).
* **M1**: For substituting their value of \(x = 0.1\) into their binomial expansion from part (a).
* **A1**: For showing the evaluated terms: \(2 - 0.225 - 0.012656... - 0.001423...\).
* **A1**: For the final answer of \(1.76092\) (must be rounded to exactly 5 decimal places).
Question 7 · Connected rates of change
6 marks
A container in the shape of an inverted right circular cone of semi-vertical angle \(\alpha\) is being filled with water at a constant rate of \(12\pi\text{ cm}^3/\text{s}\). Given that \(\tan\alpha = \frac{3}{4}\), find the rate of change of the depth of water in the container, in \(\text{cm/s}\), at the instant when the depth of the water is \(4\text{ cm}\).
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Worked solution

Let \(r\) and \(h\) be the radius of the water surface and the depth of the water at time \(t\) respectively. From the geometry of the cone, we have \(\frac{r}{h} = \tan\alpha = \frac{3}{4}\), which gives \(r = \frac{3}{4}h\). The volume \(V\) of water in the cone is given by: \(V = \frac{1}{3}\pi r^2 h\). Substituting \(r = \frac{3}{4}h\) into the volume formula: \(V = \frac{1}{3}\pi \left(\frac{3}{4}h\right)^2 h = \frac{3}{16}\pi h^3\). Differentiating \(V\) with respect to \(h\) gives: \(\frac{dV}{dh} = \frac{9}{16}\pi h^2\). By the chain rule: \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\). We are given that \(\frac{dV}{dt} = 12\pi\text{ cm}^3/\text{s}\). At the instant when \(h = 4\text{ cm}\): \(\frac{dV}{dh} = \frac{9}{16}\pi (4)^2 = 9\pi\). Substituting these values into the chain rule formula: \(12\pi = 9\pi \times \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{12\pi}{9\pi} = \frac{4}{3}\text{ cm/s}\).

Marking scheme

M1: Use \(\tan\alpha = \frac{3}{4}\) to express \(r\) in terms of \(h\) as \(r = \frac{3}{4}h\). M1: Substitute \(r\) into the volume formula to obtain \(V\) in terms of \(h\) only: \(V = \frac{3}{16}\pi h^3\). A1: Correctly differentiate to find \(\frac{dV}{dh} = \frac{9}{16}\pi h^2\). M1: State or use the chain rule relation \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\). M1: Substitute \(h = 4\) and \(\frac{dV}{dt} = 12\pi\) into their relation. A1: Obtain the correct value of \(\frac{dh}{dt} = \frac{4}{3}\) (or \(1.33\) cm/s).
Question 8 · structured
13 marks
Figure 1 shows a right pyramid \(VABCD\) with a rectangular base \(ABCD\) where \(AB = 12\text{ cm}\) and \(BC = 10\text{ cm}\). The vertex \(V\) is vertically above the centre \(O\) of the base. The length of the slant edge \(VA\) is \(14\text{ cm}\). (a) Show that the height, \(VO\), of the pyramid is \(3\sqrt{15}\text{ cm}\). (3 marks) (b) Calculate, in degrees to 1 decimal place, the angle between the plane \(VAB\) and the base \(ABCD\). (4 marks) (c) Calculate, in degrees to 1 decimal place, the angle between the line \(VA\) and the plane \(VBC\). (6 marks)
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Worked solution

(a) The base \(ABCD\) is a rectangle with \(AB = 12\text{ cm}\) and \(BC = 10\text{ cm}\). The diagonal \(AC\) is given by Pythagoras' Theorem: \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 10^2} = \sqrt{144 + 100} = \sqrt{244} = 2\sqrt{61}\text{ cm}\). Since \(O\) is the centre of the base, \(AO = \frac{1}{2}AC = \sqrt{61}\text{ cm}\). Since \(V\) is vertically above \(O\), triangle \(VOA\) is right-angled at \(O\). Using Pythagoras' Theorem: \(VO^2 + AO^2 = VA^2\) which gives \(VO^2 + 61 = 14^2 = 196\), so \(VO^2 = 135\) and thus \(VO = \sqrt{135} = 3\sqrt{15}\text{ cm}\). (b) Let \(M\) be the midpoint of \(AB\). Since \(VAB\) is an isosceles triangle with \(VA = VB\), the line \(VM\) is perpendicular to \(AB\). The line \(OM\) is also perpendicular to \(AB\), and its length is half the length of \(BC\): \(OM = \frac{1}{2}BC = 5\text{ cm}\). The angle between the plane \(VAB\) and the base \(ABCD\) is the angle \(VMO\). In the right-angled triangle \(VOM\): \(\tan(\angle VMO) = \frac{VO}{OM} = \frac{3\sqrt{15}}{5}\). Hence, \(\angle VMO = \arctan\left(\frac{3\sqrt{15}}{5}\right) \approx \arctan(2.3238) \approx 66.715^\circ \approx 66.7^\circ\) (to 1 decimal place). (c) Let \(P\) be the projection of the vertex \(A\) onto the plane \(VBC\), so that \(AP\) is perpendicular to the plane \(VBC\). The angle between the line \(VA\) and the plane \(VBC\) is the angle \(AVP\). We can find the perpendicular distance \(AP\) by considering the volume of the tetrahedron \(VABC\). The area of the right-angled triangular base \(ABC\) is: \(\text{Area}(ABC) = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 12 \times 10 = 60\text{ cm}^2\). The height of the tetrahedron with base \(ABC\) is \(VO = 3\sqrt{15}\text{ cm}\). The volume of the tetrahedron \(VABC\) is: \(\text{Volume}(VABC) = \frac{1}{3} \times \text{Area}(ABC) \times VO = \frac{1}{3} \times 60 \times 3\sqrt{15} = 60\sqrt{15}\text{ cm}^3\). Now, let \(N\) be the midpoint of \(BC\). In the isosceles triangle \(VBC\), the slant height \(VN\) is perpendicular to \(BC\). In the right-angled triangle \(VON\), \(ON = \frac{1}{2}AB = 6\text{ cm}\) and \(VO = 3\sqrt{15}\text{ cm}\). Using Pythagoras' Theorem: \(VN = \sqrt{VO^2 + ON^2} = \sqrt{135 + 36} = \sqrt{171} = 3\sqrt{19}\text{ cm}\). The area of the triangular base \(VBC\) is: \(\text{Area}(VBC) = \frac{1}{2} \times BC \times VN = \frac{1}{2} \times 10 \times 3\sqrt{19} = 15\sqrt{19}\text{ cm}^2\). The volume of the tetrahedron can also be written as: \(\text{Volume}(VABC) = \frac{1}{3} \times \text{Area}(VBC) \times AP\), which gives \(60\sqrt{15} = \frac{1}{3} \times 15\sqrt{19} \times AP\), so \(60\sqrt{15} = 5\sqrt{19} \times AP\), giving \(AP = \frac{12\sqrt{15}}{\sqrt{19}}\text{ cm}\). In the right-angled triangle \(VPA\) (with right angle at \(P\)): \(\sin(\angle AVP) = \frac{AP}{VA} = \frac{12\sqrt{15}}{14\sqrt{19}} = \frac{6\sqrt{15}}{7\sqrt{19}} \approx 0.7616\). Hence, \(\angle AVP = \arcsin(0.7616) \approx 49.605^\circ \approx 49.6^\circ\) (to 1 decimal place).

Marking scheme

(a) M1: For an attempt to use Pythagoras' Theorem to find the diagonal \(AC\) or half-diagonal \(AO\) (e.g., \(\sqrt{12^2 + 10^2}\)). M1: For setting up a correct Pythagorean equation for \(VO\), such as \(VO^2 = 14^2 - (\sqrt{61})^2\). A1: For simplifying to show \(VO = 3\sqrt{15}\) with no errors seen. (b) M1: For identifying the angle between plane \(VAB\) and base \(ABCD\) as \(\angle VMO\), where \(M\) is the midpoint of \(AB\). B1: For stating or finding that \(OM = 5\text{ cm}\). M1: For using a correct trigonometric ratio to find the angle, e.g., \(\tan(\theta) = \frac{3\sqrt{15}}{5}\). A1: For the correct angle \(66.7^\circ\) (accept answers rounding to \(66.7\)). (c) M1: For finding the length of the slant height \(VN\) of triangle \(VBC\), where \(N\) is the midpoint of \(BC\) (e.g., \(VN = \sqrt{171} = 3\sqrt{19}\)). M1: For an attempt to express the volume of the tetrahedron \(VABC\) using \(\text{Area}(ABC) \times VO\). M1: For setting up the volume equation \(\frac{1}{3} \times \text{Area}(VBC) \times AP = \text{Volume}(VABC)\) to find the perpendicular distance \(AP\). A1: For finding the correct perpendicular distance \(AP = \frac{12\sqrt{15}}{\sqrt{19}}\) (or decimal equivalent \(\approx 10.66\)). M1: For a correct trigonometric relationship to find the angle \(\angle AVP\), e.g., \(\sin(\angle AVP) = \frac{AP}{14}\). A1: For the correct angle \(49.6^\circ\) (accept answers rounding to \(49.6\)).
Question 9 · Calculus
13 marks
The curve \(C\) has equation \(y = 2x^3 - 9x^2 + 12x - 2\).

(a) Find the coordinates of the stationary points of \(C\) and determine their nature. (5 marks)

(b) Find an equation of the tangent to the curve \(C\) at the point where \(x = 0\). (3 marks)

The tangent found in part (b) intersects the curve \(C\) again at the point \(R\).

(c) Find the coordinates of \(R\). (2 marks)

(d) Find the area of the finite region bounded by the curve \(C\) and the tangent to the curve at \(x = 0\). (3 marks)
Show answer & marking scheme

Worked solution

(a) First, find the derivative of \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 6x^2 - 18x + 12\)
For stationary points, set \(\frac{dy}{dx} = 0\):
\(6(x^2 - 3x + 2) = 0\)
\(6(x - 1)(x - 2) = 0\)
So, \(x = 1\) or \(x = 2\).

When \(x = 1\):
\(y = 2(1)^3 - 9(1)^2 + 12(1) - 2 = 3\)
So, one stationary point is \((1, 3)\).

When \(x = 2\):
\(y = 2(2)^3 - 9(2)^2 + 12(2) - 2 = 16 - 36 + 24 - 2 = 2\)
So, the other stationary point is \((2, 2)\).

To determine the nature of these points, find the second derivative:
\(\frac{d^2y}{dx^2} = 12x - 18\)

At \((1, 3)\):
\(\frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0\), which means \((1, 3)\) is a local maximum.

At \((2, 2)\):
\(\frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0\), which means \((2, 2)\) is a local minimum.

(b) At \(x = 0\):
\(y = 2(0)^3 - 9(0)^2 + 12(0) - 2 = -2\)

The gradient of the tangent at this point is found by substituting \(x = 0\) into \(\frac{dy}{dx}\):
\(m = 6(0)^2 - 18(0) + 12 = 12\)

The equation of the tangent is:
\(y - (-2) = 12(x - 0)\)
\(y = 12x - 2\)

(c) To find the intersection of the tangent and the curve, set their equations equal:
\(2x^3 - 9x^2 + 12x - 2 = 12x - 2\)
\(2x^3 - 9x^2 = 0\)
\(x^2(2x - 9) = 0\)

Since \(x = 0\) is the point of tangency (which is a repeated root), the other intersection point \(R\) is at:
\(2x - 9 = 0 \implies x = 4.5\) (or \(x = \frac{9}{2}\))

Substitute \(x = 4.5\) into the tangent equation to find \(y\):
\(y = 12(4.5) - 2 = 52\)

So, the coordinates of \(R\) are \((4.5, 52)\).

(d) The finite region is bounded between \(x = 0\) and \(x = 4.5\).
On this interval, the tangent \(y = 12x - 2\) lies above the curve \(y = 2x^3 - 9x^2 + 12x - 2\).

The area \(A\) is:
\(A = \int_{0}^{4.5} \left( (12x - 2) - (2x^3 - 9x^2 + 12x - 2) \right) dx\)
\(A = \int_{0}^{4.5} (9x^2 - 2x^3) dx\)
\(A = \left[ 3x^3 - \frac{1}{2}x^4 \right]_{0}^{4.5}\)
\(A = \left( 3(4.5)^3 - \frac{1}{2}(4.5)^4 \right) - 0\)
\(A = 3(91.125) - 0.5(410.0625)\)
\(A = 273.375 - 205.03125 = 68.34375\)
In fractional form, this is \(\frac{2187}{32}\).

Marking scheme

Part (a):
- M1: Attempts to differentiate the cubic expression to find \(\frac{dy}{dx}\).
- M1: Sets their derivative equal to 0 and attempts to solve the quadratic equation for \(x\).
- A1: Obtains the correct coordinates \((1, 3)\) and \((2, 2)\).
- M1: Differentiates again to find \(\frac{d^2y}{dx^2}\) and substitutes at least one of their \(x\) values to determine the nature.
- A1: Correctly identifies \((1, 3)\) as a maximum and \((2, 2)\) as a minimum with supporting working.

Part (b):
- M1: Attempts to find the gradient of the curve at \(x = 0\) by substituting into their \(\frac{dy}{dx}\).
- M1: Uses their gradient and the coordinates of the point of tangency \((0, -2)\) to form a linear equation.
- A1: Correct equation, e.g., \(y = 12x - 2\) (any equivalent form).

Part (c):
- M1: Equates the curve and tangent equations and attempts to solve for \(x\).
- A1: Correct coordinates of \(R\): \((4.5, 52)\) or \(\left(\frac{9}{2}, 52\right)\).

Part (d):
- M1: Sets up a correct integral for the area with limits from 0 to their \(x\)-coordinate of \(R\).
- M1: Integrates the difference of the two functions correctly, obtaining \(3x^3 - \frac{1}{2}x^4\).
- A1: Correct exact area of \(\frac{2187}{32}\) or \(68.34375\) (accept \(68.3\) or better).
Question 10 · Structured
11 marks
(a) Solve the equation

\[2 \log_x 4 + 3 \log_4 x = 7\]

giving your answers in exact form. (5)

(b) Solve the simultaneous equations

\[\log_3 (y - 2) - \log_9 x = 1\]
\[2^x \times 2^y = 4096\] (6)
Show answer & marking scheme

Worked solution

**(a)**

Let \(u = \log_4 x\).

Using the change of base formula:
\[\log_x 4 = \frac{1}{\log_4 x} = \frac{1}{u}\]

Substitute this into the equation:
\[2\left(\frac{1}{u}\right) + 3u = 7\]

Multiply through by \(u\) (since \(u \neq 0\)):
\[2 + 3u^2 = 7u \implies 3u^2 - 7u + 2 = 0\]

Factor the quadratic equation:
\[(3u - 1)(u - 2) = 0\]

This gives:
\[u = \frac{1}{3} \quad \text{or} \quad u = 2\]

Now convert back to solve for \(x\):
For \(u = \frac{1}{3}\):
\[\log_4 x = \frac{1}{3} \implies x = 4^{1/3} = \sqrt[3]{4}\]

For \(u = 2\):
\[\log_4 x = 2 \implies x = 4^2 = 16\]

So, the solutions are \(x = 16\) and \(x = 4^{1/3}\).

**(b)**

First, consider the second equation:
\[2^x \times 2^y = 4096\]

Using index laws:
\[2^{x+y} = 2^{12} \implies x + y = 12 \quad \text{--- (Equation 1)}\]

Now, consider the first equation:
\[\log_3 (y - 2) - \log_9 x = 1\]

Apply the change of base rule to \\log_9 x\:
\[\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \log_3 \left(x^{1/2}\right)\]

Substitute back into the equation:
\[\log_3 (y - 2) - \log_3 \left(x^{1/2}\right) = 1\]

Using the subtraction law of logarithms:
\[\log_3 \left(\frac{y - 2}{\sqrt{x}}\right) = 1\]
\[\frac{y - 2}{\sqrt{x}} = 3^1 = 3 \implies y - 2 = 3\sqrt{x} \quad \text{--- (Equation 2)}\]

From Equation 1, express \(y\) in terms of \(x\):
\[y = 12 - x\]

Substitute \(y = 12 - x\) into Equation 2:
\[12 - x - 2 = 3\sqrt{x} \implies 10 - x = 3\sqrt{x}\]

Let \(w = \sqrt{x}\) where \(w \ge 0\). The equation becomes:
\[10 - w^2 = 3w \implies w^2 + 3w - 10 = 0\]

Factor the quadratic:
\[(w + 5)(w - 2) = 0\]

This gives \(w = 2\) or \(w = -5\).

Since \(w = \sqrt{x}\) must be non-negative, we reject \(w = -5\).

Thus:
\[\sqrt{x} = 2 \implies x = 4\]

Using \(y = 12 - x\):
\[y = 12 - 4 = 8\]

(Note: If we squared the equation \(10 - x = 3\sqrt{x}\) directly, we would get \(x^2 - 29x + 100 = 0 \implies (x-25)(x-4)=0\). But if \(x = 25\), then \(y = -13\), which makes the term \(\log_3(y - 2) = \log_3(-15)\) undefined. Therefore, \(x = 25\) is rejected as an extraneous solution.)

The unique solution is \(x = 4, y = 8\).

Marking scheme

**Part (a)**
- **M1**: Uses change of base formula correctly to express \(\log_x 4\) as \(\frac{1}{\log_4 x}\).
- **M1**: Substitutes and obtains a quadratic equation in terms of \(u\) (e.g. \(3u^2 - 7u + 2 = 0\)).
- **A1**: Correctly solves the quadratic to obtain \(u = 2\) and \(u = \frac{1}{3}\).
- **M1**: Applies the definition of logarithms to set up \(x = 4^2\) or \(x = 4^{1/3}\).
- **A1**: Both correct values: \(x = 16\) and \(x = 4^{1/3}\) (or \(\sqrt[3]{4}\)).

**Part (b)**
- **M1**: Simplifies the index equation to get \(x + y = 12\).
- **M1**: Expresses \(\log_9 x\) as \(\frac{1}{2}\log_3 x\) (or \(\log_3 \sqrt{x}\)) using change of base.
- **M1**: Converts the logarithmic equation into a non-logarithmic equation: e.g. \(y - 2 = 3\sqrt{x}\) or \((y - 2)^2 = 9x\).
- **M1**: Substitutes \(y = 12 - x\) (or equivalent) to form a quadratic equation in one variable (e.g. \(x^2 - 29x + 100 = 0\) or \(w^2 + 3w - 10 = 0\)).
- **A1**: Solves the quadratic to find possible values (e.g. \(x = 4, x = 25\)) and clearly rejects the extraneous solution (e.g. noting that \(y = -13\) is invalid as \(y > 2\) is required).
- **A1**: Correct final unique values: \(x = 4\) and \(y = 8\).
Question 11 · structured
7 marks
The curve \(C\) has the equation \(y = 2\sqrt{x} - x\), where \(x \ge 0\).

The curve intersects the \(x\)-axis at the origin and at the point \(A\).

(a) Find the coordinates of \(A\).

(b) The finite region \(R\), which is bounded by the curve \(C\) and the \(x\)-axis, is rotated through \(360^\circ\) about the \(x\)-axis.

Find the exact volume of the solid generated, giving your answer in the form \(\frac{p}{q}\pi\), where \(p\) and \(q\) are integers.
Show answer & marking scheme

Worked solution

**(a)**
To find the coordinates of \(A\), set \(y = 0\):
\[2\sqrt{x} - x = 0\]
\[\sqrt{x}(2 - \sqrt{x}) = 0\]
Since \(A\) is not the origin, we have:
\[2 - \sqrt{x} = 0 \implies \sqrt{x} = 2 \implies x = 4\]
Thus, the coordinates of \(A\) are \((4, 0)\).

**(b)**
The volume \(V\) of the solid of revolution is given by:
\[V = \pi \int_{0}^{4} y^2 \mathrm{d}x\]
First, expand the integrand \(y^2\):
\[y^2 = (2\sqrt{x} - x)^2 = 4x - 4x^{3/2} + x^2\]

Now, integrate each term with respect to \(x\):
\[\int (4x - 4x^{3/2} + x^2) \mathrm{d}x = 2x^2 - \frac{8}{5}x^{5/2} + \frac{1}{3}x^3\]

Substitute the limits of integration \(0\) and \(4\):
\[V = \pi \left[ 2x^2 - \frac{8}{5}x^{5/2} + \frac{1}{3}x^3 \right]_{0}^{4}\]
\[V = \pi \left\{ \left( 2(4)^2 - \frac{8}{5}(4)^{5/2} + \frac{1}{3}(4)^3 \right) - 0 \right\}\]
\[V = \pi \left( 32 - \frac{8}{5}(32) + \frac{64}{3} \right)\]
\[V = \pi \left( 32 - \frac{256}{5} + \frac{64}{3} \right)\]
Using a common denominator of \(15\):
\[V = \pi \left( \frac{480 - 768 + 320}{15} \right) = \frac{32}{15}\pi\]

Marking scheme

**(a)**
- **M1**: Sets \(2\sqrt{x} - x = 0\) and attempts to solve for \(x\).
- **A1**: Correct coordinates for \(A\): \((4, 0)\) (or clearly states \(x = 4, y = 0\)).

**(b)**
- **M1**: Uses the correct volume formula \(V = \pi \int y^2 \mathrm{d}x\) and attempts to expand \(y^2\).
- **A1**: Correct expansion of \(y^2\) to get \(4x - 4x^{3/2} + x^2\).
- **M1**: Integrates the expression with at least two terms integrated correctly.
- **A1**: Fully correct integrated expression: \(2x^2 - \frac{8}{5}x^{5/2} + \frac{1}{3}x^3\).
- **A1**: Correctly substitutes both limits of integration (0 and their upper limit from part a) into their integrated expression.
- **A1**: Obtains the exact volume of \(V = \frac{32}{15}\pi\) (or equivalent exact fraction).

Paper 2 (4PM1/02)

Answer all eleven questions. Write your answers in the spaces provided. Without sufficient working, correct answers may be awarded no marks.
11 Question · 100 marks
Question 1 · Trigonometric equations
5 marks
Find the values of \(\theta\) in the interval \(0 \le \theta < 2\pi\) which satisfy the equation \(3 \cos 2\theta + 5 \sin \theta - 4 = 0\). Give your answers to 3 significant figures.
Show answer & marking scheme

Worked solution

Using the trigonometric identity \(\cos 2\theta = 1 - 2\sin^2\theta\), we substitute this into the given equation: \(3(1 - 2\sin^2\theta) + 5\sin\theta - 4 = 0\). Simplifying this gives: \(3 - 6\sin^2\theta + 5\sin\theta - 4 = 0\), which rearranges to the quadratic equation: \(6\sin^2\theta - 5\sin\theta + 1 = 0\). Factorising this quadratic equation, we get: \((3\sin\theta - 1)(2\sin\theta - 1) = 0\). This yields two possible values: \(\sin\theta = \frac{1}{3}\) or \(\sin\theta = \frac{1}{2}\). For \(\sin\theta = \frac{1}{3}\), the solutions in the interval \(0 \le \theta < 2\pi\) are: \(\theta = \arcsin\left(\frac{1}{3}\right) \approx 0.340\) and \(\theta = \pi - \arcsin\left(\frac{1}{3}\right) \approx 2.80\). For \(\sin\theta = \frac{1}{2}\), the solutions in the interval are: \(\theta = \frac{\pi}{6} \approx 0.524\) and \(\theta = \frac{5\pi}{6} \approx 2.62\). Thus, the values of \(\theta\) to 3 significant figures are 0.340, 0.524, 2.62, and 2.80.

Marking scheme

M1: Substitutes \(\cos 2\theta = 1 - 2\sin^2\theta\) into the equation to obtain a quadratic equation in \(\sin\theta\). A1: Obtains the correct simplified quadratic equation \(6\sin^2\theta - 5\sin\theta + 1 = 0\) (or equivalent). M1: Solves their three-term quadratic equation to find two values for \(\sin\theta\) (namely \(\sin\theta = \frac{1}{3}\) and \(\sin\theta = \frac{1}{2}\)). A1: Finds at least two correct values of \(\theta\) in the interval (either 0.340 and 2.80, or 0.524 and 2.62). A1: Finds all four correct values to 3 significant figures: \(\theta = 0.340, 0.524, 2.62, 2.80\) (accept exact forms \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) for those respective values). Deduct 1 mark for any extra solutions within the interval.
Question 2 · short_answer
5 marks
Find the set of values of \( x \) for which both of the following inequalities are satisfied:

\[ x^2 - 2x - 15 < 0 \]
and
\[ 3(x + 1) \ge 5x - 1 \]
Show answer & marking scheme

Worked solution

First, solve the quadratic inequality:
\[ x^2 - 2x - 15 < 0 \]
Find the critical values by solving the equation:
\[ x^2 - 2x - 15 = 0 \]
\[ (x - 5)(x + 3) = 0 \]
This gives the critical values as \( x = 5 \) and \( x = -3 \).
Since the inequality is \( < 0 \), we choose the region between these values:
\[ -3 < x < 5 \]

Next, solve the linear inequality:
\[ 3(x + 1) \ge 5x - 1 \]
\[ 3x + 3 \ge 5x - 1 \]
\[ 4 \ge 2x \]
\[ x \le 2 \]

To satisfy both inequalities, we find the intersection of \( -3 < x < 5 \) and \( x \le 2 \).
This gives the final set of values as:
\[ -3 < x \le 2 \]

Marking scheme

M1: Factorises the quadratic expression \( x^2 - 2x - 15 \) or uses the quadratic formula to find the critical values \( x = 5 \) and \( x = -3 \).
A1: Correctly identifies the interval for the quadratic inequality as \( -3 < x < 5 \) (or equivalent).
B1: Correctly solves the linear inequality to obtain \( x \le 2 \) (or equivalent).
M1: Attempts to find the intersection of the two solution sets (e.g., by drawing a number line or combining the inequalities).
A1: Correct final interval \( -3 < x \le 2 \) (accept set notation \( \{x : -3 < x \le 2\} \) or interval notation \( (-3, 2] \)).
Question 3 · short_answer
5 marks
Find the set of values of \( x \) for which both of the following inequalities are satisfied:

\[ x^2 - 2x - 15 < 0 \]
and
\[ 3(x + 1) \ge 5x - 1 \]
Show answer & marking scheme

Worked solution

First, solve the quadratic inequality:
\[ x^2 - 2x - 15 < 0 \]
Find the critical values by solving the equation:
\[ x^2 - 2x - 15 = 0 \]
\[ (x - 5)(x + 3) = 0 \]
This gives the critical values as \( x = 5 \) and \( x = -3 \).
Since the inequality is \( < 0 \), we choose the region between these values:
\[ -3 < x < 5 \]

Next, solve the linear inequality:
\[ 3(x + 1) \ge 5x - 1 \]
\[ 3x + 3 \ge 5x - 1 \]
\[ 4 \ge 2x \]
\[ x \le 2 \]

To satisfy both inequalities, we find the intersection of \( -3 < x < 5 \) and \( x \le 2 \).
This gives the final set of values as:
\[ -3 < x \le 2 \]

Marking scheme

M1: Factorises the quadratic expression \( x^2 - 2x - 15 \) or uses the quadratic formula to find the critical values \( x = 5 \) and \( x = -3 \).
A1: Correctly identifies the interval for the quadratic inequality as \( -3 < x < 5 \) (or equivalent).
B1: Correctly solves the linear inequality to obtain \( x \le 2 \) (or equivalent).
M1: Attempts to find the intersection of the two solution sets (e.g., by drawing a number line or combining the inequalities).
A1: Correct final interval \( -3 < x \le 2 \) (accept set notation \( \{x : -3 < x \le 2\} \) or interval notation \( (-3, 2] \)).
Question 4 · Geometric series
6 marks
A geometric series has first term \( a \) and common ratio \( r \), where \( r > 0 \).

The sum to infinity of the series is \( S \).
The sum of the first three terms of the series is \( \frac{7}{8}S \).

(a) Find the value of \( r \).
(3 marks)

The sum of the second and third terms of the series is 12.
(b) Find the value of \( a \).
(3 marks)
Show answer & marking scheme

Worked solution

(a) The sum to infinity of a geometric series is given by:
\( S = \frac{a}{1-r} \)

The sum of the first three terms is:
\( S_3 = \frac{a(1-r^3)}{1-r} = S(1-r^3) \)

We are given that \( S_3 = \frac{7}{8}S \), so:
\( S(1-r^3) = \frac{7}{8}S \)

Since \( S \neq 0 \) (as \( a \neq 0 \)), we can divide both sides by \( S \):
\( 1-r^3 = \frac{7}{8} \)
\( r^3 = 1 - \frac{7}{8} = \frac{1}{8} \)

Taking the cube root of both sides:
\( r = \frac{1}{2} \)

(b) The \( n \)-th term of a geometric series is given by \( T_n = a r^{n-1} \).

The second term is \( T_2 = ar \) and the third term is \( T_3 = ar^2 \).

We are given that the sum of the second and third terms is 12:
\( ar + ar^2 = 12 \)
\( a(r + r^2) = 12 \)

Substituting \( r = \frac{1}{2} \) into the equation:
\( a\left(\frac{1}{2} + \left(\frac{1}{2}\right)^2\right) = 12 \)
\( a\left(\frac{1}{2} + \frac{1}{4}\right) = 12 \)
\( a\left(\frac{3}{4}\right) = 12 \)
\( a = 12 \times \frac{4}{3} = 16 \)

Marking scheme

Part (a)
- M1: Writes an equation relating \( S_3 \), \( S \), and \( r \), e.g., \( \frac{a(1-r^3)}{1-r} = \frac{7}{8}\left(\frac{a}{1-r}\right) \) or \( S(1-r^3) = \frac{7}{8}S \).
- M1: Simplifies the equation to find a value for \( r^3 \), e.g., \( r^3 = \frac{1}{8} \).
- A1: Correct value of \( r = \frac{1}{2} \) (or 0.5).

Part (b)
- M1: Formulates an equation for the sum of the second and third terms, e.g., \( ar + ar^2 = 12 \).
- M1: Substitutes their value of \( r \) from part (a) into their equation.
- A1: Correct value of \( a = 16 \).
Question 5 · free_response
8 marks
In triangle \(ABC\), \(AB = x\text{ cm}\), \(BC = (x + 3)\text{ cm}\), \(AC = 7\text{ cm}\) and angle \(ABC = 60^\circ\).

(a) Show that \(x = 5\).
(3)

(b) Find the exact area of triangle \(ABC\), giving your answer in the form \(a\sqrt{3}\) where \(a\) is an integer.
(2)

(c) Find the exact value of \(\sin(\angle ACB)\).
(3)
Show answer & marking scheme

Worked solution

(a) Using the cosine rule on triangle \(ABC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]
Substitute the given terms into the formula:
\[7^2 = x^2 + (x+3)^2 - 2x(x+3)\cos(60^\circ)\]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[49 = x^2 + x^2 + 6x + 9 - 2x(x+3)\left(\frac{1}{2}\right)\]
\[49 = 2x^2 + 6x + 9 - (x^2 + 3x)\]
\[49 = x^2 + 3x + 9\]
\[x^2 + 3x - 40 = 0\]
Factorising this quadratic equation gives:
\[(x - 5)(x + 8) = 0\]
Since \(x\) represents a length, we must have \(x > 0\). Therefore, \(x = 5\).

(b) For \(x = 5\), the side lengths are \(AB = 5\text{ cm}\) and \(BC = 5 + 3 = 8\text{ cm}\).
Using the area formula:
\[\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\]
\[\text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin(60^\circ)\]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[\text{Area} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\]
Thus, the area is \(10\sqrt{3}\text{ cm}^2\) (where \(a = 10\)).

(c) Using the sine rule on triangle \(ABC\):
\[\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle ABC)}{AC}\]
\[\frac{\sin(\angle ACB)}{5} = \frac{\sin(60^\circ)}{7}\]
\[\sin(\angle ACB) = \frac{5 \sin(60^\circ)}{7}\]
Substitute \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[\sin(\angle ACB) = \frac{5 \times \frac{\sqrt{3}}{2}}{7} = \frac{5\sqrt{3}}{14}\]

Marking scheme

(a)
- M1: Correctly substitutes the algebraic expressions and \(\cos(60^\circ)\) into the cosine rule, e.g., \(7^2 = x^2 + (x+3)^2 - 2x(x+3)\cos(60^\circ)\).
- A1: Obtains a correct simplified three-term quadratic equation, e.g., \(x^2 + 3x - 40 = 0\).
- A1: Solves the quadratic to find \(x = 5\) and explicitly rejects \(x = -8\) or states \(x > 0\) as justification.

(b)
- M1: Uses the area formula \(\frac{1}{2}ab\sin C\) with \(AB = 5\), \(BC = 8\) and \(\sin(60^\circ)\).
- A1: Obtains the correct exact area \(10\sqrt{3}\).

(c)
- M1: Attempts to use the sine rule to express \(\sin(\angle ACB)\) in terms of other known values.
- A1: Obtains a correct equation, e.g., \(\frac{\sin(\angle ACB)}{5} = \frac{\sin(60^\circ)}{7}\).
- A1: Simplifies to the final exact value \(\frac{5\sqrt{3}}{14}\).
Question 6 · structured
10 marks
A closed right circular cylinder has radius \(r\) cm and height \(h\) cm. The total surface area of the cylinder is \(600\pi\text{ cm}^2\).

(a) Show that the volume, \(V\text{ cm}^3\), of the cylinder is given by \(V = 300\pi r - \pi r^3\).

(b) Use calculus to find the maximum volume of the cylinder, leaving your answer as a multiple of \(\pi\).

(c) Prove that the volume you have found is a maximum.
Show answer & marking scheme

Worked solution

(a) The total surface area \(S\) of a closed cylinder is given by:
\(S = 2\pi r^2 + 2\pi r h\)

Given that \(S = 600\pi\\:\n\)2\\pi r^2 + 2\\pi r h = 600\\pi\)

Divide through by \(2\pi\\:\n\)r^2 + r h = 300\)

Rearrange to make \(h\) the subject:
\(h = \frac{300 - r^2}{r}\)

The volume \(V\) of a cylinder is given by:
\(V = \pi r^2 h\)

Substitute the expression for \(h\) into the volume formula:
\(V = \pi r^2 \left(\frac{300 - r^2}{r}\right)\)
\(V = \pi r (300 - r^2)\)
\(V = 300\pi r - \pi r^3\) (as required)

(b) To find the maximum volume, differentiate \(V\) with respect to \(r\):
\(\frac{\mathrm{d}V}{\mathrm{d}r} = 300\pi - 3\pi r^2\)

Set \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\):
\(300\pi - 3\pi r^2 = 0\)
\(3\pi r^2 = 300\pi\)
\(r^2 = 100\)

Since \(r > 0\), we have \(r = 10\).

Substitute \(r = 10\) back into the expression for \(V\):
\(V = 300\pi(10) - \pi(10)^3\)
\(V = 3000\pi - 1000\pi = 2000\pi\text{ cm}^3\).

(c) Differentiate \(\frac{\mathrm{d}V}{\mathrm{d}r}\) again with respect to \(r\) to find the second derivative:
\(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\)

Substitute \(r = 10\):
\(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi(10) = -60\pi\)

Since \(-60\pi < 0\), the volume is a maximum.

Marking scheme

(a)
M1: Writes down the correct formula for the total surface area of a closed cylinder and sets it equal to \(600\pi\\, e.g., \)2\\pi r^2 + 2\\pi r h = 600\\pi\).
M1: Rearranges the formula to express \(h\) in terms of \(r\\, e.g., \)h = \\frac{300-r^2}{r}\).
B1: States the formula for the volume of a cylinder, \(V = \pi r^2 h\).
A1*: Correctly substitutes \(h\) into the volume formula and simplifies to obtain the given expression \(V = 300\pi r - \pi r^3\) with no errors shown.

(b)
M1: Differentiates \(V\) with respect to \(r\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}r} = 300\pi - 3\pi r^2\).
M1: Sets their derivative equal to 0 and attempts to solve for \(r\).
A1: Obtains \(r = 10\) (rejecting the negative root \(r = -10\)).
A1: Substitutes \(r = 10\) into the volume formula and obtains \(2000\pi\) (accept with units \(\text{cm}^3\)).

(c)
M1: Differentiates again to obtain the second derivative \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\).
A1: Substitutes \(r = 10\) into their second derivative to show that \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -60\pi < 0\\, and concludes that the volume is a maximum. (Accept any valid alternative justification, e.g., testing values of \)\\frac{\\mathrm{d}V}{\\mathrm{d}r}\) on either side of \(r = 10\) with a complete table/explanation).
Question 7 · structured
10 marks
The curve \( C \) has equation \( y = \log_2(2x + 1) \).

(a) Complete the table of values below for \( y = \log_2(2x + 1) \).

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 0 & 0.5 & 1.5 & 3.5 & 7.5 \\ \hline
y & 0 & & 2 & & 4 \\ \hline
\end{array}
\]
(2)

(b) On a axes grid, plot the points from your completed table and draw the curve \( C \) for \( 0 \le x \le 7.5 \).
(2)

(c) By drawing a suitable straight line on your grid, obtain an estimate, to 1 decimal place, for the root of the equation
\[ \log_2(2x + 1) + 2x - 5 = 0 \]
(3)

(d) Show algebraically that the equation \( \log_2(2x + 1) + 2x - 5 = 0 \) can be written in the form \( 2x + 1 = 2^{5 - 2x} \).
(3)
Show answer & marking scheme

Worked solution

**Part (a)**
To find the missing \( y \)-values in the table:
When \( x = 0.5 \), \( y = \log_2(2(0.5) + 1) = \log_2(2) = 1 \).
When \( x = 3.5 \), \( y = \log_2(2(3.5) + 1) = \log_2(8) = 3 \).

**Part (b)**
Plot the points \( (0, 0) \), \( (0.5, 1) \), \( (1.5, 2) \), \( (3.5, 3) \), and \( (7.5, 4) \) and connect them with a smooth, continuous curve starting at the origin.

**Part (c)**
To solve \( \log_2(2x + 1) + 2x - 5 = 0 \), rearrange it to align with the plotted curve \( y = \log_2(2x + 1) \):
\[ \log_2(2x + 1) = 5 - 2x \]
Thus, we draw the line \( y = 5 - 2x \) on the same grid.
Two points on this line are \( (0, 5) \) and \( (2.5, 0) \).
The intersection of the line and the curve occurs at \( (1.5, 2) \).
Therefore, the estimated root is \( x = 1.5 \) (accept estimates in the range \( 1.4 \le x \le 1.6 \)).

**Part (d)**
Starting with the equation:
\[ \log_2(2x + 1) + 2x - 5 = 0 \]
Isolate the logarithmic term on one side:
\[ \log_2(2x + 1) = 5 - 2x \]
By the definition of logarithms, \( \log_b(A) = B \iff A = b^B \). With base \( b = 2 \):
\[ 2x + 1 = 2^{5 - 2x} \]
This is the required form.

Marking scheme

**Part (a)**
- **B1**: For correctly identifying \( y = 1 \) when \( x = 0.5 \).
- **B1**: For correctly identifying \( y = 3 \) when \( x = 3.5 \).

**Part (b)**
- **M1**: For plotting at least 4 of their points from the table correctly.
- **A1**: For drawing a smooth, single-valued curve passing through all 5 correct points.

**Part (c)**
- **M1**: For rearranging the equation to identify that the line to draw is \( y = 5 - 2x \).
- **M1**: For drawing the straight line \( y = 5 - 2x \) on the grid, passing through at least two correct points (e.g., \( (1.5, 2) \) and \( (2.5, 0) \)).
- **A1**: For finding the root \( x = 1.5 \) (accept range \( 1.4 \le x \le 1.6 \) from graph reading).

**Part (d)**
- **M1**: For rearranging the equation to isolate the log term: \( \log_2(2x + 1) = 5 - 2x \).
- **M1**: For correctly applying the base-2 exponential definition of logarithms.
- **A1**: For fully establishing the given result \( 2x + 1 = 2^{5 - 2x} \) with clear, logically sound steps.
Question 8 · structured
11 marks
In triangle \(OAB\), \(P\) is the point on \(OA\) such that \(OP : PA = 2:1\) and \(Q\) is the point on \(OB\) such that \(OQ : QB = 1:3\). The line \(AQ\) and the line \(BP\) intersect at the point \(X\). Let \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\).

(a) Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\), an expression for:
(i) \(\overrightarrow{AQ}\)
(ii) \(\overrightarrow{BP}\)

(b) Express \(\overrightarrow{OX}\) in two different forms in terms of \(\mathbf{a}\), \(\mathbf{b}\) and the scalar parameters \(\lambda\) and \(\mu\), where \(\overrightarrow{OX} = \overrightarrow{OA} + \mu \overrightarrow{AQ}\) and \(\overrightarrow{OX} = \overrightarrow{OB} + \lambda \overrightarrow{BP}\).

(c) Find the value of \(\lambda\) and the value of \(\mu\), and hence write down \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

(d) The point \(Y\) lies on the line \(OX\) produced such that \(OX : XY = 1 : k\). Given that \(AY\) is parallel to \(OB\), find the value of \(k\).
Show answer & marking scheme

Worked solution

(a)
(i) Since \(P\) divides \(OA\) in the ratio \(2:1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(Q\) divides \(OB\) in the ratio \(1:3\), \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\).
Therefore, \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\).
(ii) Similarly, \(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(b)
Using the first expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \mathbf{a} + \mu\left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1-\mu)\mathbf{a} + \frac{1}{4}\mu\mathbf{b}\)
Using the second expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b}\)

(c)
Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate their coefficients from the two expressions for \(\overrightarrow{OX}\):
For \(\mathbf{a}\): \(1-\mu = \frac{2}{3}\lambda\)
For \(\mathbf{b}\): \(\frac{1}{4}\mu = 1-\lambda\)
From the second equation, we get \(\lambda = 1 - \frac{1}{4}\mu\).
Substituting this into the first equation:
\(1-\mu = \frac{2}{3}\left(1 - \frac{1}{4}\mu\right) = \frac{2}{3} - \frac{1}{6}\mu\)
Rearranging terms to solve for \(\mu\):
\(1 - \frac{2}{3} = \mu - \frac{1}{6}\mu \implies \frac{1}{3} = \frac{5}{6}\mu \implies \mu = \frac{2}{5}\)
Substituting \(\mu = \frac{2}{5}\) back to find \(\lambda\):
\(\lambda = 1 - \frac{1}{4}\left(\frac{2}{5}\right) = 1 - \frac{1}{10} = \frac{9}{10}\)
Substituting either parameter back gives:
\(\overrightarrow{OX} = \left(1-\frac{2}{5}\right)\mathbf{a} + \frac{1}{4}\left(\frac{2}{5}\right)\mathbf{b} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\).

(d)
Since \(Y\) lies on \(OX\) produced, let \(\overrightarrow{OY} = m\overrightarrow{OX}\).
So, \(\overrightarrow{OY} = m\left(\frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\right)\).
Then \(\overrightarrow{AY} = \overrightarrow{OY} - \overrightarrow{OA} = \left(\frac{3}{5}m - 1\right)\mathbf{a} + \frac{1}{10}m\mathbf{b}\).
For \(AY\) to be parallel to \(OB\) (which is in the direction of \(\mathbf{b}\)), the coefficient of \(\mathbf{a}\) must be zero:
\(\frac{3}{5}m - 1 = 0 \implies m = \frac{5}{3}\).
This means \(\overrightarrow{OY} = \frac{5}{3}\overrightarrow{OX}\).
Since \(\overrightarrow{OY} = \overrightarrow{OX} + \overrightarrow{XY}\), we have \(\overrightarrow{XY} = \frac{2}{3}\overrightarrow{OX}\).
Therefore, \(OX : XY = 1 : \frac{2}{3}\), which gives \(k = \frac{2}{3}\).

Marking scheme

Part (a)
M1: For a correct attempt at vector addition for either \(\overrightarrow{AQ}\) or \(\overrightarrow{BP}\) (e.g., writing \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ}\) with at least one term correct in terms of \(\mathbf{a}\) or \(\mathbf{b}\)).
A1: Both \(\overrightarrow{AQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\) and \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\) correct.

Part (b)
B1: One correct expression for \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\), and either \(\lambda\) or \(\mu\).
B1: A second correct expression for \(\overrightarrow{OX}\) in terms of the other parameter.

Part (c)
M1: Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from their two expressions of \(\overrightarrow{OX}\) to form two linear equations.
M1: A complete algebraic method to solve the simultaneous equations for \(\lambda\) or \(\mu\).
A1: Finding one correct parameter value (either \(\mu = \frac{2}{5}\) or \(\lambda = \frac{9}{10}\)).
A1: Finding both correct parameter values.
A1: Expressing \(\overrightarrow{OX} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\) (or equivalent fractional forms).

Part (d)
M1: Writing \(\overrightarrow{OY} = m\overrightarrow{OX}\), expressing \(\overrightarrow{AY}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), and setting the coefficient of \(\mathbf{a}\) equal to \(0\) to find \(m\).
A1: Correctly identifying \(k = \frac{2}{3}\).
Question 9 · Calculus
8 marks
A curve \(C\) has equation \(y = f(x)\) such that

\[\frac{dy}{dx} = (2y + 1)e^{3x}\]

Given that \(y = 0\) when \(x = 0\):

(a) Find an expression for \(y\) in terms of \(x\).
(5 marks)

(b) Find the exact value of \(\frac{dy}{dx}\) when \(x = \ln(2)\).
(3 marks)
Show answer & marking scheme

Worked solution

(a) Separating the variables gives:
\[\int \frac{1}{2y + 1} \, dy = \int e^{3x} \, dx\]

Integrating both sides:
\[\frac{1}{2} \ln(2y + 1) = \frac{1}{3} e^{3x} + C\]

Using the boundary condition \(y = 0\) when \(x = 0\):
\[\frac{1}{2} \ln(1) = \frac{1}{3} e^0 + C \implies C = -\frac{1}{3}\]

Substituting \(C\) back into the equation:
\[\frac{1}{2} \ln(2y + 1) = \frac{1}{3} e^{3x} - \frac{1}{3}\]

Multiplying by 2:
\[\ln(2y + 1) = \frac{2}{3} (e^{3x} - 1)\]

Exponentiating both sides:
\[2y + 1 = e^{\frac{2}{3}(e^{3x} - 1)}\]

Rearranging to make \(y\) the subject:
\[y = \frac{1}{2} \left( e^{\frac{2}{3}(e^{3x} - 1)} - 1 \right)\]

(b) When \(x = \ln(2)\):
\[e^{3x} = e^{3\ln(2)} = e^{\ln(8)} = 8\]

From the expression in part (a):
\[2y + 1 = e^{\frac{2}{3}(e^{3x} - 1)}\]

Substituting \(e^{3x} = 8\):
\[2y + 1 = e^{\frac{2}{3}(8 - 1)} = e^{\frac{14}{3}}\]

Using the differential equation \(\frac{dy}{dx} = (2y + 1)e^{3x}\):
\[\frac{dy}{dx} = e^{\frac{14}{3}} \times 8 = 8e^{\frac{14}{3}}\]

Marking scheme

(a)
M1: Separation of variables to obtain \(\int \frac{1}{2y + 1} \, dy = \int e^{3x} \, dx\) (condone missing integral signs).
M1: Integrates both sides, LHS to involve \(\ln(2y+1)\) and RHS to involve \(e^{3x}\) (ignore constant of integration at this stage).
A1: Fully correct integrated expression: \(\frac{1}{2} \ln(2y+1) = \frac{1}{3}e^{3x} + C\) (or equivalent).
M1: Substitute \(x = 0, y = 0\) to find the value of the constant of integration \(C\).
A1: Correct explicit expression for \(y\): \(y = \frac{1}{2} \left( e^{\frac{2}{3}(e^{3x} - 1)} - 1 \right)\) (or any equivalent correct explicit form).

(b)
B1: Deduces that \(e^{3x} = 8\) when \(x = \ln(2)\).
M1: Substitutes their value of \(e^{3x}\) into their expression for \(2y+1\) (or \(y\)) and substitutes these into the differential equation to find \(\frac{dy}{dx}\).
A1: Correct exact value of \(8e^{\frac{14}{3}}\) (or equivalent exact expression, e.g. \(8e^{14/3}\)).
Question 10 · structured
17 marks
The curve \(C\) has the equation \(y = \frac{2x - 5}{x - 3}\), where \(x \neq 3\).

(a) Write down the equation of
(i) the asymptote to \(C\) which is parallel to the \(y\)-axis,
(ii) the asymptote to \(C\) which is parallel to the \(x\)-axis.
(2)

(b) Find the coordinates of the points where \(C\) crosses the coordinate axes.
(3)

(c) Sketch the curve \(C\), showing clearly its asymptotes and the coordinates of the points where \(C\) crosses the coordinate axes.
(4)

(d) Find an equation of the normal to \(C\) at the point \(P\) on \(C\) where \(x = 4\). Show all your working.
(5)

(e) The normal to \(C\) at \(P\) intersects \(C\) again at the point \(Q\).
Find the coordinates of \(Q\).
(3)
Show answer & marking scheme

Worked solution

**(a)**

(i) The vertical asymptote occurs where the denominator is zero:
\(x - 3 = 0 \implies x = 3\)

(ii) As \(x \to \pm \infty\), \(y \to \frac{2x}{x} = 2\).
So the horizontal asymptote is:
\(y = 2\)

**(b)**

To find the intersection with the \(y\)-axis, set \(x = 0\):
\(y = \frac{2(0) - 5}{0 - 3} = \frac{5}{3}\)
So the intersection is \(\left(0, \frac{5}{3}\right)\).

To find the intersection with the \(x\)-axis, set \(y = 0\):
\(2x - 5 = 0 \implies x = \frac{5}{2}\)
So the intersection is \(\left(\frac{5}{2}, 0\right)\).

**(c)**

To sketch the curve:
1. Draw the asymptotes \(x = 3\) and \(y = 2\) as dashed lines.
2. One branch lies in the region \(x > 3\) and \(y > 2\). Since \(P(4, 3)\) lies on this branch, it lies entirely in the upper-right region relative to the asymptotes.
3. The other branch lies in the region \(x < 3\) and \(y < 2\). It must pass through the axes at \(\left(0, \frac{5}{3}\right)\) and \(\left(\frac{5}{2}, 0\right)\).

**(d)**

We find the derivative \(\frac{dy}{dx}\) using the quotient rule on \(y = \frac{2x - 5}{x - 3}\):
Let \(u = 2x - 5 \implies \frac{du}{dx} = 2\)
Let \(v = x - 3 \implies \frac{dv}{dx} = 1\)
\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{2(x - 3) - (2x - 5)(1)}{(x - 3)^2}\)
\(\frac{dy}{dx} = \frac{2x - 6 - 2x + 5}{(x - 3)^2} = \frac{-1}{(x - 3)^2}\)

At the point \(P\) where \(x = 4\):
\(y = \frac{2(4) - 5}{4 - 3} = 3\)
So \(P\) has coordinates \((4, 3)\).

Substitute \(x = 4\) into \(\frac{dy}{dx}\) to find the gradient of the tangent, \(m_t\):
\(m_t = \frac{-1}{(4 - 3)^2} = -1\)

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

The equation of the normal at \(P(4, 3)\) is:
\(y - 3 = 1(x - 4)\)
\(y = x - 1\)

**(e)**

To find the intersection of the normal \(y = x - 1\) and the curve \(C\), we set their equations equal:
\(x - 1 = \frac{2x - 5}{x - 3}\)
\((x - 1)(x - 3) = 2x - 5\)
\(x^2 - 4x + 3 = 2x - 5\)
\(x^2 - 6x + 8 = 0\)
\((x - 2)(x - 4) = 0\)

We already know that \(x = 4\) is the point \(P\).
Thus, for the other point of intersection \(Q\), we have \(x = 2\).

Substitute \(x = 2\) into the equation of the normal to find the \(y\)-coordinate:
\(y = 2 - 1 = 1\)

So the coordinates of \(Q\) are \((2, 1)\).

Marking scheme

**(a)**
* **B1**: For \(x = 3\) (or equivalent, e.g., asymptote parallel to \(y\)-axis is \(x = 3\)).
* **B1**: For \(y = 2\) (or equivalent, e.g., asymptote parallel to \(x\)-axis is \(y = 2\)).

**(b)**
* **B1**: For setting \(x = 0\) to find \(y = \frac{5}{3}\) or writing down \(\left(0, \frac{5}{3}\right)\).
* **M1**: For setting \(y = 0\) to solve \(2x - 5 = 0\).
* **A1**: For the coordinate \(\left(\frac{5}{2}, 0\right)\) or \(x = 2.5\).

**(c)**
* **B1**: Both asymptotes correctly drawn as dashed lines and labelled with equations \(x=3\) and \(y=2\).
* **B1**: One branch correctly drawn in the region \(x > 3\) and \(y > 2\).
* **B1**: The other branch correctly drawn in the region \(x < 3\) and \(y < 2\), approaching the asymptotes.
* **B1**: Intercepts on both axes clearly marked with correct coordinates or values \(\frac{5}{3}\) on the \(y\)-axis and \(\frac{5}{2}\) on the \(x\)-axis.

**(d)**
* **M1**: For an attempt to differentiate using the quotient rule, product rule, or chain rule (after division).
* **A1**: Correct derivative: \(\frac{dy}{dx} = -\frac{1}{(x-3)^2}\).
* **M1**: For substituting \(x=4\) into their \(\frac{dy}{dx}\) to find the gradient of the tangent, and then finding the negative reciprocal to obtain the gradient of the normal.
* **M1**: Correctly finding \(y = 3\) at \(x = 4\), and setting up the straight-line equation \(y - 3 = m(x - 4)\) using their normal gradient \(m\).
* **A1**: A correct equation of the normal, e.g., \(y = x - 1\) (or \(x - y - 1 = 0\)).

**(e)**
* **M1**: For equating their line equation from part (d) with the equation of curve \(C\).
* **M1**: For multiplying out and forming a three-term quadratic equation in the form \(x^2 + ax + b = 0\) and attempting to solve it (by factoring or formula) to find \(x \neq 4\).
* **A1**: For the correct coordinates \((2, 1)\) of \(Q\) (both \(x\) and \(y\) must be correct).
Question 11 · Trig identities, equations & integration
15 marks

(a) Show that \(\cos(3\theta) = 4\cos^3 \theta - 3\cos \theta\). (4 marks)

(b) Hence, or otherwise, find the exact value of \(\int_{0}^{\frac{\pi}{3}} \cos^3 \theta \text{ d}\theta\). (4 marks)

(c) Solve the equation \(\cos(3\theta) + 2\cos\theta = 0\) for \(0 \le \theta \le \pi\). (3 marks)

(d) Find the exact coordinates of the stationary points on the curve with equation \(y = \frac{1}{3}\cos(3x) - \cos x\) in the interval \(0 < x < \pi\). (4 marks)

Show answer & marking scheme

Worked solution

(a) Use the compound angle formula for \(\cos(A + B)\) where \(A = 2\theta\) and \(B = \theta\):
\(\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos\theta - \sin(2\theta)\sin\theta\)
Substitute the double angle formulas \(\cos(2\theta) = 2\cos^2\theta - 1\) and \(\sin(2\theta) = 2\sin\theta\cos\theta\):
\(\cos(3\theta) = (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta\)
\(= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta\)
Using the Pythagorean identity \(\sin^2\theta = 1 - \cos^2\theta\):
\(\cos(3\theta) = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta\)
\(= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta\)
\(= 4\cos^3\theta - 3\cos\theta\)

(b) Rearranging the identity from part (a):
\(\cos^3\theta = \frac{1}{4}(\cos(3\theta) + 3\cos\theta)\)
Now integrate with respect to \(\theta\):
\(\int_{0}^{\frac{\pi}{3}} \cos^3\theta\text{ d}\theta = \int_{0}^{\frac{\pi}{3}} \frac{1}{4}(\cos(3\theta) + 3\cos\theta)\text{ d}\theta\)
\(= \left[ \frac{1}{12}\sin(3\theta) + \frac{3}{4}\sin\theta \right]_{0}^{\frac{\pi}{3}}\)
Substitute the upper limit \(\theta = \frac{\pi}{3}\):
\(\frac{1}{12}\sin(\pi) + \frac{3}{4}\sin\left(\frac{\pi}{3}\right) = 0 + \frac{3}{4}\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{8}\)
Substitute the lower limit \(\theta = 0\):
\(\frac{1}{12}\sin(0) + \frac{3}{4}\sin(0) = 0\)
Therefore, the exact value is \(\frac{3\sqrt{3}}{8}\).

(c) Substitute the identity from part (a) into the equation:
\(4\cos^3\theta - 3\cos\theta + 2\cos\theta = 0\)
\(4\cos^3\theta - \cos\theta = 0\)
\(\cos\theta(4\cos^2\theta - 1) = 0\)
Thus, \(\cos\theta = 0\) or \(\cos^2\theta = \frac{1}{4} \implies \cos\theta = \pm\frac{1}{2}\).
For \(0 \le \theta \le \pi\):
If \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}\)
If \(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\)
If \(\cos\theta = -\frac{1}{2} \implies \theta = \frac{\pi}{3}\text{ in second quadrant, which is } \frac{2\pi}{3}\)
So the solutions are \(\theta = \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}\).

(d) Differentiate \(y = \frac{1}{3}\cos(3x) - \cos x\) with respect to \(x\):
\(\frac{\text{d}y}{\text{d}x} = -\sin(3x) + \sin x\)
At stationary points, \(\frac{\text{d}y}{\text{d}x} = 0 \implies \sin(3x) - \sin x = 0\).
Using the sum-to-product formula:
\(2\cos(2x)\sin(-x) = 0 \implies -2\cos(2x)\sin x = 0\).
Since \(0 < x < \pi\), \(\sin x \neq 0\). Thus, we must have \(\cos(2x) = 0\).
For \(0 < x < \pi \implies 0 < 2x < 2\pi\).
\(2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}\)
\(2x = \frac{3\pi}{2} \implies x = \frac{3\pi}{4}\)
Now find the corresponding \(y\)-coordinates:
For \(x = \frac{\pi}{4}\):
\(y = \frac{1}{3}\cos\left(\frac{3\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{1}{3}\left(-\frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2} = -\frac{2\sqrt{2}}{3}\).
For \(x = \frac{3\pi}{4}\):
\(y = \frac{1}{3}\cos\left(\frac{9\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{2}}{2}\right) = \frac{2\sqrt{2}}{3}\).
So, the exact coordinates of the stationary points are \(\left(\frac{\pi}{4}, -\frac{2\sqrt{2}}{3}\right)\) and \(\left(\frac{3\pi}{4}, \frac{2\sqrt{2}}{3}\right)\).

Marking scheme

Part (a):
M1: Attempts to express \(\cos(3\theta)\) as \(\cos(2\theta + \theta)\) and expands using compound angle formula.
M1: Uses double angle formulas for both \(\cos(2\theta)\) and \(\sin(2\theta)\).
A1: Obtains correct intermediate expression \((2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta\).
A1*: Employs \(\sin^2\theta = 1 - \cos^2\theta\) and simplifies correctly to show the given identity.

Part (b):
M1: Expresses \(\cos^3\theta\) in terms of \(\cos(3\theta)\) and \(\cos\theta\).
M1: Attempts to integrate, showing at least one term of the form \(A\sin(3\theta)\) or \(B\sin\theta\).
A1: Correct integration yielding \(\frac{1}{12}\sin(3\theta) + \frac{3}{4}\sin\theta\).
A1: Substitutes limits correctly to obtain exact value \(\frac{3\sqrt{3}}{8}\).

Part (c):
M1: Uses the identity from (a) to form a cubic equation in \(\cos\theta\).
M1: Factorises or solves to find values for \(\cos\theta\) (at least \(\cos\theta = 0\) and one of \(\pm\frac{1}{2}\)).
A1: Obtains all three correct angles \(\theta = \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}\) and no others in the range.

Part (d):
M1: Differentiates the curve equation to find \(\frac{\text{d}y}{\text{d}x}\).
M1: Sets \(\frac{\text{d}y}{\text{d}x} = 0\) and solves the resulting trigonometric equation to find \(x\) in the given interval.
A1: Correct \(x\) values \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).
A1: Substitutes \(x\) back to get the correct exact \(y\)-coordinates, giving the final coordinates as \(\left(\frac{\pi}{4}, -\frac{2\sqrt{2}}{3}\right)\) and \(\left(\frac{3\pi}{4}, \frac{2\sqrt{2}}{3}\right)\).

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