Edexcel IAL · Thinka-original Practice Paper

2023 Edexcel IAL Pure Mathematics (YPM01) Practice Paper with Answers

Thinka Jan 2023 Cambridge International A Level-Style Mock — Pure Mathematics (YPM01)

300 marks360 mins2023
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Jan 2023 Cambridge International A Level Pure Mathematics (YPM01) paper. Not affiliated with or reproduced from Cambridge.

Section WMA11/01: Pure Mathematics P1

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
11 Question · 74.7 marks
Question 1 · Short Answer
4 marks
Find the exact value of

\(\int_{1}^{4} \left( \frac{6}{\sqrt{x}} - \frac{1}{2}x^2 \right) \text{d}x\)
Show answer & marking scheme

Worked solution

To find the exact value of the integral:

1. Rewrite the integrand in a form ready for integration:
\(\frac{6}{\sqrt{x}} - \frac{1}{2}x^2 = 6x^{-\frac{1}{2}} - \frac{1}{2}x^2\)

2. Integrate term by term:
\(\int \left( 6x^{-\frac{1}{2}} - \frac{1}{2}x^2 \right) \text{d}x = \left[ \frac{6x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{\frac{1}{2}x^3}{3} \right] = \left[ 12x^{\frac{1}{2}} - \frac{1}{6}x^3 \right]\)

3. Substitute the upper limit \(x = 4\):
\(12(4)^{\frac{1}{2}} - \frac{1}{6}(4)^3 = 12(2) - \frac{64}{6} = 24 - \frac{32}{3} = \frac{40}{3}\)

4. Substitute the lower limit \(x = 1\):
\(12(1)^{\frac{1}{2}} - \frac{1}{6}(1)^3 = 12 - \frac{1}{6} = \frac{71}{6}\)

5. Subtract the lower limit result from the upper limit result:
\(\frac{40}{3} - \frac{71}{6} = \frac{80}{6} - \frac{71}{6} = \frac{9}{6} = \frac{3}{2}\)

Marking scheme

M1: For an attempt to integrate at least one term (power of \(x\) increased by 1).
A1: Correct integration: \(12x^{\frac{1}{2}} - \frac{1}{6}x^3\) (ignore constant of integration).
M1: For substituting both limits 4 and 1 into their integrated expression and subtracting the correct way round.
A1: Correct exact value of \(\frac{3}{2}\) or \(1.5\) (or any equivalent fraction).
Question 2 · Short Answer
5 marks
The curve \(C\) has equation

\[y = \frac{4x^3 - 5}{2x}, \quad x > 0\]

Find the equation of the tangent to \(C\) at the point where \(x = 1\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers to be found.
Show answer & marking scheme

Worked solution

1. Express \(y\) in a form suitable for differentiation:
\[y = \frac{4x^3}{2x} - \frac{5}{2x} = 2x^2 - \frac{5}{2}x^{-1}\]

2. Find the coordinate of the point where \(x = 1\):
\[y = 2(1)^2 - \frac{5}{2}(1)^{-1} = 2 - 2.5 = -0.5 = -\frac{1}{2}\]
So the point of contact is \((1, -0.5)\).

3. Differentiate \(y\) with respect to \(x\) to find the gradient function:
\[\frac{\text{d}y}{\text{d}x} = 4x - \left(-\frac{5}{2}\right)x^{-2} = 4x + \frac{5}{2}x^{-2}\]

4. Evaluate the gradient at \(x = 1\):
\[m = 4(1) + \frac{5}{2}(1)^{-2} = 4 + 2.5 = 6.5 = \frac{13}{2}\]

5. Find the equation of the tangent line using \(y - y_1 = m(x - x_1)\):
\[y - \left(-\frac{1}{2}\right) = \frac{13}{2}(x - 1)\]
\[y + \frac{1}{2} = \frac{13}{2}x - \frac{13}{2}\]
Multiply the entire equation by 2 to clear fractions:
\[2y + 1 = 13x - 13\]
Rearrange to the form \(ax + by + c = 0\):
\[13x - 2y - 14 = 0\]

Marking scheme

M1: For attempting to expand or split the fraction to express \(y\) as separate terms in powers of \(x\).
A1: Correct simplified expression for \(y = 2x^2 - \frac{5}{2}x^{-1}\) (or equivalent index form).
M1: Differentiates their expression. Look for at least one term differentiated correctly (power multiplied by coefficient and power reduced by 1).
A1: Correct derivative \(\frac{\text{d}y}{\text{d}x} = 4x + \frac{5}{2}x^{-2}\). Finds correct gradient \(m = \frac{13}{2}\) and correct coordinate \(y = -\frac{1}{2}\).
M1: Applies the tangent formula \(y - y_1 = m(x - x_1)\) using their calculated \(y\) and \(m\) at \(x=1\).
A1: Correct equation in the form \(13x - 2y - 14 = 0\) (or any integer multiple thereof, e.g., \(26x - 4y - 28 = 0\)).
Question 3 · Short Answer
4 marks
Solve, for \(0 \le \theta < 360^\circ\), the equation

\[3\cos^2\theta - 5\sin\theta - 1 = 0\]

giving your answers to one decimal place.
Show answer & marking scheme

Worked solution

1. Use the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\):
\[3(1 - \sin^2\theta) - 5\sin\theta - 1 = 0\]
\[3 - 3\sin^2\theta - 5\sin\theta - 1 = 0\]
\[-3\sin^2\theta - 5\sin\theta + 2 = 0\]
Multiply by \(-1\) to make the leading term positive:
\[3\sin^2\theta + 5\sin\theta - 2 = 0\]

2. Factorise the quadratic equation:
\[(3\sin\theta - 1)(\sin\theta + 2) = 0\]

3. Solve for \(\sin\theta\):
- Either \(3\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{3}\)
- Or \(\sin\theta + 2 = 0 \implies \sin\theta = -2\) (which has no real solutions since \(-1 \le \sin\theta \le 1\))

4. Find the values of \(\theta\) for \(\sin\theta = \frac{1}{3}\) in the range \(0 \le \theta < 360^\circ\):
- Primary solution: \(\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\), which rounds to \(19.5^\circ\).
- Secondary solution: \(\theta = 180^\circ - 19.47^\circ \approx 160.53^\circ\), which rounds to \(160.5^\circ\).

Marking scheme

M1: Substitution of \(\cos^2\theta = 1 - \sin^2\theta\) to form an equation in \(\sin\theta\).
A1: Correct three-term quadratic equation, e.g., \(3\sin^2\theta + 5\sin\theta - 2 = 0\) (or equivalent).
M1: Solves the quadratic to find a value for \(\sin\theta\) (must be \(\frac{1}{3}\)), and finds at least one correct principal value for \(\theta\) to 1 d.p.
A1: Correct final answers of \(19.5^\circ\) and \(160.5^\circ\) only (with no extra angles in the range).
Question 4 · Short Answer
5 marks
The line \(l_1\) has equation \(2x + 3y - 6 = 0\).

The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \(P(4, 7)\).

(a) Find an equation for \(l_2\) in the form \(y = mx + c\).

(b) Find the coordinates of the point of intersection of \(l_1\) and \(l_2\).
Show answer & marking scheme

Worked solution

**(a)**
First, express the equation of \(l_1\) in gradient-intercept form:
\[2x + 3y - 6 = 0 \implies 3y = -2x + 6 \implies y = -\frac{2}{3}x + 2\]
So the gradient of \(l_1\) is \(m_1 = -\frac{2}{3}\).

Since \(l_2\) is perpendicular to \(l_1\), the gradient of \(l_2\) is:
\[m_2 = -\frac{1}{m_1} = \frac{3}{2}\]

Using the point-slope form with point \(P(4, 7)\):
\[y - 7 = \frac{3}{2}(x - 4)\]
\[y - 7 = \frac{3}{2}x - 6\]
\[y = \frac{3}{2}x + 1\]

**(b)**
To find the point of intersection, substitute \(y = \frac{3}{2}x + 1\) into the equation of \(l_1\):
\[2x + 3\left(\frac{3}{2}x + 1\right) - 6 = 0\]
\[2x + \frac{9}{2}x + 3 - 6 = 0\]
\[\frac{13}{2}x - 3 = 0\]
\[\frac{13}{2}x = 3 \implies x = \frac{6}{13}\]

Substitute \(x = \frac{6}{13}\) back into the equation of \(l_2\):
\[y = \frac{3}{2}\left(\frac{6}{13}\right) + 1 = \frac{9}{13} + 1 = \frac{22}{13}\]

So the point of intersection is \(\left(\frac{6}{13}, \frac{22}{13}\right)\).

Marking scheme

**(a)**
M1: Attempts to find the gradient of \(l_1\) and uses \(m_1 m_2 = -1\) to find the gradient of \(l_2\).
A1: Correct gradient of \(l_2\) is \(\frac{3}{2}\).
A1: Correct equation of \(l_2\) in the form \(y = mx + c\), which is \(y = \frac{3}{2}x + 1\).

**(b)**
M1: Eliminates one variable (either \(x\) or \(y\)) to find the intersection point.
A1: Correct coordinates \(\left(\frac{6}{13}, \frac{22}{13}\right)\) (both must be exact fractions or equivalent).
Question 5 · Structured
8.1 marks
The curve \(C\) has equation \(y = x^2 - 4kx + 2\), where \(k\) is a constant.
The line \(L\) has equation \(y = 2x - k\).

(a) Show that the \(x\)-coordinates of the points of intersection of \(C\) and \(L\) satisfy the equation
\[x^2 - 2(2k+1)x + k + 2 = 0\]

(b) Given that \(C\) and \(L\) do not intersect, find the set of possible values of \(k\).
Show answer & marking scheme

Worked solution

(a) Equating the equations of the curve and line:
\[x^2 - 4kx + 2 = 2x - k\]
Rearranging to make the right-hand side zero:
\[x^2 - 4kx - 2x + k + 2 = 0\]
\[x^2 - 2(2k+1)x + k + 2 = 0\] as required.

(b) For the curve and line to not intersect, this quadratic equation in \(x\) must have no real roots. Therefore, the discriminant must be less than zero:
\[b^2 - 4ac < 0\]
Here, \(a = 1\), \(b = -2(2k+1)\), and \(c = k+2\).
\[[-2(2k+1)]^2 - 4(1)(k+2) < 0\]
\[4(4k^2 + 4k + 1) - 4(k+2) < 0\]
Divide both sides by 4:
\[4k^2 + 4k + 1 - (k+2) < 0\]
\[4k^2 + 3k - 1 < 0\]
Factorising the quadratic inequality:
\[(4k-1)(k+1) < 0\]
The critical values are \(k = -1\) and \(k = 0.25\).
Since the inequality is less than zero, the solution is the region between these values:
\[-1 < k < 0.25\]

Marking scheme

(a)
M1: Equating the curve and line expressions to set up an equation.
A1: Rearranging terms to show the given equation with correct working.

(b)
M1: Attempting to use the discriminant condition \(b^2 - 4ac < 0\).
A1: Correct substitution of \(a\), \(b\), and \(c\) into the discriminant expression.
M1: Expanding and simplifying to a three-term quadratic inequality, e.g., \(4k^2 + 3k - 1 < 0\).
M1: Finding critical values of their quadratic inequality (factorising or using formula).
A1.1: Correct set of values: \(-1 < k < 0.25\) (or equivalent notation, e.g., interval notation).
Question 6 · Structured
8.1 marks
The points \(A\) and \(B\) have coordinates \((-2, 5)\) and \((6, 1)\) respectively.

(a) Find an equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ay + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers to be found.

The perpendicular bisector of \(AB\) crosses the \(x\)-axis at the point \(P\).

(b) Find the area of the triangle \(ABP\).
Show answer & marking scheme

Worked solution

(a) First, find the midpoint \(M\) of \(AB\):
\[M = \left(\frac{-2+6}{2}, \frac{5+1}{2}\right) = (2, 3)\]
Next, find the gradient of \(AB\):
\[m_{AB} = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}\]
The gradient of the perpendicular bisector is the negative reciprocal:
\[m_{\perp} = -\frac{1}{m_{AB}} = 2\]
Using the point-slope form with \(M(2,3)\) and \(m=2\):
\[y - 3 = 2(x - 2)\]
\[y - 3 = 2x - 4\]
Rearranging to the form \(ay + bx + c = 0\):
\[y - 2x + 1 = 0\] (or any integer multiple, e.g., \(2x - y - 1 = 0\)).

(b) The point \(P\) is on the \(x\)-axis, so let \(y = 0\) in the equation of the perpendicular bisector:
\[0 - 2x + 1 = 0 \implies x = 0.5\]
So, \(P\) has coordinates \((0.5, 0)\).
To find the area of triangle \(ABP\), we can use the coordinates of the vertices \(A(-2, 5)\), \(B(6, 1)\), and \(P(0.5, 0)\):
\[\text{Area} = \frac{1}{2} |x_A(y_B - y_P) + x_B(y_P - y_A) + x_P(y_A - y_B)|\]
\[\text{Area} = \frac{1}{2} |-2(1 - 0) + 6(0 - 5) + 0.5(5 - 1)|\]
\[\text{Area} = \frac{1}{2} |-2 - 30 + 2| = \frac{1}{2} |-30| = 15\]

Marking scheme

(a)
M1: Finding the midpoint of \(AB\).
M1: Finding the gradient of \(AB\).
M1: Finding the gradient of the perpendicular bisector (negative reciprocal).
M1: Applying \(y - y_1 = m(x - x_1)\) with their midpoint and perpendicular gradient.
A1: Obtaining a correct equation in the required form, e.g., \(y - 2x + 1 = 0\).

(b)
M1: Finding the coordinates of \(P\) by setting \(y = 0\).
M1: Using an appropriate method to find the area of the triangle (e.g., Shoelace formula, or finding length of base \(AB\) and perpendicular height \(PM\)).
A1.1: Correct area of \(15\).
Question 7 · Structured
8.1 marks
(a) Show that the equation
\[6\sin^2 \theta - \cos \theta - 4 = 0\]
can be written as
\[6\cos^2 \theta + \cos \theta - 2 = 0\]

(b) Hence, solve for \(0 \le x < 360^\circ\), the equation
\[6\sin^2 (2x + 10^\circ) - \cos (2x + 10^\circ) - 4 = 0\]
giving your answers to 1 decimal place.
Show answer & marking scheme

Worked solution

(a) Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[6(1 - \cos^2 \theta) - \cos \theta - 4 = 0\]
\[6 - 6\cos^2 \theta - \cos \theta - 4 = 0\]
\[-6\cos^2 \theta - \cos \theta + 2 = 0\]
Multiplying by \(-1\):
\[6\cos^2 \theta + \cos \theta - 2 = 0\] as required.

(b) Let \(\theta = 2x + 10^\circ\). Since \(0 \le x < 360^\circ\), we have:
\[0 \le 2x < 720^\circ \implies 10^\circ \le \theta < 730^\circ\]
From part (a), \(6\cos^2 \theta + \cos \theta - 2 = 0\).
Factorising this quadratic equation:
\[(3\cos \theta + 2)(2\cos \theta - 1) = 0\]
So, \(\cos \theta = -\frac{2}{3}\) or \(\cos \theta = \frac{1}{2}\).

Case 1: \(\cos \theta = \frac{1}{2}\)
Within the interval \(10^\circ \le \theta < 730^\circ\):
\[\theta = 60^\circ, 300^\circ, 420^\circ, 660^\circ\]
Solving for \(x\) where \(x = \frac{\theta - 10^\circ}{2}\):
- \(\theta = 60^\circ \implies x = 25^\circ\)
- \(\theta = 300^\circ \implies x = 145^\circ\)
- \(\theta = 420^\circ \implies x = 205^\circ\)
- \(\theta = 660^\circ \implies x = 325^\circ\)

Case 2: \(\cos \theta = -\frac{2}{3}\)
The principal value is \(\theta = \cos^{-1}(-2/3) \approx 131.81^\circ\).
Other values in the interval \(10^\circ \le \theta < 730^\circ\) are:
\[\theta \approx 131.81^\circ, 228.19^\circ, 491.81^\circ, 588.19^\circ\]
Solving for \(x\):
- \(\theta = 131.81^\circ \implies x = 60.9^\circ\)
- \(\theta = 228.19^\circ \implies x = 109.1^\circ\)
- \(\theta = 491.81^\circ \implies x = 240.9^\circ\)
- \(\theta = 588.19^\circ \implies x = 289.1^\circ\)

Marking scheme

(a)
M1: Attempting to use \(\sin^2 \theta = 1 - \cos^2 \theta\) in the equation.
A1: Correct substitution and rearrangement to show the given quadratic in \(\cos \theta\).

(b)
M1: Attempting to factorise or solve the quadratic equation to find values of \(\cos(2x+10^\circ)\).
A1: Correct values of \(\cos(2x+10^\circ) = -2/3\) and \(\cos(2x+10^\circ) = 1/2\).
M1: Finding at least two correct values of \(\theta = 2x+10^\circ\) in the appropriate range.
M1: Setting up correct equations to solve for \(x\) from their values of \(\theta\).
A1: Finding the four solutions from \(\cos(2x+10^\circ) = 1/2\): \(25^\circ, 145^\circ, 205^\circ, 325^\circ\).
A1.1: Finding the four solutions from \(\cos(2x+10^\circ) = -2/3\): \(60.9^\circ, 109.1^\circ, 240.9^\circ, 289.1^\circ\) (allow rounding errors of 0.1).
Question 8 · Structured
8.1 marks
The curve \(C\) has equation \(y = 2x^2 + \frac{32}{x} - 5\), \(x \neq 0\).

(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\).

(b) Find the equation of the tangent to \(C\) at the point \(P(4, 35)\), giving your answer in the form \(y = mx + c\).

(c) Find the coordinates of the stationary point of \(C\).
Show answer & marking scheme

Worked solution

(a) Rewrite \(y\) as: \(y = 2x^2 + 32x^{-1} - 5\).
Differentiating with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4x - 32x^{-2} = 4x - \frac{32}{x^2}\]

(b) To find the gradient of the tangent at \(P(4, 35)\), substitute \(x = 4\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[m = 4(4) - \frac{32}{4^2} = 16 - 2 = 14\]
Using the tangent line equation: \(y - 35 = 14(x - 4)\)
\[y - 35 = 14x - 56 \implies y = 14x - 21\]

(c) For stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[4x - \frac{32}{x^2} = 0\]
\[4x^3 = 32 \implies x^3 = 8 \implies x = 2\]
Substitute \(x = 2\) back into the equation of the curve to find the \(y\)-coordinate:
\[y = 2(2)^2 + \frac{32}{2} - 5 = 8 + 16 - 5 = 19\]
So the stationary point has coordinates \((2, 19)\).

Marking scheme

(a)
M1: Differentiating at least one term correctly (power decreased by 1).
A1: Fully correct derivative: \(4x - \frac{32}{x^2}\) (or equivalent).

(b)
M1: Substituting \(x=4\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of the tangent.
M1: Using \(y - 35 = m(x - 4)\) with their gradient.
A1: Obtaining \(y = 14x - 21\).

(c)
M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempting to solve for \(x\).
A1: Correct value \(x = 2\).
A1.1: Correct coordinates \((2, 19)\).
Question 9 · Structured
8.1 marks
Figure 1 shows a sketch of part of the curve \(C\) with equation \(y = 9x^{\frac{1}{2}} - 3x^{\frac{3}{2}}\), \(x \ge 0\).
The curve meets the \(x\)-axis at the origin and at the point \(A(3, 0)\).
The region \(R\) is bounded by the curve \(C\) and the \(x\)-axis.

(a) Use integration to find the exact area of the region \(R\).

(b) State, with a brief reason, whether the area of the region bounded by the curve \(y = 9(x-2)^{\frac{1}{2}} - 3(x-2)^{\frac{3}{2}}
\) and the \(x\)-axis is greater than, equal to, or less than the area of \(R\).
Show answer & marking scheme

Worked solution

(a) The area \(R\) is given by the definite integral:
\[R = \int_0^3 \left( 9x^{\frac{1}{2}} - 3x^{\frac{3}{2}} \right) \mathrm{d}x\]
Integrating each term:
\[\int 9x^{\frac{1}{2}} \mathrm{d}x = 9 \frac{x^{3/2}}{3/2} = 6x^{\frac{3}{2}}\]
\[\int -3x^{\frac{3}{2}} \mathrm{d}x = -3 \frac{x^{5/2}}{5/2} = -\frac{6}{5}x^{\frac{5/2}{}}\]
So, the integral is:
\[\left[ 6x^{\frac{3}{2}} - \frac{6}{5}x^{\frac{5/2}{}} \right]_0^3\]
Substituting the upper limit \(x = 3\):
\[6(3)^{\frac{3}{2}} - \frac{6}{5}(3)^{\frac{5/2}{}} = 6(3\sqrt{3}) - \frac{6}{5}(9\sqrt{3}) = 18\sqrt{3} - \frac{54}{5}\sqrt{3} = \frac{90\sqrt{3} - 54\sqrt{3}}{5} = \frac{36\sqrt{3}}{5}\]
Substituting the lower limit \(x = 0\) gives \(0\).
Thus, the exact area is \(\frac{36\sqrt{3}}{5}\).

(b) The curve \(y = 9(x-2)^{\frac{1}{2}} - 3(x-2)^{\frac{3}{2}}\) is obtained by translating the curve \(C\) by 2 units to the right (translation vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\)).
Since translations are rigid transformations, they preserve the size and shape of regions.
Therefore, the area of the new region is equal to the area of \(R\).

Marking scheme

(a)
M1: Attempting to integrate: raising the power of \(x\) by 1 on at least one term.
A1: Fully correct integrated expression: \(6x^{\frac{3}{2}} - \frac{6}{5}x^{\frac{5/2}{}}\).
M1: Substituting the limits 3 and 0 and subtracting.
M1: Attempting to simplify powers of 3 to surd form (e.g., \(3^{3/2} = 3\sqrt{3}\)).
A1: Correct exact area, e.g., \(\frac{36\sqrt{3}}{5}\) or \(7.2\sqrt{3}\).

(b)
M1: Explaining that the new curve represents a translation (by 2 units to the right).
A1: Stating that translation does not alter the area of the region.
A1.1: Concluding that the area is equal to the original area of \(R\).
Question 10 · Structured
8.1 marks
The curve \(C_1\) has equation \(y = (x-2)^2(x+3)\).

(a) Sketch \(C_1\), showing the coordinates of the points where the curve meets or cuts the coordinate axes.

The curve \(C_2\) has equation \(y = x^3 - 3x^2 - 4x + 12\).

(b) Show that \(C_1\) and \(C_2\) intersect at the point where \(x = 0\).

(c) Find the coordinates of the other point of intersection of \(C_1\) and \(C_2\).
Show answer & marking scheme

Worked solution

(a) For \(C_1\): \(y = (x-2)^2(x+3)\)
- The \(x\)-intercepts are at \(x = 2\) (where it touches the \(x\)-axis because of the squared factor) and \(x = -3\) (where it crosses the \(x\)-axis).
- The \(y\)-intercept is at \(x = 0\), where \(y = (-2)^2(3) = 12\).
- Since the coefficient of \(x^3\) is positive, the curve has a standard positive cubic shape.

(b) At \(x = 0\):
For \(C_1\): \(y = (0-2)^2(0+3) = 4 \times 3 = 12\).
For \(C_2\): \(y = 0^3 - 3(0)^2 - 4(0) + 12 = 12\).
Since both curves have the same \(y\)-coordinate at \(x = 0\), they intersect at this point.

(c) Equating \(C_1\) and \(C_2\):
\[(x-2)^2(x+3) = x^3 - 3x^2 - 4x + 12\]
Expand \(C_1\):
\[(x^2 - 4x + 4)(x+3) = x^3 + 3x^2 - 4x^2 - 12x + 4x + 12 = x^3 - x^2 - 8x + 12\]
Set the equations equal:
\[x^3 - x^2 - 8x + 12 = x^3 - 3x^2 - 4x + 12\]
Subtract \(x^3\) and 12 from both sides:
\[-x^2 - 8x = -3x^2 - 4x\]
Rearrange:
\[2x^2 - 4x = 0\]
\[2x(x - 2) = 0\]
The solutions are \(x = 0\) (from part b) and \(x = 2\).
Substituting \(x = 2\) back into either equation:
\[y = (2-2)^2(2+3) = 0\]
Thus, the other point of intersection is \((2, 0)\).

Marking scheme

(a)
M1: Drawing a positive cubic shape.
A1: Touching the \(x\)-axis at \(2\) and crossing at \(-3\).
A1: Labeling the \(y\)-intercept at \((0, 12)\).

(b)
M1: Finding the value of \(y\) for both curves when \(x = 0\).
A1: Showing both equal to 12 and concluding they intersect.

(c)
M1: Expanding the expression for \(C_1\) correctly to \(x^3 - x^2 - 8x + 12\).
M1: Equating the two cubic curves and simplifying to a quadratic equation.
A1.1: Solving the quadratic to find \(x = 2\) and finding the correct coordinate \((2, 0)\).
Question 11 · Structured
8.1 marks
The circle \(C\) has equation
\[x^2 + y^2 - 8x + 6y - 11 = 0\]

(a) Find
(i) the coordinates of the centre of \(C\),
(ii) the radius of \(C\).

The line \(L\) has equation \(y = 2x + k\), where \(k\) is a constant.
Given that \(L\) is a tangent to \(C\),

(b) find the two possible values of \(k\), giving your answers in the form \(p \pm q\sqrt{5}\), where \(p\) and \(q\) are integers to be found.
Show answer & marking scheme

Worked solution

(a) Complete the square for \(x\) and \(y\):
\[(x - 4)^2 - 16 + (y + 3)^2 - 9 - 11 = 0\]
\[(x - 4)^2 + (y + 3)^2 - 36 = 0\]
\[(x - 4)^2 + (y + 3)^2 = 36\]
(i) The centre of \(C\) is \((4, -3)\).
(ii) The radius of \(C\) is \(\sqrt{36} = 6\).

(b) Since \(L\) is a tangent to \(C\), the perpendicular distance from the centre of the circle \((4, -3)\) to the line \(L\) (which can be written as \(2x - y + k = 0\)) must equal the radius of the circle, \(6\).
Using the formula for the perpendicular distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\):
\[d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\]
\[6 = \frac{|2(4) - (-3) + k|}{\sqrt{2^2 + (-1)^2}}\]
\[6 = \frac{|8 + 3 + k|}{\sqrt{5}}\]
\[|11 + k| = 6\sqrt{5}\]
This gives two possible equations:
\[11 + k = 6\sqrt{5} \implies k = -11 + 6\sqrt{5}\]
\[11 + k = -6\sqrt{5} \implies k = -11 - 6\sqrt{5}\]
Thus, the two possible values of \(k\) are \(-11 \pm 6\sqrt{5}\).

Alternatively, substituting \(y = 2x + k\) into the circle's equation yields a quadratic in \(x\) with discriminant set to zero, which simplifies to \(k^2 + 22k - 59 = 0\), solving to give the same result.

Marking scheme

(a)
M1: Attempting to complete the square for both \(x\) and \(y\).
A1: Correct centre of \((4, -3)\).
A1: Correct radius of \(6\).

(b)
M1: Attempting to use the distance formula from the centre \((4, -3)\) to the line \(2x - y + k = 0\) equated to their radius, or substituting \(y = 2x + k\) into the circle's equation.
A1: Obtaining the correct equation, e.g., \(\frac{|11+k|}{\sqrt{5}} = 6\) or the quadratic \(5x^2 + (4k+4)x + k^2 + 6k - 11 = 0\).
M1: Formulating a solvability condition, e.g., setting \(|11+k| = 6\sqrt{5}\) or setting the discriminant of the quadratic to 0.
A1: Obtaining a correct simplified relation for \(k\), e.g., \(k^2 + 22k - 59 = 0\) or \(|11+k| = 6\sqrt{5}\).
A1.1: Correct values: \(k = -11 \pm 6\sqrt{5}\) (or \(p = -11\) and \(q = 6\)).

Section WMA12/01: Pure Mathematics P2

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
10 Question · 75 marks
Question 1 · Short Answer
4 marks
A geometric series has second term \( 6 \) and fifth term \( \frac{16}{9} \). Find the sum to infinity of this series.
Show answer & marking scheme

Worked solution

The \( n \)-th term of a geometric series is given by \( u_n = a r^{n-1} \). We are given \( u_2 = ar = 6 \) and \( u_5 = ar^4 = \frac{16}{9} \). Dividing the two equations: \( \frac{ar^4}{ar} = \frac{16/9}{6} \implies r^3 = \frac{16}{54} = \frac{8}{27} \). Taking the cube root gives \( r = \frac{2}{3} \). Substituting this back into the first equation: \( a \left(\frac{2}{3}\right) = 6 \implies a = 9 \). Since \( |r| < 1 \), the sum to infinity exists and is given by: \( S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27 \).

Marking scheme

M1: Attempts to write two equations in \( a \) and \( r \), and divides them to obtain an equation in \( r^3 \) or \( r \). A1: Correct value of \( r = \frac{2}{3} \) (or equivalent fraction). M1: Uses their \( r \) to find \( a \) and applies the sum to infinity formula \( S_{\infty} = \frac{a}{1-r} \). A1: Correct answer of \( 27 \).
Question 2 · Short Answer
4 marks
Solve the equation \( 2 \log_3(x - 2) - \log_3(x + 4) = 1 \).
Show answer & marking scheme

Worked solution

Using the laws of logarithms: \( \log_3(x - 2)^2 - \log_3(x + 4) = 1 \implies \log_3\left(\frac{(x - 2)^2}{x + 4}\right) = 1 \). Converting from logarithmic to exponential form: \( \frac{(x - 2)^2}{x + 4} = 3^1 = 3 \). Expanding and rearranging: \( (x - 2)^2 = 3(x + 4) \implies x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0 \). Factoring the quadratic equation: \( (x - 8)(x + 1) = 0 \), which gives \( x = 8 \) or \( x = -1 \). Since \( \log_3(x - 2) \) is only defined for \( x > 2 \), the solution \( x = -1 \) is rejected. Thus, the only valid solution is \( x = 8 \).

Marking scheme

M1: Applies log laws to express the equation in the form \( \log_3(f(x)) = 1 \) or equivalent. M1: Eliminates logarithms correctly to form a quadratic equation, e.g. \( \frac{(x-2)^2}{x+4} = 3 \). A1: Solves the quadratic to find \( x = 8 \) and \( x = -1 \). A1: Correctly identifies \( x = 8 \) as the only valid solution and rejects \( x = -1 \) with a valid reason.
Question 3 · Short Answer
4 marks
The circle \( C \) has equation \( x^2 + y^2 - 8x + 6y = 0 \). Find the equation of the tangent to \( C \) at the point \( P(7, 1) \), giving your answer in the form \( ax + by + c = 0 \), where \( a \), \( b \) and \( c \) are integers.
Show answer & marking scheme

Worked solution

Completing the square for the circle equation: \( (x - 4)^2 - 16 + (y + 3)^2 - 9 = 0 \implies (x - 4)^2 + (y + 3)^2 = 25 \). The centre of the circle is \( Q(4, -3) \). The gradient of the radius \( QP \) is \( m_r = \frac{1 - (-3)}{7 - 4} = \frac{4}{3} \). Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is \( m_t = -\frac{1}{m_r} = -\frac{3}{4} \). The equation of the tangent is \( y - 1 = -\frac{3}{4}(x - 7) \). Multiplying by 4: \( 4y - 4 = -3(x - 7) \implies 4y - 4 = -3x + 21 \). Rearranging into the required form: \( 3x + 4y - 25 = 0 \).

Marking scheme

M1: Completes the square to find the centre of the circle \( (4, -3) \). M1: Finds the gradient of the radius \( QP \) using their centre and the point \( P(7, 1) \). M1: Uses the perpendicular gradient condition \( m_t = -\frac{1}{m_r} \) and attempts the equation of the straight line. A1: Correct equation in the form \( ax + by + c = 0 \) with integer coefficients, e.g., \( 3x + 4y - 25 = 0 \) (or any non-zero integer multiple).
Question 4 · Structured
9 marks

The function \(f(x)\) is defined by \(f(x) = (2 + kx)^6\), where \(k\) is a non-zero constant.

(a) Find the first 4 terms in ascending powers of \(x\) of the binomial expansion of \((2 + kx)^6\), giving each term in its simplest form.

(b) Given that the coefficient of \(x^3\) is 5 times the coefficient of \(x^2\), find the value of \(k\).

(c) Using this value of \(k\), find the coefficient of \(x\) in the expansion.

Show answer & marking scheme

Worked solution

(a) Using the binomial theorem:

\((2+kx)^6 = 2^6 + \binom{6}{1} 2^5 (kx) + \binom{6}{2} 2^4 (kx)^2 + \binom{6}{3} 2^3 (kx)^3 + \dots\)

\(= 64 + 6(32)kx + 15(16)k^2x^2 + 20(8)k^3x^3 + \dots\)

\(= 64 + 192kx + 240k^2x^2 + 160k^3x^3 + \dots\)

(b) The coefficient of \(x^3\) is \(160k^3\) and the coefficient of \(x^2\) is \(240k^2\).

We are given: \(160k^3 = 5 \times 240k^2\)

\(160k^3 = 1200k^2\)

Since \(k \neq 0\), we can divide both sides by \(160k^2\):

\(k = \frac{1200}{160} = 7.5\)

(c) The coefficient of \(x\) is \(192k\).

Substituting \(k = 7.5\):

\(192 \times 7.5 = 1440\)

Marking scheme

(a)

M1: Attempt binomial expansion with at least 3 terms. Powers of \(2\) decreasing, powers of \(kx\) increasing.

A1: Any two terms correct and simplified.

A1: Any three terms correct and simplified.

A1: Fully correct simplified expression: \(64 + 192kx + 240k^2x^2 + 160k^3x^3\).

(b)

M1: Sets up a correct equation using their coefficients of \(x^3\) and \(x^2\): \(160k^3 = 5 \times 240k^2\).

M1: Attempts to solve the equation for \(k\) (must involve dividing by \(k^2\) or factorising).

A1: \(k = 7.5\) or \(\frac{15}{2}\).

(c)

M1: Substitutes their value of \(k\) into their coefficient of \(x\), which must be of the form \(C k\).

A1: \(1440\) (must be a single constant value).

Question 5 · Structured
9 marks

(a) Solve, for \(0 \le \theta < 360^\circ\), the equation \(4\cos^2 \theta + 7\sin \theta - 7 = 0\), giving your answers to 1 decimal place where appropriate.

(b) Hence, solve for \(0 \le x < 180^\circ\) the equation \(4\cos^2 (2x - 20^\circ) + 7\sin (2x - 20^\circ) - 7 = 0\), giving your answers to 1 decimal place where appropriate.

Show answer & marking scheme

Worked solution

(a) Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(4(1 - \sin^2 \theta) + 7\sin \theta - 7 = 0\)

\(4 - 4\sin^2 \theta + 7\sin \theta - 7 = 0\)

\(4\sin^2 \theta - 7\sin \theta + 3 = 0\)

Factorising the quadratic:

\((4\sin \theta - 3)(\sin \theta - 1) = 0\)

So \\sin \theta = \frac{3}{4}\) or \(\sin \theta = 1\).

For \(\sin \theta = 1\): \(\theta = 90^\circ\)

For \(\sin \theta = 0.75\): \(\theta = \sin^{-1}(0.75) \approx 48.59^\circ\)

Other solution: \(180^\circ - 48.59^\circ = 131.41^\circ\)

So the solutions in the interval are \(\theta = 48.6^\circ, 90^\circ, 131.4^\circ\).

(b) Let \(\theta = 2x - 20^\circ\). Since \(0 \le x < 180^\circ\), the range for \(\theta\) is \(-20^\circ \le \theta < 340^\circ\).

The solutions for \(\theta\) from part (a) all fall in this range.

Case 1: \(2x - 20^\circ = 48.59^\circ \implies 2x = 68.59^\circ \implies x \approx 34.3^\circ\)

Case 2: \(2x - 20^\circ = 90^\circ \implies 2x = 110^\circ \implies x = 55^\circ\)

Case 3: \(2x - 20^\circ = 131.41^\circ \implies 2x = 151.41^\circ \implies x \approx 75.7^\circ\)

Marking scheme

(a)

M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic equation in \(\sin \theta\).

A1: Obtains a correct quadratic: \(4\sin^2 \theta - 7\sin \theta + 3 = 0\).

M1: Attempts to solve the quadratic to find values of \(\sin \theta\).

A1: Obtains \(\theta = 90^\circ\) or both \(48.6^\circ\) and \(131.4^\circ\).

A1: All three solutions correct: \(48.6^\circ, 90^\circ, 131.4^\circ\). Deduct 1 mark for extra solutions in the range.

(b)

M1: Equates \(2x - 20^\circ\) to their values of \(\theta\).

M1: Attempts to solve for \(x\) for at least one of their values of \(\theta\).

A1: Any two of \(34.3^\circ, 55^\circ, 75.7^\circ\) correct.

A1: All three of \(34.3^\circ, 55^\circ, 75.7^\circ\) and no other values in the interval.

Question 6 · Structured
9 marks

The polynomial \(f(x)\) is given by \(f(x) = 2x^3 + ax^2 + bx - 10\), where \(a\) and \(b\) are constants.

Given that \((x-2)\) is a factor of \(f(x)\),

(a) show that \(2a + b = -3\).

Given also that when \(f(x)\) is divided by \((x+1)\), the remainder is \(-18\),

(b) find the value of \(a\) and the value of \(b\).

(c) Hence, factorise \(f(x)\) completely.

Show answer & marking scheme

Worked solution

(a) Since \((x-2)\) is a factor, \(f(2) = 0\).

\(2(2)^3 + a(2)^2 + b(2) - 10 = 0\)

\(16 + 4a + 2b - 10 = 0\)

\(4a + 2b = -6\)

Dividing by 2 gives: \(2a + b = -3\) (as required).

(b) Since the remainder when divided by \((x+1)\) is \(-18\), \(f(-1) = -18\).

\(2(-1)^3 + a(-1)^2 + b(-1) - 10 = -18\)

\(-2 + a - b - 10 = -18\)

\(a - b = -6\)

We solve the simultaneous equations:

1) \(2a + b = -3\)

2) \(a - b = -6\)

Adding 1) and 2) gives:

\(3a = -9 \implies a = -3\)

Substituting \(a = -3\) into 2):

\(-3 - b = -6 \implies b = 3\)

(c) Substituting \(a = -3\) and \(b = 3\) gives: \(f(x) = 2x^3 - 3x^2 + 3x - 10\).

Since \((x-2)\) is a factor, we can divide by \((x-2)\) to find the quadratic factor:

\(2x^3 - 3x^2 + 3x - 10 = (x-2)(2x^2 + cx + d)\)

Comparing constant terms: \(-2d = -10 \implies d = 5\).

Comparing \(x^2\) terms: \(-4x^2 + cx^2 = -3x^2 \implies c = 1\).

So the quadratic factor is \(2x^2 + x + 5\).

For \(2x^2 + x + 5\), the discriminant is \(1^2 - 4(2)(5) = -39 < 0\), so it does not factorise further.

Thus, \(f(x) = (x-2)(2x^2 + x + 5)\).

Marking scheme

(a)

M1: Attempts to evaluate \(f(2) = 0\) with substitution.

A1*: Fully correct proof with no errors showing the intermediate step \(4a + 2b = -6\) leading to \(2a + b = -3\).

(b)

M1: Attempts to evaluate \(f(-1) = -18\).

A1: Obtains a correct simplified equation in \(a\) and \(b\), e.g. \(a - b = -6\).

M1: Eliminates one variable from their two equations to solve for \(a\) or \(b\).

A1: Correct values: \(a = -3\) and \(b = 3\).

(c)

M1: Attempts algebraic division or equating coefficients to find the quadratic factor of \(f(x)\).

A1: Correct quadratic factor: \(2x^2 + x + 5\).

A1: Fully factorised form: \((x-2)(2x^2 + x + 5)\).

Question 7 · Structured
9 marks

(a) Find the exact value of \(x\) for which \(\log_3(x + 5) - \log_3(x - 1) = 2\).

(b) Solve the equation \(3^{2y + 1} - 10(3^y) + 3 = 0\), giving your answers in exact form.

Show answer & marking scheme

Worked solution

(a) Using the subtraction law of logarithms:

\(\log_3\left(\frac{x+5}{x-1}\right) = 2\)

Removing the logarithm:

\(\frac{x+5}{x-1} = 3^2 = 9\)

\(x + 5 = 9(x - 1)\)

\(x + 5 = 9x - 9\)

\(8x = 14 \implies x = 1.75\) (or \(\frac{7}{4}\))

(b) Using index laws, write \(3^{2y+1}\) as \(3 \times 3^{2y} = 3(3^y)^2\).

Let \(u = 3^y\).

The equation becomes:

\(3u^2 - 10u + 3 = 0\)

Factorising the quadratic:

\((3u - 1)(u - 3) = 0\)

So \(u = \frac{1}{3}\) or \(u = 3\).

If \(3^y = \frac{1}{3} \implies y = -1\).

If \(3^y = 3 \implies y = 1\).

Marking scheme

(a)

M1: Uses the subtraction law of logarithms correctly: \(\log_3\left(\frac{x+5}{x-1}\right)\).

M1: Converts the logarithmic equation to exponential form: \(\frac{x+5}{x-1} = 3^2\).

M1: Solves the linear equation for \(x\).

A1: \(x = 1.75\) (or equivalent fraction, e.g., \(\frac{7}{4}\)).

(b)

M1: Recognises that \(3^{2y+1} = 3 \times (3^y)^2\).

M1: Substitutes \(u = 3^y\) to obtain a quadratic equation in \(u\).

A1: Correct quadratic: \(3u^2 - 10u + 3 = 0\).

M1: Solves the quadratic to find two values for \(3^y\).

A1: Both \(y = 1\) and \(y = -1\) (must be exact values).

Question 8 · Structured
9 marks

The curve \(C\) has equation \(y = 2x^2 - 8x + 10\) and the line \(L\) has equation \(y = x + 6\).

(a) Show that the line \(L\) and the curve \(C\) intersect at the points with \(x\)-coordinates \(x = \frac{1}{2}\) and \(x = 4\).

(b) Find, using calculus, the exact area of the finite region bounded by the curve \(C\) and the line \(L\).

Show answer & marking scheme

Worked solution

(a) To find points of intersection, set \(y_C = y_L\):

\(2x^2 - 8x + 10 = x + 6\)

\(2x^2 - 9x + 4 = 0\)

Factorising the quadratic:

\((2x - 1)(x - 4) = 0\)

This gives solutions \(x = \frac{1}{2}\) and \(x = 4\), as required.

(b) The finite region is bounded between \(x = \frac{1}{2}\) and \(x = 4\). In this interval, the line is above the curve.

\(\text{Area} = \int_{1/2}^4 (y_L - y_C) \, dx\)

\(= \int_{1/2}^4 ((x+6) - (2x^2 - 8x + 10)) \, dx\)

\(= \int_{1/2}^4 (-2x^2 + 9x - 4) \, dx\)

Integrating each term:

\(= \left[ -\frac{2}{3}x^3 + \frac{9}{2}x^2 - 4x \right]_{1/2}^4\)

Evaluating at the upper limit \(x = 4\):

\(-\frac{2}{3}(64) + \frac{9}{2}(16) - 4(4) = -\frac{128}{3} + 72 - 16 = \frac{40}{3}\)

Evaluating at the lower limit \(x = \frac{1}{2}\):

\(-\frac{2}{3}\left(\frac{1}{8}\right) + \frac{9}{2}\left(\frac{1}{4}\right) - 4\left(\frac{1}{2}\right) = -\frac{1}{12} + \frac{9}{8} - 2 = -\frac{23}{24}\)

Finding the difference:

\(\text{Area} = \frac{40}{3} - \left(-\frac{23}{24}\right) = \frac{320}{24} + \frac{23}{24} = \frac{343}{24}\)

Marking scheme

(a)

M1: Equates the equations of the curve and the line.

A1: Obtains a correct simplified quadratic equation: \(2x^2 - 9x + 4 = 0\).

A1*: Factorises or uses the quadratic formula to show both roots are \(x = \frac{1}{2}\) and \(x = 4\).

(b)

M1: Formulates a correct integral with appropriate limits: \(\int_{1/2}^4 (x+6 - (2x^2 - 8x + 10)) \, dx\).

M1: Integrates a quadratic expression of the form \(Ax^2 + Bx + C\), increasing the power of at least one term.

A1: Correct integrated expression: \(-\frac{2}{3}x^3 + \frac{9}{2}x^2 - 4x\) (ignore constant of integration).

M1: Substitutes both limits \(4\) and \(\frac{1}{2}\) into their integrated expression.

A1: Correct intermediate values: \(\frac{40}{3}\) and \(-\frac{23}{24}\) (or equivalent values if integrated separately).

A1: \(\frac{343}{24}\) (or exact equivalent, e.g., \(14 \frac{7}{24}\)).

Question 9 · Structured
9 marks

A circle \(C\) has equation \(x^2 + y^2 - 6x + 8y = 0\).

(a) Find:

(i) the coordinates of the centre of \(C\),

(ii) the radius of \(C\).

The line \(L\) has equation \(y = \frac{3}{4}x\).

(b) Show that the line \(L\) is a tangent to the circle \(C\).

(c) Find the coordinates of the point of contact between \(L\) and \(C\).

Show answer & marking scheme

Worked solution

(a) Rearranging the equation by completing the square:

\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 0\)

\((x - 3)^2 + (y + 4)^2 = 25\)

(i) Centre of \(C\) is \((3, -4)\).

(ii) Radius is \(\sqrt{25} = 5\).

(b) Method 1: Algebraic Substitution

Substitute \(y = \frac{3}{4}x\) into the circle equation:

\(x^2 + \left(\frac{3}{4}x\right)^2 - 6x + 8\left(\frac{3}{4}x\right) = 0\)

\(x^2 + \frac{9}{16}x^2 - 6x + 6x = 0\)

\(\frac{25}{16}x^2 = 0\)

This quadratic equation has a single repeated root, \(x = 0\), which indicates that the line touches the circle at exactly one point, meaning \(L\) is a tangent to \(C\).

Method 2: Distance from centre

The equation of \(L\) is \(3x - 4y = 0\). The perpendicular distance from the centre \((3, -4)\) to the line is:

\(d = \frac{|3(3) - 4(-4)|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 16|}{\sqrt{25}} = \frac{25}{5} = 5\)

Since the perpendicular distance from the centre to the line is equal to the radius of the circle, the line is a tangent.

(c) Using Method 1, the repeated root occurs at \(x = 0\). Substituting \(x = 0\) into the line equation: \(y = \frac{3}{4}(0) = 0\).

So the point of contact is \((0, 0)\).

Marking scheme

(a)

M1: Completes the square for both \(x\) and \(y\) terms.

A1: Centre is \((3, -4)\).

A1: Radius is \(5\).

(b)

M1: Substitutes the line equation into the circle equation, OR attempts to find the perpendicular distance from the centre to the line.

A1: Obtains the correct simplified quadratic equation \(\frac{25}{16}x^2 = 0\) OR calculates distance as \(\frac{25}{5}\).

M1: Identifies that there is a repeated root \(x = 0\) OR compares the distance \(5\) with the radius \(5\).

A1: Clear conclusion stating that since there is a repeated root OR distance equals radius, \(L\) is a tangent.

(c)

M1: Solves for the coordinates of the intersection point.

A1: Point of contact is \((0, 0)\).

Question 10 · Structured
9 marks

An open-topped rectangular box is to be made from thin sheet metal. The box has a rectangular base of length \(2x\) cm and width \(x\) cm, and a height of \(h\) cm. Given that the volume of the box is \(36\text{ cm}^3\):

(a) Show that the surface area of the box, \(A\text{ cm}^2\), is given by \(A = 2x^2 + \frac{108}{x}\).

(b) Use calculus to find the value of \(x\) for which \(A\) is a minimum, justifying that your value of \(x\) gives a minimum.

Show answer & marking scheme

Worked solution

(a) The volume of the box is given by:

\(V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2h\)

We are given \(V = 36\), so:

\(2x^2h = 36 \implies h = \frac{18}{x^2}\)

The box has no top. The surface area \(A\) consists of the base and four vertical sides:

\(A = \text{Base area} + 2(\text{front/back area}) + 2(\text{side area})\)

\(A = (2x)(x) + 2(2xh) + 2(xh)\)

\(A = 2x^2 + 6xh\)

Substitute \(h = \frac{18}{x^2}\) into the equation for \(A\):

\(A = 2x^2 + 6x\left(\frac{18}{x^2}\right)\)

\(A = 2x^2 + \frac{108}{x}\) (as required).

(b) To find stationary points, differentiate \(A\) with respect to \(x\):

\(A = 2x^2 + 108x^{-1}\)

\(\frac{dA}{dx} = 4x - 108x^{-2} = 4x - \frac{108}{x^2}\)

Set \(\frac{dA}{dx} = 0\):

\(4x - \frac{108}{x^2} = 0\)

\(4x^3 = 108\)

\(x^3 = 27 \implies x = 3\)

To justify that this gives a minimum, find the second derivative:

\(\frac{d^2A}{dx^2} = 4 + 216x^{-3} = 4 + \frac{216}{x^3}\)

When \(x = 3\):

\(\frac{d^2A}{dx^2} = 4 + \frac{216}{27} = 4 + 8 = 12\)

Since \(\frac{d^2A}{dx^2} = 12 > 0\), the value of \(x = 3\) gives a minimum value for the surface area.

Marking scheme

(a)

M1: Expresses the volume in terms of \(x\) and \(h\): \(V = 2x^2h\).

M1: Uses \(V = 36\) to express \(h\) in terms of \(x\): \(h = \frac{18}{x^2}\).

M1: Formulates a correct expression for the total surface area of an open-topped box: \(A = 2x^2 + 6xh\).

A1*: Substitutes \(h\) and simplifies to obtain the given formula \(A = 2x^2 + \frac{108}{x}\) with no errors.

(b)

M1: Differentiates \(A\) with respect to \(x\) (at least one term correct).

A1: Correct derivative: \(4x - \frac{108}{x^2}\).

M1: Sets their \(\frac{dA}{dx} = 0\) and solves for \(x\).

A1: \(x = 3\).

M1: Finds the second derivative and evaluates it at \(x = 3\) (or uses another valid method) to show that it is a minimum.

Section WMA13/01: Pure Mathematics P3

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
10 Question · 75 marks
Question 1 · Short Answer
4 marks
The function \( f(x) = \ln(3x + 1) - x^2 + 2 \) has a single positive root \(\alpha\). (a) Show that \(\alpha\) lies in the interval \([1.5, 2.0]\). (b) Use the iteration formula \( x_{n+1} = \sqrt{\ln(3x_n + 1) + 2} \) with \( x_0 = 1.7 \) to find the value of \( x_3 \) to 3 decimal places.
Show answer & marking scheme

Worked solution

(a) Evaluating the function at the boundaries: \( f(1.5) = \ln(3(1.5) + 1) - (1.5)^2 + 2 = \ln(5.5) - 2.25 + 2 \approx 1.7047 - 0.25 = 1.455 \) (to 3 d.p.). \( f(2.0) = \ln(3(2) + 1) - (2)^2 + 2 = \ln(7) - 4 + 2 \approx 1.9459 - 2 = -0.054 \) (to 3 d.p.). Since there is a change of sign and \( f(x) \) is continuous in the interval \([1.5, 2.0]\), a root \(\alpha\) lies in this interval. (b) Using the iteration formula: \( x_1 = \sqrt{\ln(3(1.7) + 1) + 2} = \sqrt{\ln(6.1) + 2} \approx 1.95148 \), \( x_2 = \sqrt{\ln(3(1.95148) + 1) + 2} = \sqrt{\ln(6.85444) + 2} \approx 1.98114 \), \( x_3 = \sqrt{\ln(3(1.98114) + 1) + 2} = \sqrt{\ln(6.94342) + 2} \approx 1.98439 \). Thus, to 3 decimal places, \( x_3 = 1.984 \).

Marking scheme

(a) M1: Attempts to evaluate both \( f(1.5) \) and \( f(2.0) \) (at least one correct value to 1 d.p.). A1: Finds correct values of \( f(1.5) \approx 1.45 \) and \( f(2.0) \approx -0.05 \), and concludes with a reason (change of sign and continuity). (b) M1: Attempts to find \( x_1 \) using the iteration formula. A1: Correctly calculates \( x_3 = 1.984 \) to 3 decimal places.
Question 2 · Short Answer
4 marks
The function \( \mathrm{g} \) is defined by \( \mathrm{g}(x) = \frac{3e^x - 2}{e^x + 4} \), \( x \in \mathbb{R} \). (a) Find an expression for \( \mathrm{g}^{-1}(x) \). (b) State the domain of \( \mathrm{g}^{-1} \).
Show answer & marking scheme

Worked solution

(a) Let \( y = \frac{3e^x - 2}{e^x + 4} \). Rearranging to make \( e^x \) the subject: \( y(e^x + 4) = 3e^x - 2 \implies y e^x + 4y = 3e^x - 2 \implies 4y + 2 = e^x(3 - y) \implies e^x = \frac{4y+2}{3-y} \). Taking the natural logarithm of both sides: \( x = \ln\left(\frac{4y+2}{3-y}\right) \). Replacing \( y \) with \( x \), we obtain: \( \mathrm{g}^{-1}(x) = \ln\left(\frac{4x+2}{3-x}\right) \). (b) The domain of \( \mathrm{g}^{-1} \) is the range of \( \mathrm{g} \). Since \( e^x > 0 \), \( \mathrm{g}(x) = 3 - \frac{14}{e^x + 4} \). As \( x \to -\infty \), \( e^x \to 0 \), so \( \mathrm{g}(x) \to -\frac{2}{4} = -\frac{1}{2} \). As \( x \to \infty \), \( \mathrm{g}(x) \to 3 \). Thus, the range of \( \mathrm{g} \) is \( -\frac{1}{2} < \mathrm{g}(x) < 3 \). Therefore, the domain of \( \mathrm{g}^{-1} \) is \( -\frac{1}{2} < x < 3 \).

Marking scheme

(a) M1: Starts process to make \( e^x \) the subject by multiplying out and collecting terms. M1: Correctly isolates \( e^x = \frac{4y+2}{3-y} \) or equivalent. A1: Correct expression \( \mathrm{g}^{-1}(x) = \ln\left(\frac{4x+2}{3-x}\right) \) or equivalent, with correct notation. (b) B1: Correctly states \( -\frac{1}{2} < x < 3 \) (or equivalent interval notation).
Question 3 · Short Answer
4 marks
The curve \( C \) has equation \( y = \frac{\ln(2x - 1)}{x} \), \( x > \frac{1}{2} \). Find the equation of the tangent to \( C \) at the point where \( x = 1 \).
Show answer & marking scheme

Worked solution

To find the equation of the tangent, we first find the coordinates of the point of contact and the gradient at that point. At \( x = 1 \), \( y = \frac{\ln(2(1) - 1)}{1} = \frac{\ln(1)}{1} = 0 \). So the point is \( (1, 0) \). To find the gradient, we use the quotient rule: let \( u = \ln(2x-1) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{2x-1} \) and \( v = x \implies \frac{\mathrm{d}v}{\mathrm{d}x} = 1 \). Then \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x \left(\frac{2}{2x-1}\right) - \ln(2x-1)}{x^2} \). Substituting \( x = 1 \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 \left(\frac{2}{1}\right) - \ln(1)}{1^2} = 2 \). The equation of the tangent is given by \( y - y_1 = m(x - x_1) \), which becomes \( y - 0 = 2(x - 1) \implies y = 2x - 2 \).

Marking scheme

M1: Applies the quotient rule (or product rule) to differentiate the function, with correct structure. A1: Correct derivative \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x \left(\frac{2}{2x-1}\right) - \ln(2x-1)}{x^2} \) or equivalent. M1: Finds the value of \( y \) at \( x = 1 \) (which is 0) and attempts to evaluate their \( \frac{\mathrm{d}y}{\mathrm{d}x} \) at \( x = 1 \). A1: Correct equation of the tangent, e.g., \( y = 2x - 2 \) or equivalent.
Question 4 · Structured
9 marks
The function \( f \) is defined by
\[ f(x) = 3|x - 2| - 4, \quad x \in \mathbb{R} \]

(a) Sketch the graph of \( y = f(x) \), showing the coordinates of the vertex and the \( y \)-intercept. (3)

(b) Solve the equation \( f(x) = 2x + 1 \). (4)

(c) State the range of \( f(x) \). (2)
Show answer & marking scheme

Worked solution

(a) The graph of \( y = 3|x - 2| - 4 \) is a V-shape.
The vertex is located at \( (2, -4) \).
The \( y \)-intercept is found by setting \( x = 0 \):
\( y = 3|0 - 2| - 4 = 3(2) - 4 = 2 \). So the \( y \)-intercept is at \( (0, 2) \).

(b) To solve \( 3|x - 2| - 4 = 2x + 1 \):
Case 1: \( x \ge 2 \)
\( 3(x - 2) - 4 = 2x + 1 \Rightarrow 3x - 10 = 2x + 1 \Rightarrow x = 11 \)
Since \( 11 \ge 2 \), this is a valid solution.

Case 2: \( x < 2 \)
\( -3(x - 2) - 4 = 2x + 1 \Rightarrow -3x + 2 = 2x + 1 \Rightarrow 5x = 1 \Rightarrow x = 0.2 \)
Since \( 0.2 < 2 \), this is a valid solution.
So the solutions are \( x = 11 \) and \( x = 0.2 \).

(c) The minimum value of \( f(x) \) is \( -4 \), and it goes to positive infinity.
So the range is \( f(x) \ge -4 \).

Marking scheme

(a)
M1: For a V-shaped graph in the correct orientation.
A1: Vertex at \( (2, -4) \) correctly labelled or coordinates stated.
A1: \( y \)-intercept at \( (0, 2) \) correctly labelled or coordinates stated.

(b)
M1: Formulates the equation for \( x \ge 2 \): \( 3(x-2) - 4 = 2x + 1 \).
A1: Obtains \( x = 11 \).
M1: Formulates the equation for \( x < 2 \): \( -3(x-2) - 4 = 2x + 1 \).
A1: Obtains \( x = 0.2 \) (or \( \frac{1}{5} \)).

(c)
M1: Realises that the range is dependent on the y-coordinate of the vertex.
A1: Correct range: \( f(x) \ge -4 \) (accept \( y \ge -4 \), do not accept \( x \ge -4 \)).
Question 5 · Structured
9 marks
(a) Express \( 5\cos\theta - 12\sin\theta \) in the form \( R\cos(\theta + \alpha) \), where \( R > 0 \) and \( 0 < \alpha < \frac{\pi}{2} \). Give the value of \( \alpha \) in radians to 4 decimal places. (3)

(b) Hence, solve for \( 0 \le \theta < 2\pi \), the equation
\[ 5\cos\theta - 12\sin\theta = 6.5 \]
giving your answers to 3 significant figures. (4)

(c) Find the minimum value of
\[ \frac{30}{17 + (5\cos\theta - 12\sin\theta)^2} \] (2)
Show answer & marking scheme

Worked solution

(a) Let \( 5\cos\theta - 12\sin\theta = R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha \).
Comparing coefficients:
\( R\cos\alpha = 5 \)
\( R\sin\alpha = 12 \)
\( R^2 = 5^2 + 12^2 = 169 \Rightarrow R = 13 \).
\( \tan\alpha = \frac{12}{5} \Rightarrow \alpha = \arctan(2.4) \approx 1.1760 \) radians.
So, \( 13\cos(\theta + 1.1760) \).

(b) \( 13\cos(\theta + 1.1760) = 6.5 \Rightarrow \cos(\theta + 1.1760) = 0.5 \).
Let \( X = \theta + 1.1760 \). Since \( 0 \le \theta < 2\pi \), we have \( 1.1760 \le X < 2\pi + 1.1760 \approx 7.4592 \).
The solutions to \( \cos X = 0.5 \) in this range are:
\( X = \frac{5\pi}{3} \approx 5.2360 \)
\( X = \frac{7\pi}{3} \approx 7.3304 \)
So:
\( \theta = 5.2360 - 1.1760 = 4.0600 \approx 4.06 \)
\( \theta = 7.3304 - 1.1760 = 6.1544 \approx 6.15 \)

(c) The expression is \( \frac{30}{17 + [13\cos(\theta + \alpha)]^2} = \frac{30}{17 + 169\cos^2(\theta + \alpha)} \).
To minimize this expression, we maximize the denominator.
The maximum value of \( \cos^2(\theta + \alpha) \) is 1.
Maximum denominator \( = 17 + 169(1) = 186 \).
So the minimum value is \( \frac{30}{186} = \frac{5}{31} \).

Marking scheme

(a)
B1: \( R = 13 \).
M1: \( \tan\alpha = \pm \frac{12}{5} \) or \( \pm \frac{5}{12} \).
A1: \( \alpha \approx 1.1760 \) (must be 4 d.p.).

(b)
M1: Sets \( \cos(\theta + \alpha) = \frac{6.5}{13} = 0.5 \).
M1: Finds at least one correct value for \( \theta + \alpha \) (e.g., \( \frac{5\pi}{3} \) or \( 5.24 \)).
A1: One correct solution, \( \theta \approx 4.06 \).
A1: Second correct solution, \( \theta \approx 6.15 \). (Accept answers that round to 4.06 and 6.15).

(c)
M1: Recognises that the minimum value of the expression occurs when \( \cos^2(\theta + \alpha) = 1 \).
A1: \( \frac{5}{31} \) (or exact equivalent).
Question 6 · Structured
9 marks
The mass, \( M \) grams, of a radioactive substance \( t \) years after it was first observed is modeled by the equation
\[ M = A \mathrm{e}^{-kt} \]
where \( A \) and \( k \) are positive constants.

Given that the initial mass of the substance was 80 grams,
(a) write down the value of \( A \). (1)

Given also that the mass of the substance was 50 grams after 6 years,
(b) find the value of \( k \), giving your answer to 4 decimal places. (3)

(c) Find the rate of decrease of the mass of the substance when \( t = 10 \), giving your answer in grams per year to 3 significant figures. (3)

(d) Find the value of \( t \) when the mass is 10 grams, giving your answer to 1 decimal place. (2)
Show answer & marking scheme

Worked solution

(a) At \( t = 0 \), \( M = 80 \Rightarrow A \mathrm{e}^0 = 80 \Rightarrow A = 80 \).

(b) When \( t = 6 \), \( M = 50 \).
\( 50 = 80 \mathrm{e}^{-6k} \Rightarrow \mathrm{e}^{-6k} = 0.625 \)
\( -6k = \ln(0.625) \Rightarrow k = -\frac{\ln(0.625)}{6} \approx 0.07833 \approx 0.0783 \).

(c) The rate of decrease is given by \( -\frac{\mathrm{d}M}{\mathrm{d}t} \).
\( \frac{\mathrm{d}M}{\mathrm{d}t} = -k A \mathrm{e}^{-kt} = -0.07833 \times 80 \mathrm{e}^{-0.07833 t} \).
When \( t = 10 \):
Rate of decrease \( = 0.07833 \times 80 \mathrm{e}^{-0.07833 \times 10} \approx 6.2664 \mathrm{e}^{-0.7833} \approx 2.86 \) grams per year.

(d) When \( M = 10 \):
\( 10 = 80 \mathrm{e}^{-0.07833 t} \Rightarrow \mathrm{e}^{-0.07833 t} = 0.125 \)
\( -0.07833 t = \ln(0.125) \Rightarrow t = \frac{\ln(8)}{0.07833} \approx 26.5 \) years.

Marking scheme

(a)
B1: \( A = 80 \).

(b)
M1: Substitutes \( M = 50 \), \( t = 6 \), and their \( A \) into the model.
M1: Correct use of logarithms to solve for \( k \).
A1: \( k = 0.0783 \) (must be 4 d.p.).

(c)
M1: Differentiates \( M \) with respect to \( t \) to find \( \frac{\mathrm{d}M}{\mathrm{d}t} = -k A \mathrm{e}^{-kt} \).
M1: Substitutes \( t = 10 \) into their derivative.
A1: \( 2.86 \) (accept \( 2.86 \) or \( 2.87 \) depending on rounding of \( k \)).

(d)
M1: Sets \( M = 10 \) and solves for \( t \) using their \( A \) and \( k \).
A1: \( t = 26.5 \) (accept 26.5 or 26.6).
Question 7 · Structured
9 marks
A curve \( C \) has equation
\[ y = \frac{x^2 - 3}{2x + 5}, \quad x \neq -2.5 \]

(a) Show that \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 + 10x + 6}{(2x + 5)^2} \). (4)

(b) Find the exact coordinates of the stationary points of \( C \). (5)
Show answer & marking scheme

Worked solution

(a) Let \( u = x^2 - 3 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \).
Let \( v = 2x + 5 \Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = 2 \).
Using the quotient rule:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x + 5)(2x) - (x^2 - 3)(2)}{(2x + 5)^2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x^2 + 10x - 2x^2 + 6}{(2x + 5)^2} = \frac{2x^2 + 10x + 6}{(2x + 5)^2} \). (Shown)

(b) At stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 2x^2 + 10x + 6 = 0 \Rightarrow x^2 + 5x + 3 = 0 \).
Using the quadratic formula:
\( x = \frac{-5 \pm \sqrt{25 - 12}}{2} = \frac{-5 \pm \sqrt{13}}{2} \).
To find the y-coordinates, we can use \( x^2 = -5x - 3 \):
\( y = \frac{x^2 - 3}{2x + 5} = \frac{-5x - 6}{2x + 5} \).
For \( x = \frac{-5 + \sqrt{13}}{2} \):
\( 2x + 5 = \sqrt{13} \).
\( -5x - 6 = \frac{25 - 5\sqrt{13}}{2} - 6 = \frac{13 - 5\sqrt{13}}{2} \).
\( y = \frac{13 - 5\sqrt{13}}{2\sqrt{13}} = \frac{\sqrt{13} - 5}{2} \).
For \( x = \frac{-5 - \sqrt{13}}{2} \):
\( 2x + 5 = -\sqrt{13} \).
\( -5x - 6 = \frac{25 + 5\sqrt{13}}{2} - 6 = \frac{13 + 5\sqrt{13}}{2} \).
\( y = \frac{13 + 5\sqrt{13}}{-2\sqrt{13}} = \frac{-\sqrt{13} - 5}{2} \).
So the exact coordinates are:
\( \left(\frac{-5 + \sqrt{13}}{2}, \frac{\sqrt{13} - 5}{2}\right) \) and \( \left(\frac{-5 - \sqrt{13}}{2}, \frac{-\sqrt{13} - 5}{2}\right) \).

Marking scheme

(a)
M1: Applies quotient rule with \( u = x^2 - 3 \) and \( v = 2x + 5 \).
A1: Correct derivative of \( u \) is \( 2x \), and of \( v \) is \( 2 \).
M1: Correct structure of quotient rule: \( \frac{v u' - u v'}{v^2} \).
A1*: Simplifies numerator to \( 2x^2 + 10x + 6 \) with no errors seen.

(b)
M1: Sets \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to get \( 2x^2 + 10x + 6 = 0 \).
A1: Solves to find \( x = \frac{-5 \pm \sqrt{13}}{2} \) (or exact equivalents).
M1: Substitutes at least one exact \( x \)-value back into the original curve equation \( y = \frac{x^2 - 3}{2x + 5} \).
A1: One correct exact \( y \)-coordinate: \( y = \frac{\sqrt{13} - 5}{2} \) (or equivalent).
A1: Second correct exact \( y \)-coordinate: \( y = \frac{-\sqrt{13} - 5}{2} \) (or equivalent).
Question 8 · Structured
9 marks
A curve \( C \) has equation
\[ y = \mathrm{e}^{2x} \sin(3x) \]

(a) Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \). (3)

(b) Show that the \( x \)-coordinates of the stationary points of \( C \) satisfy the equation \( \tan(3x) = -1.5 \). (3)

(c) Find the equation of the tangent to \( C \) at the point where \( x = \pi \). Give your answer in the form \( y = mx + c \), where \( m \) and \( c \) are exact constants. (3)
Show answer & marking scheme

Worked solution

(a) Using the product rule:
\( u = \mathrm{e}^{2x} \Rightarrow u' = 2\mathrm{e}^{2x} \)
\( v = \sin(3x) \Rightarrow v' = 3\cos(3x) \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = u'v + uv' = 2\mathrm{e}^{2x}\sin(3x) + 3\mathrm{e}^{2x}\cos(3x) = \mathrm{e}^{2x}(2\sin(3x) + 3\cos(3x)) \).

(b) At stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
Since \( \mathrm{e}^{2x} \neq 0 \) for all \( x \):
\( 2\sin(3x) + 3\cos(3x) = 0 \).
Divide both sides by \( \cos(3x) \):
\( 2\tan(3x) + 3 = 0 \Rightarrow \tan(3x) = -1.5 \). (Shown)

(c) When \( x = \pi \):
\( y = \mathrm{e}^{2\pi}\sin(3\pi) = 0 \).
The gradient of the tangent is:
\( m = \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\pi} = \mathrm{e}^{2\pi}(2\sin(3\pi) + 3\cos(3\pi)) = \mathrm{e}^{2\pi}(0 - 3) = -3\mathrm{e}^{2\pi} \).
The equation of the tangent is:
\( y - 0 = -3\mathrm{e}^{2\pi}(x - \pi) \Rightarrow y = -3\mathrm{e}^{2\pi}x + 3\pi\mathrm{e}^{2\pi} \).

Marking scheme

(a)
M1: Correctly applies product rule: \( \frac{\mathrm{d}y}{\mathrm{d}x} = A\mathrm{e}^{2x}\sin(3x) + B\mathrm{e}^{2x}\cos(3x) \).
A1: One term correct: \( 2\mathrm{e}^{2x}\sin(3x) \) or \( 3\mathrm{e}^{2x}\cos(3x) \).
A1: Fully correct derivative: \( \mathrm{e}^{2x}(2\sin(3x) + 3\cos(3x)) \).

(b)
M1: Sets their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
M1: Recognises \( \mathrm{e}^{2x} \neq 0 \) and uses \( \tan(3x) = \frac{\sin(3x)}{\cos(3x)} \).
A1*: Fully correct proof of \( \tan(3x) = -1.5 \) with no errors.

(c)
M1: Finds the value of \( y \) and \( \frac{\mathrm{d}y}{\mathrm{d}x} \) at \( x = \pi \).
M1: Uses \( y - y_1 = m(x - x_1) \) with their coordinates and gradient.
A1: \( y = -3\mathrm{e}^{2\pi}x + 3\pi\mathrm{e}^{2\pi} \) (or exact equivalent).
Question 9 · Structured
9 marks
Find:

(a) \( \int (4x + 1) \mathrm{e}^{2x^2 + x} \mathrm{d}x \) (3)

(b) the exact value of \( \int_{0}^{\pi/6} (\sin(3x) + \cos(2x)) \mathrm{d}x \) (3)

(c) Show that \( \int_{2}^{4} \frac{8}{2x - 3} \mathrm{d}x = \ln(k) \), where \( k \) is a constant to be found. (3)
Show answer & marking scheme

Worked solution

(a) Let \( u = 2x^2 + x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 4x + 1 \).
So the integral becomes:
\( \int \mathrm{e}^u \mathrm{d}u = \mathrm{e}^u + C = \mathrm{e}^{2x^2 + x} + C \).

(b) \( \int_{0}^{\pi/6} (\sin(3x) + \cos(2x)) \mathrm{d}x = \left[ -\frac{1}{3}\cos(3x) + \frac{1}{2}\sin(2x) \right]_{0}^{\pi/6} \)
Evaluating at upper bound \( \pi/6 \):
\( -\frac{1}{3}\cos(\pi/2) + \frac{1}{2}\sin(\pi/3) = 0 + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} \).
Evaluating at lower bound \( 0 \):
\( -\frac{1}{3}\cos(0) + \frac{1}{2}\sin(0) = -\frac{1}{3} \).
Subtracting:
\( \frac{\sqrt{3}}{4} - \left(-\frac{1}{3}\right) = \frac{3\sqrt{3} + 4}{12} \).

(c) \( \int_{2}^{4} \frac{8}{2x - 3} \mathrm{d}x = \left[ 4\ln|2x - 3| \right]_{2}^{4} \)
Evaluating at 4: \( 4\ln|5| = 4\ln(5) \).
Evaluating at 2: \( 4\ln|1| = 0 \).
So the integral value is \( 4\ln(5) = \ln(5^4) = \ln(625) \).
Thus \( k = 625 \).

Marking scheme

(a)
M1: Recognises reverse chain rule or substitutes \( u = 2x^2 + x \).
A1: \( \mathrm{e}^{2x^2 + x} \).
A1: Correct constant of integration \( + C \).

(b)
M1: Integrates to find \( -A\cos(3x) + B\sin(2x) \).
A1: Correct integration: \( -\frac{1}{3}\cos(3x) + \frac{1}{2}\sin(2x) \).
A1: Evaluates and obtains \( \frac{3\sqrt{3} + 4}{12} \) (or exact equivalent).

(c)
M1: Integrates to find \( A\ln|2x - 3| \).
A1: Correct integration: \( 4\ln|2x - 3| \).
A1: Correctly applies laws of logs to show \( k = 625 \).
Question 10 · Structured
9 marks
The curve \( C \) has equation \( y = x^3 - 5x + 1 \).

(a) Show that \( C \) has a root \( \alpha \) in the interval \( [2, 2.5] \). (2)

(b) Show that the equation \( x^3 - 5x + 1 = 0 \) can be rearranged into the iterative formula
\[ x_{n+1} = \sqrt{5 - \frac{1}{x_n}} \] (2)

(c) Using the iterative formula with \( x_0 = 2 \), find the values of \( x_1 \), \( x_2 \), and \( x_3 \), giving each answer to 4 decimal places. (3)

(d) By choosing a suitable interval, prove that \( \alpha = 2.128 \) to 3 decimal places. (2)
Show answer & marking scheme

Worked solution

(a) Let \( f(x) = x^3 - 5x + 1 \).
\( f(2) = 2^3 - 5(2) + 1 = 8 - 10 + 1 = -1 \).
\( f(2.5) = (2.5)^3 - 5(2.5) + 1 = 15.625 - 12.5 + 1 = 4.125 \).
Since \( f(x) \) is continuous and there is a change of sign between \( f(2) < 0 \) and \( f(2.5) > 0 \), a root \( \alpha \) lies in the interval \( [2, 2.5] \).

(b) \( x^3 - 5x + 1 = 0 \Rightarrow x^3 = 5x - 1 \).
Since \( x \neq 0 \) in the interval, dividing both sides by \( x \):
\( x^2 = 5 - \frac{1}{x} \).
Taking the positive square root (since \( x > 0 \)):
\( x = \sqrt{5 - \frac{1}{x}} \).
This yields \( x_{n+1} = \sqrt{5 - \frac{1}{x_n}} \).

(c) With \( x_0 = 2 \):
\( x_1 = \sqrt{5 - \frac{1}{2}} = \sqrt{4.5} \approx 2.1213 \) (4 d.p.)
\( x_2 = \sqrt{5 - \frac{1}{2.12132}} \approx 2.1280 \) (4 d.p.)
\( x_3 = \sqrt{5 - \frac{1}{2.12805}} \approx 2.1284 \) (4 d.p.)

(d) To show \( \alpha = 2.128 \) to 3 d.p., evaluate \( f(x) \) at the bounds \( 2.1275 \) and \( 2.1285 \):
\( f(2.1275) = (2.1275)^3 - 5(2.1275) + 1 \approx -0.0083 \)
\( f(2.1285) = (2.1285)^3 - 5(2.1285) + 1 \approx 0.0003 \)
Since there is a sign change and \( f(x) \) is continuous, the root lies in \( (2.1275, 2.1285) \), so \( \alpha = 2.128 \) to 3 d.p.

Marking scheme

(a)
M1: Evaluates \( f(2) \) and \( f(2.5) \) with at least one correct calculation.
A1: Both \( f(2) = -1 \) and \( f(2.5) = 4.125 \) correct, with a statement mentioning sign change and continuity.

(b)
M1: Realises division by \( x \) is required or makes \( x^2 \) the subject first.
A1*: Fully correct rearrangement showing all intermediate steps without errors.

(c)
M1: Correct substitution of \( x_0 = 2 \) into the formula.
A1: \( x_1 = 2.1213 \) and \( x_2 = 2.1280 \).
A1: \( x_3 = 2.1284 \).

(d)
M1: Selects the correct interval bounds: \( [2.1275, 2.1285] \).
A1: Obtains \( f(2.1275) \approx -0.0083 \) and \( f(2.1285) \approx 0.0003 \), and draws the correct conclusion based on the sign change.

Section WMA14/01: Pure Mathematics P4

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
8 Question · 75.10000000000001 marks
Question 1 · Short Answer
4.7 marks
Find the coefficient of \(x^2\) in the binomial expansion of \(\frac{1+3x}{\sqrt{4-x}}\), in ascending powers of \(x\).
Show answer & marking scheme

Worked solution

First, rewrite the term as \((1 + 3x)(4 - x)^{-\frac{1}{2}}\). Next, expand \((4 - x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\). Using the binomial expansion formula: \(\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2 + \dots = 1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots\). Multiplying by \(\frac{1}{2}\) gives \((4 - x)^{-\frac{1}{2}} = \frac{1}{2} + \frac{x}{16} + \frac{3}{256}x^2 + \dots\). Now expand \((1+3x)\left(\frac{1}{2} + \frac{x}{16} + \frac{3}{256}x^2 + \dots\)\). The term in \(x^2\) is given by \(1 \cdot \frac{3}{256}x^2 + 3x \cdot \frac{x}{16} = \left(\frac{3}{256} + \frac{3}{16}\right)x^2 = \left(\frac{3}{256} + \frac{48}{256}\right)x^2 = \frac{51}{256}x^2\). Thus, the coefficient of \(x^2\) is \(\frac{51}{256}\).

Marking scheme

M1: Attempts to write as \(k\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\) with a correct value of \(k = \frac{1}{2}\). A1: Correct simplified term in \(x\) or \(x^2\) of the expansion of \(\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\), e.g., \(\frac{x}{8}\) or \(\frac{3}{128}x^2\). M1: Multiplies their expansion by \((1+3x)\) and identifies the relevant terms to find the coefficient of \(x^2\). A1.7: Correct final coefficient of \(\frac{51}{256}\) (or exact equivalent).
Question 2 · Short Answer
4.7 marks
A curve \(C\) has equation \(2x^2 - 3xy + y^2 + 4x = 11\). Find the value of \(\frac{dy}{dx}\) at the point \((2, 1)\) on \(C\).
Show answer & marking scheme

Worked solution

Differentiate the equation implicitly with respect to \(x\). Differentiating term by term: \(\frac{d}{dx}(2x^2) = 4x\). Using the product rule for \(-3xy\): \(\frac{d}{dx}(-3xy) = -3y - 3x\frac{dy}{dx}\). Differentiating the remaining terms: \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) and \(\frac{d}{dx}(4x) = 4\), and \(\frac{d}{dx}(11) = 0\). Putting it all together: \(4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0\). Substitute \(x = 2\) and \(y = 1\) into the differentiated equation: \(4(2) - 3(1) - 3(2)\frac{dy}{dx} + 2(1)\frac{dy}{dx} + 4 = 0\). Simplify to get: \(8 - 3 - 6\frac{dy}{dx} + 2\frac{dy}{dx} + 4 = 0\), which gives \(9 - 4\frac{dy}{dx} = 0\). Rearranging gives \(\frac{dy}{dx} = \frac{9}{4}\).

Marking scheme

M1: Attempts implicit differentiation with at least one term correct involving \(\frac{dy}{dx}\), such as \(2y\frac{dy}{dx}\) or via product rule on \(-3xy\). A1: Completely correct differentiation: \(4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0\). M1: Substitutes \(x = 2\) and \(y = 1\) into their differentiated expression to solve for \(\frac{dy}{dx}\). A1.7: Correct value of \(\frac{9}{4}\) or \(2.25\).
Question 3 · Short Answer
4.7 marks
Find the exact value of \(\int_{0}^{\ln 2} \frac{e^{2x}}{e^x + 1} \, dx\).
Show answer & marking scheme

Worked solution

We can write the integrand as \(\frac{e^{2x}}{e^x + 1} = \frac{e^x(e^x+1) - e^x}{e^x + 1} = e^x - \frac{e^x}{e^x + 1}\). Alternatively, using the substitution \(u = e^x + 1\) with \(du = e^x \, dx\). The integral of \(e^x\) is \(e^x\) and the integral of \(\frac{e^x}{e^x + 1}\) is \(\ln(e^x + 1)\). Thus, the indefinite integral is \(e^x - \ln(e^x + 1) + C\). Evaluating this between the limits of \(0\) and \(\ln 2\): at the upper limit \(x = \ln 2\), we have \(e^{\ln 2} - \ln(e^{\ln 2} + 1) = 2 - \ln(2+1) = 2 - \ln 3\). At the lower limit \(x = 0\), we have \(e^0 - \ln(e^0 + 1) = 1 - \ln(1+1) = 1 - \ln 2\). Subtracting the lower limit value from the upper limit value: \((2 - \ln 3) - (1 - \ln 2) = 1 - \ln 3 + \ln 2 = 1 - \ln\left(\frac{3}{2}\right)\).

Marking scheme

M1: Realises the need to split the fraction or use a suitable substitution like \(u = e^x + 1\) or \(u = e^x\). A1: Obtains a correct integrated expression, such as \(e^x - \ln(e^x + 1)\) or \(u - \ln u\). M1: Applies the limits of \(0\) and \(\ln 2\) (or corresponding limits \(u = 2\) and \(u = 3\) if using substitution) to their integrated function. A1.7: Correct exact answer in the form \(1 - \ln\left(\frac{3}{2}\right)\) or \(1 + \ln\left(\frac{2}{3}\right)\) or equivalent.
Question 4 · Structured
12.2 marks
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

(a) Find the binomial expansion of \((4 + 3x)^{-\frac{1}{2}}\), up to and including the term in \(x^3\), simplifying each coefficient. (5 marks)

(b) Hence, or otherwise, find the expansion of \(\frac{2-x}{\sqrt{4+3x}}\) in ascending powers of \(x\) up to and including the term in \(x^2\). (5 marks)

(c) Find the set of values of \(x\) for which this expansion is valid. (2.2 marks)
Show answer & marking scheme

Worked solution

(a) Rewrite the expression:
\[ (4+3x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} \]
Apply the binomial theorem:
\[ \left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{3}{4}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\frac{3}{4}x\right)^3 + \dots \]
\[ = 1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{135}{1024}x^3 + \dots \]
Multiply by the constant term \(\frac{1}{2}\):
\[ \frac{1}{2}\left(1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{135}{1024}x^3\right) = \frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2 - \frac{135}{2048}x^3 \]

(b) Multiply \((2-x)\) by the expansion found in (a):
\[ \frac{2-x}{\sqrt{4+3x}} = (2-x)\left(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2 + \dots\right) \]
Expand up to the term in \(x^2\):
\[ = 2\left(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2\right) - x\left(\frac{1}{2} - \frac{3}{16}x\right) \]
\[ = 1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{1}{2}x + \frac{3}{16}x^2 \]
Combine like terms:
\[ = 1 - \left(\frac{3}{8} + \frac{1}{2}\right)x + \left(\frac{27}{128} + \frac{24}{128}\right)x^2 \]
\[ = 1 - \frac{7}{8}x + \frac{51}{128}x^2 \]

(c) The expansion is valid when:
\[ \left|\frac{3}{4}x\right| < 1 \implies |x| < \frac{4}{3} \]

Marking scheme

(a)
B1: For factorising out \(4^{-1/2}\) or \(\frac{1}{2}\).
M1: For a correct attempt at the binomial expansion with fractional power \(-\frac{1}{2}\) showing at least two terms containing \(\frac{3}{4}x\).
A1: For correct simplified first and second terms: \(\frac{1}{2} - \frac{3}{16}x\).
A1: For the third term: \(\frac{27}{256}x^2\).
A1: For the fourth term: \(-\frac{135}{2048}x^3\).

(b)
M1: For an attempt to multiply their expansion by \((2-x)\).
M1: For collecting terms up to \(x^2\) (must see at least two multiplication steps correctly handled).
A1: For \(1 - \frac{7}{8}x\).
A1: For \(\frac{51}{128}x^2\).

(c)
M1: Recognises that the validity requires \(| \frac{3}{4}x | < 1\) or equivalent.
A1.2: Correct range of validity, written as \(|x| < \frac{4}{3}\) or \(-\frac{4}{3} < x < \frac{4}{3}\).
Question 5 · Structured
12.2 marks
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A curve \(C\) has parametric equations
\[x = 3t^2 + 2, \quad y = 2t^3 - 3t\]

(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\). (3 marks)

(b) Find the equation of the normal to \(C\) at the point where \(t = 2\). (5.2 marks)

(c) Find the coordinates of the points on \(C\) where the tangent is parallel to the x-axis. (4 marks)
Show answer & marking scheme

Worked solution

(a) First, find the derivatives of \(x\) and \(y\) with respect to \(t\):
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = 6t \]
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = 6t^2 - 3 \]
Using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{6t^2 - 3}{6t} = \frac{2t^2 - 1}{2t} \]

(b) At \(t = 2\):
\[ x = 3(2)^2 + 2 = 14 \]
\[ y = 2(2)^3 - 3(2) = 10 \]
The gradient of the tangent at \(t = 2\) is:
\[ m_t = \frac{2(2)^2 - 1}{2(2)} = \frac{7}{4} \]
Therefore, the gradient of the normal is:
\[ m_n = -\frac{1}{m_t} = -\frac{4}{7} \]
The equation of the normal is:
\[ y - 10 = -\frac{4}{7}(x - 14) \]
\[ 7y - 70 = -4x + 56 \implies 4x + 7y - 126 = 0 \]

(c) The tangent is parallel to the x-axis when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ \frac{2t^2 - 1}{2t} = 0 \implies 2t^2 - 1 = 0 \implies t^2 = \frac{1}{2} \implies t = \pm\frac{1}{\sqrt{2}} \]
When \(t = \frac{1}{\sqrt{2}}\):
\[ x = 3\left(\frac{1}{2}\right) + 2 = 3.5 \]
\[ y = 2\left(\frac{1}{2\sqrt{2}}\right) - \frac{3}{\sqrt{2}} = \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}} = -\sqrt{2} \]
When \(t = -\frac{1}{\sqrt{2}}\):
\[ x = 3\left(\frac{1}{2}\right) + 2 = 3.5 \]
\[ y = 2\left(-\frac{1}{2\sqrt{2}}\right) + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \sqrt{2} \]
So the coordinates are \((3.5, -\sqrt{2})\) and \((3.5, \sqrt{2})\).

Marking scheme

(a)
M1: Differentiates \(x\) and \(y\) with respect to \(t\) correctly to find \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\).
M1: Applies the chain rule correctly to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Correct expression \(\frac{2t^2 - 1}{2t}\) or equivalent.

(b)
M1: Substitutes \(t = 2\) to find coordinates \((14, 10)\).
M1: Substitutes \(t = 2\) into their expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the tangent gradient.
M1: Uses the negative reciprocal of their tangent gradient to obtain the normal gradient.
M1: Applies the straight line equation using their point and normal gradient.
A1.2: Correct equation of the normal, e.g., \(4x + 7y - 126 = 0\) or \(y = -\frac{4}{7}x + 18\).

(c)
M1: Equates the numerator of their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to zero and solves for \(t\).
A1: Obtains \(t = \pm \frac{1}{\sqrt{2}}\) or equivalent.
M1: Substitutes at least one \(t\) value back to find both \(x\) and \(y\) coordinates.
A1: Correct coordinates: \((3.5, -\sqrt{2})\) and \((3.5, \sqrt{2})\) (or exact equivalents).
Question 6 · Structured
12.2 marks
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A container is being filled with water. The volume of water in the container, \(V\text{ cm}^3\), at time \(t\text{ seconds}\) satisfies the differential equation
\[\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{k(200 - V)}{t+4}\]
where \(k\) is a positive constant.
Given that when \(t = 0\), \(V = 50\), and when \(t = 4\), \(V = 125\):

(a) Show that \(V = 200 - 150\left(\frac{4}{t+4}\right)^k\). (7.2 marks)

(b) Find the value of \(k\). (3 marks)

(c) Find the rate at which the volume of water is increasing when \(t = 12\). (2 marks)
Show answer & marking scheme

Worked solution

(a) Separating the variables in the differential equation:
\[ \int \frac{1}{200 - V}\,\mathrm{d}V = \int \frac{k}{t+4}\,\mathrm{d}t \]
Integrating both sides yields:
\[ -\ln|200 - V| = k\ln|t+4| + C \]
\[ \ln|200 - V| = -k\ln(t+4) - C \]
Exponentiate both sides:
\[ 200 - V = A(t+4)^{-k} \quad \text{where } A = \mathrm{e}^{-C} \]
Using the boundary condition \(t = 0\), \(V = 50\):
\[ 200 - 50 = A(4)^{-k} \implies 150 = A \cdot 4^{-k} \implies A = 150 \cdot 4^k \]
Substitute \(A\) back into the equation:
\[ 200 - V = 150 \cdot 4^k \cdot (t+4)^{-k} = 150 \left(\frac{4}{t+4}\right)^k \]
\[ V = 200 - 150 \left(\frac{4}{t+4}\right)^k \]

(b) Use the condition \(t = 4\), \(V = 125\):
\[ 125 = 200 - 150 \left(\frac{4}{4+4}\right)^k \]
\[ 125 = 200 - 150 \left(\frac{1}{2}\right)^k \]
\[ 150 \left(\frac{1}{2}\right)^k = 75 \implies \left(\frac{1}{2}\right)^k = 0.5 \implies k = 1 \]

(c) With \(k = 1\), the volume equation is:
\[ V = 200 - 150\left(\frac{4}{t+4}\right) \]
When \(t = 12\):
\[ V = 200 - 150\left(\frac{4}{16}\right) = 200 - 37.5 = 162.5 \]
Substitute \(t = 12\), \(V = 162.5\), and \(k = 1\) into the differential equation:
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1(200 - 162.5)}{12+4} = \frac{37.5}{16} = 2.34375 \text{ cm}^3/\text{s} \text{ (or } \frac{75}{32}\text{)} \]

Marking scheme

(a)
M1: Separates variables correctly.
M1: Integrates both sides to obtain log terms, allowing for a constant of integration.
A1: Correct integration: \(-\ln|200 - V| = k\ln(t+4) + C\) or equivalent.
M1: Applies exponential rules correctly to express \(200 - V\) as a function of \(t\).
M1: Substitutes boundary condition \(t = 0\), \(V = 50\) to evaluate their constant of integration.
A1: Obtains the correct expression for \(A = 150 \cdot 4^k\).
A1.2: Completes the algebraic steps to show the given answer cleanly.

(b)
M1: Substitutes \(t = 4\), \(V = 125\) into the given equation.
M1: Simplifies to find \((0.5)^k = 0.5\) or uses logarithms to solve for \(k\).
A1: Correct value \(k = 1\).

(c)
M1: Substitutes \(t = 12\) into the equation to find \(V\) (or substitutes \(k = 1\) and \(t = 12\) into an expression for \(\frac{\mathrm{d}V}{\mathrm{d}t}\)).
A1: Correct rate \(2.34\) (or exact fraction \(\frac{75}{32}\)).
Question 7 · Structured
12.2 marks
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

The line \(l_1\) has vector equation
\[\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\]
and the line \(l_2\) has vector equation
\[\mathbf{r} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}\]

(a) Show that the lines \(l_1\) and \(l_2\) intersect, and find the coordinates of their point of intersection, \(P\). (5 marks)

(b) The point \(A\) lies on \(l_1\) where \(l_1\) has parameter \(\lambda = 4\). The point \(B\) lies on \(l_2\) where \(l_2\) has parameter \(\mu = 1\). Find the cosine of the angle \(APB\). (4.2 marks)

(c) Find the exact area of the triangle \(APB\). (3 marks)
Show answer & marking scheme

Worked solution

(a) Equate components of \(l_1\) and \(l_2\):
1) \(1 + 2\lambda = 4 - \mu \implies 2\lambda + \mu = 3\)
2) \(-2 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 4\)
3) \(3 - \lambda = 4 + 3\mu \implies \lambda + 3\mu = -1\)

From equation 1), \\mu = 3 - 2\lambda\). Substitute this into 2):
\[ \lambda - 2(3 - 2\lambda) = 4 \implies \lambda - 6 + 4\lambda = 4 \implies 5\lambda = 10 \implies \lambda = 2 \]
Then, \(\mu = 3 - 2(2) = -1\).
Substitute \(\lambda = 2\) and \(\mu = -1\) into equation 3) to verify consistency:
\[ \text{LHS} = 2 + 3(-1) = -1 \quad \text{RHS} = -1 \]
Since the equations are consistent, the lines intersect.
Using \(\lambda = 2\) in \(l_1\), the intersection point \(P\) is:
\[ P = \begin{pmatrix} 1 + 2(2) \\ -2 + 2 \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix} \]
So \(P\) is \((5, 0, 1)\).

(b) Find coordinates of \(A\) (where \(\lambda = 4\)):
\[ A = \begin{pmatrix} 1 + 2(4) \\ -2 + 4 \\ 3 - 4 \end{pmatrix} = \begin{pmatrix} 9 \\ 2 \\ -1 \end{pmatrix} \]
Find coordinates of \(B\) (where \(\mu = 1\)):
\[ B = \begin{pmatrix} 4 - 1 \\ 2 + 2(1) \\ 4 + 3(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix} \]
Now, find vectors \(\vec{PA}\) and \(\vec{PB}\):
\[ \vec{PA} = A - P = \begin{pmatrix} 9 - 5 \\ 2 - 0 \\ -1 - 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \]
\[ \vec{PB} = B - P = \begin{pmatrix} 3 - 5 \\ 4 - 0 \\ 7 - 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 6 \end{pmatrix} \]
Find magnitudes and dot product:
\[ |\vec{PA}| = \sqrt{4^2 + 2^2 + (-2)^2} = \sqrt{24} = 2\sqrt{6} \]
\[ |\vec{PB}| = \sqrt{(-2)^2 + 4^2 + 6^2} = \sqrt{56} = 2\sqrt{14} \]
\[ \vec{PA} \cdot \vec{PB} = 4(-2) + 2(4) + (-2)(6) = -8 + 8 - 12 = -12 \]
Using the scalar product:
\[ \cos(\angle APB) = \frac{\vec{PA} \cdot \vec{PB}}{|\vec{PA}| |\vec{PB}|} = \frac{-12}{2\sqrt{6} \cdot 2\sqrt{14}} = \frac{-12}{4\sqrt{84}} = -\frac{3}{2\sqrt{21}} = -\frac{\sqrt{21}}{14} \]

(c) First find \\sin(\angle APB)\):
\[ \sin^2(\angle APB) = 1 - \cos^2(\angle APB) = 1 - \frac{21}{196} = \frac{175}{196} \implies \sin(\angle APB) = \frac{5\sqrt{7}}{14} \]
Now, compute area:
\[ \text{Area} = \frac{1}{2} |\vec{PA}| |\vec{PB}| \sin(\angle APB) = \frac{1}{2} (2\sqrt{6})(2\sqrt{14}) \left(\frac{5\sqrt{7}}{14}\right) \]
\[ = 2\sqrt{84} \cdot \frac{5\sqrt{7}}{14} = 4\sqrt{21} \cdot \frac{5\sqrt{7}}{14} = \frac{20\sqrt{147}}{14} = \frac{140\sqrt{3}}{14} = 10\sqrt{3} \]

Marking scheme

(a)
M1: Equates components of both lines to form a system of equations in terms of \(\lambda\) and \(\mu\).
M1: Solves a pair of equations to find values for \(\lambda\) and \(\mu\).
A1: Obtains \(\lambda = 2\) and \(\mu = -1\).
M1: Checks values in the third equation to confirm intersection.
A1: Correct coordinates of the intersection point \(P(5, 0, 1)\).

(b)
M1: Finds coordinates of \(A\) and \(B\) and forms vectors \(\vec{PA}\) and \(\vec{PB}\).
M1: Finds the dot product \(\vec{PA} \cdot \vec{PB}\) and magnitudes of both vectors.
M1: Applies the cosine formula correctly.
A1.2: Correct simplified exact value of \(\cos(\angle APB) = -\frac{\sqrt{21}}{14}\).

(c)
M1: Finds \(\sin(\angle APB)\) using trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\).
M1: Applies the triangle area formula \(\frac{1}{2} ab \sin C\).
A1: Correct exact area of \(10\sqrt{3}\).
Question 8 · Structured
12.2 marks
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A curve \(C\) has parametric equations
\[x = \ln(t + 2), \quad y = \frac{t}{t+1}, \quad t > -1\]
The region \(R\) is bounded by the curve \(C\), the x-axis, and the lines \(x = \ln(2)\) and \(x = \ln(3)\).

(a) Show that the area of \(R\) is given by \(\int_0^1 \frac{t}{(t+1)(t+2)}\,\mathrm{d}t\). (4 marks)

(b) Using partial fractions, find the exact area of \(R\), giving your answer in the form \(\ln(a)\) where \(a\) is a rational number to be found. (6.2 marks)

(c) Find a cartesian equation of the curve \(C\) in the form \(y = f(x)\). (2 marks)
Show answer & marking scheme

Worked solution

(a) The area of the region \(R\) is given by:
\[ \text{Area} = \int y\,\mathrm{d}x \]
Since \(x = \ln(t + 2)\), we have:
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t+2} \implies \mathrm{d}x = \frac{1}{t+2}\,\mathrm{d}t \]
Now find the limits of integration in terms of \(t\):
When \(x = \ln(2)\):
\[ \ln(t+2) = \ln(2) \implies t + 2 = 2 \implies t = 0 \]
When \(x = \ln(3)\):
\[ \ln(t+2) = \ln(3) \implies t + 2 = 3 \implies t = 1 \]
Substitute \(y\), \(\mathrm{d}x\), and limits into the integral:
\[ \text{Area} = \int_0^1 \frac{t}{t+1} \cdot \frac{1}{t+2}\,\mathrm{d}t = \int_0^1 \frac{t}{(t+1)(t+2)}\,\mathrm{d}t \]

(b) Express the integrand in partial fractions:
\[ \frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \]
\[ t = A(t+2) + B(t+1) \]
Let \(t = -1\):
\[ -1 = A(1) \implies A = -1 \]
Let \(t = -2\):
\[ -2 = B(-1) \implies B = 2 \]
Thus:
\[ \int_0^1 \left( \frac{2}{t+2} - \frac{1}{t+1} \right)\,\mathrm{d}t = \Big[ 2\ln(t+2) - \ln(t+1) \Big]_0^1 \]
Evaluate at the limits:
\[ = \Big( 2\ln(3) - \ln(2) \Big) - \Big( 2\ln(2) - \ln(1) \Big) \]
\[ = 2\ln(3) - 3\ln(2) \]
\[ = \ln(3^2) - \ln(2^3) = \ln(9) - \ln(8) = \ln\left(\frac{9}{8}\right) \]
So \(a = \frac{9}{8}\).

(c) We have:
\[ x = \ln(t+2) \implies \mathrm{e}^x = t+2 \implies t = \mathrm{e}^x - 2 \]
Substitute into the equation for \(y\):
\[ y = \frac{\mathrm{e}^x - 2}{(\mathrm{e}^x - 2) + 1} = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1} \]

Marking scheme

(a)
M1: Correct expression for \(\frac{\mathrm{d}x}{\mathrm{d}t}\).
M1: Finding correct \(t\) limits corresponding to \(x = \ln(2)\) and \(x = \ln(3)\).
M1: Integrates using \(\int y \frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\).
A1: Correctly derives the given integral with clear working.

(b)
M1: Correct form for partial fractions and finds values for \(A\) and \(B\).
A1: Correct partial fractions: \(\frac{2}{t+2} - \frac{1}{t+1}\).
M1: Integrates partial fractions to logarithmic terms.
A1: Correct integration: \(2\ln(t+2) - \ln(t+1)\).
M1: Substitutes limits 0 and 1, applying logarithmic subtraction/addition laws.
A1.2: Finds \(a = \frac{9}{8}\) in the final form \(\ln\left(\frac{9}{8}\right)\).

(c)
M1: Rearranges \(x = \ln(t+2)\) to make \(t\) the subject.
A1: Correct Cartesian equation: \(y = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1}\).

Wondering how well you actually know this?

Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Start Practising FreeSee pricing

Want more questions like this? Practice unlimited on Thinka — instant answers included.

Start Practising Free