1. [Maximum mark: 3]
Given that \( y = \frac{\ln x}{x^2} \), find the exact value of \( \frac{dy}{dx} \) when \( x = e \).
▶️Answer/Explanation
Solution:
Differentiate using the quotient rule (or product rule):
\( \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} – 2x \ln x}{x^4} = \frac{1 – 2 \ln x}{x^3} \).
Substitute \( x = e \):
\( \frac{dy}{dx} = \frac{1 – 2 \ln e}{e^3} = \frac{1 – 2}{e^3} = -\frac{1}{e^3} \).
2. [Maximum mark: 4]
(a) Sketch, on the same diagram, the graphs of \( y = |2x – 9| \) and \( y = 5x – 3 \).
(b) Solve the equation \( |2x – 9| = 5x – 3 \).
▶️Answer/Explanation
(a)
– Draw a V-shaped graph for \( y = |2x – 9| \) with vertex at \( x = 4.5 \).
– Draw a straight line for \( y = 5x – 3 \) with a steeper gradient, intersecting the \( x \)-axis between the origin and the vertex of the first graph.
(b)
Solve \( -2x + 9 = 5x – 3 \):
\( 12 = 7x \) ⇒ \( x = \frac{12}{7} \approx 1.71 \).
(No other solution is valid as the other case \( 2x – 9 = 5x – 3 \) leads to \( x = -2 \), which does not satisfy the original equation.)
3. [Maximum mark: 5]
A curve has equation \( e^{2x} \cos 2y + \sin y = 1 \). Find the exact gradient of the curve at the point \( \left(0, \frac{\pi}{6}\right) \).
▶️Answer/Explanation
Solution:
Differentiate implicitly with respect to \( x \):
\( 2e^{2x} \cos 2y – 2e^{2x} \sin 2y \cdot \frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = 0 \).
Substitute \( x = 0 \) and \( y = \frac{\pi}{6} \):
\( 2 \cos \frac{\pi}{3} – 2 \sin \frac{\pi}{3} \cdot \frac{dy}{dx} + \cos \frac{\pi}{6} \cdot \frac{dy}{dx} = 0 \).
Simplify using \( \cos \frac{\pi}{3} = \frac{1}{2} \), \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), and \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \):
\( 1 – \sqrt{3} \cdot \frac{dy}{dx} + \frac{\sqrt{3}}{2} \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{2}{\sqrt{3}} \).
4. [Maximum mark: 6]
(a) Use the trapezium rule with three intervals to show that the value of \( \int_{1}^{4} \ln x \, dx \) is approximately \( \ln 12 \).
(b) Use a graph of \( y = \ln x \) to show that \( \ln 12 \) is an under-estimate of the true value of \( \int_{1}^{4} \ln x \, dx \).
▶️Answer/Explanation
(a)
Using the trapezium rule with \( h = 1 \) and \( y \)-values \( \ln 1, \ln 2, \ln 3, \ln 4 \):
\( \int_{1}^{4} \ln x \, dx \approx \frac{1}{2} [\ln 1 + 2 \ln 2 + 2 \ln 3 + \ln 4] \).
Simplify using logarithm properties:
\( \frac{1}{2} [\ln 1 + \ln 4 + 2 \ln 6] = \frac{1}{2} [\ln (1 \cdot 4 \cdot 6^2)] = \ln 12 \).
(b)
The graph of \( y = \ln x \) is concave down, so the tops of the trapeziums lie below the curve. Hence, the trapezium rule under-estimates the true area.
5. [Maximum mark: 7]
The polynomial \( p(x) \) is defined by \( p(x) = 2x^3 + ax^2 – 3x – 4 \), where \( a \) is a constant. It is given that \( (x – 4) \) is a factor of \( p(x) \).
(a) Find the value of \( a \) and hence factorise \( p(x) \). [4]
(b) Show that the equation \( p(e^{3y}) = 0 \) has only one real root and find its exact value. [3]
▶️Answer/Explanation
(a)
Substitute \( x = 4 \) into \( p(x) = 0 \):
\( 128 + 16a – 12 – 4 = 0 \) ⇒ \( 16a = -112 \) ⇒ \( a = -7 \).
Factorise \( p(x) = (x – 4)(2x^2 + x + 1) \).
(b)
Solve \( p(e^{3y}) = 0 \):
\( e^{3y} = 4 \) (since \( 2x^2 + x + 1 = 0 \) has no real roots).
Take natural logarithms: \( 3y = \ln 4 \) ⇒ \( y = \frac{1}{3} \ln 4 \).
6. [Maximum mark: 8]
The diagram shows the curve \( y = 3e^{2x-1} \). The shaded region is bounded by the curve and the lines \( x = a \), \( x = a + 1 \), and \( y = 0 \), where \( a \) is a constant. It is given that the area of the shaded region is 120 square units.
(a) Show that \( a = \frac{1}{2} \ln(80 + e^{2a-1}) – \frac{1}{2} \). [5]
(b) Use an iterative formula, based on the equation in part (a), to find the value of \( a \) correct to 3 significant figures. [3]
▶️Answer/Explanation
(a)
Integrate \( y = 3e^{2x-1} \) from \( x = a \) to \( x = a + 1 \):
\( \int_{a}^{a+1} 3e^{2x-1} \, dx = \frac{3}{2} e^{2x-1} \Big|_{a}^{a+1} = \frac{3}{2} (e^{2a+1} – e^{2a-1}) \).
Set the area equal to 120 and simplify:
\( \frac{3}{2} e^{2a-1} (e^2 – 1) = 120 \) ⇒ \( e^{2a-1} = \frac{80}{e^2 – 1} \).
Take natural logarithms and rearrange to obtain the given expression.
(b)
Using iteration \( a_{n+1} = \frac{1}{2} \ln(80 + e^{2a_n – 1}) – \frac{1}{2} \), starting with \( a_0 = 1 \):
Converges to \( a \approx 1.76 \) (correct to 3 s.f.).
7. [Maximum mark: 8]
The diagram shows the curves \( y = \sqrt{2\pi – 2x} \) and \( y = \sin^2 x \) for \( 0 \leq x \leq \pi \). The shaded region is bounded by the two curves and the line \( x = 0 \). Find the exact area of the shaded region.
▶️Answer/Explanation
Solution:
Integrate \( \sqrt{2\pi – 2x} \) from \( x = 0 \) to \( x = \pi \):
\( \int_{0}^{\pi} (2\pi – 2x)^{1/2} \, dx = -\frac{1}{3} (2\pi – 2x)^{3/2} \Big|_{0}^{\pi} = \frac{(2\pi)^{3/2}}{3} \).
Integrate \( \sin^2 x \) using identity \( \sin^2 x = \frac{1}{2} – \frac{1}{2} \cos 2x \):
\( \int_{0}^{\pi} \sin^2 x \, dx = \frac{\pi}{2} \).
Subtract the second integral from the first to find the shaded area:
\( \frac{(2\pi)^{3/2}}{3} – \frac{\pi}{2} \).
8. [Maximum mark: 9]
(a) Express \( 3 \sin 2\theta \sec \theta + 10 \cos(\theta – 30^\circ) \) in the form \( R \sin(\theta + \alpha) \), where \( R > 0 \) and \( 0^\circ < \alpha < 90^\circ \). Give the value of \( \alpha \) correct to 2 decimal places. [6]
(b) Hence solve the equation \( 3 \sin 4\theta \sec 2\theta + 10 \cos(2\theta – 30^\circ) = 2 \) for \( 0^\circ < \theta < 90^\circ \). [3]
▶️Answer/Explanation
(a)
Simplify \( 3 \sin 2\theta \sec \theta \) to \( 6 \sin \theta \).
Expand \( 10 \cos(\theta – 30^\circ) \) to \( 5\sqrt{3} \cos \theta + 5 \sin \theta \).
Combine terms: \( 11 \sin \theta + 5\sqrt{3} \cos \theta \).
Express as \( R \sin(\theta + \alpha) \) where \( R = 14 \) and \( \alpha = \arctan\left(\frac{5\sqrt{3}}{11}\right) \approx 38.21^\circ \).
(b)
Substitute \( \theta \) with \( 2\theta \) in part (a) to get \( 14 \sin(2\theta + 38.21^\circ) = 2 \).
Solve \( \sin(2\theta + 38.21^\circ) = \frac{1}{7} \):
\( 2\theta + 38.21^\circ = \arcsin\left(\frac{1}{7}\right) \) ⇒ \( \theta \approx 66.8^\circ \).