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:
\( \frac{dy}{dx} = \frac{\frac{1}{x} \cdot x^2 – \ln x \cdot 2x}{x^4} = \frac{x – 2x \ln x}{x^4} \).
Simplify and substitute \( x = e \):
\( \frac{dy}{dx} = \frac{e – 2e \ln e}{e^4} = \frac{e – 2e}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} \).
2. [Maximum mark: 4]
(a) Sketch, on the same diagram, the graphs of \( y = |2x – 9| \) and \( y = 5x – 3 \). [2]
(b) Solve the equation \( |2x – 9| = 5x – 3 \). [2]
▶️Answer/Explanation
(a)
– Draw a V-shaped graph for \( y = |2x – 9| \) with vertex at \( x = 4.5 \).
– Draw a straight line for \( y = 5x – 3 \) intersecting the V-shaped graph.
(b)
Solve \( -2x + 9 = 5x – 3 \):
\( 9 + 3 = 7x \) → \( x = \frac{12}{7} \).
Verify \( x = \frac{12}{7} \) satisfies 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 \( \left(0, \frac{1}{6}\pi\right) \).
▶️Answer/Explanation
Solution:
Differentiate implicitly with respect to \( x \):
\( 2e^{2x} \cos 2y – 2e^{2x} \sin 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0 \).
Substitute \( x = 0 \) and \( y = \frac{\pi}{6} \):
\( 2 \cos \frac{\pi}{3} – 2 \sin \frac{\pi}{3} \frac{dy}{dx} + \cos \frac{\pi}{6} \frac{dy}{dx} = 0 \).
Simplify and solve for \( \frac{dy}{dx} \):
\( 1 – \sqrt{3} \frac{dy}{dx} + \frac{\sqrt{3}}{2} \frac{dy}{dx} = 0 \) → \( \frac{dy}{dx} = \frac{2}{\sqrt{3}} \).
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 \( \ln 12 \). [4]
(b) Use a graph of \( y = \ln x \) to show that \( \ln 12 \) is an under-estimate of the true value of \( \int_{1}^{4} \ln x \, dx \). [2]
▶️Answer/Explanation
(a)
Using trapezium rule with \( h = 1 \):
\( \frac{1}{2} [\ln 1 + 2 \ln 2 + 2 \ln 3 + \ln 4] = \frac{1}{2} \ln (1 \cdot 2^2 \cdot 3^2 \cdot 4) = \ln 12 \).
(b)
The graph of \( y = \ln x \) is concave down, so the trapezium rule under-estimates the 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 \( p(x) = 0 \):
\( 128 + 16a – 12 – 4 = 0 \) → \( a = -7 \).
Factorise: \( p(x) = (x – 4)(2x^2 + x + 1) \).
(b)
Solve \( e^{3y} = 4 \):
\( y = \frac{1}{3} \ln 4 \).
The quadratic \( 2x^2 + x + 1 \) has no real roots (discriminant \( < 0 \)), so only one real root exists.
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. Give the result of each iteration to 5 significant figures. [3]
▶️Answer/Explanation
(a)
Integrate \( y \): \( \int_{a}^{a+1} 3e^{2x-1} \, dx = \frac{3}{2} e^{2x-1} \Big|_{a}^{a+1} = \frac{3}{2} (e^{2a+1} – e^{2a-1}) \).
Set equal to 120 and solve for \( a \).
(b)
Iterative formula: \( a_{n+1} = \frac{1}{2} \ln(80 + e^{2a_n-1}) – \frac{1}{2} \).
Converges to \( a \approx 1.76 \).
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 \( \sqrt{2\pi – 2x} \): \( -\frac{1}{3} (2\pi – 2x)^{3/2} \Big|_{0}^{\pi} = \frac{1}{3} (2\pi)^{3/2} \).
Integrate \( \sin^2 x \): \( \frac{1}{2}x – \frac{1}{4} \sin 2x \Big|_{0}^{\pi} = \frac{\pi}{2} \).
Subtract the second integral from the first: \( \frac{1}{3} (2\pi)^{3/2} – \frac{\pi}{2} \).
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: \( 6 \sin \theta + 10 \cos(\theta – 30^\circ) = 11 \sin \theta + 5\sqrt{3} \cos \theta \).
Express as \( R \sin(\theta + \alpha) \): \( R = 14 \), \( \alpha = \arctan\left(\frac{5\sqrt{3}}{11}\right) \approx 38.21^\circ \).
(b)
Substitute \( \theta \) with \( 2\theta \): \( 14 \sin(2\theta + 38.21^\circ) = 2 \).
Solve: \( 2\theta + 38.21^\circ = \arcsin\left(\frac{2}{14}\right) \) → \( \theta \approx 66.8^\circ \).