1. [Maximum mark: 4]
The variables \(x\) and \(y\) satisfy the equation \(y=4^{2x-a}\), where \(a\) is an integer. As shown in the diagram, the graph of \(\ln y\) against \(x\) is a straight line passing through the point \((0,-20.8)\), where the second coordinate is given correct to 3 significant figures.
(a) Show that the gradient of the straight line is \(\ln 16\).
(b) Determine the value of \(a\).
▶️Answer/Explanation
(a) Starting with \( y = 4^{2x – a} \), take the natural logarithm of both sides:
\(\ln y = (2x – a)\ln 4\).
This is of the form \(\ln y = m x + c\), where \(m = 2\ln 4\).
Since \(2\ln 4 = \ln 4^2 = \ln 16\), the gradient is \(\ln 16\).
(b) Substitute the point \((0, -20.8)\) into the equation \(\ln y = 2x\ln 4 – a\ln 4\):
\(-20.8 = 0 – a\ln 4\).
Solve for \(a\):
\(a = \frac{20.8}{\ln 4} \approx 15\).
Thus, the value of \(a\) is 15.
2. [Maximum mark: 6]
(a) Express the equation \(7 \tan \theta + 4 \cot \theta – 13 \sec \theta = 0\) in terms of \(\sin \theta\) only.
(b) Hence solve the equation \(7 \tan \theta + 4 \cot \theta – 13 \sec \theta = 0\) for \(0^\circ < \theta < 360^\circ\).
▶️Answer/Explanation
(a) Rewrite the equation using trigonometric identities:
\(7 \tan \theta = \frac{7\sin \theta}{\cos \theta}\), \(4 \cot \theta = \frac{4\cos \theta}{\sin \theta}\), and \(13 \sec \theta = \frac{13}{\cos \theta}\).
Multiply through by \(\sin \theta \cos \theta\) to eliminate denominators:
\(7\sin^2 \theta + 4\cos^2 \theta – 13\sin \theta = 0\).
Use \(\cos^2 \theta = 1 – \sin^2 \theta\):
\(7\sin^2 \theta + 4(1 – \sin^2 \theta) – 13\sin \theta = 0\).
Simplify to \(3\sin^2 \theta – 13\sin \theta + 4 = 0\).
(b) Solve the quadratic equation \(3\sin^2 \theta – 13\sin \theta + 4 = 0\):
Let \(u = \sin \theta\), then \(3u^2 – 13u + 4 = 0\).
Solutions: \(u = \frac{1}{3}\) or \(u = 4\) (invalid).
Thus, \(\sin \theta = \frac{1}{3}\), giving \(\theta = 19.5^\circ\) or \(160.5^\circ\).
3. [Maximum mark: 7]
The diagram shows the curve with equation \(y=3\sin x-3\sin 2x\) for \(0 \leq x \leq \pi\). The curve meets the \(x\)-axis at the origin and at the points with \(x\)-coordinates \(a\) and \(\pi\).
(a) Find the exact value of \(a\).
(b) Find the area of the shaded region.
▶️Answer/Explanation
(a) Set \(y = 0\):
\(3\sin x – 3\sin 2x = 0\).
Factor out 3: \(\sin x = \sin 2x\).
Use double-angle identity: \(\sin x = 2\sin x \cos x\).
Solutions: \(\sin x = 0\) or \(\cos x = \frac{1}{2}\).
For \(0 \leq x \leq \pi\), \(x = 0\), \(x = \frac{\pi}{3}\), or \(x = \pi\).
Thus, \(a = \frac{\pi}{3}\).
(b) Integrate \(y\) from \(0\) to \(\frac{\pi}{3}\):
\(\int_{0}^{\frac{\pi}{3}} (3\sin x – 3\sin 2x) \, dx = \left[ -3\cos x + \frac{3}{2}\cos 2x \right]_{0}^{\frac{\pi}{3}}\).
Evaluate:
At \(x = \frac{\pi}{3}\): \(-3\cos \frac{\pi}{3} + \frac{3}{2}\cos \frac{2\pi}{3} = -1.5 – 0.75 = -2.25\).
At \(x = 0\): \(-3\cos 0 + \frac{3}{2}\cos 0 = -3 + 1.5 = -1.5\).
Area = \(-2.25 – (-1.5) = -0.75\) (take absolute value): \(0.75\) or \(\frac{3}{4}\).
4. [Maximum mark: 7]
A curve has equation \(x^2y + 2y^3 = 48\). Find the equation of the normal to the curve at the point \((4, 2)\), giving your answer in the form \(ax + by + c = 0\) where \(a\), \(b\), and \(c\) are integers.
▶️Answer/Explanation
Differentiate implicitly:
\(x^2 \frac{dy}{dx} + 2xy + 6y^2 \frac{dy}{dx} = 0\).
Substitute \((4, 2)\):
\(16 \frac{dy}{dx} + 16 + 24 \frac{dy}{dx} = 0\).
Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{16}{40} = -\frac{2}{5}\).
Slope of normal: \(\frac{5}{2}\).
Equation of normal: \(y – 2 = \frac{5}{2}(x – 4)\).
Simplify to \(5x – 2y – 16 = 0\).
5. [Maximum mark: 9]
(a) By sketching the graphs of \(y = |5 – 2x|\) and \(y = 3 \ln x\) on the same diagram, show that the equation \(|5 – 2x| = 3 \ln x\) has exactly two roots. [3]
(b) Show that the value of the larger root satisfies the equation \(x = 2.5 + 1.5 \ln x\). [1]
(c) Show by calculation that the value of the larger root lies between 4.5 and 5.0. [2]
(d) Use an iterative formula, based on the equation in part (b), to find the value of the larger root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
▶️Answer/Explanation
(a) The graph of \(y = |5 – 2x|\) is a V-shape with vertex at \(x = 2.5\). The graph of \(y = 3 \ln x\) is a logarithmic curve increasing for \(x > 0\). The two graphs intersect at exactly two points, indicating two roots.
(b) For the larger root, \(5 – 2x < 0\), so \(|5 – 2x| = 2x – 5\).
Thus, \(2x – 5 = 3 \ln x\), which rearranges to \(x = 2.5 + 1.5 \ln x\).
(c) Let \(f(x) = x – 2.5 – 1.5 \ln x\).
At \(x = 4.5\): \(f(4.5) \approx -0.25\).
At \(x = 5.0\): \(f(5.0) \approx 0.08\).
Since \(f(x)\) changes sign, the root lies between 4.5 and 5.0.
(d) Use the iterative formula \(x_{n+1} = 2.5 + 1.5 \ln x_n\) with \(x_0 = 4.5\):
\(x_1 \approx 4.7354\),
\(x_2 \approx 4.8276\),
\(x_3 \approx 4.8649\),
\(x_4 \approx 4.8801\),
\(x_5 \approx 4.8864\).
The root is approximately 4.88 (to 3 s.f.).
6. [Maximum mark: 8]
(a) Show that the \(x\)-coordinate of any stationary point on the curve \(y = \frac{9e^{2x} + 16}{e^x – 1}\) satisfies the equation \(e^x (3e^x – 8)(3e^x + 2) = 0\). [4]
(b) Hence show that the curve has only one stationary point and find its exact coordinates. [4]
▶️Answer/Explanation
(a) Differentiate \(y\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(18e^{2x})(e^x – 1) – (9e^{2x} + 16)(e^x)}{(e^x – 1)^2}\).
Set \(\frac{dy}{dx} = 0\):
\(18e^{2x}(e^x – 1) – e^x(9e^{2x} + 16) = 0\).
Simplify to \(9e^{3x} – 18e^{2x} – 16e^x = 0\).
Factorize: \(e^x (9e^{2x} – 18e^x – 16) = 0\).
Further factorize: \(e^x (3e^x – 8)(3e^x + 2) = 0\).
(b) Solutions: \(e^x = 0\) (invalid), \(3e^x – 8 = 0\) (valid), or \(3e^x + 2 = 0\) (invalid).
Thus, \(e^x = \frac{8}{3}\), so \(x = \ln \left(\frac{8}{3}\right)\).
Substitute back to find \(y = \frac{9 \left(\frac{8}{3}\right)^2 + 16}{\frac{8}{3} – 1} = 48\).
The only stationary point is \(\left(\ln \left(\frac{8}{3}\right), 48\right)\).
7. [Maximum mark: 9]
(a) Find, in terms of \(x\) and \(a\), the quotient when \(p(x) = 2x^3 + 5x^2 + ax + 2a\) is divided by \((x + 2)\), and show that the remainder is 4. [3]
(b) It is given that \(\int_{-1}^{1} \frac{p(x)}{x+2} \, dx = \frac{22}{3} + \ln b\), where \(b\) is an integer. Find the values of \(a\) and \(b\). [6]
▶️Answer/Explanation
(a) Perform polynomial division of \(p(x)\) by \((x + 2)\):
Quotient: \(2x^2 + x + (a – 2)\).
Remainder: Substitute \(x = -2\) into \(p(x)\):
\(p(-2) = -16 + 20 – 2a + 2a = 4\).
(b) Rewrite the integrand using the quotient and remainder:
\(\frac{p(x)}{x+2} = 2x^2 + x + (a – 2) + \frac{4}{x+2}\).
Integrate term by term:
\(\int_{-1}^{1} \left(2x^2 + x + (a – 2) + \frac{4}{x+2}\right) dx = \left[\frac{2}{3}x^3 + \frac{1}{2}x^2 + (a – 2)x + 4\ln|x+2|\right]_{-1}^{1}\).
Evaluate at limits:
At \(x = 1\): \(\frac{2}{3} + \frac{1}{2} + (a – 2) + 4\ln 3\).
At \(x = -1\): \(-\frac{2}{3} + \frac{1}{2} – (a – 2) + 4\ln 1\).
Subtract: \(\frac{4}{3} + 2(a – 2) + 4\ln 3 = \frac{22}{3} + \ln b\).
Solve for \(a\): \(2a – 4 = 6 \Rightarrow a = 5\).
Then \(4\ln 3 = \ln b \Rightarrow b = 81\).