1. [Maximum mark: 4]
The equation of a curve is such that \(\frac{dy}{dx} = \frac{3}{x^4} + 32x^3\). It is given that the curve passes through the point \(\left(\frac{1}{2}, 4\right)\).
Find the equation of the curve.
▶️Answer/Explanation
Integrate \(\frac{dy}{dx}\) to find \(y\):
\(y = \int \left(\frac{3}{x^4} + 32x^3\right) dx = -\frac{1}{x^3} + 8x^4 + c\)
Substitute \(\left(\frac{1}{2}, 4\right)\) into the equation:
\(4 = -8 + \frac{1}{2} + c \Rightarrow c = \frac{23}{2}\)
Thus, the equation of the curve is:
\(y = -\frac{1}{x^3} + 8x^4 + \frac{23}{2}\)
2. [Maximum mark: 5]
The sum of the first 20 terms of an arithmetic progression is 405 and the sum of the first 40 terms is 1410.
Find the 60th term of the progression.
▶️Answer/Explanation
Using the sum formula for an arithmetic progression:
\(S_{20} = 10(2a + 19d) = 405\)
\(S_{40} = 20(2a + 39d) = 1410\)
Solve simultaneously to find \(a = 6\) and \(d = 1.5\).
The 60th term is:
\(a + 59d = 6 + 59 \times 1.5 = 94.5\)
3. [Maximum mark: 6]
(a) Find the first three terms in the expansion of \((3 – 2x)^5\) in ascending powers of \(x\).
(b) Hence find the coefficient of \(x^2\) in the expansion of \((4 + x)^2(3 – 2x)^5\).
▶️Answer/Explanation
(a) Using the binomial expansion:
\((3 – 2x)^5 = 243 – 810x + 1080x^2 – \dots\)
First three terms: \(243, -810x, 1080x^2\)
(b) Expand \((4 + x)^2 = 16 + 8x + x^2\).
Multiply by the expansion from (a):
Coefficient of \(x^2\) is \(16 \times 1080 + 8 \times (-810) + 1 \times 243 = 11043\)
4. [Maximum mark: 3]
The diagram shows part of the graph of \(y = a\tan(x – b) + c\).
Given that \(0 < b < \pi\), state the values of the constants \(a\), \(b\) and \(c\).
▶️Answer/Explanation
From the graph analysis:
\(a = 2\) ,
\(b = \frac{\pi}{4}\) (phase shift), \(c = 1\) (vertical shift).
5. [Maximum mark: 4]
The fifth, sixth and seventh terms of a geometric progression are \(8k\), \(-12\) and \(2k\) respectively.
Given that \(k\) is negative, find the sum to infinity of the progression.
▶️Answer/Explanation
Using the property of a geometric progression:
\((-12)^2 = 8k \times 2k \Rightarrow 144 = 16k^2 \Rightarrow k = -3\) (since \(k\) is negative).
Common ratio \(r = \frac{-12}{8k} = 0.5\).
First term \(a = \frac{8k}{r^4} = -384\).
Sum to infinity: \(S_\infty = \frac{a}{1 – r} = \frac{-384}{0.5} = -768\).
6. [Maximum mark: 5]
The equation of a curve is \(y = (2k – 3)x^2 – kx – (k – 2)\), where \(k\) is a constant. The line \(y = 3x – 4\) is a tangent to the curve
. Find the value of \(k\).
▶️Answer/Explanation
Set the curve equal to the line:
\((2k – 3)x^2 – (k + 3)x – (k – 6) = 0\).
For tangency, the discriminant must be zero:
\((k + 3)^2 + 4(2k – 3)(k – 6) = 0\).
Simplify to \(9k^2 – 54k + 81 = 0 \Rightarrow k = 3\).
7. [Maximum mark: 5]
(a) Prove the identity \(\frac{1 – 2\sin^2 \theta}{1 – \sin^2 \theta} \equiv 1 – \tan^2 \theta\).
(b) Hence solve the equation \(\frac{1 – 2\sin^2 \theta}{1 – \sin^2 \theta} = 2\tan^4 \theta\) for \(0^\circ \leq \theta \leq 180^\circ\).
▶️Answer/Explanation
(a) Simplify the left-hand side:
\(\frac{\cos^2 \theta – \sin^2 \theta}{\cos^2 \theta} = 1 – \tan^2 \theta\).
(b) Using the identity from (a):
\(1 – \tan^2 \theta = 2\tan^4 \theta\).
Let \(u = \tan^2 \theta\): \(2u^2 + u – 1 = 0 \Rightarrow u = 0.5\) or \(-1\).
Solutions: \(\theta = 35.3^\circ\) and \(144.7^\circ\).
8. [Maximum mark: 10]
The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre C. The boundary of the plate consists of two arcs PS and QR of the original circle and two semicircles with PQ and RS as diameters. The radius of the circle with centre C is 4 cm, and PQ = RS = 4 cm also.
(a) Show that angle \(PCS = \frac{2}{3}\pi\) radians.
(b) Find the exact perimeter of the plate.
(c) Show that the area of the plate is \(\left(\frac{20}{3}\pi + 8\sqrt{3}\right) \text{cm}^2\).
▶️Answer/Explanation
(a) Using geometry, angle \(PCS = \pi – 2 \times \frac{\pi}{6} = \frac{2}{3}\pi\).
(b) Perimeter includes two arcs and two semicircles:
\(2 \times 4 \times \frac{2\pi}{3} + 2 \times \pi \times 2 = \frac{28\pi}{3}\).
(c) Calculate the area of the large circle, subtract the area of two semicircles and two segments:
\(\pi \times 4^2 – \pi \times 2^2 – 2\left(\frac{8\pi}{3} – 4\sqrt{3}\right) = \frac{20\pi}{3} + 8\sqrt{3}\).
9. [Maximum mark: 11]
Functions \(f\) and \(g\) are defined as follows:
\(f(x) = (x – 2)^2 – 4\) for \(x \geq 2\),
\(g(x) = ax + 2\) for \(x \in \mathbb{R}\),
where \(a\) is a constant.
(a) State the range of \(f\).
(b) Find \(f^{-1}(x)\).
(c) Given that \(a = -\frac{5}{3}\), solve the equation \(f(x) = g(x)\).
(d) Given instead that \(g(g(f^{-1}(12))) = 62\), find the possible values of \(a\).
▶️Answer/Explanation
(a) Range of \(f\) is f(x) ⩾−4
(b) Find the inverse:
\(y = (x – 2)^2 – 4 \Rightarrow x = \sqrt{y + 4} + 2\).
Thus, \(f^{-1}(x) = \sqrt{x + 4} + 2\).
(c) Solve \((x – 2)^2 – 4 = -\frac{5}{3}x + 2\):
Quadratic solution gives \(x = 3\) (only valid solution).
(d) Compute \(f^{-1}(12) = 6\), then \(g(g(6)) = 62\):
Solve \(6a^2 + 2a – 60 = 0\) to find \(a = 3\) or \(-\frac{10}{3}\).
10. [Maximum mark: 8]
The equation of a circle is \(x^2 + y^2 – 4x + 6y – 77 = 0\).
(a) Find the \(x\)-coordinates of the points \(A\) and \(B\) where the circle intersects the \(x\)-axis.
(b) Find the point of intersection of the tangents to the circle at \(A\) and \(B\).
▶️Answer/Explanation
(a) Set \(y = 0\): \(x^2 – 4x – 77 = 0\).
Solutions: \(x = -7\) and \(x = 11\).
(b) Find gradients of radii:
Gradient of \(CA\) is \(\frac{3}{9} = \frac{1}{3}\), so tangent gradient at \(A\) is \(-3\).
Gradient of \(CB\) is \(\frac{3}{-9} = -\frac{1}{3}\), so tangent gradient at \(B\) is \(-3\).
Equations: \(y = 3x + 21\) and \(y = -3x + 33\).
Intersection at \((2, 27)\).
11. [Maximum mark: 14]
The equation of a curve is \(y = 2\sqrt{3x + 4} – x\).
(a) Find the equation of the normal to the curve at the point \((4, 4)\), giving your answer in the form \(y = mx + c\).
(b) Find the coordinates of the stationary point.
(c) Determine the nature of the stationary point.
(d) Find the exact area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 4\).
▶️Answer/Explanation
(a) Differentiate \(y\) to find \(\frac{dy}{dx} = 3(3x + 4)^{-0.5} – 1\).
At \(x = 4\), gradient of tangent is \(\frac{1}{4}\), so normal gradient is \(4\).
Equation: \(y – 4 = -4(x – 4) \Rightarrow y = 4x – 12\).
(b) Set \(\frac{dy}{dx} = 0\): \(3(3x + 4)^{-0.5} – 1 = 0\).
Solve to find stationary point at \(\left(\frac{5}{3}, \frac{13}{3}\right)\).
(c) Second derivative is negative, so it’s a maximum point.
$\frac{d^2y}{dx^2} = -\frac{9}{2}(3x+4)^{-1.5}$
$\text{At } x = \frac{5}{3}, \frac{d^2y}{dx^2} \text{ is negative so the point is a maximum}$
(d) Integrate \(y\) from \(0\) to \(4\):
$\text{Area} = \int [2(3x+4)^{0.5} – x] dx = \frac{4}{9}(3x+4)^{1.5} – \frac{1}{2}x^2$
$\left[ \frac{4}{9}(3x+4)^{1.5} – \frac{1}{2}x^2 \right]_0^4 = \left( \frac{4}{9}(16)^{1.5} – \frac{1}{2}(4)^2 \right) – \left( \frac{4}{9}(4)^{1.5} – \frac{1}{2}(0)^2 \right) = \frac{256}{9} – 8 – \frac{32}{9}$
$16\frac{8}{9}$