1. [Maximum mark: 4]
(a) Express \(16x^2 – 24x + 10\) in the form \((4x + a)^2 + b\).
(b) It is given that the equation \(16x^2 – 24x + 10 = k\), where \(k\) is a constant, has exactly one root. Find the value of this root.
▶️Answer/Explanation
(a) \((4x – 3)^2 + 1\) or \(a = -3\), \(b = 1\).
(b) For one root, \(k = 1\). The root is \(\frac{3}{4}\) or \(0.75\).
2. [Maximum mark: 5]
(a) The graph of \(y = f(x)\) is transformed to the graph of \(y = 2f(x – 1)\). Describe fully the two single transformations which have been combined to give the resulting transformation.
(b) The curve \(y = \sin 2x – 5x\) is reflected in the \(y\)-axis and then stretched by scale factor \(\frac{1}{3}\) in the \(x\)-direction. Write down the equation of the transformed curve.
▶️Answer/Explanation
(a) Translation by 1 unit in the \(x\)-direction, followed by a stretch with scale factor 2 in the \(y\)-direction.
(b) \(y = -\sin 6x + 15x\) or \(y = \sin(-6x) + 15x\).
3. [Maximum mark: 4]
The equation of a curve is \(y = (x – 3)\sqrt{x + 1} + 3\). The following points lie on the curve. Non-exact values are rounded to 4 decimal places
\(A(2, k)\), \(B(2.9, 2.8025)\), \(C(2.99, 2.9800)\), \(D(2.999, 2.9980)\), and \(E(3, 3)\)
(a) Find \(k\), giving your answer correct to 4 decimal places.
(b) Find the gradient of \(AE\), giving your answer correct to 4 decimal places.
The gradients of BE, CE and DE, rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997
respectively.
(c) State what the gradients of \(BE\), \(CE\), and \(DE\) suggest about the gradient of the curve at point \(E\).
▶️Answer/Explanation
(a) \(k = 1.2679\).
(b) Gradient of \(AE = 1.7321\).
(c) The gradients approach 2, suggesting the gradient of the curve at \(E\) is 2.
4. [Maximum mark: 5]
The coefficient of \(x\) in the expansion of \((4x + \frac{10}{x})^3\) is \(p\). The coefficient of \(\frac{1}{x}\) in the expansion of \((2x + \frac{k}{x^2})^5\) is \(q\).
Given that \(p = 6q\), find the possible values of \(k\).
▶️Answer/Explanation
Solution:
– \(p = 480\) (coefficient of \(x\)).
– \(q = 80k^2\) (coefficient of \(\frac{1}{x}\)).
– Solving \(480 = 6 \times 80k^2\) gives \(k^2 = 1\), so \(k = \pm 1\).
5. [Maximum mark: 6]
The function \(f\) is defined by \(f(x) = 2x^2 + 3\) for \(x \geq 0\).
(a) Find and simplify an expression for \(ff(x)\).
(b) Solve the equation \(ff(x) = 34x^2 + 19\).
▶️Answer/Explanation
(a) \(ff(x) = 8x^4 + 24x^2 + 21\).
(b) Solving \(8x^4 – 10x^2 + 2 = 0\) gives \(x = \pm 1\) or \(x = \pm \frac{1}{2}\).
6. [Maximum mark: 4]
Points \(A(8, 3)\) and \(B(p, q)\) lie on a line. The equation of the perpendicular bisector of \(AB\) is \(y = -2x + 4\). Find the values of \(p\) and \(q\).
▶️Answer/Explanation
Solution:
– Gradient of \(AB = \frac{1}{2}\).
– Midpoint of \(AB\) is \((2, 0)\).
– Solving gives \(p = -4\) and \(q = -3\).
7. [Maximum mark: 5]
The point \(A\) has coordinates \((1, 5)\), and the line \(l\) has gradient \(-\frac{2}{3}\) and passes through \(A\). A circle has center \((5, 11)\) and radius \(\sqrt{52}\).
(a) Show that \(l\) is the tangent to the circle at \(A\).
(b) Find the equation of the other circle of radius \(\sqrt{52}\) for which \(l\) is also the tangent at \(A\).
▶️Answer/Explanation
(a) The distance from the center to \(A\) is \(\sqrt{52}\), and the gradient of \(l\) is perpendicular to the radius at \(A\).
Showing gradient of circle at A is = $-\frac{2}{3}$
(b) The other circle has center \((-3, -1)\) and equation \((x + 3)^2 + (y + 1)^2 = 52\).
8. [Maximum mark: 8]
The first, second, and third terms of an arithmetic progression are \(a\), \(\frac{3}{2}a\), and \(b\) respectively. The first, second, and third terms of a geometric progression are \(a\), 18, and \(b + 3\) respectively.
(a) Find the values of \(a\) and \(b\).
(b) Find the sum of the first 20 terms of the arithmetic progression.
▶️Answer/Explanation
(a) Solving gives \(a = 12\) and \(b = 24\).
(b) Common difference \(d = 6\), sum of first 20 terms is \(1380\).
9. [Maximum mark: 6]
The diagram shows part of the curve \(y^2 = x – 2\) and the lines \(x = 5\) and \(y = 1\). The shaded region enclosed by the curve and the lines is rotated through \(360^\circ\) about the \(x\)-axis. Find the volume obtained.
▶️Answer/Explanation
Solution:
$\text{Curve intersects } y = 1 \text{ at } (3, 1)$
$\text{Volume} = \int \pi (x-2)^2 dx$
$\left[ \pi \left( \frac{1}{2}x^2 – 2x \right) \right] \text{ or } \left[ \pi \frac{1}{2}(x-2)^2 \right]$
$= [\pi] \left[ \left( \frac{5^2}{2} – 2 \times 5 \right) – \left( \frac{\text{their } 3^2}{2} – 2 \times \text{their } 3 \right) \right]$
$= [\pi] \left[ \frac{5}{2} + \frac{3}{2} \right] \text{ as a minimum requirement for their values}$
– Volume = \(\pi \int_{3}^{5} (x – 2) \, dx – \pi \int_{3}^{5} 1 \, dx = 2\pi\) or \(6.28\).
10. [Maximum mark: 7]
(a) Prove the identity \(\frac{1 + \sin x}{1 – \sin x} – \frac{1 – \sin x}{1 + \sin x} \equiv \frac{4 \tan x}{\cos x}\).
(b) Hence solve the equation \(\frac{1 + \sin x}{1 – \sin x} – \frac{1 – \sin x}{1 + \sin x} = 8 \tan x\) for \(0 \leq x \leq \frac{1}{2}\pi\).
▶️Answer/Explanation
(a) Combine fractions and simplify to \(\frac{4 \sin x}{\cos^2 x} = \frac{4 \tan x}{\cos x}\).
(b) Solving gives \(x = \frac{\pi}{3}\) or \(x = 0\).
11. [Maximum mark: 10]
The gradient of a curve is given by \(\frac{dy}{dx} = 6(3x – 5)^3 – kx^2\), where \(k\) is a constant. The curve has a stationary point at \((2, -3.5)\).
(a) Find the value of \(k\).
(b) Find the equation of the curve.
(c) Find \(\frac{d^2y}{dx^2}\).
(d) Determine the nature of the stationary point at \((2, -3.5)\).
▶️Answer/Explanation
(a) Substituting \(x = 2\) gives \(k = \frac{3}{2}\).
(b) Integrating gives \(y = \frac{1}{2}(3x – 5)^4 – \frac{1}{2}x^3\).
(c) \(\frac{d^2y}{dx^2} = 54(3x – 5)^2 – 3x\).
(d) At \(x = 2\), \(\frac{d^2y}{dx^2} > 0\), so it is a minimum point.
12. [Maximum mark: 11]
The diagram shows a cross-section of seven cylindrical pipes, each of radius 20 cm, held together by a thin rope wrapped tightly around them. The centers of the six outer pipes are \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\). Points \(P\) and \(Q\) are where straight sections of the rope meet the pipe with center \(A\).
(a) Show that angle \(PAQ = \frac{1}{3}\pi\) radians.
(b) Find the length of the rope.
(c) Find the area of the hexagon \(ABCDEF\), giving your answer in terms of \(\sqrt{3}\).
(d) Find the area of the complete region enclosed by the rope.
▶️Answer/Explanation
(a) By symmetry, \(PAQ = \frac{2\pi}{6} = \frac{\pi}{3}\).
(b) Total length = \(6 \times (40 + \frac{20\pi}{3}) = 240 + 40\pi\) cm.
(c) Area of hexagon = \(2400\sqrt{3}\) cm².
(d) Total enclosed area = \(2400\sqrt{3} + 4800 + 400\pi\) cm².
$\text{Each straight section of rope has length } 40 \text{ cm}$
$\text{Each curved section round each pipe has length } r\theta = 20 \times \frac{\pi}{3}$
$\text{Total length} = 6 \times (\text{(their } 40) + k\pi)$
$240 + 40\pi \text{ or } 366 \text{ (AWRT) (cm)}$