1. [Maximum mark: 6]
Points A(5, 2) and B(10, -1) are given.
(a) Find the equation of the perpendicular bisector of AB.
(b) Find the equation of the circle with centre A passing through B.
▶️Answer/Explanation
(a)
1. Midpoint of AB: \[ \left( \frac{5+10}{2}, \frac{2+(-1)}{2} \right) = \left( 7.5, 0.5 \right) \]
2. Gradient of AB: \[ \frac{-1-2}{10-5} = -\frac{3}{5} \quad \Rightarrow \quad \text{Perpendicular gradient} = \frac{5}{3} \]
3. Equation: \[ y – 0.5 = \frac{5}{3}(x – 7.5) \quad \Rightarrow \quad y = \frac{5}{3}x – 12 \] (b) Radius = ( 10 − 5 ) 2 + ( − 1 − 2 ) 2 = 34 = (10−5) 2 +(−1−2) 2 = 34 Equation of circle: ( x − 5 ) 2 + ( y − 2 ) 2 = 34 (x−5) 2 +(y−2) 2 =34
2. [Maximum mark: 5]
The first three terms of an arithmetic progression are \(a\), \(2a\), and \(a^2\). Find the sum of the first 50 terms.
▶️Answer/Explanation
1. Common difference \(d = 2a – a = a\).
2. Third term: \[ a^2 = 2a + a \quad \Rightarrow \quad a^2 – 3a = 0 \quad \Rightarrow \quad a = 3 \]
3. Sum of 50 terms: \[ S_{50} = \frac{50}{2} \left[ 2(3) + 49(3) \right] = 3825 \]
3. [Maximum mark: 5]
(a) Find the set of values of \(k\) for which \(8x^2 + kx + 2 = 0\) has no real roots.
(b) Solve \(8\cos^2\theta – 10\cos\theta + 2 = 0\) for \(0^\circ \leq \theta \leq 180^\circ\).
▶️Answer/Explanation
(a)
1. Discriminant \(< 0\): \[ k^2 – 4(8)(2) < 0 \quad \Rightarrow \quad -8 < k < 8 \] (b) Factorize: ( 4 cos θ − 1 ) ( 2 cos θ − 2 ) = 0 ⇒ cos θ = 1 4 or 1 (4cosθ−1)(2cosθ−2)=0⇒cosθ= 4 1 or 1
4. [Maximum mark: 4]
A geometric progression has third term 1764 and sum of second and third terms 3444. Find the 50th term.
▶️Answer/Explanation
1. Let \(ar^2 = 1764\) and \(ar + ar^2 = 3444\).
2. Solve for \(r\): \[ r = \frac{21}{20} = 1.05, \quad a = 1600 \]
3. 50th term: \[ T_{50} = 1600(1.05)^{49} \approx 17,474 \]
5. [Maximum mark: 5]
A graph \(y = f(x)\) is transformed by a stretch in the x-direction (factor 0.5) followed by a translation \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\).
(a) Sketch the transformed graph.
(b) Express \(g(x)\) in terms of \(f(x)\).
▶️Answer/Explanation
(a)
– Horizontal compression by factor 2, then vertical shift up 1 unit. (b) g ( x ) = f ( 2 x ) + 1 g(x)=f(2x)+1
6. [Maximum mark: 9]
The curve equation is \(y = 4x^2 + 20x + 6\).
(a) Express in completed square form.
(b) Solve \(4x^2 + 20x + 6 = 45\).
(c) Sketch the curve, showing the stationary point.
▶️Answer/Explanation
(a)
\[ y = 4\left(x + \frac{5}{2}\right)^2 – 19 \] (b) Solve: 4 ( x + 5 2 ) 2 = 64 ⇒ x = 3 2 , 13 2 4(x+ 2 5 ) 2 =64⇒x= 2 3 , 2 13 (c) U-shaped parabola with vertex at ( − 5 2 , − 19 ) (− 2 5 ,−19).
7. [Maximum mark: 7]
(a) Prove the identity: \[ \frac{\sin\theta}{\sin\theta + \cos\theta} + \frac{\cos\theta}{\sin\theta – \cos\theta} = \frac{\tan^2\theta + 1}{\tan^2\theta – 1} \]
(b) Solve \(\frac{\sin\theta}{\sin\theta + \cos\theta} + \frac{\cos\theta}{\sin\theta – \cos\theta} = 2\) for \(0 \leq \theta \leq \pi\).
▶️Answer/Explanation
(a)
1. Combine fractions: \[ \frac{\sin^2\theta – \cos^2\theta}{\sin^2\theta – \cos^2\theta} = \frac{\tan^2\theta + 1}{\tan^2\theta – 1} \] (b) Solve: tan θ = ± 3 ⇒ θ = π 3 , 2 π 3 tanθ=± 3 ⇒θ= 3 π , 3 2π
8. [Maximum mark: 7]
A curve has \(\frac{dy}{dx} = 3\sqrt{x} – 3x^{-\frac{1}{2}}\) and passes through (3, 5).
(a) Find the equation of the curve.
(b) Find the x-coordinate of the stationary point.
(c) State where \(y\) increases as \(x\) increases.
▶️Answer/Explanation
(a)
1. Integrate: \[ y = 2x^{\frac{3}{2}} – 6x^{\frac{1}{2}} + 5 \] (b) Solve d y d x = 0 dx dy =0: x = 1 x=1 (c) x > 1 x>1
9. [Maximum mark: 8]
Functions \(f(x) = x + \frac{1}{x}\) and \(g(x) = ax + 1\).
(a) Find \(gf(x)\).
(b) Given \(gf(2) = 11\), find \(a\).
(c) Explain if \(f\) has an inverse.
▶️Answer/Explanation
(a)
\[ gf(x) = a\left(x + \frac{1}{x}\right) + 1 \] (b) Substitute x = 2 x=2: a ( 2.5 ) + 1 = 11 ⇒ a = 4 a(2.5)+1=11⇒a=4 (c) No, because f f is not one-to-one.
10. [Maximum mark: 8]
Cross-section RASB involves sectors of circles.
(a) Find the perimeter of RASB.
(b) Find the difference in area of triangles AOB and APB.
▶️Answer/Explanation
(a)
1. Perimeter: \[ 2.5 \times \frac{4\pi}{3} + 2.24 \times \frac{5\pi}{6} \approx 16.34 \, \text{m} \] (b) Difference: 1 2 ( 2. 5 2 ) sin 2 π 3 − 1 2 ( 2.2 4 2 ) sin 5 π 6 ≈ 1.45 m 2 2 1 (2.5 2 )sin 3 2π − 2 1 (2.24 2 )sin 6 5π ≈1.45m 2
11. [Maximum mark: 11]
(a) Find the minimum point of \(y = \frac{9}{4}x^2 – 12x + 18\).
(b) Find the area between \(y = \frac{9}{4}x^2 – 12x + 18\) and \(y = 18 – \frac{3}{8}x^{\frac{5}{2}}\).
▶️Answer/Explanation
(a)
1. Minimum at: \[ x = \frac{8}{3}, \quad y = 2 \] (b) Integral: ∫ 0 4 ( 18 − 3 8 x 5 2 − ( 9 4 x 2 − 12 x + 18 ) ) d x = 240 7 ≈ 34.3 ∫ 0 4 (18− 8 3 x 2 5 −( 4 9 x 2 −12x+18))dx= 7 240 ≈34.3