1. [Maximum mark: 4]
(a) Express \( x^2 – 8x + 11 \) in the form \( (x + p)^2 + q \) where \( p \) and \( q \) are constants.
(b) Hence find the exact solutions of the equation \( x^2 – 8x + 11 = 1 \).
▶️Answer/Explanation
(a) \( x^2 – 8x + 11 = (x – 4)^2 – 5 \)
(b) \((x – 4)^2 – 5 = 1 \) leads to \((x – 4)^2 = 6 \).
Solutions: \( x = 4 \pm \sqrt{6} \).
2. [Maximum mark: 5]
The thirteenth term of an arithmetic progression is 12 and the sum of the first 30 terms is −15. Find the sum of the first 50 terms of the progression.
▶️Answer/Explanation
Using \( a + 12d = 12 \) and \( \frac{30}{2}(2a + 29d) = -15 \), solve for \( a \) and \( d \):
\( a = 72 \), \( d = -5 \).
Sum of first 50 terms: \( S_{50} = \frac{50}{2}(2 \times 72 + 49 \times -5) = -2525 \).
3. [Maximum mark: 6]
(a) The coefficient of \( x^4 \) in the expansion of \( \left(2x^2 + \frac{k^2}{x}\right)^5 \) is \( a \). The coefficient of \( x^2 \) in the expansion of \( (2kx – 1)^4 \) is \( b \). Find \( a \) and \( b \) in terms of the constant \( k \).
(b) Given that \( a + b = 216 \), find the possible values of \( k \).
▶️Answer/Explanation
(a) \( a = 80k^4 \), \( b = 24k^2 \).
(b) Solve \( 80k^4 + 24k^2 – 216 = 0 \):
\( k = \pm \sqrt{\frac{3}{2}} \) (exact values).
4. [Maximum mark: 6]
(a) Prove the identity:
\[ \frac{\sin^3 \theta}{\sin \theta – 1} – \frac{\sin^2 \theta}{1 + \sin \theta} \equiv -\tan^2 \theta (1 + \sin^2 \theta). \]
(b) Hence solve the equation:
\[ \frac{\sin^3 \theta}{\sin \theta – 1} – \frac{\sin^2 \theta}{1 + \sin \theta} = \tan^2 \theta (1 – \sin^2 \theta) \] for \( 0 < \theta < 2\pi \).
▶️Answer/Explanation
(a) Combine fractions and simplify using \( \sin^2 \theta + \cos^2 \theta = 1 \) to reach the given identity.
(b) Solutions: \( \theta = \pi \) (only valid solution in the interval).
5. [Maximum mark: 7]
The diagram shows a sector \( ABC \) of a circle with centre \( A \) and radius \( r \). The line \( BD \) is perpendicular to \( AC \). Angle \( CAB \) is \( \theta \) radians.
(a) Given that \( \theta = \frac{1}{6}\pi \), find the exact area of \( BCD \) in terms of \( r \).
(b) Given instead that the length of \( BD \) is \( \frac{\sqrt{3}}{2} r \), find the exact perimeter of \( BCD \) in terms of \( r \).
▶️Answer/Explanation\\
(a) Area of \( BCD = \frac{\pi}{12} r^2 – \frac{\sqrt{3}}{8} r^2 \).
(b) Perimeter of \( BCD = \frac{\sqrt{3}}{2} r + \frac{1}{2} r + \frac{\pi}{3} r \).
6. [Maximum mark: 8]
The function \( f \) is defined as follows:
\[ f(x) = \frac{x^2 – 4}{x^2 + 4} \text{ for } x > 2. \]
(a) Find an expression for \( f^{-1}(x) \).
(b) Show that \( 1 – \frac{8}{x^2 + 4} \) can be expressed as \( \frac{x^2 – 4}{x^2 + 4} \) and hence state the range of \( f \).
(c) Explain why the composite function \( ff \) cannot be formed.
▶️Answer/Explanation
(a) \( f^{-1}(x) = \sqrt{\frac{4x + 4}{1 – x}} \).
(b) Range of \( f \): \( 0 < f(x) < 1 \).
(c) The range of \( f \) does not include the entire domain of \( f \).
7. [Maximum mark: 9]
The diagram shows the curve with equation \( y = (3x – 2)^{\frac{1}{2}} \) and the line \( y = \frac{1}{2}x + 1 \). The curve and the line intersect at points \( A \) and \( B \).
(a) Find the coordinates of \( A \) and \( B \).
(b) Hence find the area of the region enclosed between the curve and the line.
▶️Answer/Explanation
(a) Coordinates: \( A(2, 2) \) and \( B(6, 4) \).
(b) Area: \( \frac{4}{9} \) (exact value).
8. [Maximum mark: 8]
(a) The curve \( y = \sin x \) is transformed to the curve \( y = 4 \sin\left(\frac{1}{2}x – 30^\circ\right) \). Describe fully a sequence of transformations that have been combined, making clear the order in which the transformations are applied. [5]
(b) Find the exact solutions of the equation \( 4 \sin\left(\frac{1}{2}x – 30^\circ\right) = 2\sqrt{2} \) for \( 0^\circ \leq x \leq 360^\circ \). [3]
▶️Answer/Explanation
(a) Sequence:
1. Translation \( \begin{pmatrix} 60^\circ \\ 0 \end{pmatrix} \).
2. Stretch factor 2 in \( x \)-direction.
3. Stretch factor 4 in \( y \)-direction.
(b) Solutions: \( x = 150^\circ, 330^\circ \).
9. [Maximum mark: 10]
The equation of a circle is \( x^2 + y^2 + 6x – 2y – 26 = 0 \).
(a) Find the coordinates of the centre of the circle and the radius. Hence find the coordinates of the lowest point on the circle. [4]
(b) Find the set of values of the constant \( k \) for which the line with equation \( y = kx – 5 \) intersects the circle at two distinct points. [6]
▶️Answer/Explanation
(a) Centre: \( (-3, 1) \), Radius: 6. Lowest point: \( (-3, -5) \).
(b) Values of \( k \): \( k < 0 \) or \( k > \frac{4}{3} \).
10. [Maximum mark: 12]
The equation of a curve is such that \( \frac{d^2y}{dx^2} = 6x^2 – \frac{4}{x^3} \). The curve has a stationary point at \( \left(-1, \frac{9}{2}\right) \).
(a) Determine the nature of the stationary point at \( \left(-1, \frac{9}{2}\right) \).
(b) Find the equation of the curve.
(c) Show that the curve has no other stationary points.
(d) A point \( A \) is moving along the curve and the \( y \)-coordinate of \( A \) is increasing at a rate of 5 units per second. Find the rate of increase of the \( x \)-coordinate of \( A \) at the point where \( x = 1 \).
▶️Answer/Explanation
(a) Minimum point (since \( \frac{d^2y}{dx^2} > 0 \)).
(b) Equation: \( y = \frac{1}{2}x^4 – \frac{2}{x} + 2 \).
(c) Only stationary point is at \( x = -1 \).
(d) Rate of increase: \( \frac{5}{4} \) units per second.