1. [Maximum mark: 3]
A curve with equation \( y = f(x) \) is such that \( f'(x) = 6x^2 – \frac{8}{x^2} \). It is given that the curve passes through the point \( (2, 7) \).
Find \( f(x) \).
▶️Answer/Explanation
Solution:
Integrate \( f'(x) \):
\( f(x) = \int (6x^2 – \frac{8}{x^2}) \, dx = 2x^3 + \frac{8}{x} + c \)
Substitute \( (2, 7) \) to find \( c \):
\( 7 = 2(2)^3 + \frac{8}{2} + c \)
\( 7 = 16 + 4 + c \)
\( c = -13 \)
Thus, \( f(x) = 2x^3 + \frac{8}{x} – 13 \).
2. [Maximum mark: 4]
The function \( f \) is defined by \( f(x) = \frac{1}{3}(2x – 1)^{\frac{3}{2}} – 2x \) for \( \frac{1}{2} < x < a \). It is given that \( f \) is a decreasing function. Find the maximum possible value of the constant \( a \).
▶️Answer/Explanation
Solution:
Find \( f'(x) \):
\( f'(x) = (2x – 1)^{\frac{1}{2}} – 2 \)
For \( f \) to be decreasing, \( f'(x) \leq 0 \):
\( (2x – 1)^{\frac{1}{2}} – 2 \leq 0 \)
\( (2x – 1)^{\frac{1}{2}} \leq 2 \)
Square both sides:
\( 2x – 1 \leq 4 \)
\( x \leq \frac{5}{2} \)
Thus, the maximum value of \( a \) is \( \frac{5}{2} \).
3. [Maximum mark: 6]
A line with equation \( y = mx – 6 \) is a tangent to the curve with equation \( y = x^2 – 4x + 3 \). Find the possible values of the constant \( m \), and the corresponding coordinates of the points at which the line touches the curve.
▶️Answer/Explanation
Solution:
Set the equations equal for tangency condition:
\( x^2 – 4x + 3 = mx – 6 \)
Rearrange:
\( x^2 – (4 + m)x + 9 = 0 \)
For tangency, discriminant \( D = 0 \):
\( (4 + m)^2 – 36 = 0 \)
Solve for \( m \):
\( 4 + m = \pm 6 \)
\( m = 2 \) or \( m = -10 \)
Find corresponding \( x \)-coordinates:
For \( m = 2 \): \( x = 3 \), \( y = 0 \) → Point \( (3, 0) \)
For \( m = -10 \): \( x = -3 \), \( y = 24 \) → Point \( (-3, 24) \).
4. [Maximum mark: 6]
(a) Show that the equation \( \frac{\tan x + \sin x}{\tan x – \sin x} = k \) may be expressed as \( \frac{1 + \cos x}{1 – \cos x} = k \).
(b) Hence express \( \cos x \) in terms of \( k \).
(c) Hence solve the equation \( \frac{\tan x + \sin x}{\tan x – \sin x} = 4 \) for \( -\pi < x < \pi \).
▶️Answer/Explanation
(a) Solution:
Multiply numerator and denominator by \( \cos x \):
\( \frac{\sin x + \sin x \cos x}{\sin x – \sin x \cos x} = \frac{1 + \cos x}{1 – \cos x} = k \).
(b) Solution:
Rearrange \( \frac{1 + \cos x}{1 – \cos x} = k \):
\( 1 + \cos x = k – k \cos x \)
\( \cos x (1 + k) = k – 1 \)
\( \cos x = \frac{k – 1}{k + 1} \).
(c) Solution:
Substitute \( k = 4 \):
\( \cos x = \frac{3}{5} \)
Solutions in \( -\pi < x < \pi \):
\( x = \pm 0.927 \) radians.
5. [Maximum mark: 6]
The diagram shows a triangle \( ABC \), in which angle \( ABC = 90^\circ \) and \( AB = 4 \, \text{cm} \). The sector \( ABD \) is part of a circle with centre \( A \).
The area of the sector is \( 10 \, \text{cm}^2 \).
(a) Find angle \( BAD \) in radians.
(b) Find the perimeter of the shaded region.
▶️Answer/Explanation
(a) Solution:
Use sector area formula:
\( \frac{1}{2} \times 4^2 \times \theta = 10 \)
\( \theta = 1.25 \) radians.
(b) Solution:
Arc length \( BD = 4 \times 1.25 = 5 \, \text{cm} \).
Find \( BC \):
\( BC = 4 \tan(1.25) \approx 12.04 \, \text{cm} \).
Find \( CD \):
\( CD = \frac{4}{\cos(1.25)} – 4 \approx 8.69 \, \text{cm} \).
Perimeter: \( 5 + 12.04 + 8.69 \approx 25.7 \, \text{cm} \).
6. [Maximum mark: 6]
Functions \( f \) and \( g \) are both defined for \( x \in \mathbb{R} \) and are given by
\( f(x) = x^2 – 2x + 5 \),
\( g(x) = x^2 + 4x + 13 \).
(a) By first expressing each of \( f(x) \) and \( g(x) \) in completed square form, express \( g(x) \) in the form \( f(x + p) + q \), where \( p \) and \( q \) are constants.
(b) Describe fully the transformation which transforms the graph of \( y = f(x) \) to the graph of \( y = g(x) \)
▶️Answer/Explanation
(a) Solution:
Complete the square for \( f(x) \):
\( f(x) = (x – 1)^2 + 4 \).
Complete the square for \( g(x) \):
\( g(x) = (x + 2)^2 + 9 \).
Express \( g(x) \) in terms of \( f(x + p) + q \):
\( g(x) = f(x + 3) + 5 \).
(b) Solution:
The transformation is a translation by \( \begin{pmatrix} -3 \\ 5 \end{pmatrix} \).
7. [Maximum mark: 7]
(a) Write down the first four terms of the expansion, in ascending powers of \( x \), of \( (a – x)^6 \).
(b) Given that the coefficient of \( x^2 \) in the expansion of \( \left(1 + \frac{2}{a}x\right)(a – x)^6 \) is \( -20 \), find in exact form the possible values of the constant \( a \).
▶️Answer/Explanation
(a) Solution:
First four terms:
\( a^6 – 6a^5x + 15a^4x^2 – 20a^3x^3 \).
(b) Solution:
Multiply \( \left(1 + \frac{2}{a}x\right) \) by the expansion of \( (a – x)^6 \):
Coefficient of \( x^2 \): \( 15a^4 – 40a^2 = -20 \).
Solve:
\( 15a^4 – 40a^2 + 20 = 0 \).
Factorize:
\( (5a^2 – 10)(3a^2 – 2) = 0 \).
Solutions: \( a = \pm \sqrt{2}, \pm \sqrt{\frac{2}{3}} \).
8. [Maximum mark: 7]
Functions \( f \) and \( g \) are defined as follows:
\( f: x \mapsto x^2 – 1 \) for \( x < 0 \),
\( g: x \mapsto \frac{1}{2x + 1} \) for \( x < -\frac{1}{2} \).
(a) Solve the equation \( fg(x) = 3 \).
(b) Find an expression for \( (fg)^{-1}(x) \).
▶️Answer/Explanation
(a) Solution:
Compute \( fg(x) \):
\( fg(x) = \left(\frac{1}{2x + 1}\right)^2 – 1 = 3 \).
Solve:
\( \left(\frac{1}{2x + 1}\right)^2 = 4 \).
\( \frac{1}{2x + 1} = \pm 2 \).
Only valid solution: \( x = -\frac{3}{4} \).
(b) Solution:
Let \( y = \left(\frac{1}{2x + 1}\right)^2 – 1 \).
Rearrange for \( x \):
\( x = -\frac{1}{2\sqrt{y + 1}} – \frac{1}{2} \).
Thus, \( (fg)^{-1}(x) = -\frac{1}{2\sqrt{x + 1}} – \frac{1}{2} \).
9. [Maximum mark: 9]
(a) A geometric progression is such that the second term is equal to 24% of the sum to infinity. Find the possible values of the common ratio.
(b) An arithmetic progression \( P \) has first term \( a \) and common difference \( d \). An arithmetic progression \( Q \) has first term \( 2(a + 1) \) and common difference \( (d + 1) \). It is given that:
$\frac{\text{5th term of } P}{\text{12th term of } Q} = \frac{1}{3} \quad \text{and}$ $\quad \frac{\text{Sum of first 5 terms of } P}{\text{Sum of first 5 terms of } Q} = \frac{2}{3}.$
Find the value of \( a \) and the value of \( d \).
▶️Answer/Explanation
(a) Solution:
Given \( ar = 0.24 \times \frac{a}{1 – r} \).
Simplify:
\( r(1 – r) = 0.24 \).
Solve:
\( r = 0.6 \) or \( r = 0.4 \).
(b) Solution:
For the 5th term ratio:
\( \frac{a + 4d}{2(a + 1) + 11(d + 1)} = \frac{1}{3} \).
Simplify to \( a + d = 13 \).
For the sum ratio:
\( \frac{\frac{5}{2}(2a + 4d)}{\frac{5}{2}(4(a + 1) + 4(d + 1))} = \frac{2}{3} \).
Simplify to \( -a + 2d = 8 \).
Solve system:
\( a = 6 \), \( d = 7 \).
10. [Maximum mark: 10]
Points \( A(-2, 3) \), \( B(3, 0) \), and \( C(6, 5) \) lie on the circumference of a circle with centre \( D \).
(a) Show that angle \( ABC = 90^\circ \).
(b) Hence state the coordinates of \( D \).
(c) Find an equation of the circle.
(d) The point \( E \) lies on the circumference of the circle such that \( BE \) is a diameter. Find an equation of the tangent to the circle at \( E \).
▶️Answer/Explanation
(a) Solution:
Find gradients:
\( m_{AB} = \frac{0 – 3}{3 – (-2)} = -\frac{3}{5} \).
\( m_{BC} = \frac{5 – 0}{6 – 3} = \frac{5}{3} \).
Product: \( m_{AB} \times m_{BC} = -1 \), so \( ABC = 90^\circ \).
(b) Solution:
Centre \( D \) is midpoint of \( AC \): \( (2, 4) \).
(c) Solution:
Radius \( AD = \sqrt{(2 + 2)^2 + (4 – 3)^2} = \sqrt{17} \).
Equation: \( (x – 2)^2 + (y – 4)^2 = 17 \).
(d) Solution:
Find \( E \):
\( E \) is reflection of \( B \) over \( D \): \( (1, 8) \).
Gradient of tangent at \( E \):
\( m = \frac{1}{4} \) (negative reciprocal of \( BE \) gradient).
Equation: \( y – 8 = \frac{1}{4}(x – 1) \).
11. [Maximum mark: 11]
The diagram shows part of the curve with equation \( y = x^{\frac{1}{2}} + k^2x^{-\frac{1}{2}} \), where \( k \) is a positive constant.
(a) Find the coordinates of the minimum point of the curve, giving your answer in terms of \( k \).
(b) The tangent at the point on the curve where \( x = 4k^2 \) intersects the \( y \)-axis at \( P \). Find the \( y \)-coordinate of \( P \) in terms of \( k \).
(c) The shaded region is bounded by the curve, the \( x \)-axis, and the lines \( x = \frac{9}{4}k^2 \) and \( x = 4k^2 \). Find the area of the shaded region in terms of \( k \).
▶️Answer/Explanation
(a) Solution:
Find derivative:
\( \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} – \frac{1}{2}k^2x^{-\frac{3}{2}} \).
Set to zero for minimum:
\( \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2}k^2x^{-\frac{3}{2}} \).
Solve: \( x = k^2 \).
Coordinates: \( (k^2, 2k) \).
(b) Solution:
At \( x = 4k^2 \):
\( y = 2k + \frac{k^2}{2k} = \frac{5k}{2} \).
Gradient: \( \frac{3}{16k} \).
Equation of tangent:
\( y – \frac{5k}{2} = \frac{3}{16k}(x – 4k^2) \).
At \( x = 0 \), \( y = \frac{7k}{4} \).
(c) Solution:
Integrate \( y \):
\( \int \left(x^{\frac{1}{2}} + k^2x^{-\frac{1}{2}}\right) dx = \frac{2}{3}x^{\frac{3}{2}} + 2k^2x^{\frac{1}{2}} \).
Evaluate from \( \frac{9}{4}k^2 \) to \( 4k^2 \):
Area: \( \frac{49k^3}{12} \).