1. [Maximum mark: 5]
Consider the functions \( f(x) = x – 3 \) and \( g(x) = x^2 + k^2 \), where \( k \) is a real constant.
(a) Write down an expression for \( (g \circ f)(x) \). [2]
(b) Given that \( (g \circ f)(2) = 10 \), find the possible values of \( k \). [3]
▶️Answer/Explanation
(a) \( (g \circ f)(x) = (x – 3)^2 + k^2 \)
(b) Substituting \( x = 2 \) into \( (g \circ f)(x) \):
\( (2 – 3)^2 + k^2 = 10 \)
\( 1 + k^2 = 10 \)
\( k^2 = 9 \)
\( k = \pm 3 \)
2. [Maximum mark: 4]
Events A and B are such that \( P(A) = 0.65 \), \( P(B) = 0.75 \), and \( P(A \cap B) = 0.6 \).
(a) Find \( P(A \cup B) \). [2]
(b) Hence, or otherwise, find \( P(A’ \cap B’) \). [2]
▶️Answer/Explanation
(a) \( P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.65 + 0.75 – 0.6 = 0.8 \)
(b) \( P(A’ \cap B’) = 1 – P(A \cup B) = 1 – 0.8 = 0.2 \)
3. [Maximum mark: 7]
The sum of the first \( n \) terms of an arithmetic sequence is given by \( S_n = pn^2 – qn \), where \( p \) and \( q \) are positive constants. It is given that \( S_4 = 40 \) and \( S_5 = 65 \).
(a) Find the value of \( p \) and the value of \( q \). [5]
(b) Find the value of \( u_5 \). [2]
▶️Answer/Explanation
(a) Using \( S_4 = 40 \) and \( S_5 = 65 \):
\( 40 = 16p – 4q \)
\( 65 = 25p – 5q \)
Solving the equations, \( p = 3 \) and \( q = 2 \).
(b) \( u_5 = S_5 – S_4 = 65 – 40 = 25 \)
4. [Maximum mark: 6]
In the following triangle ABC, \( AB = 6 \, \text{cm} \), \( AC = 10 \, \text{cm} \), and \( \cos BAC = \frac{1}{5} \). Find the area of triangle ABC.
▶️Answer/Explanation
Using the formula for the area of a triangle:
\( \text{Area} = \frac{1}{2} \times AB \times AC \times \sin BAC \)
First, find \( \sin BAC \):
\( \sin BAC = \sqrt{1 – \cos^2 BAC} = \sqrt{1 – \left(\frac{1}{5}\right)^2} = \frac{\sqrt{24}}{5} \)
Then, \( \text{Area} = \frac{1}{2} \times 6 \times 10 \times \frac{\sqrt{24}}{5} = 12 \, \text{cm}^2 \)
5. [Maximum mark: 6]
The binomial expansion of \( (1 + kx)^n \) is given by \( 1 + 12x + 28k^2x^2 + \ldots + k^nx^n \), where \( n \in \mathbb{Z}^+ \) and \( k \in \mathbb{R} \). Find the value of \( n \) and the value of \( k \).
▶️Answer/Explanation
Using the binomial expansion:
\( (1 + kx)^n = 1 + nkx + \frac{n(n-1)}{2}k^2x^2 + \ldots \)
Comparing coefficients:
\( nk = 12 \) and \( \frac{n(n-1)}{2}k^2 = 28 \)
Solving, \( n = 8 \) and \( k = \frac{3}{2} \)
6. [Maximum mark: 7]
Prove by mathematical induction that \( 5^{2n} – 2^{3n} \) is divisible by 17 for all \( n \in \mathbb{Z}^+ \).
▶️Answer/Explanation
Base case: For \( n = 1 \), \( 5^2 – 2^3 = 25 – 8 = 17 \), which is divisible by 17.
Inductive step: Assume \( 5^{2k} – 2^{3k} \) is divisible by 17.
For \( n = k + 1 \):
\( 5^{2(k+1)} – 2^{3(k+1)} = 25 \cdot 5^{2k} – 8 \cdot 2^{3k} \)
\( = 25(5^{2k} – 2^{3k}) + 17 \cdot 2^{3k} \), which is divisible by 17.
Hence, by induction, \( 5^{2n} – 2^{3n} \) is divisible by 17 for all \( n \in \mathbb{Z}^+ \).
7. [Maximum mark: 5]
It is given that \( z = 5 + qi \) satisfies the equation \( z^2 + iz = -p + 25i \), where \( p, q \in \mathbb{R} \). Find the value of \( p \) and the value of \( q \).
▶️Answer/Explanation
Substituting \( z = 5 + qi \) into the equation:
\( (5 + qi)^2 + i(5 + qi) = -p + 25i \)
Expanding and equating real and imaginary parts:
\( 25 – q^2 + p – q = 0 \) and \( 10q – 20 = 0 \)
Solving, \( q = 2 \) and \( p = -19 \)
8. [Maximum mark: 9]
(a) Find \( \int x(\ln x)^2 dx \). [6]
(b) Hence, show that \( \int_1^4 x(\ln x)^2 dx = 32 (\ln 2)^2 – 16 \ln 2 + \frac{15}{4} \). [3]
▶️Answer/Explanation
(a) Using integration by parts:
\( \int x(\ln x)^2 dx = \frac{x^2 (\ln x)^2}{2} – \int x \ln x dx \)
Then, \( \int x \ln x dx = \frac{x^2 \ln x}{2} – \frac{x^2}{4} \)
Combining, \( \int x(\ln x)^2 dx = \frac{x^2 (\ln x)^2}{2} – \frac{x^2 \ln x}{2} + \frac{x^2}{4} + C \)
(b) Substituting limits \( x = 1 \) to \( x = 4 \):
\( \int_1^4 x(\ln x)^2 dx = 32 (\ln 2)^2 – 16 \ln 2 + \frac{15}{4} \)
9. [Maximum mark: 8]
Consider the function \( f(x) = \frac{\sin^2(kx)}{x^2} \), where \( x \neq 0 \) and \( k \in \mathbb{R}^+ \).
(a) Show that \( f \) is an even function. [2]
(b) Given that \( \lim_{x \to 0} f(x) = 16 \), find the value of \( k \). [6]
▶️Answer/Explanation
(a) \( f(-x) = \frac{\sin^2(-kx)}{(-x)^2} = \frac{\sin^2(kx)}{x^2} = f(x) \), so \( f \) is even.
(b) Using L’Hôpital’s rule:
\( \lim_{x \to 0} \frac{\sin^2(kx)}{x^2} = k^2 \)
Given \( k^2 = 16 \), so \( k = 4 \)
10. [Maximum mark: 15]
The functions \( f \) and \( g \) are defined by \( f(x) = \ln(2x – 9) \), where \( x > \frac{9}{2} \), and \( g(x) = 2\ln x – \ln d \), where \( x > 0 \), \( d \in \mathbb{R}^+ \).
(a) State the equation of the vertical asymptote to the graph of \( y = g(x) \). [1]
(b) (i) Show that, at the points of intersection, \( x^2 – 2dx + 9d = 0 \). [3]
(ii) Hence show that \( d^2 – 9d > 0 \). [2]
(iii) Find the range of possible values of \( d \). [3]
(c) In the case where \( d = 10 \), find the value of \( q – p \), where \( p \) and \( q \) are the points of intersection. [6]
▶️Answer/Explanation
(a) The vertical asymptote is \( x = 0 \).
(b)(i) Setting \( f(x) = g(x) \):
\( \ln(2x – 9) = 2\ln x – \ln d \)
Simplifying, \( x^2 – 2dx + 9d = 0 \).
(b)(ii) For two distinct solutions, the discriminant \( d^2 – 9d > 0 \).
(b)(iii) Solving \( d^2 – 9d > 0 \), \( d > 9 \).
(c) For \( d = 10 \), solving \( x^2 – 20x + 90 = 0 \):
\( x = 10 \pm \sqrt{10} \), so \( q – p = 2\sqrt{10} \)
11. [Maximum mark: 21]
Consider the function \( f(x) = e^{\cos 2x} \), where \( -\frac{\pi}{4} \leq x \leq \frac{5\pi}{4} \).
(a) Find the coordinates of the points on the curve \( y = f(x) \) where the gradient is zero. [5]
(b) Using the second derivative at each point found in part (a), show that the curve \( y = f(x) \) has two local maximum points and one local minimum point. [4]
(c) Sketch the curve of \( y = f(x) \) for \( 0 \leq x \leq \pi \), taking into consideration the relative values of the second derivative found in part (b). [3]
(d) (i) Find the Maclaurin series for \( \cos 2x \), up to and including the term in \( x^4 \). [2]
(ii) Hence, find the Maclaurin series for \( e^{\cos 2x – 1} \), up to and including the term in \( x^4 \). [2]
(iii) Hence, write down the Maclaurin series for \( f(x) \), up to and including the term in \( x^4 \). [2]
(e) Use the first two non-zero terms in the Maclaurin series for \( f(x) \) to show that \( \int_{0}^{1/10} e^{\cos 2x} dx \approx \frac{149e}{1500} \). [3]
▶️Answer/Explanation
(a) The gradient is zero when \( f'(x) = -2\sin 2x e^{\cos 2x} = 0 \), so \( x = 0, \frac{\pi}{2}, \pi \).
Coordinates: \( (0, e) \), \( \left(\frac{\pi}{2}, \frac{1}{e}\right) \), \( (\pi, e) \).
(b) Using the second derivative \( f”(x) \):
At \( x = 0 \) and \( x = \pi \), \( f”(x) < 0 \) (local maxima).
At \( x = \frac{\pi}{2} \), \( f”(x) > 0 \) (local minimum).
(c) Sketch the curve with maxima at \( (0, e) \) and \( (\pi, e) \), and minimum at \( \left(\frac{\pi}{2}, \frac{1}{e}\right) \).
(d)(i) \( \cos 2x = 1 – 2x^2 + \frac{2x^4}{3} + \ldots \)
(d)(ii) \( e^{\cos 2x – 1} = 1 – 2x^2 + \frac{8x^4}{3} + \ldots \)
(d)(iii) \( f(x) = e \left(1 – 2x^2 + \frac{8x^4}{3} + \ldots \right) \)
(e) Using the first two terms:
\( \int_{0}^{1/10} e^{\cos 2x} dx \approx \int_{0}^{1/10} e(1 – 2x^2) dx = \frac{149e}{1500} \)
12. [Maximum mark: 17]
(a) Find the binomial expansion of \( (\cos \theta + i \sin \theta)^5 \). Give your answer in the form \( a + bi \), where \( a \) and \( b \) are expressed in terms of \( \sin \theta \) and \( \cos \theta \). [4]
(b) By using De Moivre’s theorem and your answer to part (a), show that \( \sin 5\theta = 16\sin^5 \theta – 20\sin^3 \theta + 5\sin \theta \). [6]
(c) (i) Hence, show that \( \theta = \frac{\pi}{5} \) and \( \theta = \frac{3\pi}{5} \) are solutions of the equation \( 16\sin^4 \theta – 20\sin^2 \theta + 5 = 0 \). [4]
(ii) Hence, show that \( \sin \frac{\pi}{5} \sin \frac{3\pi}{5} = \frac{\sqrt{5}}{4} \). [3]
▶️Answer/Explanation
(a) Using the binomial expansion:
\( (\cos \theta + i \sin \theta)^5 = \cos^5 \theta + 5i \cos^4 \theta \sin \theta – 10 \cos^3 \theta \sin^2 \theta – 10i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta \)
Grouping real and imaginary parts:
Real part: \( \cos^5 \theta – 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta \)
Imaginary part: \( 5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta \)
Thus, \( (\cos \theta + i \sin \theta)^5 = (\cos^5 \theta – 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta) + i (5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta) \).
(b) By De Moivre’s theorem:
\( (\cos \theta + i \sin \theta)^5 = \cos 5\theta + i \sin 5\theta \)
Equating the imaginary parts:
\( \sin 5\theta = 5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta \)
Substitute \( \cos^2 \theta = 1 – \sin^2 \theta \):
\( \sin 5\theta = 5(1 – \sin^2 \theta)^2 \sin \theta – 10(1 – \sin^2 \theta) \sin^3 \theta + \sin^5 \theta \)
Simplify:
\( \sin 5\theta = 16\sin^5 \theta – 20\sin^3 \theta + 5\sin \theta \).
(c)(i) From part (b), \( \sin 5\theta = 0 \) when \( \theta = \frac{\pi}{5} \) and \( \theta = \frac{3\pi}{5} \).
Thus, \( 16\sin^4 \theta – 20\sin^2 \theta + 5 = 0 \) for these values of \( \theta \).
(c)(ii) Let \( \alpha = \sin^2 \frac{\pi}{5} \) and \( \beta = \sin^2 \frac{3\pi}{5} \).
From the quadratic equation \( 16x^2 – 20x + 5 = 0 \), the product of the roots is \( \alpha \beta = \frac{5}{16} \).
Thus, \( \sin \frac{\pi}{5} \sin \frac{3\pi}{5} = \sqrt{\alpha \beta} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4} \).