1. [Maximum mark: 4]
Solve the equation \( 2 \cdot 3^{2x – 1} = 4^{x + 1} \), giving your answer correct to 2 decimal places.
▶️Answer/Explanation
Solution:
Use the logarithm of a product or power to linearize the equation:
\( \ln 2 + (2x – 1)\ln 3 = (x + 1)\ln 4 \)
Solve for \( x \):
\( x = \frac{\ln 4 + \ln 3 – \ln 2}{2\ln 3 – \ln 4} \)
Obtain answer \( x = 2.21 \) (to 2 decimal places).
2. [Maximum mark: 5]
(a) Expand \( (2 – x^2)^{-2} \) in ascending powers of \( x \), up to and including the term in \( x^4 \), simplifying the coefficients. [4]
(b) State the set of values of \( x \) for which the expansion is valid. [1]
▶️Answer/Explanation
(a) The expansion is:
\( \frac{1}{4} + \frac{1}{4}x^2 + \frac{3}{16}x^4 \) (simplified coefficients).
(b) The expansion is valid for \( |x| < \sqrt{2} \).
3. [Maximum mark: 6]
Solve the equation \( 2 \cot 2x + 3 \cot x = 5 \), for \( 0^\circ < x < 180^\circ \).
▶️Answer/Explanation
Solution:
Use trigonometric identities to rewrite the equation:
\( 2 \left( \frac{\cot^2 x – 1}{2 \cot x} \right) + 3 \cot x = 5 \).
Simplify to a quadratic in \( \cot x \):
\( \cot^2 x + 5 \cot x – 4 = 0 \).
Solve for \( \cot x \), then find \( x \):
\( x = 35.1^\circ \) and \( x = 99.9^\circ \).
4. [Maximum mark: 7]
The variables \( x \) and \( y \) satisfy the differential equation \( \frac{dy}{dx} = \frac{xy}{1 + x^2} \), and \( y = 2 \) when \( x = 0 \). Solve the differential equation, obtaining a simplified expression for \( y \) in terms of \( x \).
▶️Answer/Explanation
Solution:
Separate variables and integrate:
\( \int \frac{dy}{y} = \int \frac{x}{1 + x^2} dx \).
Obtain \( \ln y = \frac{1}{2} \ln(1 + x^2) + C \).
Use initial condition \( y = 2 \) when \( x = 0 \) to find \( C = \ln 2 \).
Simplify to \( y = 2 \sqrt{1 + x^2} \).
5. [Maximum mark: 8]
The polynomial \( p(x) = ax^3 – 10x^2 + bx + 8 \), where \( a \) and \( b \) are constants, has \( (x – 2) \) as a factor of both \( p(x) \) and \( p'(x) \).
(a) Find the values of \( a \) and \( b \). [5]
(b) When \( a \) and \( b \) have these values, factorise \( p(x) \) completely. [3]
▶️Answer/Explanation
(a) Substitute \( x = 2 \) into \( p(x) \) and \( p'(x) \):
\( 8a – 40 + 2b + 8 = 0 \) and \( 12a – 40 + b = 0 \).
Solve to find \( a = 3 \) and \( b = 4 \).
(b) Factorise \( p(x) \) completely:
\( p(x) = (x – 2)^2 (3x + 2) \).
6. [Maximum mark: 8]
Let \( I = \int_0^3 \frac{27}{(9 + x^2)^2} dx \).
(a) Using the substitution \( x = 3 \tan \theta \), show that \( I = \int_0^{\frac{\pi}{4}} \cos^2 \theta d\theta \). [4]
(b) Hence find the exact value of \( I \). [4]
▶️Answer/Explanation
(a) Substitute \( x = 3 \tan \theta \), \( dx = 3 \sec^2 \theta d\theta \):
Transform the integral and simplify to \( \int \cos^2 \theta d\theta \).
(b) Integrate \( \cos^2 \theta \) and evaluate:
\( I = \frac{1}{8} (\pi + 2) \).
7. [Maximum mark: 8]
The complex number \( u \) is defined by \( u = \frac{\sqrt{2} – a \sqrt{2}i}{1 + 2i} \), where \( a \) is a positive integer.
(a) Express \( u \) in terms of \( a \), in the form \( x + iy \), where \( x \) and \( y \) are real and exact. [3]
(b) When \( a = 3 \), express \( u \) in the form \( re^{i\theta} \), giving exact values of \( r \) and \( \theta \). [2]
(c) Find the two square roots of \( u \), giving answers in the form \( re^{i\theta} \). [3]
▶️Answer/Explanation
(a) Multiply numerator and denominator by \( 1 – 2i \):
\( u = \frac{(1 – 2a)\sqrt{2}}{5} – \frac{(2 + a)\sqrt{2}}{5}i \).
(b) For \( a = 3 \):
\( r = 2 \), \( \theta = \frac{3\pi}{4} \).
(c) Square roots:
\( \sqrt{2} e^{i \frac{3\pi}{8}} \) and \( \sqrt{2} e^{i \frac{11\pi}{8}} \).
8. [Maximum mark: 9]
The equation of a curve is \( x^3 + y^3 + 2xy + 8 = 0 \).
(a) Express \( \frac{dy}{dx} \) in terms of \( x \) and \( y \). [4]
(b) The tangent to the curve at the point where \( x = 0 \) and the tangent at the point where \( y = 0 \) intersect at the acute angle \( \alpha \). Find the exact value of \( \tan \alpha \). [5]
▶️Answer/Explanation
(a) Differentiate implicitly:
\( \frac{dy}{dx} = -\frac{3x^2 + 2y}{3y^2 + 2x} \).
(b) Find gradients at \( x = 0 \) and \( y = 0 \):
Gradients are \( \frac{1}{3} \) and \( 3 \).
Use \( \tan \alpha = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right| \) to find \( \tan \alpha = \frac{4}{3} \).
9. [Maximum mark: 9]
In the diagram, \( OABCDEFG \) is a cuboid with \( OA = 2 \), \( OC = 4 \), and \( OG = 2 \). The point \( M \) is the midpoint of \( DF \), and \( N \) is on \( AB \) such that \( AN = 3NB \).
(a) Express the vectors \( \overrightarrow{OM} \) and \( \overrightarrow{MN} \) in terms of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). [3]
(b) Find a vector equation for the line through \( M \) and \( N \). [2]
(c) Show that the length of the perpendicular from \( O \) to the line through \( M \) and \( N \) is \( \frac{\sqrt{53}}{6} \). [4]
▶️Answer/Explanation
(a) \( \overrightarrow{OM} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \), \( \overrightarrow{MN} = \mathbf{i} + \mathbf{j} – 2\mathbf{k} \).
(b) Vector equation: \( \mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda (\mathbf{i} + \mathbf{j} – 2\mathbf{k}) \).
(c) Find perpendicular distance using projection:
Solve \( \overrightarrow{OP} \cdot \overrightarrow{MN} = 0 \) to find \( \lambda \), then compute distance.
10. [Maximum mark: 11]
The curve \( y = x \sqrt{\sin x} \) has one stationary point in the interval \( 0 < x < \pi \), where \( x = a \).
(a) Show that \( \tan a = -\frac{1}{2}a \). [4]
(b) Verify by calculation that \( a \) lies between 2 and 2.5. [2]
(c) Show that the iterative formula \( x_{n+1} = \pi – \tan^{-1}\left(\frac{1}{2}x_n\right) \) converges to \( a \). [2]
(d) Use the iterative formula to determine \( a \) correct to 2 decimal places. [3]
▶️Answer/Explanation
(a) Differentiate \( y \) and set \( \frac{dy}{dx} = 0 \):
Obtain \( \tan a = -\frac{1}{2}a \).
(b) Evaluate \( \tan 2 + 1 \) and \( \tan 2.5 + 1.25 \):
Sign change confirms \( a \in (2, 2.5) \).
(c) Rearrange \( \tan a = -\frac{1}{2}a \) to show convergence.
(d) Iterate to find \( a \approx 2.29 \).