1. [Maximum mark: 4]
Expand \( (1 + 3x)^{\frac{2}{3}} \) in ascending powers of \( x \), up to and including the term in \( x^3 \), simplifying the coefficients.
▶️Answer/Explanation
Solution:
Using the binomial expansion for \( (1 + 3x)^{\frac{2}{3}} \):
\( 1 + \frac{2}{3}(3x) + \frac{\frac{2}{3} \cdot (-\frac{1}{3})}{2!}(3x)^2 + \frac{\frac{2}{3} \cdot (-\frac{1}{3}) \cdot (-\frac{4}{3})}{3!}(3x)^3 \)
Simplify each term:
\( 1 + 2x – x^2 + \frac{4}{3}x^3 \)
2. [Maximum mark: 5]
Solve the equation \( 4^x = 3 + 4^{-x} \). Give your answer correct to 3 decimal places.
▶️Answer/Explanation
Solution:
Let \( u = 4^x \), then the equation becomes:
\( u = 3 + \frac{1}{u} \)
Multiply through by \( u \): \( u^2 – 3u – 1 = 0 \)
Solve the quadratic: \( u = \frac{3 \pm \sqrt{13}}{2} \)
Take the positive root: \( u = \frac{3 + \sqrt{13}}{2} \)
Convert back to \( x \): \( 4^x = \frac{3 + \sqrt{13}}{2} \)
Take logs: \( x = \frac{\ln(\frac{3 + \sqrt{13}}{2})}{\ln 4} \approx 0.862 \)
3. [Maximum mark: 7]
The parametric equations of a curve are \( x = t + \ln(t + 2) \), \( y = (t – 1)e^{-2t} \), where \( t > -2 \).
(a) Express \( \frac{dy}{dx} \) in terms of \( t \), simplifying your answer.
(b) Find the exact y-coordinate of the stationary point of the curve.
▶️Answer/Explanation
(a) Solution:
Find \( \frac{dx}{dt} = 1 + \frac{1}{t+2} \) and \( \frac{dy}{dt} = e^{-2t} – 2(t-1)e^{-2t} \)
Then \( \frac{dy}{dx} = \frac{e^{-2t}(3 – 2t)}{1 + \frac{1}{t+2}} = \frac{(3-2t)(t+2)}{t+3}e^{-2t} \)
(b) Solution:
Set \( \frac{dy}{dx} = 0 \Rightarrow 3 – 2t = 0 \Rightarrow t = \frac{3}{2} \)
Substitute into y: \( y = (\frac{1}{2})e^{-3} = \frac{1}{2}e^{-3} \)
4. [Maximum mark: 7]
Let \( f(x) = \frac{15 – 6x}{(1 + 2x)(4 – x)} \).
(a) Express \( f(x) \) in partial fractions.
(b) Hence find \( \int_1^2 f(x)dx \), giving your answer in the form \( \ln\left(\frac{a}{b}\right) \), where \( a \) and \( b \) are integers.
▶️Answer/Explanation
(a) Solution:
Partial fractions form: \( \frac{A}{1+2x} + \frac{B}{4-x} \)
Solve to get \( A = 4 \), \( B = -1 \)
(b) Solution:
Integrate: \( \int ( \frac{4}{1+2x} – \frac{1}{4-x} ) dx = 2\ln|1+2x| + \ln|4-x| \)
Evaluate from 1 to 2: \( (2\ln5 + \ln2) – (2\ln3 + \ln3) = \ln\left(\frac{50}{27}\right) \)
5. [Maximum mark: 7]
(a) By first expanding \( \tan(2θ + 2θ) \), show that the equation \( \tan 4θ = \frac{1}{2} \tan θ \) may be expressed as \( \tan^4 θ + 2\tan^2 θ – 7 = 0 \).
(b) Hence solve the equation \( \tan 4θ = \frac{1}{2} \tan θ \), for \( 0° < θ < 180° \).
▶️Answer/Explanation
(a) Solution:
Using double angle formula twice:
\( \tan 4θ = \frac{2\tan 2θ}{1 – \tan^2 2θ} = \frac{2(\frac{2\tan θ}{1-\tan^2 θ})}{1 – (\frac{2\tan θ}{1-\tan^2 θ})^2} \)
Simplify and equate to \( \frac{1}{2}\tan θ \) to get the required equation.
(b) Solution:
Let \( u = \tan^2 θ \): \( u^2 + 2u – 7 = 0 \)
Solutions: \( u = -1 \pm 2\sqrt{2} \) (take positive root)
\( \tan θ = \pm\sqrt{2\sqrt{2}-1} \) ⇒ \( θ ≈ 53.5°, 126.5° \)
6. [Maximum mark: 7]
(a) By sketching a suitable pair of graphs, show that the equation \( \cot\frac{1}{2}x = 1 + e^{-x} \) has exactly one root in the interval \( 0 < x \leq \pi \).
(b) Verify by calculation that this root lies between 1 and 1.5.
(c) Use the iterative formula \( x_{n+1} = 2\tan^{-1}\left(\frac{1}{1 + e^{-x_n}}\right) \) to determine the root correct to 2 decimal places.
▶️Answer/Explanation
(a) Solution:
Sketch \( y = \cot(x/2) \) (decreasing from ∞ to 0) and \( y = 1 + e^{-x} \) (decreasing from 2 to 1).
They intersect exactly once in \( (0, \pi] \).
(b) Solution:
At \( x=1 \): LHS ≈ 1.19, RHS ≈ 1.37 ⇒ LHS < RHS
At \( x=1.5 \): LHS ≈ 0.82, RHS ≈ 1.22 ⇒ LHS < RHS
But at \( x=0.5 \): LHS > RHS, so root is between 1 and 1.5.
(c) Solution:
Iterations converge to \( x ≈ 1.34 \) (show sufficient iterations to justify).
7. [Maximum mark: 9]
For the curve shown in the diagram, the normal to the curve at point \( P(x,y) \) meets the x-axis at \( N \). \( M \) is the foot of the perpendicular from \( P \) to the x-axis.
The curve is such that for all values of x in the interval 0 ≤ x <1/2 π, the area of triangle PMN is equal to tan
(a)(i) Show that \( \frac{MN}{y} = \frac{dy}{dx} \).
(a)(ii) Hence show that \( \frac{1}{2}y^2\frac{dy}{dx} = \tan x \).
(b) Given that \( y = 1 \) when \( x = 0 \), solve the differential equation to find \( y \) in terms of \( x \).
▶️Answer/Explanation
(a)(i) Solution:
Gradient of normal is \( -\frac{dx}{dy} \), so equation is \( Y – y = -\frac{dx}{dy}(X – x) \).
At \( N \), \( Y=0 \), so \( MN = x + y\frac{dy}{dx} – x = y\frac{dy}{dx} \).
(a)(ii) Solution:
Area of triangle \( PMN = \frac{1}{2}y \cdot MN = \frac{1}{2}y^2\frac{dy}{dx} = \tan x \).
(b) Solution:
Separate variables: \( \frac{1}{2}y^2 dy = \tan x dx \)
Integrate: \( \frac{1}{6}y^3 = -\ln \cos x + C \)
Use initial condition: \( C = \frac{1}{6} \)
Final solution: \( y = \sqrt[3]{1 – 6\ln \cos x} \)
8. [Maximum mark: 10]
The diagram shows the curve \( y = \frac{\ln x}{x^4} \) and its maximum point \( M \).
(a) Find the exact coordinates of \( M \).
(b) By using integration by parts, show that for all \( a > 1 \), \( \int_1^a \frac{\ln x}{x^4} dx < \frac{1}{9} \).
▶️Answer/Explanation
(a) Solution:
Find derivative: \( \frac{dy}{dx} = \frac{1 – 4\ln x}{x^5} \)
Set to zero: \( 1 – 4\ln x = 0 \) ⇒ \( x = e^{1/4} \)
\( y \)-coordinate: \( y = \frac{1/4}{e} = \frac{1}{4e} \)
(b) Solution:
Integrate by parts: \( \int \frac{\ln x}{x^4} dx = -\frac{\ln x}{3x^3} + \frac{1}{9x^3} + C \)
Evaluate from 1 to \( a \): \( -\frac{\ln a}{3a^3} + \frac{1}{9a^3} + \frac{1}{9} \)
As \( a \to \infty \), the expression approaches \( \frac{1}{9} \) from below.
9. [Maximum mark: 9]
The quadrilateral \( ABCD \) is a trapezium with \( AB \parallel DC \). The position vectors are \( \vec{OA} = -i + 2j + 3k \), \( \vec{OB} = i + 3j + k \), \( \vec{OC} = 2i + 2j – 3k \).
(a) Given that \( \vec{DC} = 3\vec{AB} \), find \( \vec{OD} \).
(b) State a vector equation for the line through \( A \) and \( B \).
(c) Find the distance between the parallel sides and hence find the area of the trapezium.
▶️Answer/Explanation
(a) Solution:
\( \vec{AB} = (2i + j – 2k) \)
\( \vec{DC} = 3\vec{AB} = 6i + 3j – 6k \)
\( \vec{OD} = \vec{OC} – \vec{DC} = -4i – j + 3k \)
(b) Solution:
\( r = -i + 2j + 3k + \lambda(2i + j – 2k) \)
(c) Solution:
Shortest distance is perpendicular distance from \( C \) to \( AB \):
Find point \( P \) on \( AB \) where \( CP \) is perpendicular to \( AB \) ⇒ distance is 3.
Area = \( \frac{1}{2}(|AB| + |DC|) \times \) distance = \( \frac{1}{2}(3 + 9) \times 3 = 18 \)
10. [Maximum mark: 10]
(a) Verify that \( -1 + \sqrt{2}i \) is a root of \( z^4 + 3z^2 + 2z + 12 = 0 \).
(b) Find the other roots of this equation.
▶️Answer/Explanation
(a) Solution:
Substitute \( z = -1 + \sqrt{2}i \) into the equation and show it equals zero.
(b) Solution:
Complex roots come in conjugate pairs, so \( -1 – \sqrt{2}i \) is also a root.
Find quadratic factor: \( (z – (-1+\sqrt{2}i))(z – (-1-\sqrt{2}i)) = z^2 + 2z + 3 \)
Polynomial division gives: \( (z^2 + 2z + 3)(z^2 – 2z + 4) = 0 \)
Solve \( z^2 – 2z + 4 = 0 \) to get \( z = 1 \pm \sqrt{3}i \)