1. [Maximum mark: 4]
Find the set of values of \( x \) satisfying the inequality \( |2^{x+1} – 2| < 0.5 \), giving your answer to 3 significant figures.
▶️Answer/Explanation
Remove the modulus and rewrite the inequality as \(-0.5 < 2^{x+1} – 2 < 0.5\). Simplify to \(1.5 < 2^{x+1} < 2.5\). Take logarithms (base 2) to solve for \(x\): \(\log_2 1.5 < x + 1 < \log_2 2.5\). Subtract 1 to find the range \(-0.415 < x < 0.322\) (to 3 significant figures).
2. [Maximum mark: 5]
On an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z – 1 + 2i| \leq |z| \) and \( |z – 2| \leq 1 \).
▶️Answer/Explanation
The inequality \( |z – 1 + 2i| \leq |z| \) represents the perpendicular bisector of the line joining \( 1 – 2i \) and the origin. The inequality \( |z – 2| \leq 1 \) describes a circle centered at \( 2 \) with radius \( 1 \). Shade the intersection of the half-plane (from the first inequality) and the circle (from the second inequality).
3. [Maximum mark: 5]
The polynomial \( 2x^3 + ax^2 + bx + 6 \), where \( a \) and \( b \) are constants, is denoted by \( p(x) \). When \( p(x) \) is divided by \( (x + 2) \), the remainder is \(-38\), and when \( p(x) \) is divided by \( (2x – 1) \), the remainder is \( \frac{19}{2} \). Find the values of \( a \) and \( b \).
▶️Answer/Explanation
Substitute \( x = -2 \) into \( p(x) \) to get \( 4a – 2b = -28 \). Substitute \( x = \frac{1}{2} \) into \( p(x) \) to get \( a + 2b = 13 \). Solve the system of equations to find \( a = -3 \) and \( b = 8 \).
4. [Maximum mark: 5]
Solve the quadratic equation \( (3 + i)w^2 – 2w + 3 – i = 0 \), giving your answers in the form \( x + iy \), where \( x \) and \( y \) are real.
▶️Answer/Explanation
Use the quadratic formula: \( w = \frac{2 \pm \sqrt{(-2)^2 – 4(3 + i)(3 – i)}}{2(3 + i)} \).
Simplify the discriminant: \( \sqrt{4 – 4(9 + 1)} = \sqrt{-36} = 6i \).
Substitute back: \( w = \frac{2 \pm 6i}{6 + 2i} \).
Rationalize the denominators:
For \( + \): \( \frac{2 + 6i}{6 + 2i} \times \frac{6 – 2i}{6 – 2i} = \frac{12 + 32i}{40} = \frac{3}{5} + \frac{4}{5}i \).
For \( – \): \( \frac{2 – 6i}{6 + 2i} \times \frac{6 – 2i}{6 – 2i} = \frac{-12i}{40} = -i \).
Final Answer: \( w = \frac{3}{5} + \frac{4}{5}i \) and \( w = -i \).
5. [Maximum mark: 6]
Find the exact coordinates of the stationary points of the curve \( y = \frac{e^{3x^2 – 1}}{1 – x^2} \).
▶️Answer/Explanation
Use the quotient rule to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{(6x(1 – x^2) + 2x e^{3x^2 – 1}) e^{3x^2 – 1}}{(1 – x^2)^2} \).
Set \( \frac{dy}{dx} = 0 \): \( 6x(1 – x^2) + 2x = 0 \).
Solve for \( x \): \( x = 0 \), \( x = \pm \frac{2\sqrt{5}}{3} \).
Find corresponding \( y \)-values:
For \( x = 0 \): \( y = e^{-1} \).
For \( x = \pm \frac{2\sqrt{5}}{3} \): \( y = -3e^3 \).
Final Answer: Stationary points at \( (0, e^{-1}) \), \( \left(\frac{2\sqrt{5}}{3}, -3e^3\right) \), and \( \left(-\frac{2\sqrt{5}}{3}, -3e^3\right) \).
6. [Maximum mark: 6]
(a) Show that the equation \( \cot^2 \theta + 2 \cos 2\theta = 4 \) can be written in the form \( 4 \sin^4 \theta + 3 \sin^2 \theta – 1 = 0 \).
(b) Hence solve the equation \( \cot^2 \theta + 2 \cos 2\theta = 4 \), for \( 0^\circ < \theta < 360^\circ \).
▶️Answer/Explanation
Part (a):
Rewrite \( \cot^2 \theta \) as \( \frac{\cos^2 \theta}{\sin^2 \theta} \).
Use \( \cos 2\theta = 1 – 2 \sin^2 \theta \).
Substitute and simplify to \( 4 \sin^4 \theta + 3 \sin^2 \theta – 1 = 0 \).
Part (b):
Let \( u = \sin^2 \theta \) and solve \( 4u^2 + 3u – 1 = 0 \).
Solutions: \( u = \frac{1}{4} \) (valid) and \( u = -1 \) (invalid).
Solve \( \sin \theta = \pm \frac{1}{2} \) for \( \theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ \).
Final Answer: \( \theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ \).
7. [Maximum mark: 8]
The equation of a curve is \( x^3 + y^2 + 3x^2 + 3y = 4 \).
(a) Show that \( \frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3} \).
(b) Hence find the coordinates of the points on the curve at which the tangent is parallel to the \( x \)-axis.
▶️Answer/Explanation
Part (a):
Differentiate implicitly: \( 3x^2 + 2y \frac{dy}{dx} + 6x + 3 \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3} \).
Part (b):
Set numerator \( 3x^2 + 6x = 0 \): \( x = 0 \) or \( x = -2 \).
Substitute \( x = 0 \) into the curve equation: \( y^2 + 3y – 4 = 0 \), giving \( y = 1 \) or \( y = -4 \).
Substitute \( x = -2 \) into the curve equation: \( y^2 + 3y = 0 \), giving \( y = 0 \) or \( y = -3 \).
Final Answer: Points are \( (0, 1) \), \( (0, -4) \), \( (-2, 0) \), and \( (-2, -3) \).
8. [Maximum mark: 7]
The variables \( x \) and \( y \) satisfy the differential equation \( e^{4x} \frac{dy}{dx} = \cos^2 3y \). It is given that \( y = 0 \) when \( x = 2 \). Solve the differential equation, obtaining an expression for \( y \) in terms of \( x \).
▶️Answer/Explanation
Separate variables: \( \frac{dy}{\cos^2 3y} = e^{-4x} dx \).
Integrate both sides: \( \frac{1}{3} \tan 3y = -\frac{1}{4} e^{-4x} + C \).
Use initial condition \( y = 0 \) when \( x = 2 \): \( C = \frac{1}{4} e^{-8} \).
Solve for \( y \): \( y = \frac{1}{3} \tan^{-1} \left( \frac{3}{4} e^{-8} – \frac{3}{4} e^{-4x} \right) \).
Final Answer: \( y = \frac{1}{3} \tan^{-1} \left( \frac{3}{4} e^{-8} – \frac{3}{4} e^{-4x} \right) \).
9. [Maximum mark: 11]
Let \( f(x) = \frac{17x^2 – 7x + 16}{(2 + 3x^2)(2 – x)} \).
(a) Express \( f(x) \) in partial fractions.
(b) Hence obtain the expansion of \( f(x) \) in ascending powers of \( x \), up to and including the term in \( x^3 \).
(c) State the set of values of \( x \) for which the expansion in (b) is valid. Give your answer in an exact form.
▶️Answer/Explanation
Part (a):
Assume partial fractions form: \( \frac{Ax + B}{2 + 3x^2} + \frac{C}{2 – x} \).
Solve for \( A, B, C \): \( A = -2 \), \( B = 3 \), \( C = 5 \).
Final Answer: \( f(x) = \frac{-2x + 3}{2 + 3x^2} + \frac{5}{2 – x} \).
Part (b):
Expand \( \frac{5}{2 – x} \) as \( \frac{5}{2} \left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8}\right) \).
Expand \( \frac{-2x + 3}{2 + 3x^2} \) as \( \frac{3}{2} – x – \frac{9x^2}{4} + \frac{9x^3}{4} \).
Final Answer: \( 4 + \frac{1}{4}x – \frac{13}{8}x^2 + \frac{29}{16}x^3 \).
Part (c):
Final Answer: \( |x| < \sqrt{\frac{2}{3}} \).
10. [Maximum mark: 9]
The diagram shows the curve \( y = x \cos 2x \), for \( x \geq 0 \).
(a) Find the equation of the tangent to the curve at the point where \( x = \frac{1}{2}\pi \).
(b) Find the exact area of the shaded region shown in the diagram, bounded by the curve and the \( x \)-axis.
▶️Answer/Explanation
Part (a):
Find \( y \) at \( x = \frac{\pi}{2} \): \( y = -\frac{\pi}{2} \).
Find \( \frac{dy}{dx} = \cos 2x – 2x \sin 2x \), evaluate at \( x = \frac{\pi}{2} \): \( \frac{dy}{dx} = -1 \).
Final Answer: Tangent equation: \( x + y = 0 \).
Part (b):
Integrate by parts: \( \int x \cos 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \).
Evaluate from \( 0 \) to \( \frac{\pi}{4} \): \( \frac{\pi}{8} \).
Final Answer: Area \( = \frac{\pi}{8} \).
11. [Maximum mark: 9]
The line \( l \) has equation \( \mathbf{r} = \mathbf{i} – 2\mathbf{j} – 3\mathbf{k} + \lambda (-\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \). The points \( A \) and \( B \) have position vectors \( -2\mathbf{i} + 2\mathbf{j} – \mathbf{k} \) and \( 3\mathbf{i} – \mathbf{j} + \mathbf{k} \) respectively.
(a) Find a unit vector in the direction of \( l \).
(b) Find a vector equation for the line \( m \) passing through \( A \) and \( B \).
(c) Determine whether lines \( l \) and \( m \) are parallel, intersect, or are skew.
▶️Answer/Explanation
Part (a):
Find the direction vector of \( l \): \( -\mathbf{i} + \mathbf{j} + 2\mathbf{k} \).
Normalize: \( \frac{1}{\sqrt{6}} (-\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).
Final Answer: Unit vector \( = \frac{1}{\sqrt{6}} (-\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \).
Part (b):
Direction vector of \( m \): \( \mathbf{AB} = 5\mathbf{i} – 3\mathbf{j} + 2\mathbf{k} \).
Final Answer: \( \mathbf{r} = -2\mathbf{i} + 2\mathbf{j} – \mathbf{k} + \mu (5\mathbf{i} – 3\mathbf{j} + 2\mathbf{k}) \).
Part (c):
Check if direction vectors are scalar multiples (they are not).
Solve for intersection: no solution exists.
Final Answer: The lines are skew.