1. [Maximum mark: 5]
Find the exact coordinates of the points on the curve \( y = \frac{x^2}{1 – 3x} \) at which the gradient of the tangent is equal to 8.
▶️Answer/Explanation
Use the quotient rule to find the derivative of \( y \):
\( \frac{dy}{dx} = \frac{(1-3x)(2x) – x^2(-3)}{(1-3x)^2} = \frac{2x – 3x^2}{(1-3x)^2} \).
Set the derivative equal to 8 and solve for \( x \):
\( \frac{2x – 3x^2}{(1-3x)^2} = 8 \).
\( 2x – 3x^2 = 8(1 – 6x + 9x^2) \).
\( 2x – 3x^2 = 8 – 48x + 72x^2 \).
\( 75x^2 – 50x + 8 = 0 \).
\( (15x – 4)(5x – 2) = 0 \).
\( x = \frac{2}{5} \) or \( x = \frac{4}{15} \).
Find the corresponding \( y \)-coordinates:
For \( x = \frac{2}{5} \), \( y = \frac{\left(\frac{2}{5}\right)^2}{1 – 3 \times \frac{2}{5}} = \frac{\frac{4}{25}}{-\frac{1}{5}} = -\frac{4}{5} \).
For \( x = \frac{4}{15} \), \( y = \frac{\left(\frac{4}{15}\right)^2}{1 – 3 \times \frac{4}{15}} = \frac{\frac{16}{225}}{\frac{3}{15}} = \frac{16}{45} \).
Final Answer: The points are \( \left(\frac{2}{5}, -\frac{4}{5}\right) \) and \( \left(\frac{4}{15}, \frac{16}{45}\right) \).
2. [Maximum mark: 4]
On an Argand diagram, shade the region whose points represent complex numbers \( z \) satisfying the inequalities \( |z – 2i| \leq |z + 2 – i| \) and \( 0 \leq \arg(z + 1) \leq \frac{1}{4}\pi \).
▶️Answer/Explanation
Interpret the first inequality \( |z – 2i| \leq |z + 2 – i| \):
This represents all points \( z \) closer to \( 2i \) than to \( -2 + i \). The boundary is the perpendicular bisector of the line segment joining \( 2i \) and \( -2 + i \).
Interpret the second inequality \( 0 \leq \arg(z + 1) \leq \frac{\pi}{4} \):
This represents all points \( z \) such that the argument of \( z + 1 \) lies between 0 and \( \frac{\pi}{4} \). The boundary is a half-line starting at \( -1 \) and making an angle of \( \frac{\pi}{4} \) with the positive real axis.
Shade the region that satisfies both conditions:
The shaded region is the intersection of the area below the perpendicular bisector (from Step 1) and the sector defined by the argument condition (from Step 2).
3. [Maximum mark: 4]
The variables \( x \) and \( y \) are related by the equation \( y = ab^x \), where \( a \) and \( b \) are constants. The diagram shows the result of plotting \( \ln y \) against \( x \) for two pairs of values of \( x \) and \( y \). The coordinates of these points are \( (1, 3.7) \) and \( (2.2, 6.46) \). Use this information to find the values of \( a \) and \( b \).
▶️Answer/Explanation
Take the natural logarithm of both sides of the equation \( y = ab^x \):
\( \ln y = \ln a + x \ln b \).
Set up equations using the given points:
For \( (1, 3.7) \): \( 3.7 = \ln a + \ln b \).
For \( (2.2, 6.46) \): \( 6.46 = \ln a + 2.2 \ln b \).
Solve the system of equations:
Subtract the first equation from the second: \( 2.76 = 1.2 \ln b \).
\( \ln b = \frac{2.76}{1.2} = 2.3 \).
Substitute back to find \( \ln a = 3.7 – 2.3 = 1.4 \).
Exponentiate to find \( a \) and \( b \):
\( a = e^{1.4} \approx 4.06 \).
\( b = e^{2.3} \approx 9.97 \).
Final Answer: \( a \approx 4.06 \), \( b \approx 9.97 \).
4. [Maximum mark: 5]
The complex number \( u \) is defined by \( u = \frac{3 + 2i}{a – 5i} \), where \( a \) is real.
(a) Express \( u \) in the Cartesian form \( x + iy \), where \( x \) and \( y \) are in terms of \( a \).
(b) Given that \( \arg u = \frac{1}{4}\pi \), find the value of \( a \).
▶️Answer/Explanation
(a)
Multiply numerator and denominator by the conjugate of the denominator:
\( u = \frac{(3 + 2i)(a + 5i)}{(a – 5i)(a + 5i)} \).
Expand and simplify:
Numerator: \( 3a + 15i + 2ai + 10i^2 = 3a + (15 + 2a)i – 10 \).
Denominator: \( a^2 + 25 \).
\( u = \frac{(3a – 10) + (2a + 15)i}{a^2 + 25} \).
Final Answer for (a): \( u = \frac{3a – 10}{a^2 + 25} + \frac{2a + 15}{a^2 + 25}i \).
(b)
For \( \arg u = \frac{\pi}{4} \), the real and imaginary parts must be equal and positive:
\( \frac{3a – 10}{a^2 + 25} = \frac{2a + 15}{a^2 + 25} \).
\( 3a – 10 = 2a + 15 \).
\( a = 25 \).
Final Answer for (b): \( a = 25 \).
5. [Maximum mark: 6]
(a) Given that \( \sin\left(x + \frac{\pi}{6}\right) – \sin\left(x – \frac{\pi}{6}\right) = \cos\left(x + \frac{\pi}{3}\right) – \cos\left(x – \frac{\pi}{3}\right) \), find the exact value of \( \tan x \).
(b) Hence find the exact roots of the equation for \( 0 \leq x \leq 2\pi \).
▶️Answer/Explanation
(a)
Use trigonometric identities to simplify both sides:
LHS: \( \sin\left(x + \frac{\pi}{6}\right) – \sin\left(x – \frac{\pi}{6}\right) = 2\cos x \sin \frac{\pi}{6} \).
RHS: \( \cos\left(x + \frac{\pi}{3}\right) – \cos\left(x – \frac{\pi}{3}\right) = -2\sin x \sin \frac{\pi}{3} \).
Substitute known values:
\( 2\cos x \left(\frac{1}{2}\right) = -2\sin x \left(\frac{\sqrt{3}}{2}\right) \).
\( \cos x = -\sqrt{3} \sin x \).
Solve for \( \tan x \):
\( \tan x = -\frac{1}{\sqrt{3}} \).
Final Answer for (a): \( \tan x = -\frac{1}{\sqrt{3}} \).
(b)
Solve \( \tan x = -\frac{1}{\sqrt{3}} \) in the interval \( [0, 2\pi] \):
\( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \).
Final Answer for (b): \( x = \frac{5\pi}{6}, \frac{11\pi}{6} \).
6. [Maximum mark: 6]
The parametric equations of a curve are \( x = \sqrt{t + 3} \), \( y = \ln t \), for \( t > 0 \).
(a) Obtain a simplified expression for \( \frac{dy}{dx} \) in terms of \( t \).
(b) Hence find the exact coordinates of the point on the curve at which the gradient of the normal is \(-2\).
▶️Answer/Explanation
(a)
Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\( \frac{dx}{dt} = \frac{1}{2}(t + 3)^{-\frac{1}{2}} \).
\( \frac{dy}{dt} = \frac{1}{t} \).
Use the chain rule to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{1/t}{1/(2\sqrt{t + 3})} = \frac{2\sqrt{t + 3}}{t} \).
Final Answer for (a): \( \frac{dy}{dx} = \frac{2\sqrt{t + 3}}{t} \).
(b)
The gradient of the normal is \(-2\), so the gradient of the tangent is \( \frac{1}{2} \):
\( \frac{2\sqrt{t + 3}}{t} = \frac{1}{2} \).
Solve for \( t \):
\( 4\sqrt{t + 3} = t \).
Square both sides: \( 16(t + 3) = t^2 \).
\( t^2 – 16t – 48 = 0 \).
\( t = 8 \pm \sqrt{112} \). Only \( t = 8 + 4\sqrt{7} \) is valid since \( t > 0 \).
Find \( x \) and \( y \):
\( x = \sqrt{8 + 4\sqrt{7} + 3} = \sqrt{11 + 4\sqrt{7}} \).
\( y = \ln(8 + 4\sqrt{7}) \).
Final Answer for (b): The point is \( \left( \sqrt{11 + 4\sqrt{7}}, \ln(8 + 4\sqrt{7}) \right) \).
7. [Maximum mark: 7]
The variables \( x \) and \( \theta \) satisfy the differential equation \( \frac{x}{\tan \theta} \frac{dx}{d\theta} = x^2 + 3 \). It is given that \( x = 1 \) when \( \theta = 0 \). Solve the differential equation, obtaining an expression for \( x^2 \) in terms of \( \theta \).
▶️Answer/Explanation
Separate the variables:
\( \frac{x}{x^2 + 3} dx = \tan \theta \, d\theta \).
Integrate both sides:
\( \int \frac{x}{x^2 + 3} dx = \int \tan \theta \, d\theta \).
\( \frac{1}{2} \ln(x^2 + 3) = -\ln(\cos \theta) + C \).
Use the initial condition \( x = 1 \), \( \theta = 0 \) to find \( C \):
\( \frac{1}{2} \ln(1 + 3) = -\ln(1) + C \).
\( \frac{1}{2} \ln 4 = C \).
\( C = \ln 2 \).
Solve for \( x^2 \):
\( \frac{1}{2} \ln(x^2 + 3) = -\ln(\cos \theta) + \ln 2 \).
\( \ln(x^2 + 3) = -2\ln(\cos \theta) + 2\ln 2 \).
\( x^2 + 3 = \frac{4}{\cos^2 \theta} \).
Final Answer: \( x^2 = \frac{4}{\cos^2 \theta} – 3 \).
8. [Maximum mark: 8]
(a) By sketching a suitable pair of graphs, show that the equation \( \sqrt{x} = e^x – 3 \) has only one root.
(b) Show by calculation that this root lies between 1 and 2.
(c) Show that, if a sequence of values given by the iterative formula \( x_{n+1} = \ln(3 + \sqrt{x_n}) \) converges, then it converges to the root of the equation in (a).
(d) Use the iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
(a)
Sketch \( y = \sqrt{x} \) and \( y = e^x – 3 \):
– \( y = \sqrt{x} \) starts at \( (0, 0) \) and increases.
– \( y = e^x – 3 \) crosses the y-axis at \( (0, -2) \) and increases rapidly.
The graphs intersect only once.
Final Answer for (a): The equation has only one root.
(b)
Evaluate \( f(x) = \sqrt{x} – e^x + 3 \) at \( x = 1 \) and \( x = 2 \):
\( f(1) = 1 – e + 3 \approx 1.28 > 0 \).
\( f(2) = \sqrt{2} – e^2 + 3 \approx -2.98 < 0 \).
By the Intermediate Value Theorem, the root lies between 1 and 2.
Final Answer for (b): The root lies in \( (1, 2) \).
(c)
Rearrange \( \sqrt{x} = e^x – 3 \) to \( x = \ln(3 + \sqrt{x}) \).
The iterative formula \( x_{n+1} = \ln(3 + \sqrt{x_n}) \) converges to the root.
Final Answer for (c): The iterative formula converges to the root.
(d)
Perform iterations starting with \( x_0 = 1 \):
\( x_1 = \ln(3 + \sqrt{1}) \approx 1.3863 \).
\( x_2 = \ln(3 + \sqrt{1.3863}) \approx 1.4297 \).
\( x_3 = \ln(3 + \sqrt{1.4297}) \approx 1.4341 \).
\( x_4 = \ln(3 + \sqrt{1.4341}) \approx 1.4345 \).
Final Answer for (d): The root is \( 1.43 \) (to 2 decimal places).
9. [Maximum mark: 9]
The diagram shows the curve \( y = xe^{-\frac{1}{4}x^2} \), for \( x \geq 0 \), and its maximum point \( M \).
(a) Find the exact coordinates of \( M \).
(b) Using the substitution \( x = \sqrt{u} \), or otherwise, find by integration the exact area of the shaded region bounded by the curve, the x-axis, and the line \( x = 3 \).
▶️Answer/Explanation
(a)
Find the derivative of \( y \):
\( \frac{dy}{dx} = e^{-\frac{1}{4}x^2} + x \left( -\frac{1}{2}x e^{-\frac{1}{4}x^2} \right) = e^{-\frac{1}{4}x^2} \left( 1 – \frac{1}{2}x^2 \right) \).
Set \( \frac{dy}{dx} = 0 \) and solve for \( x \):
\( 1 – \frac{1}{2}x^2 = 0 \).
\( x = \sqrt{2} \) (since \( x \geq 0 \)).
Find the corresponding \( y \)-coordinate:
\( y = \sqrt{2} e^{-\frac{1}{4} \times 2} = \sqrt{2} e^{-\frac{1}{2}} \).
Final Answer for (a): \( M \left( \sqrt{2}, \sqrt{2} e^{-\frac{1}{2}} \right) \).
(b)
Use the substitution \( x = \sqrt{u} \), \( dx = \frac{1}{2\sqrt{u}} du \):
The integral becomes \( \int_{0}^{3} x e^{-\frac{1}{4}x^2} dx = \int_{0}^{9} \sqrt{u} e^{-\frac{1}{4}u} \cdot \frac{1}{2\sqrt{u}} du = \frac{1}{2} \int_{0}^{9} e^{-\frac{1}{4}u} du \).
Integrate:
\( \frac{1}{2} \left[ -4 e^{-\frac{1}{4}u} \right]_{0}^{9} = -2 \left( e^{-\frac{9}{4}} – 1 \right) = 2 – 2 e^{-\frac{9}{4}} \).
Final Answer for (b): The area is \( 2 – 2 e^{-\frac{9}{4}} \).
10. [Maximum mark: 11]
Let \( f(x) = \frac{24x + 13}{(1 – 2x)(2 + x)^2} \).
(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 \( x^2 \).
(c) State the set of values of \( x \) for which the expansion in (b) is valid.
▶️Answer/Explanation
(a)
Assume partial fractions of the form:
\( \frac{A}{1 – 2x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2} \).
Multiply through by the denominator and solve for \( A \), \( B \), and \( C \):
\( 24x + 13 = A(2 + x)^2 + B(1 – 2x)(2 + x) + C(1 – 2x) \).
Solving gives \( A = 4 \), \( B = 2 \), \( C = -7 \).
Final Answer for (a): \( f(x) = \frac{4}{1 – 2x} + \frac{2}{2 + x} – \frac{7}{(2 + x)^2} \).
(b)
Expand each term as a binomial series:
\( \frac{4}{1 – 2x} = 4(1 + 2x + 4x^2 + \cdots) \).
\( \frac{2}{2 + x} = 1 – \frac{x}{2} + \frac{x^2}{4} – \cdots \).
\( -\frac{7}{(2 + x)^2} = -\frac{7}{4} \left( 1 – x + \frac{3x^2}{4} – \cdots \right) \).
Combine the expansions:
\( f(x) = 4 + 8x + 16x^2 + 1 – \frac{x}{2} + \frac{x^2}{4} – \frac{7}{4} + \frac{7x}{4} – \frac{21x^2}{16} + \cdots \).
Simplify: \( \frac{13}{4} + \frac{37x}{4} + \frac{239x^2}{16} \).
Final Answer for (b): \( \frac{13}{4} + \frac{37}{4}x + \frac{239}{16}x^2 \).
(c)
Final Answer for (c): The expansion is valid for \( |x| < \frac{1}{2} \).
11. [Maximum mark: 10]
In the diagram, OABCDEFG is a cuboid in which \( OA = 3 \) units, \( OC = 2 \) units, and \( OD = 2 \) units. Unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are parallel to \( OA \), \( OD \), and \( OC \) respectively. \( M \) is the midpoint of \( EF \).
(a) Find the position vector of \( M \).
(b) The position vector of \( P \) is \( \mathbf{i} + \mathbf{j} + 2\mathbf{k} \). Calculate angle \( PAM \).
(c) Find the exact length of the perpendicular from \( P \) to the line passing through \( O \) and \( M \).
▶️Answer/Explanation
(a)
The position vector of \( M \) is \( 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \).
Final Answer for (a): \( \mathbf{M} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \).
(b)
Find vectors \( \mathbf{PA} \) and \( \mathbf{PM} \):
\( \mathbf{PA} = \mathbf{A} – \mathbf{P} = 3\mathbf{i} – (\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} – \mathbf{j} – 2\mathbf{k} \).
\( \mathbf{PM} = \mathbf{M} – \mathbf{P} = 2\mathbf{i} + \mathbf{j} – \mathbf{k} \).
Calculate the dot product and magnitudes:
\( \mathbf{PA} \cdot \mathbf{PM} = 4 – 1 + 2 = 5 \).
\( |\mathbf{PA}| = \sqrt{4 + 1 + 4} = 3 \).
\( |\mathbf{PM}| = \sqrt{4 + 1 + 1} = \sqrt{6} \).
Find the angle using \( \cos \theta = \frac{\mathbf{PA} \cdot \mathbf{PM}}{|\mathbf{PA}| |\mathbf{PM}|} \):
\( \theta = \cos^{-1}\left( \frac{5}{3\sqrt{6}} \right) \approx 53.4^\circ \).
Final Answer for (b): \( \theta \approx 53.4^\circ \).
(c)
The line through \( O \) and \( M \) has direction vector \( \mathbf{M} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \).
The perpendicular distance from \( P \) to this line is given by:
\( \frac{|\mathbf{OP} \times \mathbf{M}|}{|\mathbf{M}|} \).
Compute \( \mathbf{OP} \times \mathbf{M} \):
\( \mathbf{OP} \times \mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{vmatrix} = (-3)\mathbf{i} + 5\mathbf{j} – \mathbf{k} \).
Compute the magnitude:
\( |\mathbf{OP} \times \mathbf{M}| = \sqrt{9 + 25 + 1} = \sqrt{35} \).
\( |\mathbf{M}| = \sqrt{9 + 4 + 1} = \sqrt{14} \).
Final Answer for (c): The perpendicular distance is \( \frac{\sqrt{35}}{\sqrt{14}} = \frac{\sqrt{10}}{2} \).