1. [Maximum mark: 4]
Solve the equation \(\ln(e^{2x} + 3) = 2x + \ln 3\), giving your answer correct to 3 decimal places.
▶️Answer/Explanation
Solution:
Use the logarithm property \(\ln A – \ln B = \ln\left(\frac{A}{B}\right)\):
\(\ln(e^{2x} + 3) – \ln 3 = 2x\)
\(\ln\left(\frac{e^{2x} + 3}{3}\right) = 2x\)
Exponentiate both sides:
\(\frac{e^{2x} + 3}{3} = e^{2x}\)
Solve for \(e^{2x}\):
\(e^{2x} + 3 = 3e^{2x}\)
\(3 = 2e^{2x}\)
\(e^{2x} = \frac{3}{2}\)
Take the natural logarithm:
\(2x = \ln\left(\frac{3}{2}\right)\)
\(x = \frac{1}{2}\ln\left(\frac{3}{2}\right) \approx 0.203\)
Final answer: \(x = 0.203\) (to 3 decimal places).
2. [Maximum mark: 5]
Solve the equation \(3 \cos 2\theta = 3 \cos \theta + 2\), for \(0^\circ \leq \theta \leq 360^\circ\).
▶️Answer/Explanation
Solution:
Use the double-angle identity \(\cos 2\theta = 2\cos^2 \theta – 1\):
\(3(2\cos^2 \theta – 1) = 3\cos \theta + 2\)
Simplify:
\(6\cos^2 \theta – 3 = 3\cos \theta + 2\)
Rearrange to quadratic form:
\(6\cos^2 \theta – 3\cos \theta – 5 = 0\)
Solve the quadratic equation for \(\cos \theta\):
\(\cos \theta = \frac{3 \pm \sqrt{9 + 120}}{12} = \frac{3 \pm \sqrt{129}}{12}\)
Only the positive root is valid (\(\cos \theta \in [-1, 1]\)):
\(\cos \theta \approx -0.696\)
Find \(\theta\):
\(\theta = 134.1^\circ, 225.9^\circ\) (to 1 decimal place).
3. [Maximum mark: 5]
The polynomial \(ax^3 + x^2 + bx + 3\) is denoted by \(p(x)\). It is given that \(p(x)\) is divisible by \((2x – 1)\) and that when \(p(x)\) is divided by \((x + 2)\) the remainder is 5. Find the values of \(a\) and \(b\).
▶️Answer/Explanation
Solution:
Since \(p(x)\) is divisible by \((2x – 1)\), \(p\left(\frac{1}{2}\right) = 0\):
\(a\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 3 = 0\)
Simplify:
\(\frac{a}{8} + \frac{1}{4} + \frac{b}{2} + 3 = 0\)
Multiply by 8:
\(a + 2 + 4b + 24 = 0\)
\(a + 4b = -26\) (Equation 1).
When \(p(x)\) is divided by \((x + 2)\), the remainder is \(p(-2) = 5\):
\(a(-2)^3 + (-2)^2 + b(-2) + 3 = 5\)
Simplify:
\(-8a + 4 – 2b + 3 = 5\)
\(-8a – 2b = -2\)
\(4a + b = 1\) (Equation 2).
Solve the system:
From Equation 2: \(b = 1 – 4a\).
Substitute into Equation 1:
\(a + 4(1 – 4a) = -26\)
\(a + 4 – 16a = -26\)
\(-15a = -30\)
\(a = 2\).
Then \(b = 1 – 4(2) = -7\).
Final answer: \(a = 2\), \(b = -7\).
4. [Maximum mark: 6]
The equation of a curve is \(y = \cos^3 x \sqrt{\sin x}\). It is given that the curve has one stationary point in the interval \(0 < x < \frac{1}{2}\pi\). Find the \(x\)-coordinate of this stationary point, giving your answer correct to 3 significant figures.
▶️Answer/Explanation
Solution:
Differentiate \(y = \cos^3 x \sqrt{\sin x}\) using the product rule:
\(\frac{dy}{dx} = -3\cos^2 x \sin x \sqrt{\sin x} + \cos^3 x \cdot \frac{\cos x}{2\sqrt{\sin x}}\).
Set \(\frac{dy}{dx} = 0\) for stationary points:
\(-3\cos^2 x \sin x \sqrt{\sin x} + \frac{\cos^4 x}{2\sqrt{\sin x}} = 0\).
Multiply through by \(2\sqrt{\sin x}\):
\(-6\cos^2 x \sin^2 x + \cos^4 x = 0\).
Factor out \(\cos^2 x\):
\(\cos^2 x (-6\sin^2 x + \cos^2 x) = 0\).
Since \(0 < x < \frac{\pi}{2}\), \(\cos x \neq 0\), so:
\(-6\sin^2 x + \cos^2 x = 0\).
Use \(\cos^2 x = 1 – \sin^2 x\):
\(-6\sin^2 x + 1 – \sin^2 x = 0\)
\(-7\sin^2 x = -1\)
\(\sin^2 x = \frac{1}{7}\)
\(\sin x = \frac{1}{\sqrt{7}}\)
\(x = \sin^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 0.388\) radians.
Final answer: \(x = 0.388\) (to 3 significant figures).
5. [Maximum mark: 7]
(a) By sketching a suitable pair of graphs, show that the equation \(\ln x = 3x – x^2\) has one real root.
(b) Verify by calculation that the root lies between 2 and 2.8.
(c) Use the iterative formula \(x_{n+1} = \sqrt{3x_n – \ln x_n}\) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
(a) Solution:
Sketch \(y = \ln x\) (increasing, concave down, passes through (1,0)) and \(y = 3x – x^2\) (parabola opening downward, roots at \(x = 0\) and \(x = 3\)). The graphs intersect once in the first quadrant, confirming one real root.
(b) Solution:
Evaluate at \(x = 2\): \(\ln 2 \approx 0.693\), \(3(2) – 2^2 = 2\). Since \(0.693 < 2\), the root is to the right of 2.
Evaluate at \(x = 2.8\): \(\ln 2.8 \approx 1.030\), \(3(2.8) – (2.8)^2 \approx 0.56\). Since \(1.030 > 0.56\), the root is to the left of 2.8.
(c) Solution:
Start with \(x_0 = 2.4\):
\(x_1 = \sqrt{3(2.4) – \ln 2.4} \approx 2.5346\)
\(x_2 = \sqrt{3(2.5346) – \ln 2.5346} \approx 2.6069\)
\(x_3 = \sqrt{3(2.6069) – \ln 2.6069} \approx 2.6275\)
\(x_4 = \sqrt{3(2.6275) – \ln 2.6275} \approx 2.6326\)
\(x_5 = \sqrt{3(2.6326) – \ln 2.6326} \approx 2.6339\)
The root converges to \(2.63\) (to 2 decimal places).
6. [Maximum mark: 8]
The variables \(x\) and \(y\) satisfy the differential equation \(\frac{dy}{dx} = xe^{y – x}\), and \(y = 0\) when \(x = 0\).
(a) Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
(b) Find the value of \(y\) when \(x = 1\), giving your answer in the form \(a – \ln b\), where \(a\) and \(b\) are integers.
▶️Answer/Explanation
(a) Solution:
Separate variables:
\(\frac{dy}{e^y} = xe^{-x}dx\).
Integrate both sides:
\(\int e^{-y} dy = \int xe^{-x} dx\).
Left side: \(-e^{-y} + C_1\).
Right side (integration by parts): \(-xe^{-x} – e^{-x} + C_2\).
Combine constants and solve for \(y\):
\(-e^{-y} = -xe^{-x} – e^{-x} + C\).
Use initial condition \(y(0) = 0\):
\(-1 = -0 – 1 + C \Rightarrow C = 0\).
Thus:
\(-e^{-y} = -e^{-x}(x + 1)\).
Take natural logarithm:
\(-y = \ln(e^{-x}(x + 1))\).
Simplify:
\(y = -\ln(e^{-x}(x + 1)) = x – \ln(x + 1)\).
(b) Solution:
Substitute \(x = 1\):
\(y = 1 – \ln 2\).
7. [Maximum mark: 9]
The equation of a curve is \(x^3 + 3x^2y – y^3 = 3\).
(a) Show that \(\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 – x^2}\).
(b) Find the coordinates of the points on the curve where the tangent is parallel to the \(x\)-axis.
▶️Answer/Explanation
(a) Solution:
Differentiate implicitly:
\(3x^2 + 6xy + 3x^2\frac{dy}{dx} – 3y^2\frac{dy}{dx} = 0\).
Collect terms involving \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(3x^2 – 3y^2) = -3x^2 – 6xy\).
Factor and solve:
\(\frac{dy}{dx} = \frac{-3x^2 – 6xy}{3x^2 – 3y^2} = \frac{x^2 + 2xy}{y^2 – x^2}\).
(b) Solution:
For horizontal tangent, set numerator of \(\frac{dy}{dx}\) to zero:
\(x^2 + 2xy = 0 \Rightarrow x(x + 2y) = 0\).
Case 1: \(x = 0\). Substitute into curve equation:
\(0 + 0 – y^3 = 3 \Rightarrow y = -\sqrt[3]{3}\). Point: \((0, -\sqrt[3]{3})\).
Case 2: \(x = -2y\). Substitute into curve equation:
\((-2y)^3 + 3(-2y)^2y – y^3 = 3\)
\(-8y^3 + 12y^3 – y^3 = 3\)
\(3y^3 = 3 \Rightarrow y = 1\), \(x = -2\). Point: \((-2, 1)\).
8. [Maximum mark: 10]
Let \(f(x) = \frac{x^2 + 9x}{(3x – 1)(x^2 + 3)}\).
(a) Express \(f(x)\) in partial fractions.
(b) Hence find \(\int_1^3 f(x) dx\), giving your answer in a simplified exact form.
▶️Answer/Explanation
(a) Solution:
Assume partial fractions form:
\(\frac{x^2 + 9x}{(3x – 1)(x^2 + 3)} = \frac{A}{3x – 1} + \frac{Bx + C}{x^2 + 3}\).
Multiply through by denominator:
\(x^2 + 9x = A(x^2 + 3) + (Bx + C)(3x – 1)\).
Solve for \(A\), \(B\), \(C\):
Let \(x = \frac{1}{3}\): \(\frac{1}{9} + 3 = A\left(\frac{1}{9} + 3\right) \Rightarrow A = 1\).
Compare coefficients:
\(x^2\): \(1 = A + 3B \Rightarrow B = 0\).
\(x\): \(9 = -B + 3C \Rightarrow C = 3\).
Thus:
\(f(x) = \frac{1}{3x – 1} + \frac{3}{x^2 + 3}\).
(b) Solution:
Integrate term by term:
\(\int_1^3 \left(\frac{1}{3x – 1} + \frac{3}{x^2 + 3}\right) dx\).
First term: \(\frac{1}{3}\ln|3x – 1|\).
Second term: \(\sqrt{3}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)\).
Evaluate from 1 to 3:
\(\left[\frac{1}{3}\ln(8) + \sqrt{3}\tan^{-1}\left(\sqrt{3}\right)\right] – \left[\frac{1}{3}\ln(2) + \sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]\).
Simplify:
\(\frac{1}{3}\ln 4 + \sqrt{3}\left(\frac{\pi}{3} – \frac{\pi}{6}\right) = \frac{2}{3}\ln 2 + \frac{\sqrt{3}\pi}{6}\).
9. [Maximum mark: 10]
The lines \(l\) and \(m\) have vector equations:
\(l: \mathbf{r} = -\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} + \lambda(2\mathbf{i} – \mathbf{j} – \mathbf{k})\),
\(m: \mathbf{r} = 5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} + \mu(a\mathbf{i} + b\mathbf{j} + \mathbf{k})\).
(a) Given that \(l\) and \(m\) intersect, show that \(2b – a = 4\).
(b) Given also that \(l\) and \(m\) are perpendicular, find the values of \(a\) and \(b\).
(c) When \(a\) and \(b\) have these values, find the position vector of the point of intersection of \(l\) and \(m\).
▶️Answer/Explanation
(a) Solution:
Equate components for intersection:
\(-1 + 2\lambda = 5 + a\mu\),
\(3 – \lambda = 4 + b\mu\),
\(4 – \lambda = 3 + \mu\).
Solve third equation for \(\mu\): \(\mu = 1 – \lambda\).
Substitute into first equation: \(-1 + 2\lambda = 5 + a(1 – \lambda)\).
Simplify: \(2\lambda + a\lambda = 6 + a\).
\(\lambda = \frac{6 + a}{2 + a}\).
Substitute \(\mu = 1 – \lambda\) into second equation: \(3 – \lambda = 4 + b(1 – \lambda)\).
Simplify: \(-1 – \lambda = b – b\lambda\).
Substitute \(\lambda\) and solve: \(2b – a = 4\).
(b) Solution:
Direction vectors: \(\mathbf{d}_l = 2\mathbf{i} – \mathbf{j} – \mathbf{k}\), \(\mathbf{d}_m = a\mathbf{i} + b\mathbf{j} + \mathbf{k}\).
Dot product must be zero for perpendicularity:
\(2a – b – 1 = 0\).
Solve system with \(2b – a = 4\):
From \(2a – b = 1\) and \(2b – a = 4\), solve to get \(a = 2\), \(b = 3\).
(c) Solution:
Substitute \(a = 2\), \(b = 3\) into intersection equations.
From \(\mu = 1 – \lambda\) and \(-1 + 2\lambda = 5 + 2(1 – \lambda)\):
\(-1 + 2\lambda = 7 – 2\lambda\)
\(4\lambda = 8 \Rightarrow \lambda = 2\).
Substitute back into \(l\):
\(\mathbf{r} = -\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} + 2(2\mathbf{i} – \mathbf{j} – \mathbf{k}) = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).
10. [Maximum mark: 11]
The complex number \(-1 + \sqrt{7}i\) is denoted by \(u\). It is given that \(u\) is a root of the equation \(2x^3 + 3x^2 + 14x + k = 0\), where \(k\) is a real constant.
(a) Find the value of \(k\).
(b) Find the other two roots of the equation.
(c) On an Argand diagram, sketch the locus of points representing complex numbers \(z\) satisfying the equation \(|z – u| = 2\).
(d) Determine the greatest value of \(\arg z\) for points on this locus, giving your answer in radians.
▶️Answer/Explanation
(a) Solution:
Substitute \(x = -1 + \sqrt{7}i\) into the equation:
\(2(-1 + \sqrt{7}i)^3 + 3(-1 + \sqrt{7}i)^2 + 14(-1 + \sqrt{7}i) + k = 0\).
Calculate powers:
\((-1 + \sqrt{7}i)^2 = 1 – 2\sqrt{7}i – 7 = -6 – 2\sqrt{7}i\).
\((-1 + \sqrt{7}i)^3 = (-1)(-6 – 2\sqrt{7}i) + \sqrt{7}i(-6 – 2\sqrt{7}i) = 6 + 2\sqrt{7}i – 6\sqrt{7}i + 14 = 20 – 4\sqrt{7}i\).
Substitute back:
\(2(20 – 4\sqrt{7}i) + 3(-6 – 2\sqrt{7}i) + 14(-1 + \sqrt{7}i) + k = 0\).
Simplify:
\(40 – 8\sqrt{7}i – 18 – 6\sqrt{7}i – 14 + 14\sqrt{7}i + k = 0\).
\(8 + k = 0 \Rightarrow k = -8\).
(b) Solution:
Since coefficients are real, the other roots are \(-1 – \sqrt{7}i\) and a real root.
Let the real root be \(r\). Sum of roots:
\(-1 + \sqrt{7}i – 1 – \sqrt{7}i + r = -\frac{3}{2} \Rightarrow r = \frac{1}{2}\).
Alternatively, factor as \((x + 1 – \sqrt{7}i)(x + 1 + \sqrt{7}i)(2x – 1)\).
(c) Solution:
The locus is a circle centered at \((-1, \sqrt{7})\) with radius 2 in the Argand diagram.
(d) Solution:
The greatest argument occurs at the tangent point above the center. Calculate the angle:
\(\theta = \tan^{-1}\left(\frac{\sqrt{7}}{-1}\right) + \sin^{-1}\left(\frac{2}{\sqrt{8}}\right) \approx 2.72\) radians.