1. [Maximum mark: 4]
Solve the inequality \( |3x – 7| < |4x + 5| \).
▶️Answer/Explanation
State or imply the non-modulus inequality \( (3x – 7)^2 < (4x + 5)^2 \) or corresponding equation.
Attempt to solve the quadratic inequality or pair of linear equations.
Obtain critical values \( x = -12 \) and \( x = \frac{2}{7} \).
State the final answer: \( x < -12 \) or \( x > \frac{2}{7} \).
2. [Maximum mark: 6]
By first expanding \( \sin(\theta + 30^\circ) \), solve the equation \( \sin(\theta + 30^\circ) \csc \theta = 2 \) for \( 0^\circ < \theta < 360^\circ \).
▶️Answer/Explanation
Expand \( \sin(\theta + 30^\circ) \) as \( \sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ \).
Use \( \csc \theta = \frac{1}{\sin \theta} \) to simplify the equation.
Obtain a linear equation in \( \tan \theta \) or \( \cot \theta \).
Solve for \( \tan \theta \) to get \( \tan \theta = \frac{4 + \sqrt{3}}{13} \).
Find the first solution: \( \theta = 23.8^\circ \).
Find the second solution: \( \theta = 203.8^\circ \).
3. [Maximum mark: 6]
(a) Show that \( (\sec x + \cos x)^2 \) can be expressed as \( \sec^2 x + a + b \cos 2x \), where \( a \) and \( b \) are constants to be determined.
(b) Hence find the exact value of \( \int_{0}^{\frac{1}{4}\pi} (\sec x + \cos x)^2 \, dx \).
▶️Answer/Explanation
(a)
Expand \( (\sec x + \cos x)^2 \) to \( \sec^2 x + 2 + \cos^2 x \).
Express \( \cos^2 x \) as \( \frac{1 + \cos 2x}{2} \) and simplify to \( \sec^2 x + \frac{5}{2} + \frac{1}{2} \cos 2x \).
(b)
Integrate \( \sec^2 x + \frac{5}{2} + \frac{1}{2} \cos 2x \) to get \( \tan x + \frac{5}{2}x + \frac{1}{4} \sin 2x \).
Apply the limits \( 0 \) to \( \frac{\pi}{4} \).
Substitute and simplify to obtain the exact value \( \frac{5}{4} + \frac{5}{8}\pi \).
4. [Maximum mark: 8]
A curve has parametric equations \( x = \ln(2t + 6) – \ln t \), \( y = t \ln t \).
(a) Find the value of \( t \) at the point \( P \) on the curve for which \( x = \ln 4 \).
(b) Find the exact gradient of the curve at \( P \).
▶️Answer/Explanation
(a)
Set \( x = \ln 4 \) and simplify to \( \frac{2t + 6}{t} = 4 \).
Solve for \( t \) to get \( t = 3 \).
(b)
Find \( \frac{dx}{dt} = \frac{2}{2t + 6} – \frac{1}{t} \).
Find \( \frac{dy}{dt} = \ln t + 1 \) using the product rule.
Compute \( \frac{dy}{dx} = \frac{\ln t + 1}{\frac{2}{2t + 6} – \frac{1}{t}} \).
Substitute \( t = 3 \) to get the gradient \( -6(\ln 3 + 1) \).
5. [Maximum mark: 8]
The diagram shows the curve with equation \( y = \frac{3x + 2}{\ln x} \). The curve has a minimum point \( M \).
(a) Find an expression for \( \frac{dy}{dx} \) and show that the \( x \)-coordinate of \( M \) satisfies the equation \( x = \frac{3x + 2}{3 \ln x} \).
(b) Use the equation in part (a) to show by calculation that the \( x \)-coordinate of \( M \) lies between 3 and 4.
(c) Use an iterative formula, based on the equation in part (a), to find the \( x \)-coordinate of \( M \) correct to 5 significant figures. Give the result of each iteration to 7 significant figures.
▶️Answer/Explanation
(a)
Use the quotient rule to find \( \frac{dy}{dx} = \frac{3 \ln x – \frac{3x + 2}{x}}{(\ln x)^2} \).
Set \( \frac{dy}{dx} = 0 \) and simplify to \( x = \frac{3x + 2}{3 \ln x} \).
(b)
Evaluate the expression for \( x = 3 \) and \( x = 4 \).
Show sign change to confirm the root lies between 3 and 4.
(c)
Use the iterative formula \( x_{n+1} = \frac{3x_n + 2}{3 \ln x_n} \).
Perform iterations to obtain the final answer \( x \approx 3.3223 \).
6. [Maximum mark: 9]
(a) Use the trapezium rule with three intervals to approximate \( \int_{1}^{4} \frac{6}{1 + \sqrt{x}} \, dx \). Give your answer correct to 5 significant figures.
(b) Find the exact value of \( \int_{1}^{4} 2e^{\frac{1}{2}x – 2} \, dx \).
(c)
The diagram shows the curves \( y = \frac{6}{1 + \sqrt{x}} \) and \( y = 2e^{\frac{1}{2}x – 2} \), which meet at \( x = 4 \). The shaded region is bounded by the two curves and the line \( x = 1 \). Use your answers to parts (a) and (b) to approximate the area of the shaded region. Give your answer correct to 3 significant figures.
(d) State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate.of the exact area of the shaded region.
▶️Answer/Explanation
(a)
Calculate \( y \)-values at \( x = 1, 2, 3, 4 \).
Apply the trapezium rule to obtain \( 7.1814 \).
(b)
Integrate to get \( 4e^{\frac{1}{2}x – 2} \).
Evaluate from 1 to 4 to obtain \( 4 – 4e^{-\frac{3}{2}} \).
(c)
Subtract part (b) result from part (a) result.
Obtain the approximate area \( 4.07 \).
(d)
State “over-estimate” because the tops of the trapezia lie above the curve.
7. [Maximum mark: 9]
The polynomial \( p(x) \) is defined by \( p(x) = ax^3 – 11x^2 – 19x – a \), where \( a \) is a constant. It is given that \( (x – 3) \) is a factor of \( p(x) \).
(a) Find the value of \( a \).
(b) When \( a \) has this value, factorise \( p(x) \) completely.
(c) Hence find the exact values of \( y \) that satisfy the equation \( p(e^y + e^{-y}) = 0 \).
▶️Answer/Explanation
(a)
Substitute \( x = 3 \) into \( p(x) = 0 \) and solve for \( a \).
Obtain \( a = 6 \).
(b)
Perform polynomial division or inspection to factorise \( p(x) \).
Obtain \( (x – 3)(3x + 2)(2x + 1) \).
(c)
Set \( e^y + e^{-y} \) equal to the roots \( 3, -\frac{2}{3}, -\frac{1}{2} \).
Solve \( e^y + e^{-y} = 3 \) using the quadratic formula to get \( y = \ln \left( \frac{3 \pm \sqrt{5}}{2} \right) \).
Reject negative roots as \( e^y + e^{-y} > 0 \).