1. [Maximum mark: 2]
When the polynomial \( ax^3 + 4ax^2 – 7x – 5 \) is divided by \( (x + 2) \), the remainder is 33. Find the value of the constant \( a \).
▶️Answer/Explanation
Solution:
Using the Remainder Theorem, substitute \( x = -2 \) into the polynomial and set it equal to 33:
\[ a(-2)^3 + 4a(-2)^2 – 7(-2) – 5 = 33 \]
\[ -8a + 16a + 14 – 5 = 33 \]
\[ 8a + 9 = 33 \]
\[ 8a = 24 \]
\[ a = 3 \]
2. [Maximum mark: 5]
Solve the equation \( \sec \theta \cos(\theta – 60^\circ) = 4 \) for \( -180^\circ < \theta < 180^\circ \).
▶️Answer/Explanation
Solution:
Rewrite the equation using trigonometric identities:
\[ \sec \theta \cos(\theta – 60^\circ) = \frac{\cos(\theta – 60^\circ)}{\cos \theta} = 4 \]
Using the cosine subtraction formula:
\[ \frac{\cos \theta \cos 60^\circ + \sin \theta \sin 60^\circ}{\cos \theta} = 4 \]
\[ \cos 60^\circ + \tan \theta \sin 60^\circ = 4 \]
Substitute known values \( \cos 60^\circ = 0.5 \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ 0.5 + \tan \theta \left(\frac{\sqrt{3}}{2}\right) = 4 \]
\[ \tan \theta \left(\frac{\sqrt{3}}{2}\right) = 3.5 \]
\[ \tan \theta = \frac{7}{\sqrt{3}} \]
Find the reference angle:
\[ \theta = \tan^{-1}\left(\frac{7}{\sqrt{3}}\right) \approx 76.1^\circ \]
The solutions in the given range are:
\[ \theta \approx 76.1^\circ \text{ and } \theta \approx -103.9^\circ \]
3. [Maximum mark: 5]
The diagram shows the curve with equation \( y = 6e^{-\frac{1}{2}x} \). The points on the curve with \( x \)-coordinates 0 and 2 are denoted by \( A \) and \( B \) respectively.
(a) Find the exact gradient of the curve at \( B \). [2]
The diagram shows the curve with equation y = 4 − 2 cos 2x, for 0 < x < 2π. At the point A, the
gradient of the curve is 4. The point B is a minimum point. The x-coordinates of A and B are
a and b respectively.
(b) Find the exact area of the shaded region enclosed by the curve, the line through \( A \) parallel to the \( x \)-axis and the line through \( B \) parallel to the \( y \)-axis. [3]
▶️Answer/Explanation
(a) Solution:
Differentiate and evaluate at \( x = 2 \):
\[ \frac{dy}{dx} = -3e^{-\frac{1}{2}x} \]
At \( x = 2 \): \( -\frac{3}{e} \) (b) Solution:
Area calculation:
\[ \int_{0}^{2} \left(6 – 6e^{-\frac{1}{2}x}\right) dx \]
\[ = \left[6x + 12e^{-\frac{1}{2}x}\right]_{0}^{2} \]
\[ = \frac{12}{e} \]
4. [Maximum mark: 9]
(a) Sketch, on the same diagram, the graphs of \( y = |3 – x| \) and \( y = 9 – 2x \). [2]
(b) Solve the inequality \( |3 – x| > 9 – 2x \). [3]
(c) Use logarithms to solve \( 2^{3x-10} < 500 \), giving answer as \( x < a \) (3 s.f.). [3]
(d) List integers satisfying both inequalities from (b) and (c). [1]
▶️Answer/Explanation
(a) Solution:
– V-shaped graph for \( y = |3 – x| \) (vertex at \( x = 3 \))
– Straight line for \( y = 9 – 2x \) (slope -2, y-intercept 9) (b) Solution:
Case analysis gives solution \( x > 4 \) (c) Solution:
\[ (3x-10)\ln2 < \ln500 \]
\[ x < 6.32 \] (to 3 s.f.) (d) Solution:
Integers satisfying \( 4 < x < 6.32 \): 5 and 6
5. [Maximum mark: 8]
(a) Find the quotient when \( 6x^3 – 5x^2 – 24x – 4 \) is divided by \( (2x + 1) \), showing remainder is 6. [3]
(b) Hence find \( \int_{2}^{7} \frac{6x^3 – 5x^2 – 24x – 4}{2x + 1} dx \) in form \( a + \ln b \). [5]
▶️Answer/Explanation
(a) Solution:
Quotient: \( 3x^2 – 4x – 10 \)
Remainder: 6 (verified) (b) Solution:
\[ \int \left(3x^2 – 4x – 10 + \frac{6}{2x+1}\right) dx \]
\[ = x^3 – 2x^2 – 10x + 3\ln|2x+1| \]
Evaluated from 2 to 7:
\[ = 195 + \ln 27 \]
6. [Maximum mark: 10]
The curve has parametric equations \( x = 3\ln(2t-3) \), \( y = 4t\ln t \).
(a) Find the exact gradient at point \( A \) where curve crosses y-axis. [5]
(b) Show that parameter \( t \) at point \( B \) (gradient=12) satisfies \( t = \frac{9}{1+\ln t} + \frac{3}{2} \). [2]
(c) Use iteration to find \( t \) at \( B \) (3 s.f.). [3]
▶️Answer/Explanation
(a) Solution:
\[ \frac{dx}{dt} = \frac{6}{2t-3}, \frac{dy}{dt} = 4\ln t + 4 \]
At \( A \), \( t = 2 \):
Gradient \( = \frac{2}{3}(\ln 2 + 1) \) (b) Solution:
Set \( \frac{dy}{dx} = 12 \) and rearrange to given form (c) Solution:
Iteration with \( t_0 = 5 \):
\( t_1 = 4.9626 \), \( t_2 = 4.9612 \), \( t_3 = 4.9611 \)
Final answer: 4.96 (3 s.f.)
7. [Maximum mark: 11]
(a) Prove \( \sin 2x(\cot x + 3\tan x) \equiv 4 – 2\cos 2x \). [4]
(b) Hence find exact value of \( \cot \frac{\pi}{12} + 3\tan \frac{\pi}{12} \). [2]
(c) For curve \( y = 4 – 2\cos 2x \), show \( \int_{a}^{b} (4-2\cos 2x) dx = 3\pi + 1 \). [5]
▶️Answer/Explanation
(a) Solution:
Convert to \( \sin \) and \( \cos \):
\[ \sin 2x\left(\frac{\cos x}{\sin x} + 3\frac{\sin x}{\cos x}\right) \]
Simplify to \( 2\cos^2x + 6\sin^2x \) and use identities to get RHS (b) Solution:
Using part (a) with \( x = \frac{\pi}{12} \):
\[ 8 – 2\sqrt{3} \] (c) Solution:
Find \( a = \frac{\pi}{2} \), \( b = \pi \) where gradient=4 and minimum
Integrate and evaluate:
\[ [4x – \sin 2x]_{\pi/2}^{\pi} = 3\pi + 1 \]