1. [Maximum mark: 8]
The polynomial \( p(x) \) is defined by \( p(x) = ax^3 + bx – 10 \), where \( a \) and \( b \) are constants. It is given that \( (x + 2) \) is a factor of \( p(x) \) and that the remainder is \(-55\) when \( p(x) \) is divided by \( (x + 3) \).
(a) Find the values of \( a \) and \( b \).
(b) Hence factorise \( p(x) \) completely.
▶️Answer/Explanation
(a) Substitute \( x = -2 \) into \( p(x) \):
\( p(-2) = a(-2)^3 + b(-2) – 10 = -8a – 2b – 10 = 0 \)
Substitute \( x = -3 \) into \( p(x) \):
\( p(-3) = a(-3)^3 + b(-3) – 10 = -27a – 3b – 10 = -55 \)
Solve the system:
1. \(-8a – 2b = 10\)
2. \(-27a – 3b = -45\) → \( 9a + b = 15 \)
From equation 2: \( b = 15 – 9a \). Substitute into equation 1:
\(-8a – 2(15 – 9a) = 10 \) → \(-8a – 30 + 18a = 10 \) → \( 10a = 40 \) → \( a = 4 \).
Then \( b = 15 – 9(4) = -21 \).
Answer: \( a = 4 \), \( b = -21 \).
(b) Using \( a = 4 \) and \( b = -21 \), \( p(x) = 4x^3 – 21x – 10 \).
Since \( (x + 2) \) is a factor, perform polynomial division or use inspection:
\( p(x) = (x + 2)(4x^2 – 8x – 5) \).
Factorise the quadratic: \( 4x^2 – 8x – 5 = (2x + 1)(2x – 5) \).
Answer: \( p(x) = (x + 2)(2x + 1)(2x – 5) \).
2. [Maximum mark: 7]
(a) Sketch, on the same diagram, the graphs of \( y = x + 3 \) and \( y = |2x – 1| \).
(b) Solve the equation \( x + 3 = |2x – 1| \).
(c) Find the value of \( y \) such that \( 5^{xy} + 3 = |2 \times 5^{xy} – 1| \). Give your answer correct to 3 significant figures.
▶️Answer/Explanation
(a) The graph of \( y = x + 3 \) is a straight line with gradient 1 and y-intercept 3.
The graph of \( y = |2x – 1| \) is V-shaped with its vertex at \( x = 0.5 \).
(b) Solve \( x + 3 = 2x – 1 \) → \( x = 4 \).
Solve \( x + 3 = -(2x – 1) \) → \( x + 3 = -2x + 1 \) → \( 3x = -2 \) → \( x = -\frac{2}{3} \).
Answer: \( x = 4 \) or \( x = -\frac{2}{3} \).
(c) Let \( k = 5^{xy} \). The equation becomes \( k + 3 = |2k – 1| \).
Using the positive root from part (b), \( k = 4 \).
Thus, \( 5^{xy} = 4 \) → \( xy = \log_5 4 \).
For \( x = 1 \), \( y = \log_5 4 \approx 1.72 \).
Answer: \( y \approx 1.72 \).
3. [Maximum mark: 6]
The curve with equation \( y = 5x – 2 \tan 2x \) has exactly one stationary point in the interval \( 0 \leq x < \frac{1}{4}\pi \). Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures.
▶️Answer/Explanation
Find the derivative of \( y \):
\( \frac{dy}{dx} = 5 – 4 \sec^2 2x \).
Set the derivative to zero for stationary points:
\( 5 – 4 \sec^2 2x = 0 \) → \( \sec^2 2x = \frac{5}{4} \) → \( \cos^2 2x = \frac{4}{5} \).
Thus, \( \cos 2x = \frac{2}{\sqrt{5}} \) (since \( \cos 2x > 0 \) in the given interval).
Solve for \( x \): \( 2x = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \) → \( x \approx 0.232 \) radians.
Substitute \( x \) back into \( y \): \( y \approx 5(0.232) – 2 \tan(2 \times 0.232) \approx 0.159 \).
Answer: Coordinates are \( (0.232, 0.159) \).
4. [Maximum mark: 5]
Given that \( \int_{a}^{a+14} \frac{1}{3x} \, dx = \ln 2 \), find the value of the positive constant \( a \).
▶️Answer/Explanation
Integrate \( \frac{1}{3x} \):
\( \int \frac{1}{3x} \, dx = \frac{1}{3} \ln x + C \).
Apply the limits:
\( \frac{1}{3} \ln(a + 14) – \frac{1}{3} \ln a = \ln 2 \).
Simplify: \( \ln\left(\frac{a + 14}{a}\right) = 3 \ln 2 \) → \( \frac{a + 14}{a} = 2^3 = 8 \).
Solve for \( a \): \( a + 14 = 8a \) → \( 7a = 14 \) → \( a = 2 \).
Answer: \( a = 2 \).
5. [Maximum mark: 6]
A curve has equation \( x^2 + 4x \cos 3y = 6 \). Find the exact value of the gradient of the normal to the curve at the point \( \left(\sqrt{2}, \frac{1}{12}\pi\right) \).
▶️Answer/Explanation
Step 1: Implicit Differentiation
Differentiate both sides with respect to \( x \): \[ 2x + 4 \cos 3y – 12x \sin 3y \frac{dy}{dx} = 0. \] Step 2: Substitute the Point
At \( (\sqrt{2}, \frac{1}{12}\pi) \): \[ 2\sqrt{2} + 4 \cos\left(\frac{\pi}{4}\right) – 12\sqrt{2} \sin\left(\frac{\pi}{4}\right) \frac{dy}{dx} = 0. \] Simplify using \( \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \): \[ 2\sqrt{2} + 2\sqrt{2} – 12\sqrt{2} \cdot \frac{\sqrt{2}}{2} \frac{dy}{dx} = 0 \implies 4\sqrt{2} – 12 \frac{dy}{dx} = 0. \] Step 3: Solve for \( \frac{dy}{dx} \)
\[ \frac{dy}{dx} = \frac{4\sqrt{2}}{12} = \frac{\sqrt{2}}{3}. \] Step 4: Gradient of the Normal
Negative reciprocal: \[ m_{\text{normal}} = -\frac{1}{\frac{dy}{dx}} = -\frac{3}{\sqrt{2}}. \] Rationalize: \[ m_{\text{normal}} = -\frac{3\sqrt{2}}{2}. \] Answer: The exact gradient of the normal is \( -\frac{3\sqrt{2}}{2} \).
6. [Maximum mark: 8]
(a) By sketching a suitable pair of graphs on the same diagram, show that the equation \( \ln x = 2e^{-x} \) has exactly one root.
(b) Verify by calculation that the root lies between 1.5 and 1.6.
(c) Show that if a sequence of values given by the iterative formula \( x_{n+1} = e^{2e^{-x_n}} \) converges, then it converges to the root of the equation in part (a).
(d) Use the iterative formula in part (c) to determine the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
▶️Answer/Explanation
(a) Sketch \( y = \ln x \) (increasing, passes through \( (1, 0) \)) and \( y = 2e^{-x} \) (decreasing, passes through \( (0, 2) \)). The graphs intersect once, so there is exactly one root.
(b) Let \( f(x) = \ln x – 2e^{-x} \).
\( f(1.5) \approx \ln 1.5 – 2e^{-1.5} \approx 0.4055 – 0.4463 = -0.0408 \).
\( f(1.6) \approx \ln 1.6 – 2e^{-1.6} \approx 0.4700 – 0.4066 = 0.0634 \).
Since \( f(1.5) < 0 \) and \( f(1.6) > 0 \), the root lies in \( (1.5, 1.6) \).
(c) At convergence, \( x_{n+1} = x_n = x \). Thus, \( x = e^{2e^{-x}} \). Taking natural logs: \( \ln x = 2e^{-x} \), which is the original equation.
(d) Using \( x_0 = 1.5 \):
\( x_1 = e^{2e^{-1.5}} \approx 1.5308 \).
\( x_2 \approx e^{2e^{-1.5308}} \approx 1.5389 \).
\( x_3 \approx e^{2e^{-1.5389}} \approx 1.5416 \).
\( x_4 \approx e^{2e^{-1.5416}} \approx 1.5426 \).
The root converges to \( 1.54 \) (3 s.f.).
7. [Maximum mark: 10]
(a) Prove that \( 4 \sin x \sin\left(x + \frac{1}{6}\pi\right) = \sqrt{3} – \sqrt{3} \cos 2x + \sin 2x \).
(b) Find the exact value of \( \int_{0}^{\frac{5}{6}\pi} 4 \sin x \sin\left(x + \frac{1}{6}\pi\right) \, dx \).
(c) Find the smallest positive value of \( y \) satisfying the equation \( 4 \sin(2y) \sin\left(2y + \frac{1}{6}\pi\right) = \sqrt{3} \). Give your answer in an exact form.
▶️Answer/Explanation
(a) Expand \( \sin\left(x + \frac{\pi}{6}\right) \):
\( 4 \sin x \left(\sin x \cos\frac{\pi}{6} + \cos x \sin\frac{\pi}{6}\right) = 4 \sin x \left(\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x\right) \).
Simplify: \( 2\sqrt{3} \sin^2 x + 2 \sin x \cos x \).
Use identities \( \sin^2 x = \frac{1 – \cos 2x}{2} \) and \( \sin x \cos x = \frac{\sin 2x}{2} \):
\( 2\sqrt{3} \left(\frac{1 – \cos 2x}{2}\right) + 2 \left(\frac{\sin 2x}{2}\right) = \sqrt{3} – \sqrt{3} \cos 2x + \sin 2x \).
(b) Using part (a), the integral becomes:
\( \int_{0}^{\frac{5}{6}\pi} \left(\sqrt{3} – \sqrt{3} \cos 2x + \sin 2x\right) \, dx \).
Integrate term by term:
\( \sqrt{3}x – \frac{\sqrt{3}}{2} \sin 2x – \frac{1}{2} \cos 2x \) evaluated from \( 0 \) to \( \frac{5}{6}\pi \).
Substitute limits:
\( \sqrt{3} \left(\frac{5}{6}\pi\right) – \frac{\sqrt{3}}{2} \sin\left(\frac{5}{3}\pi\right) – \frac{1}{2} \cos\left(\frac{5}{3}\pi\right) – \left(0 – 0 – \frac{1}{2}\right) \).
Simplify: \( \frac{5\sqrt{3}}{6}\pi – \frac{\sqrt{3}}{2} \left(-\frac{\sqrt{3}}{2}\right) – \frac{1}{2} \left(\frac{1}{2}\right) + \frac{1}{2} \).
Final result: \( \frac{5\sqrt{3}}{6}\pi + \frac{3}{4} – \frac{1}{4} + \frac{1}{2} = \frac{5\sqrt{3}}{6}\pi + 1 \).
(c) Using the identity from part (a):
\( 4 \sin(2y) \sin\left(2y + \frac{\pi}{6}\right) = \sqrt{3} – \sqrt{3} \cos 4y + \sin 4y = \sqrt{3} \).
Thus, \( -\sqrt{3} \cos 4y + \sin 4y = 0 \) → \( \tan 4y = \sqrt{3} \).
Solve: \( 4y = \frac{\pi}{3} + k\pi \) → \( y = \frac{\pi}{12} + \frac{k\pi}{4} \).
The smallest positive solution is \( y = \frac{\pi}{12} \).