1. [Maximum mark: 4]
The coefficient of \( x^3 \) in the expansion of \( \left( p + \frac{1}{px} \right)^4 \) is 144. Find the possible values of the constant \( p \).
▶️Answer/Explanation
The general term in the expansion is \( \binom{4}{k} p^{4-k} \left( \frac{1}{px} \right)^k \). For the coefficient of \( x^3 \), set \( -k = 3 \) ⇒ \( k = -3 \) (invalid). Alternatively, recognize the term for \( x^3 \) arises when \( k = 1 \):
\( \binom{4}{1} p^{3} \left( \frac{1}{p} \right)^1 = 4p^2 \).
Given \( 4p^2 = 144 \) ⇒ \( p^2 = 36 \) ⇒ \( p = \pm 6 \).
2. [Maximum mark: 3]
The diagram shows part of the curve with equation \( y = p \sin(q\theta) + r \), where \( p, q \) and \( r \) are constants.
(a) State the value of \( p \).
(b) State the value of \( q \).
(c) State the value of \( r \).
▶️Answer/Explanation
(a) Amplitude \( p = 3 \).
(b) Period \( \frac{2\pi}{q} = 4\pi \) ⇒ \( q = \frac{1}{2} \).
(c) Vertical shift \( r = -2 \).
3. [Maximum mark: 5]
An arithmetic progression has first term 4 and common difference \( d \). The sum of the first \( n \) terms of the progression is 5863.
(a) Show that \( (n – 1)d = \frac{11726}{n} – 8 \).
(b) Given that the \( n \)th term is 139, find the values of \( n \) and \( d \), giving the value of \( d \) as a fraction.
▶️Answer/Explanation
(a) Sum formula: \( S_n = \frac{n}{2} [8 + (n-1)d] = 5863 \) ⇒ \( (n-1)d = \frac{11726}{n} – 8 \).
(b) \( n \)th term: \( 4 + (n-1)d = 139 \) ⇒ \( (n-1)d = 135 \). Substitute into (a):
\( \frac{11726}{n} – 8 = 135 \) ⇒ \( n = 82 \).
Then \( d = \frac{135}{81} = \frac{5}{3} \).
4. [Maximum mark: 5]
(a) The curve with equation \( y = x^2 + 2x – 5 \) is translated by \( \begin{pmatrix} -1 \\ 3 \end{pmatrix} \). Find the equation of the translated curve, giving your answer in the form \( y = ax^2 + bx + c \). [3]
(b) The curve with equation \( y = x^2 + 2x – 5 \) is transformed to a curve with equation \( y = 4x^2 + 4x – 5 \). Describe fully the single transformation that has been applied. [2]
▶️Answer/Explanation
(a) Substitute \( x \rightarrow x + 1 \) and \( y \rightarrow y – 3 \):
\( y – 3 = (x+1)^2 + 2(x+1) – 5 \) ⇒ \( y = x^2 + 4x + 1 \).
(b) The transformation is a horizontal stretch with scale factor \( \frac{1}{2} \) (or stretch parallel to the \( x \)-axis).
5. [Maximum mark: 7]
(a) Solve the equation \( 6\sqrt{y} + \frac{2}{\sqrt{y}} – 7 = 0 \). [4]
(b) Hence solve the equation \( 6\sqrt{\tan x} + \frac{2}{\sqrt{\tan x}} – 7 = 0 \) for \( 0^\circ \leq x \leq 360^\circ \). [3]
▶️Answer/Explanation
(a) Let \( u = \sqrt{y} \): \( 6u + \frac{2}{u} – 7 = 0 \) ⇒ \( 6u^2 – 7u + 2 = 0 \).
Solutions: \( u = \frac{1}{2}, \frac{2}{3} \) ⇒ \( y = \frac{1}{4}, \frac{4}{9} \).
(b) \( \tan x = \frac{1}{4}, \frac{4}{9} \). Solutions: \( x = 14.0^\circ, 194.0^\circ, 24.0^\circ, 204.0^\circ \).
6. [Maximum mark: 8]
The function \( f \) is defined by \( f(x) = 2x^2 – 16x + 23 \) for \( x < 3 \).
(a) Express \( f(x) \) in the form \( 2(x + a)^2 + b \).
(b) Find the range of \( f \).
(c) Find an expression for \( f^{-1}(x) \).
(d) The function \( g \) is defined by \( g(x) = 2x + 4 \) for \( x < -1 \). Find and simplify an expression for \( fg(x) \).
▶️Answer/Explanation
(a) Complete the square: \( f(x) = 2(x – 4)^2 – 9 \).
(b) Range: \( f(x) > -9 \).
(c) Inverse: \( y = 2(x – 4)^2 – 9 \) ⇒ \( x = 4 \pm \sqrt{\frac{y + 9}{2}} \). Since \( x < 3 \), \( f^{-1}(x) = 4 – \sqrt{\frac{x + 9}{2}} \).
(d) \( fg(x) = f(2x + 4) = 2(2x + 4 – 4)^2 – 9 = 8x^2 – 9 \).
7. [Maximum mark: 7]
The diagram shows the circle with equation \( (x-2)^2 + (y+4)^2 = 20 \) and with centre \( C \). The point \( B \) has coordinates \( (0, 2) \) and the line segment \( BC \) intersects the circle at \( P \).
(a) Find the equation of \( BC \).
(b) Hence find the coordinates of \( P \), giving your answer in exact form.
▶️Answer/Explanation
(a) Centre \( C(2, -4) \). Gradient of \( BC \): \( \frac{2 – (-4)}{0 – 2} = -3 \). Equation: \( y = -3x + 2 \).
(b) Substitute \( y = -3x + 2 \) into the circle equation:
\( (x-2)^2 + (-3x + 6)^2 = 20 \) ⇒ \( 10x^2 – 40x + 20 = 0 \) ⇒ \( x = 2 \pm \sqrt{2} \).
Since \( P \) is between \( B \) and \( C \), \( x = 2 – \sqrt{2} \), \( y = 3\sqrt{2} – 4 \).
8. [Maximum mark: 8]
The diagram shows the curve with equation \( y = \sqrt{x} + \frac{4}{\sqrt{x}} \). The line \( y = 5 \) intersects the curve at the points \( A(1, 5) \) and \( B(16, 5) \).
(a) Find the equation of the tangent to the curve at the point \( A \).
(b) Calculate the area of the shaded region.
▶️Answer/Explanation
(a) Derivative: \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} – \frac{2}{x^{3/2}} \). At \( A(1, 5) \), gradient \( = -\frac{3}{2} \).
Tangent equation: \( y – 5 = -\frac{3}{2}(x – 1) \).
(b) Area under curve: \( \int_{1}^{16} \left( \sqrt{x} + \frac{4}{\sqrt{x}} \right) dx = \left[ \frac{2}{3}x^{3/2} + 8x^{1/2} \right]_{1}^{16} = 66 \).
Area under line: \( 5 \times 15 = 75 \). Shaded area: \( 75 – 66 = 9 \).
9. [Maximum mark: 8]
The diagram shows triangle \( ABC \) with \( AB = BC = 6 \) cm and angle \( ABC = 1.8 \) radians. The arc \( CD \) is part of a circle with centre \( A \) and \( ABD \) is a straight line.
(a) Find the perimeter of the shaded region.
(b) Find the area of the shaded region.
▶️Answer/Explanation
(a) Using cosine rule: \( AC = \sqrt{6^2 + 6^2 – 2 \times 6 \times 6 \cos 1.8} \approx 9.40 \) cm.
Angle \( CAD = \pi – 1.8 \) radians. Arc length \( CD = 9.40 \times (\pi – 1.8) \approx 6.31 \) cm.
Perimeter: \( 6 + (9.40 – 6) + 6.31 \approx 15.7 \) cm.
(b) Area of sector \( CAD = \frac{1}{2} \times 9.40^2 \times (\pi – 1.8) \approx 29.64 \) cm².
Area of triangle \( ABC = \frac{1}{2} \times 6 \times 6 \times \sin 1.8 \approx 17.53 \) cm².
Shaded area: \( 29.64 – 17.53 \approx 12.1 \) cm².
10. [Maximum mark: 10]
The function \( f \) is defined by \( f(x) = (4x + 2)^{-2} \) for \( x > -\frac{1}{2} \).
(a) Find \( \int_{1}^{\infty} f(x) \, dx \).
(b) A point is moving along the curve \( y = f(x) \) in such a way that, as it passes through the point \( A \), its \( y \)-coordinate is decreasing at the rate of \( k \) units per second and its \( x \)-coordinate is increasing at the rate of \( k \) units per second. Find the coordinates of \( A \).
▶️Answer/Explanation
(a) \( \int (4x + 2)^{-2} \, dx = -\frac{1}{4(4x + 2)} \). Evaluating from 1 to \( \infty \): \( 0 – \left( -\frac{1}{24} \right) = \frac{1}{24} \).
(b) Given \( \frac{dy}{dt} = -k \) and \( \frac{dx}{dt} = k \), then \( \frac{dy}{dx} = -1 \).
Derivative: \( \frac{dy}{dx} = -8(4x + 2)^{-3} \). Set \( -8(4x + 2)^{-3} = -1 \) ⇒ \( x = 0 \).
Substituting \( x = 0 \) into \( f(x) \): \( y = \frac{1}{4} \). Coordinates of \( A \): \( (0, \frac{1}{4}) \).
11. [Maximum mark: 10]
The point \( P \) lies on the line with equation \( y = mx + c \), where \( m \) and \( c \) are positive constants. A curve has equation \( y = -\frac{m}{x} \). There is a single point \( P \) on the curve such that the straight line is a tangent to the curve at \( P \).
(a) Find the coordinates of \( P \), giving the \( y \)-coordinate in terms of \( m \).
(b) The normal to the curve at \( P \) intersects the curve again at the point \( Q \). Find the coordinates of \( Q \) in terms of \( m \).
▶️Answer/Explanation
(a) For tangency, solve \( mx + c = -\frac{m}{x} \) with discriminant zero: \( c^2 – 4m^2 = 0 \) ⇒ \( c = 2m \).
Substituting back: \( x = -1 \), \( y = m \). Coordinates of \( P \): \( (-1, m) \).
(b) Gradient of normal: \( \frac{1}{m} \). Equation: \( y – m = \frac{1}{m}(x + 1) \).
Intersection with \( y = -\frac{m}{x} \): Solve \( -\frac{m}{x} = \frac{x + 1}{m} + m \) ⇒ \( x = m^2 \), \( y = -\frac{1}{m} \).
Coordinates of \( Q \): \( \left( m^2, -\frac{1}{m} \right) \).