1. [Maximum mark: 4]
A curve is such that its gradient at a point \((x, y)\) is given by \(\frac{dy}{dx} = x – 3x^{-\frac{1}{2}}\). It is given that the curve passes through the point \((4, 1)\).
Find the equation of the curve.
▶️Answer/Explanation
Solution:
Integrate \(\frac{dy}{dx} = x – 3x^{-\frac{1}{2}}\):
\( y = \int \left(x – 3x^{-\frac{1}{2}}\right) dx = \frac{x^2}{2} – 6x^{\frac{1}{2}} + c \).
Substitute \((4, 1)\) to find \(c\):
\( 1 = \frac{16}{2} – 6 \times 2 + c \) → \( 1 = 8 – 12 + c \) → \( c = 5 \).
Thus, the equation is \( y = \frac{1}{2}x^2 – 6x^{\frac{1}{2}} + 5 \).
2. [Maximum mark: 4]
The circle with equation \((x – 3)^2 + (y – 5)^2 = 40\) intersects the \(y\)-axis at points \(A\) and \(B\).
(a) Find the \(y\)-coordinates of \(A\) and \(B\), expressing your answers in terms of surds.
(b) Find the equation of the circle which has \(AB\) as its diameter.
▶️Answer/Explanation
(a) Substitute \(x = 0\) into the equation:
\((0 – 3)^2 + (y – 5)^2 = 40\) → \(9 + (y – 5)^2 = 40\) → \((y – 5)^2 = 31\).
Thus, \(y = 5 \pm \sqrt{31}\).
(b) The midpoint of \(AB\) is \((0, 5)\), and the radius is \(\sqrt{31}\).
The equation is \(x^2 + (y – 5)^2 = 31\).
3. [Maximum mark: 7]
(a) Show that the equation \(5 \cos \theta – \sin \theta \tan \theta + 1 = 0\) may be expressed in the form \(a \cos^2 \theta + b \cos \theta + c = 0\), where \(a\), \(b\), and \(c\) are constants to be found.
(b) Hence solve the equation \(5 \cos \theta – \sin \theta \tan \theta + 1 = 0\) for \(0 < \theta < 2\pi\).
▶️Answer/Explanation
(a) Rewrite \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\):
\(5 \cos \theta – \sin \theta \cdot \frac{\sin \theta}{\cos \theta} + 1 = 0\) → \(5 \cos^2 \theta – \sin^2 \theta + \cos \theta = 0\).
Use \(\sin^2 \theta = 1 – \cos^2 \theta\):
\(5 \cos^2 \theta – (1 – \cos^2 \theta) + \cos \theta = 0\) → \(6 \cos^2 \theta + \cos \theta – 1 = 0\).
Thus, \(a = 6\), \(b = 1\), \(c = -1\).
(b) Solve \(6 \cos^2 \theta + \cos \theta – 1 = 0\):
Let \(u = \cos \theta\): \(6u^2 + u – 1 = 0\) → \((3u – 1)(2u + 1) = 0\).
Solutions: \(\cos \theta = \frac{1}{3}\) → \(\theta \approx 1.23, 5.05\) radians;
\(\cos \theta = -\frac{1}{2}\) → \(\theta \approx \frac{2\pi}{3}, \frac{4\pi}{3}\) radians.
4. [Maximum mark: 7]
(a) Expand the following in ascending powers of \(x\) up to and including the term in \(x^2\):
(i) \((1 + 2x)^5\).
(ii) \((1 – ax)^6\), where \(a\) is a constant.
(b) In the expansion of \((1 + 2x)^5(1 – ax)^6\), the coefficient of \(x^2\) is \(-5\). Find the possible values of \(a\).
▶️Answer/Explanation
(a)(i) \((1 + 2x)^5 = 1 + 10x + 40x^2 + \dots\)
(a)(ii) \((1 – ax)^6 = 1 – 6ax + 15a^2x^2 + \dots\)
(b) Multiply the expansions:
Coefficient of \(x^2\): \(15a^2 – 60a + 40 = -5\).
Simplify: \(15a^2 – 60a + 45 = 0\) → \(a^2 – 4a + 3 = 0\).
Solutions: \(a = 1\) or \(a = 3\).
5. [Maximum mark: 7]
The first, second, and third terms of a geometric progression are \(2p + 6\), \(5p\), and \(8p + 2\) respectively.
(a) Find the possible values of the constant \(p\).
(b) One of the values of \(p\) found in (a) is a negative fraction. Use this value of \(p\) to find the sum to infinity of this progression.
▶️Answer/Explanation
(a) For a geometric progression, \(\frac{5p}{2p + 6} = \frac{8p + 2}{5p}\):
Cross-multiply: \(25p^2 = (2p + 6)(8p + 2)\).
Simplify: \(9p^2 – 52p – 12 = 0\).
Solutions: \(p = 6\) or \(p = -\frac{2}{9}\).
(b) Use \(p = -\frac{2}{9}\):
First term \(a = 2(-\frac{2}{9}) + 6 = \frac{50}{9}\).
Common ratio \(r = \frac{5(-\frac{2}{9})}{\frac{50}{9}} = -\frac{1}{5}\).
Sum to infinity: \(S_\infty = \frac{a}{1 – r} = \frac{\frac{50}{9}}{1 – (-\frac{1}{5})} = \frac{125}{27}\).
6. [Maximum mark: 7]
A line has equation \(y = 6x – c\) and a curve has equation \(y = cx^2 + 2x – 3\), where \(c\) is a constant. The line is a tangent to the curve at point \(P\).
Find the possible values of \(c\) and the corresponding coordinates of \(P\).
▶️Answer/Explanation
Solution:
Set the line and curve equal: \(6x – c = cx^2 + 2x – 3\).
Rearrange: \(cx^2 – 4x + (c – 3) = 0\).
For tangency, discriminant \(D = 0\):
\(16 – 4c(c – 3) = 0\) → \(4c^2 – 12c – 16 = 0\).
Solutions: \(c = 4\) or \(c = -1\).
For \(c = 4\): \(4x^2 – 4x + 1 = 0\) → \(x = \frac{1}{2}\), \(y = -1\).
For \(c = -1\): \(-x^2 – 4x – 4 = 0\) → \(x = -2\), \(y = -11\).
Thus, the possible points are \((\frac{1}{2}, -1)\) and \((-2, -11)\).
7. [Maximum mark: 7]
The function \( f \) is defined by \( f(x) = 1 + \frac{3}{x-2} \) for \( x > 2 \).
(a) State the range of \( f \).
(b) Obtain an expression for \( f^{-1}(x) \) and state the domain of \( f^{-1} \).
(c) Obtain a simplified expression for \( gf(x) \), where \( g(x) = 2x – 2 \) for \( x > 0 \).
▶️Answer/Explanation
(a) Range: \( y > 1 \).
(b) Let \( y = 1 + \frac{3}{x-2} \). Swap \( x \) and \( y \):
\( x = 1 + \frac{3}{y-2} \) → \( x – 1 = \frac{3}{y-2} \) → \( y – 2 = \frac{3}{x-1} \).
Thus, \( f^{-1}(x) = 2 + \frac{3}{x-1} \). Domain: \( x > 1 \).
(c) \( gf(x) = g\left(1 + \frac{3}{x-2}\right) = 2\left(1 + \frac{3}{x-2}\right) – 2 = \frac{6}{x-2} \).
8. [Maximum mark: 5]
The diagram shows part of the graph of \( y = \sin(a(x + b)) \), where \( a \) and \( b \) are positive constants.
(a) State the value of \( a \) and one possible value of \( b \).
(b) Another curve, with equation \( y = f(x) \), has a single stationary point at \( (p, q) \). This curve is transformed to a curve with equation \( y = -3f\left(\frac{1}{4}(x + 8)\right) \). Find the coordinates of the stationary point of the transformed curve, in terms of \( p \) and \( q \).
▶️Answer/Explanation
(a) From the graph, the period is \( 4\pi \), so \( a = \frac{1}{2} \).
A possible phase shift gives \( b = \frac{\pi}{3} \).
(b) The transformation scales \( x \) by 4 and shifts left by 8, then scales \( y \) by -3.
Thus, the new stationary point is \( (4p – 8, -3q) \).
9. [Maximum mark: 8]
A curve has equation \( y = 2x^{\frac{1}{2}} – 1 \).
(a) Find the equation of the normal to the curve at the point \( A(4, 3) \), giving your answer in the form \( y = mx + c \).
(b) A point is moving along the curve such that at \( A \), the rate of increase of the \( x \)-coordinate is \( 3 \, \text{cm/s} \). Find the rate of increase of the \( y \)-coordinate at \( A \).
(c) At \( A \), the point moves down the normal with a constant rate of decrease of the \( y \)-coordinate at \( 5 \, \text{cm/s} \). Find the rate of change of its \( x \)-coordinate.
▶️Answer/Explanation
(a) Differentiate: \( \frac{dy}{dx} = x^{-\frac{1}{2}} \). At \( x = 4 \), gradient \( m = \frac{1}{2} \).
Normal gradient: \( -2 \). Equation: \( y – 3 = -2(x – 4) \) → \( y = -2x + 11 \).
(b) Chain rule: \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{1}{2} \times 3 = \frac{3}{2} \, \text{cm/s} \).
(c) On the normal, \( \frac{dy}{dx} = -2 \). Given \( \frac{dy}{dt} = -5 \),
\( \frac{dx}{dt} = \frac{dy}{dt} / \frac{dy}{dx} = \frac{-5}{-2} = \frac{5}{2} \, \text{cm/s} \).
10. [Maximum mark: 9]
The diagram shows points \( A \), \( B \), and \( C \) on a circle with centre \( O \) and radius \( r \). Angle \( AOB \) is \( 2.8 \) radians. The shaded region is bounded by arcs of two circles: one with centre \( O \) and radius \( r \), and another with centre \( C \) and radius \( R \).
(a) State the size of angle \( ACO \) in radians.
(b) Find \( R \) in terms of \( r \).
(c) Find the area of the shaded region in terms of \( r \).
▶️Answer/Explanation
(a) Angle \( ACO = 0.7 \) radians (half of \( \pi – 2.8 \)).
(b) Using the sine rule: \( R = 2r \sin(0.7) \approx 1.53r \).
(c) Sector \( AOB \): \( \frac{1}{2} r^2 \times 2.8 = 1.4r^2 \).
Sector \( ACB \): \( \frac{1}{2} R^2 \times 1.4 \approx 1.638r^2 \).
Triangles: \( 2 \times \frac{1}{2} r^2 \sin(1.4) \approx 0.985r^2 \).
Shaded area: \( 1.4r^2 – (1.638r^2 – 0.985r^2) \approx 0.747r^2 \).
11. [Maximum mark: 10]
The curve \( y = x + \frac{2}{(2x-1)^2} \) has stationary point \( R \). The lines \( x = 1 \) and \( x = 2 \) intersect the curve at \( P \) and \( Q \) respectively.
(a) Verify that the \( x \)-coordinate of \( R \) is \( \frac{3}{2} \) and find the \( y \)-coordinate of \( R \).
(b) Find the exact value of the area of the region bounded by the curve, the lines \( x = 1 \), \( x = 2 \), and the line segment \( PQ \).
▶️Answer/Explanation
(a) Differentiate: \( \frac{dy}{dx} = 1 – \frac{8}{(2x-1)^3} \).
Set \( \frac{dy}{dx} = 0 \): \( 1 = \frac{8}{(2x-1)^3} \) → \( x = \frac{3}{2} \).
Substitute \( x = \frac{3}{2} \) into \( y \): \( y = \frac{3}{2} + \frac{2}{4} = 2 \).
(b) Coordinates: \( P(1, 3) \), \( Q(2, \frac{20}{9}) \).
Area under curve: \( \int_{1}^{2} \left(x + \frac{2}{(2x-1)^2}\right) dx = \left[\frac{x^2}{2} – \frac{1}{2x-1}\right]_{1}^{2} = \frac{13}{6} \).
Area of trapezium \( PQ \): \( \frac{1}{2} \left(3 + \frac{20}{9}\right) = \frac{47}{18} \).
Shaded area: \( \frac{47}{18} – \frac{13}{6} = \frac{4}{9} \).