1. [Maximum mark: 4]
A curve with equation \( y = f(x) \) is such that \( f'(x) = 2x – \frac{1}{3}x^{\frac{1}{3}} \). It is given that \( f(8) = 5 \). Find \( f(x) \).
▶️Answer/Explanation
Solution:
Integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int \left(2x – \frac{1}{3}x^{\frac{1}{3}}\right) dx = x^2 – \frac{1}{4}x^{\frac{4}{3}} + c \]
Use the given point \( f(8) = 5 \) to find \( c \):
\[ 5 = 8^2 – \frac{1}{4}(8)^{\frac{4}{3}} + c \]
\[ 5 = 64 – 8 + c \implies c = -51 \]
Thus, the function is:
\[ f(x) = x^2 – \frac{1}{4}x^{\frac{4}{3}} – 51 \]
2. [Maximum mark: 5]
A curve has equation \( y = x^2 + 2cx + 4 \) and a straight line has equation \( y = 4x + c \), where \( c \) is a constant. Find the set of values of \( c \) for which the curve and line intersect at two distinct points.
▶️Answer/Explanation
Solution:
Set the equations equal for intersection points:
\[ x^2 + 2cx + 4 = 4x + c \]
Rearrange to form a quadratic:
\[ x^2 + (2c – 4)x + (4 – c) = 0 \]
For two distinct real roots, the discriminant must be positive:
\[ (2c – 4)^2 – 4(1)(4 – c) > 0 \]
Simplify:
\[ 4c^2 – 16c + 16 – 16 + 4c > 0 \]
\[ 4c^2 – 12c > 0 \]
Factorize:
\[ 4c(c – 3) > 0 \]
The solution is:
\[ c < 0 \text{ or } c > 3 \]
3. [Maximum mark: 6]
(a) Find the term independent of \( x \) in the expansion of \( \left(3x + \frac{2}{x^2}\right)^6 \).
(b) Find the term independent of \( x \) in the expansion of \( \left(3x + \frac{2}{x^2}\right)^6 \left(1 – x^3\right) \).
▶️Answer/Explanation
(a) Solution:
The general term in the expansion is:
\[ \binom{6}{r} (3x)^{6-r} \left(\frac{2}{x^2}\right)^r = \binom{6}{r} 3^{6-r} 2^r x^{6-3r} \]
For the term independent of \( x \), set the exponent to zero:
\[ 6 – 3r = 0 \implies r = 2 \]
Substitute \( r = 2 \):
\[ \binom{6}{2} 3^4 2^2 = 15 \times 81 \times 4 = 4860 \]
(b) Solution:
Multiply the expansion from part (a) by \( (1 – x^3) \). The term independent of \( x \) comes from:
\[ 4860 \times 1 – \text{(term with } x^3 \text{ in part (a))} \]
For the term with \( x^3 \), set \( 6 – 3r = 3 \implies r = 1 \):
\[ \binom{6}{1} 3^5 2^1 = 6 \times 243 \times 2 = 2916 \]
Thus, the independent term is:
\[ 4860 – 2916 = 1944 \]
4. [Maximum mark: 6]
The first term of a geometric progression and the first term of an arithmetic progression are both equal to \( a \). The third term of the geometric progression is equal to the second term of the arithmetic progression. The fifth term of the geometric progression is equal to the sixth term of the arithmetic progression. Given that the terms are all positive and not all equal, find the sum of the first twenty terms of the arithmetic progression in terms of \( a \).
▶️Answer/Explanation
Solution:
For the geometric progression (GP):
\[ \text{Third term} = ar^2 \]
For the arithmetic progression (AP):
\[ \text{Second term} = a + d \]
Given \( ar^2 = a + d \) (1).
Fifth term of GP: \( ar^4 \).
Sixth term of AP: \( a + 5d \).
Given \( ar^4 = a + 5d \) (2).
Subtract (1) from (2):
\[ ar^4 – ar^2 = 4d \implies ar^2(r^2 – 1) = 4d \]
From (1): \( d = ar^2 – a \). Substitute into the above:
\[ ar^2(r^2 – 1) = 4(ar^2 – a) \]
Simplify:
\[ r^2(r^2 – 1) = 4(r^2 – 1) \implies r^2 = 4 \text{ (since terms are not all equal)} \]
Thus, \( r = 2 \) and \( d = 3a \).
Sum of first 20 terms of AP:
\[ S_{20} = \frac{20}{2} [2a + 19d] = 10[2a + 57a] = 590a \]
5. [Maximum mark: 6]
(a) Express \( 2x^2 – 8x + 14 \) in the form \( 2[(x – a)^2 + b] \).
(b) The functions \( f \) and \( g \) are defined by \( f(x) = x^2 \) for \( x \in \mathbb{R} \), and \( g(x) = 2x^2 – 8x + 14 \) for \( x \in \mathbb{R} \). Describe fully a sequence of transformations that maps the graph of \( y = f(x) \) onto the graph of \( y = g(x) \), making clear the order in which the transformations are applied.
▶️Answer/Explanation
(a) Solution:
Complete the square:
\[ 2x^2 – 8x + 14 = 2(x^2 – 4x) + 14 \]
\[ = 2[(x – 2)^2 – 4] + 14 = 2(x – 2)^2 + 6 \]
Thus, the form is:
\[ 2[(x – 2)^2 + 3] \]
(b) Solution:
The transformations are:
1. Vertical stretch by a factor of 2.
2. Horizontal translation right by 2 units.
3. Vertical translation up by 6 units.
The order must be: stretch → horizontal translation → vertical translation.
6. [Maximum mark: 8]
The circle with equation \((x + 1)^2 + (y – 2)^2 = 85\) and the straight line with equation \(y = 3x – 20\) intersect at points \(A\) and \(B\). The center of the circle is at \(C\).
(a) Find, by calculation, the coordinates of \(A\) and \(B\).
(b) Find an equation of the circle which has its center at \(C\) and for which the line \(y = 3x – 20\) is tangent to the circle.
▶️Answer/Explanation
(a) Solution:
Substitute \(y = 3x – 20\) into the circle’s equation:
\[(x + 1)^2 + (3x – 20 – 2)^2 = 85\]
\[(x + 1)^2 + (3x – 22)^2 = 85\]
Expand and simplify:
\[x^2 + 2x + 1 + 9x^2 – 132x + 484 = 85\]
\[10x^2 – 130x + 400 = 0\]
\[x^2 – 13x + 40 = 0\]
Solve the quadratic:
\[x = 5 \text{ or } x = 8\]
Find corresponding \(y\)-values:
\[A(5, -5) \text{ and } B(8, 4)\]
(b) Solution:
Center \(C(-1, 2)\). Find perpendicular distance from \(C\) to line \(y = 3x – 20\):
\[3x – y – 20 = 0\]
\[r = \frac{|3(-1) – 1(2) – 20|}{\sqrt{3^2 + (-1)^2}} = \frac{25}{\sqrt{10}}\]
Equation of new circle:
\[(x + 1)^2 + (y – 2)^2 = \left(\frac{25}{\sqrt{10}}\right)^2\]
\[(x + 1)^2 + (y – 2)^2 = \frac{625}{10} = 62.5\]
7. [Maximum mark: 7]
(a) Show that \(\frac{\sin \theta + 2\cos \theta}{\cos \theta – 2\sin \theta} – \frac{\sin \theta – 2\cos \theta}{\cos \theta + 2\sin \theta} = \frac{4}{5\cos^2 \theta – 4}\).
(b) Hence solve the equation \(\frac{\sin \theta + 2\cos \theta}{\cos \theta – 2\sin \theta} – \frac{\sin \theta – 2\cos \theta}{\cos \theta + 2\sin \theta} = 5\) for \(0^\circ < \theta < 180^\circ\).
▶️Answer/Explanation
(a) Solution:
Combine the fractions:
\[\frac{(\sin \theta + 2\cos \theta)(\cos \theta + 2\sin \theta) – (\sin \theta – 2\cos \theta)(\cos \theta – 2\sin \theta)}{(\cos \theta – 2\sin \theta)(\cos \theta + 2\sin \theta)}\]
Expand numerator:
\[\sin \theta \cos \theta + 2\sin^2 \theta + 2\cos^2 \theta + 4\sin \theta \cos \theta – [\sin \theta \cos \theta – 2\sin^2 \theta – 2\cos^2 \theta + 4\sin \theta \cos \theta]\]
Simplify:
\[4\sin^2 \theta + 4\cos^2 \theta = 4(\sin^2 \theta + \cos^2 \theta) = 4\]
Denominator is \(\cos^2 \theta – 4\sin^2 \theta\). Using \(\sin^2 \theta = 1 – \cos^2 \theta\):
\[\cos^2 \theta – 4(1 – \cos^2 \theta) = 5\cos^2 \theta – 4\]
Thus, the expression simplifies to \(\frac{4}{5\cos^2 \theta – 4}\).
(b) Solution:
Set the expression equal to 5:
\[\frac{4}{5\cos^2 \theta – 4} = 5\]
\[4 = 25\cos^2 \theta – 20\]
\[25\cos^2 \theta = 24\]
\[\cos \theta = \pm \sqrt{\frac{24}{25}} = \pm \frac{2\sqrt{6}}{5}\]
Solutions in \(0^\circ < \theta < 180^\circ\):
\[\theta \approx 11.5^\circ \text{ and } 168.5^\circ\]
8. [Maximum mark: 8]
The diagram shows the circle with equation \((x – 2)^2 + y^2 = 8\). The chord \(AB\) of the circle intersects the positive \(y\)-axis at \(A\) and is parallel to the \(x\)-axis.
(a) Find, by calculation, the coordinates of \(A\) and \(B\).
(b) Find the volume of revolution when the shaded segment, bounded by the circle and the chord \(AB\), is rotated through \(360^\circ\) about the \(x\)-axis.
▶️Answer/Explanation
(a) Solution:
Since \(AB\) is parallel to the \(x\)-axis and passes through \(A\) on the \(y\)-axis, let \(A(0, y)\). Substitute into the circle’s equation:
\[(0 – 2)^2 + y^2 = 8 \implies y = 2\]
Thus, \(A(0, 2)\). For \(B\), since \(AB\) is horizontal, \(B(4, 2)\).
(b) Solution:
Volume of revolution:
\[V = \pi \int_0^4 (8 – (x – 2)^2) dx – \pi \int_0^4 (2)^2 dx\]
\[V = \pi \left[8x – \frac{(x – 2)^3}{3}\right]_0^4 – 16\pi\]
Evaluate:
\[\pi \left(32 – \frac{8}{3} – (0 + \frac{8}{3})\right) – 16\pi = \pi \left(32 – \frac{16}{3}\right) – 16\pi\]
\[V = \frac{80}{3}\pi – 16\pi = \frac{32}{3}\pi\]
9. [Maximum mark: 8]
Functions \(f\), \(g\), and \(h\) are defined as follows:
\(f: x \mapsto x – 4x^{\frac{1}{2}} + 1\) for \(x \geq 0\),
\(g: x \mapsto mx^2 + n\) for \(x \geq -2\), where \(m\) and \(n\) are constants,
\(h: x \mapsto x^{\frac{1}{2}} – 2\) for \(x \geq 0\).
(a) Solve the equation \(f(x) = 0\), giving your solutions in the form \(x = a + b\sqrt{c}\), where \(a\), \(b\), and \(c\) are integers.
(b) Given that \(f(x) = g(h(x))\), find the values of \(m\) and \(n\).
▶️Answer/Explanation
(a) Solution:
Let \(u = x^{\frac{1}{2}}\):
\[u^2 – 4u + 1 = 0\]
Solve for \(u\):
\[u = 2 \pm \sqrt{3}\]
Thus:
\[x = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}\]
(b) Solution:
Compute \(g(h(x))\):
\[g(h(x)) = m(x^{\frac{1}{2}} – 2)^2 + n = m(x – 4x^{\frac{1}{2}} + 4) + n\]
Set equal to \(f(x)\):
\[mx – 4mx^{\frac{1}{2}} + 4m + n = x – 4x^{\frac{1}{2}} + 1\]
Compare coefficients:
\[m = 1, -4m = -4 \implies m = 1\]
\[4m + n = 1 \implies n = -3\]
10. [Maximum mark: 8]
The diagram shows a circle with center \(A\) of radius 5 cm and a circle with center \(B\) of radius 8 cm. The circles touch at point \(C\) so that \(ACB\) is a straight line. The tangent at point \(D\) on the smaller circle intersects the larger circle at \(E\) and passes through \(B\).
(a) Find the perimeter of the shaded region.
(b) Find the area of the shaded region.
▶️Answer/Explanation
(a) Solution:
Let \(\theta = \angle BAD\). Using right triangle \(ABD\):
\[\cos \theta = \frac{5}{13}, \sin \theta = \frac{12}{13}\]
Arc lengths:
\[\text{Smaller circle: } 5 \times 2\theta = 10\theta\]
\[\text{Larger circle: } 8 \times (\pi – 2\theta)\]
Length \(DE = \sqrt{13^2 – 5^2} = 12\) cm.
Total perimeter:
\[10\theta + 8(\pi – 2\theta) + 12 = 8\pi – 6\theta + 12\]
Substitute \(\theta = \cos^{-1}\left(\frac{5}{13}\right)\):
\[\text{Perimeter} \approx 13.0 \text{ cm}\]
(b) Solution:
Area of sector \(BAD\): \(\frac{1}{2} \times 5^2 \times 2\theta = 25\theta\).
Area of sector \(BAE\): \(\frac{1}{2} \times 8^2 \times (\pi – 2\theta) = 32(\pi – 2\theta)\).
Subtract triangle areas:
\[\text{Shaded area} = 25\theta + 32(\pi – 2\theta) – \frac{1}{2} \times 5 \times 12 – \frac{1}{2} \times 8 \times 12\]
Simplify:
\[\text{Area} \approx 2.67 \text{ cm}^2\]
11. [Maximum mark: 9]
It is given that a curve has equation \(y = k(3x – k)^{-1} + 3x\), where \(k\) is a constant.
(a) Find, in terms of \(k\), the values of \(x\) at which there is a stationary point.
(b) The function \(f\) has a stationary value at \(x = a\) and is defined by \(f(x) = 4(3x – 4)^{-1} + 3x\) for \(x \geq \frac{3}{2}\). Find the value of \(a\) and determine the nature of the stationary value.
(c) The function \(g\) is defined by \(g(x) = -(3x + 1)^{-1} + 3x\) for \(x \geq 0\). Determine, making your reasoning clear, whether \(g\) is an increasing function, a decreasing function, or neither.
▶️Answer/Explanation
(a) Solution:
Differentiate \(y\):
\[\frac{dy}{dx} = -3k(3x – k)^{-2} + 3\]
Set \(\frac{dy}{dx} = 0\):
\[-3k(3x – k)^{-2} + 3 = 0 \implies (3x – k)^2 = k\]
Solve for \(x\):
\[3x – k = \pm \sqrt{k} \implies x = \frac{k \pm \sqrt{k}}{3}\]
(b) Solution:
Substitute \(k = 4\) into part (a):
\[x = \frac{4 \pm 2}{3} \implies a = 2\]
Second derivative:
\[f”(x) = 72(3x – 4)^{-3}\]
Evaluate at \(x = 2\):
\[f”(2) = 72(2)^{-3} = 9 > 0 \implies \text{minimum}\]
(c) Solution:
Differentiate \(g(x)\):
\[g'(x) = 3(3x + 1)^{-2} + 3 > 0 \text{ for all } x \geq 0\]
Thus, \(g\) is strictly increasing.