1. [Maximum mark: 4]
The coefficient of \( x^4 \) in the expansion of \( (3 + x)^5 \) is equal to the coefficient of \( x^2 \) in the expansion of \( \left(2x + \frac{a}{x}\right)^6 \). Find the value of the positive constant \( a \).
▶️Answer/Explanation
Solution:
– Coefficient of \( x^4 \) in \( (3 + x)^5 \): \( \binom{5}{4} \cdot 3^1 = 15 \).
– Coefficient of \( x^2 \) in \( \left(2x + \frac{a}{x}\right)^6 \): \( \binom{6}{2} \cdot (2x)^4 \cdot \left(\frac{a}{x}\right)^2 = 240a^2 \).
– Equating the coefficients: \( 240a^2 = 15 \).
– Solving for \( a \): \( a = \frac{1}{4} \) (positive constant).
2. [Maximum mark: 4]
The second and third terms of a geometric progression are 10 and 8 respectively. Find the sum to infinity.
▶️Answer/Explanation
Solution:
– Common ratio \( r = \frac{8}{10} = 0.8 \).
– First term \( a = \frac{10}{r} = \frac{10}{0.8} = 12.5 \).
– Sum to infinity: \( S_\infty = \frac{a}{1 – r} = \frac{12.5}{1 – 0.8} = 62.5 \).
3. [Maximum mark: 4]
The equation of a curve is such that \( \frac{dy}{dx} = 3(4x – 7)^{\frac{1}{2}} – 4x^{-\frac{1}{2}} \). It is given that the curve passes through the point \( \left(4, \frac{5}{2}\right) \). Find the equation of the curve.
▶️Answer/Explanation
Solution:
– Integrate \( \frac{dy}{dx} \): \( y = \frac{3}{6}(4x – 7)^{\frac{3}{2}} – 8x^{\frac{1}{2}} + c \).
– Substitute \( (4, \frac{5}{2}) \): \( \frac{5}{2} = \frac{1}{2}(9)^{\frac{3}{2}} – 8(2) + c \).
– Solve for \( c \): \( c = 5 \).
– Final equation: \( y = \frac{1}{2}(4x – 7)^{\frac{3}{2}} – 8x^{\frac{1}{2}} + 5 \).
4. [Maximum mark: 5]
The first, second and third terms of an arithmetic progression are \( k \), \( 6k \), and \( k + 6 \) respectively.
(a) Find the value of the constant \( k \). [2]
(b) Find the sum of the first 30 terms of the progression. [3]
▶️Answer/Explanation
(a) Solution:
– Using the property of AP: \( 2 \times 6k = k + (k + 6) \).
– Solve for \( k \): \( 12k = 2k + 6 \Rightarrow k = 0.6 \).
(b) Solution:
– Common difference \( d = 6k – k = 5k = 3 \).
– Sum of first 30 terms: \( S_{30} = \frac{30}{2} [2(0.6) + 29(3)] = 1323 \).
5. [Maximum mark: 9]
The equation of a curve is \( y = 4x^2 – kx + \frac{1}{2}k^2 \) and the equation of a line is \( y = x – a \), where \( k \) and \( a \) are constants.
(a) Given that the curve and the line intersect at the points with \( x \)-coordinates 0 and \( \frac{3}{4} \), find the values of \( k \) and \( a \). [4]
(b) Given instead that \( a = -\frac{7}{2} \), find the values of \( k \) for which the line is a tangent to the curve. [5]
▶️Answer/Explanation
(a) Solution:
– Substitute \( x = 0 \): \( \frac{1}{2}k^2 = -a \).
– Substitute \( x = \frac{3}{4} \): \( 4\left(\frac{3}{4}\right)^2 – k\left(\frac{3}{4}\right) + \frac{1}{2}k^2 = \frac{3}{4} – a \).
– Solve the system: \( k = 2 \), \( a = -2 \).
(b) Solution:
– Substitute \( a = -\frac{7}{2} \) and set discriminant to zero: \( (k + 1)^2 – 16\left(\frac{1}{2}k^2 + \frac{7}{2}\right) = 0 \).
– Solve: \( 7k^2 – 2k – 57 = 0 \Rightarrow k = 3 \) or \( k = -\frac{19}{7} \).
6. [Maximum mark: 5]
The diagram shows the curve with equation \( y = 5x^{\frac{1}{2}} \) and the line with equation \( y = 2x + 2 \). Find the exact area of the shaded region which is bounded by the line and the curve.
▶️Answer/Explanation
Solution:
– Find intersection points: \( 5x^{\frac{1}{2}} = 2x + 2 \Rightarrow x = \frac{1}{4}, 4 \).
– Integrate difference: \( \int_{\frac{1}{4}}^{4} (5x^{\frac{1}{2}} – 2x – 2) \, dx \).
– Result: \( \left[ \frac{10}{3}x^{\frac{3}{2}} – x^2 – 2x \right]_{\frac{1}{4}}^{4} = \frac{45}{16} \).
7. [Maximum mark: 6]
The diagram shows a sector \(OBAC\) of a circle with centre \(O\) and radius \(10\) cm. The point \(P\) lies on \(OC\) and \(BP\) is perpendicular to \(OC\). Angle \(AOC = \frac{1}{6}\pi\) and the length of the arc \(AB\) is \(2\) cm.
(a) Find the angle \(BOC\). [2]
(b) Hence find the area of the shaded region \(BPC\) giving your answer correct to 3 significant figures. [4]
▶️Answer/Explanation
(a) Solution:
– Use arc length formula: \( r\theta = 2 \) cm \(\Rightarrow \theta = 0.2\) radians.
– Angle \(BOC = \frac{\pi}{6} – 0.2\) radians.
(b) Solution:
– Find lengths: \(BP = 10\sin(\theta)\) and \(OP = 10\cos(\theta)\).
– Area of triangle \(OBP\): \(\frac{1}{2} \times BP \times OP\).
– Area of sector \(BOC\): \(\frac{1}{2} \times 10^2 \times \theta\).
– Shaded area: Sector area \(-\) Triangle area \(= 11.4\) cm² (3 s.f.).
8. [Maximum mark: 8]
The equation of a circle is \(x^2 + y^2 + ax + by – 12 = 0\). The points \(A(1, 1)\) and \(B(2, -6)\) lie on the circle.
(a) Find the values of \(a\) and \(b\) and hence find the coordinates of the centre of the circle. [4]
(b) Find the equation of the tangent to the circle at the point \(A\), giving your answer in the form \(px + qy = k\), where \(p, q\) and \(k\) are integers. [4]
▶️Answer/Explanation
(a) Solution:
– Substitute \(A(1,1)\) and \(B(2,-6)\) into the circle equation to form two equations:
\(1 + 1 + a + b – 12 = 0\) and \(4 + 36 + 2a – 6b – 12 = 0\).
– Solve: \(a = 4\), \(b = 6\).
– Centre: \((-2, -3)\).
(b) Solution:
– Gradient of radius \(CA\): \(\frac{1 – (-3)}{1 – (-2)} = \frac{4}{3}\).
– Gradient of tangent: \(-\frac{3}{4}\).
– Equation: \(y – 1 = -\frac{3}{4}(x – 1) \Rightarrow 3x + 4y = 7\).
9. [Maximum mark: 7]
The equation of a curve is \(y = 3x + 1 – 4(3x + 1)^{\frac{1}{2}}\) for \(x > -\frac{1}{3}\).
(a) Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\). [3]
(b) Find the coordinates of the stationary point of the curve and determine its nature. [4]
▶️Answer/Explanation
(a) Solution:
– \(\frac{dy}{dx} = 3 – 6(3x + 1)^{-\frac{1}{2}}\).
– \(\frac{d^2y}{dx^2} = 9(3x + 1)^{-\frac{3}{2}}\).
(b) Solution:
– Set \(\frac{dy}{dx} = 0\): \(3 = 6(3x + 1)^{-\frac{1}{2}} \Rightarrow x = 1\).
– Substitute \(x = 1\) into \(y\): \(y = -4\).
– Stationary point: \((1, -4)\).
– Nature: \(\frac{d^2y}{dx^2} > 0\) \(\Rightarrow\) minimum point.
10. [Maximum mark: 13]
Functions \(f\) and \(g\) are defined as follows:
\[f(x) = \frac{2x + 1}{2x – 1} \quad \text{for } x \neq \frac{1}{2},\]
\[g(x) = x^2 + 4 \quad \text{for } x \in \mathbb{R}.\]
(a) State the domain of \(f^{-1}\).
(b) Find an expression for \(f^{-1}(x)\).
(c) Find \(gf^{-1}(3)\).
(d) Explain why \(g^{-1}(x)\) cannot be found.
(e) Show that \(1 + \frac{2}{2x – 1}\) can be expressed as \(\frac{2x + 1}{2x – 1}\). Hence find the area of the triangle enclosed by the tangent to the curve \(y = f(x)\) at the point where \(x = 1\) and the \(x\)- and \(y\)-axes.
▶️Answer/Explanation
(a) Solution:
– Domain of \(f^{-1}\): \(x \neq 1\).
(b) Solution:
– Let \(y = \frac{2x + 1}{2x – 1}\), solve for \(x\): \(x = \frac{y + 1}{2(y – 1)}\).
– \(f^{-1}(x) = \frac{x + 1}{2(x – 1)}\).
(c) Solution:
– \(f^{-1}(3) = 1\), \(g(1) = 5\).
(d) Solution:
– \(g\) is not one-to-one (fails horizontal line test).
(e) Solution:
– Simplify: \(1 + \frac{2}{2x – 1} = \frac{2x + 1}{2x – 1}\).
– Find tangent at \(x = 1\): \(y – 3 = -4(x – 1)\).
– Intercepts: \(\left(\frac{7}{4}, 0\right)\) and \((0, 7)\).
– Area: \(\frac{49}{8}\).
11. [Maximum mark: 10]
The function \(f\) is given by \(f(x) = 4\cos^4x + \cos^2x – k\) for \(0 \leq x \leq 2\pi\), where \(k\) is a constant.
(a) Given that \(k = 3\), find the exact solutions of the equation \(f(x) = 0\).
(b) Use the quadratic formula to show that, when \(k > 5\), the equation \(f(x) = 0\) has no solutions.
▶️Answer/Explanation
(a) Solution:
– Let \(u = \cos^2x\): \(4u^2 + u – 3 = 0\).
– Solve: \(u = \frac{3}{4}\) or \(-1\) (discarded).
– Solutions: \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).
(b) Solution:
– Quadratic in \(u\): \(4u^2 + u – k = 0\).
– Discriminant: \(1 + 16k\).
– For \(k > 5\), \(\cos^2x = \frac{-1 \pm \sqrt{1 + 16k}}{8}\) has no valid solutions (either negative or > 1).