A Level H2 Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Pure Mathematics) Version: 1 of 5 Total Marks: 100

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Question 1 (8 marks)

(a) Show that $gf$ exists and find $gf(x)$. [4 marks]

Solution:

• Domain of $f$: $x \in \mathbb{R}, x \neq 3$
• Range of $f$: Let $y = \frac{2x+1}{x-3}$. As $x \to 3^+$, $y \to +\infty$; as $x \to 3^-$, $y \to -\infty$; as $x \to \pm\infty$, $y \to 2$. So $R_f = \mathbb{R} \setminus \{2\}$.
• Domain of $g$: $x \geq -4$
• For $gf$ to exist, $R_f \subseteq D_g$: Need $\frac{2x+1}{x-3} \geq -4$ for all $x \neq 3$. $\frac{2x+1}{x-3} \geq -4 \implies \frac{2x+1}{x-3} + 4 \geq 0 \implies \frac{2x+1+4x-12}{x-3} \geq 0 \implies \frac{6x-11}{x-3} \geq 0$ Critical values: $x = \frac{11}{6}, 3$. Sign analysis shows this holds for $x \leq \frac{11}{6}$ or $x > 3$. So $gf$ exists for $x \in (-\infty, \frac{11}{6}] \cup (3, \infty)$. [2 marks]
• $gf(x) = g(f(x)) = \sqrt{\frac{2x+1}{x-3} + 4} = \sqrt{\frac{2x+1+4x-12}{x-3}} = \sqrt{\frac{6x-11}{x-3}}$ [2 marks]

(b) State domain and range of $gf$. [2 marks]

Solution:

• Domain: $x \in (-\infty, \frac{11}{6}] \cup (3, \infty)$ [1 mark]
• Range: For $x \leq \frac{11}{6}$, $\frac{6x-11}{x-3} \geq 0$ and as $x \to -\infty$, expression $\to 6$, so $gf(x) \to \sqrt{6}$. As $x \to 3^+$, expression $\to +\infty$, so $gf(x) \to \infty$. Range: $[0, \infty)$. [1 mark]

(c) Determine whether $gf$ has an inverse. [2 marks]

Solution:

• $gf$ is not one-to-one on its domain. For example, $gf(0) = \sqrt{\frac{-11}{-3}} = \sqrt{\frac{11}{3}}$ and $gf(1) = \sqrt{\frac{-5}{-2}} = \sqrt{\frac{5}{2}}$, but also for $x > 3$, $gf(4) = \sqrt{\frac{13}{1}} = \sqrt{13}$. The function takes the same value for different $x$ values (e.g., check if $\frac{6x-11}{x-3} = k$ has multiple solutions). Since $gf$ is not injective, it does not have an inverse. [2 marks]

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Question 2 (9 marks)

(a) Find the cartesian equation of $C$. [4 marks]

Solution:

• $x = t^2 - 2t$, $y = t^2 + 2t$
• $y - x = (t^2 + 2t) - (t^2 - 2t) = 4t$, so $t = \frac{y-x}{4}$ [1 mark]
• $x + y = (t^2 - 2t) + (t^2 + 2t) = 2t^2$ [1 mark]
• Substitute $t$: $x + y = 2\left(\frac{y-x}{4}\right)^2 = 2 \cdot \frac{(y-x)^2}{16} = \frac{(y-x)^2}{8}$ [1 mark]
• Therefore $(y-x)^2 = 8(x+y)$, so $k = 8$. [1 mark]

(b) Sketch the curve $C$. [3 marks]

Solution:

• The equation $(y-x)^2 = 8(x+y)$ represents a parabola.
• When $x = 0$: $y^2 = 8y \implies y(y-8) = 0$, so $(0,0)$ and $(0,8)$.
• When $y = 0$: $x^2 = 8x \implies x(x-8) = 0$, so $(0,0)$ and $(8,0)$.
• Axis of symmetry: $y = x$ (since equation is symmetric in swapping $x$ and $y$).
• Vertex: At $t = 0$: $(0,0)$. At $t = 1$: $(-1, 3)$. At $t = -1$: $(3, -1)$.
• Sketch shows parabola opening towards first quadrant, symmetric about $y = x$, passing through $(0,0)$, $(0,8)$, $(8,0)$. [3 marks]

(c) Find the exact volume of the solid formed. [2 marks]

Solution:

• The curve crosses the $x$-axis at $(0,0)$ and $(8,0)$.
• From cartesian equation: $y^2 - 2xy + x^2 = 8x + 8y \implies y^2 - 2xy - 8y + x^2 - 8x = 0$ This is quadratic in $y$: $y^2 - (2x+8)y + (x^2-8x) = 0$ $y = \frac{2x+8 \pm \sqrt{(2x+8)^2 - 4(x^2-8x)}}{2} = x+4 \pm \sqrt{4x^2+32x+64 - 4x^2+32x} = x+4 \pm \sqrt{64x+64}$ $y = x+4 \pm 8\sqrt{x+1}$
• For $0 \leq x \leq 8$, the upper branch is $y = x+4+8\sqrt{x+1}$ and lower branch is $y = x+4-8\sqrt{x+1}$.
• Volume $= \pi \int_0^8 (y_{\text{upper}}^2 - y_{\text{lower}}^2) \, dx$ $y_{\text{upper}}^2 - y_{\text{lower}}^2 = (y_{\text{upper}} - y_{\text{lower}})(y_{\text{upper}} + y_{\text{lower}}) = (16\sqrt{x+1})(2x+8) = 32(x+4)\sqrt{x+1}$
• $V = \pi \int_0^8 32(x+4)\sqrt{x+1} \, dx$. Let $u = x+1$, $du = dx$, $x = u-1$, when $x=0, u=1$; $x=8, u=9$. $V = 32\pi \int_1^9 (u-1+4)\sqrt{u} \, du = 32\pi \int_1^9 (u+3)u^{1/2} \, du = 32\pi \int_1^9 (u^{3/2} + 3u^{1/2}) \, du$ $= 32\pi \left[\frac{2}{5}u^{5/2} + 2u^{3/2}\right]_1^9 = 32\pi \left[\left(\frac{2}{5}(243) + 2(27)\right) - \left(\frac{2}{5} + 2\right)\right]$ $= 32\pi \left[\frac{486}{5} + 54 - \frac{2}{5} - 2\right] = 32\pi \left[\frac{484}{5} + 52\right] = 32\pi \cdot \frac{484+260}{5} = 32\pi \cdot \frac{744}{5} = \frac{23808\pi}{5}$ [2 marks]

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Question 3 (8 marks)

(a) Solve $\frac{x^2 - 5x + 6}{x + 1} \leq 0$. [4 marks]

Solution:

• Factorise numerator: $(x-2)(x-3)$
• Expression: $\frac{(x-2)(x-3)}{x+1} \leq 0$
• Critical values: $x = -1, 2, 3$
• Sign analysis:
  • $x < -1$: $(-)(-)/(-) = -$, negative
  • $-1 < x < 2$: $(-)(-)/(+) = +$, positive
  • $2 < x < 3$: $(+)(-)/(+) = -$, negative
  • $x > 3$: $(+)(+)/(+) = +$, positive
• At $x = 2$: numerator = 0, expression = 0 ✓
• At $x = 3$: numerator = 0, expression = 0 ✓
• At $x = -1$: undefined ✗
• Solution: $x \in (-\infty, -1) \cup [2, 3]$ [4 marks]

(b) Solve $\frac{|x|^2 - 5|x| + 6}{|x| + 1} \leq 0$. [4 marks]

Solution:

• Let $u = |x| \geq 0$. Then inequality becomes $\frac{u^2 - 5u + 6}{u + 1} \leq 0$, $u \geq 0$.
• From part (a), solution for $u$ is $u \in [0, 1) \cup [2, 3]$ (since $u \geq 0$, we take intersection with $(-\infty, -1) \cup [2, 3]$, noting $u = -1$ is not in domain).
• Wait, recalculate: For $u \geq 0$, critical values are $u = 2, 3$ (since $u = -1$ is not in domain).
  • $0 \leq u < 2$: $(u-2)(u-3) > 0$, $u+1 > 0$, so expression > 0
  • $2 \leq u \leq 3$: $(u-2)(u-3) \leq 0$, $u+1 > 0$, so expression $\leq 0$
  • $u > 3$: $(u-2)(u-3) > 0$, expression > 0
• So $u \in [2, 3]$, i.e., $2 \leq |x| \leq 3$
• This gives $x \in [-3, -2] \cup [2, 3]$ [4 marks]

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Question 4 (10 marks)

(a) Find $a$ and $r$. [4 marks]

Solution:

• $S_\infty = \frac{a}{1-r} = 24$ ... (1)
• $S_2 = a + ar = a(1+r) = 18$ ... (2)
• From (1): $a = 24(1-r)$
• Substitute into (2): $24(1-r)(1+r) = 18 \implies 24(1-r^2) = 18 \implies 1-r^2 = \frac{3}{4} \implies r^2 = \frac{1}{4}$
• Since $0 < r < 1$, $r = \frac{1}{2}$ [2 marks]
• $a = 24(1-\frac{1}{2}) = 12$ [2 marks]

(b) Find least $n$ such that $S_n > 23.5$. [3 marks]

Solution:

• $S_n = \frac{a(1-r^n)}{1-r} = \frac{12(1-0.5^n)}{0.5} = 24(1-0.5^n)$
• Need $24(1-0.5^n) > 23.5 \implies 1-0.5^n > \frac{23.5}{24} = \frac{47}{48}$
• $0.5^n < 1 - \frac{47}{48} = \frac{1}{48}$
• $n \ln 0.5 < \ln(1/48) \implies n > \frac{\ln(1/48)}{\ln 0.5} = \frac{-\ln 48}{-\ln 2} = \frac{\ln 48}{\ln 2} \approx 5.58$
• Least integer $n = 6$ [3 marks]

(c) Does sum to infinity exist for GP with first term $a$ and common ratio $2r$? [3 marks]

Solution:

• $2r = 2 \times \frac{1}{2} = 1$
• For sum to infinity to exist, we need $|2r| < 1$
• Here $|2r| = 1$, which is not less than 1.
• Therefore the sum to infinity does not exist (the series diverges). [3 marks]

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Question 5 (9 marks)

(a) Find acute angle between $l$ and $\Pi$. [3 marks]

Solution:

• Direction vector of $l$: $\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$
• Normal vector of $\Pi$: $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$
• $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{|2(2) + 1(-1) + (-1)(3)|}{\sqrt{4+1+1}\sqrt{4+1+9}} = \frac{|4-1-3|}{\sqrt{6}\sqrt{14}} = \frac{0}{\sqrt{84}} = 0$
• $\theta = 0^\circ$
• The line is parallel to the plane. [3 marks]

(b) Find point of intersection of $l$ and $\Pi$. [3 marks]

Solution:

• Parametric point on $l$: $(1+2\lambda, \lambda, 2-\lambda)$
• Substitute into plane equation: $2(1+2\lambda) - \lambda + 3(2-\lambda) = 7$
• $2 + 4\lambda - \lambda + 6 - 3\lambda = 7 \implies 8 + 0\lambda = 7 \implies 8 = 7$
• This is a contradiction, so the line does not intersect the plane.
• The line is parallel to the plane and does not lie in it. [3 marks]

(c) Find perpendicular distance from $A(3, -1, 4)$ to $\Pi$. [3 marks]

Solution:

• Distance $= \frac{|2(3) - (-1) + 3(4) - 7|}{\sqrt{2^2 + (-1)^2 + 3^2}} = \frac{|6 + 1 + 12 - 7|}{\sqrt{4+1+9}} = \frac{|12|}{\sqrt{14}} = \frac{12}{\sqrt{14}}$
• $= \frac{12\sqrt{14}}{14} = \frac{6\sqrt{14}}{7}$ [3 marks]

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Question 6 (9 marks)

(a) Express $z = 1 - i\sqrt{3}$ in modulus-argument form. [2 marks]

Solution:

• $|z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$
• $\arg(z) = -\tan^{-1}(\sqrt{3}/1) = -\frac{\pi}{3}$ (since in 4th quadrant)
• $z = 2e^{-i\pi/3}$ or $2(\cos(-\pi/3) + i\sin(-\pi/3))$ [2 marks]

(b) Find the three cube roots of $8i$ in cartesian form. [5 marks]

Solution:

• $8i = 8e^{i\pi/2}$ (since $|8i| = 8$, $\arg(8i) = \pi/2$)
• Cube roots: $w_k = 8^{1/3} e^{i(\pi/2 + 2\pi k)/3} = 2e^{i(\pi/6 + 2\pi k/3)}$ for $k = 0, 1, 2$
• $k = 0$: $w_0 = 2e^{i\pi/6} = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \sqrt{3} + i$
• $k = 1$: $w_1 = 2e^{i(\pi/6 + 2\pi/3)} = 2e^{i5\pi/6} = 2(\cos(5\pi/6) + i\sin(5\pi/6)) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i$
• $k = 2$: $w_2 = 2e^{i(\pi/6 + 4\pi/3)} = 2e^{i3\pi/2} = 2(\cos(3\pi/2) + i\sin(3\pi/2)) = 2(0 - i) = -2i$
• Roots: $\sqrt{3} + i$, $-\sqrt{3} + i$, $-2i$ [5 marks]

(c) Sketch the three cube roots on an Argand diagram. [2 marks]

Solution:

• Points: $(\sqrt{3}, 1)$, $(-\sqrt{3}, 1)$, $(0, -2)$
• These form an equilateral triangle centred at the origin.
• Sketch shows three points correctly plotted with axes labelled. [2 marks]

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Question 7 (10 marks)

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$. [3 marks]

Solution:

• Differentiate implicitly: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$
• $(x + 2y)\frac{dy}{dx} = -2x - y$
• $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$ [3 marks]

(b) Find coordinates of stationary points. [4 marks]

Solution:

• Stationary points when $\frac{dy}{dx} = 0 \implies -2x - y = 0 \implies y = -2x$
• Substitute into curve equation: $x^2 + x(-2x) + (-2x)^2 = 12$
• $x^2 - 2x^2 + 4x^2 = 12 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2$
• When $x = 2$, $y = -4$. When $x = -2$, $y = 4$.
• Stationary points: $(2, -4)$ and $(-2, 4)$ [4 marks]

(c) Determine nature of each stationary point. [3 marks]

Solution:

• Second derivative or first derivative test.
• Using first derivative test: Check sign of $\frac{dy}{dx}$ on either side.
• For $(2, -4)$: $x+2y = 2+2(-4) = -6 < 0$. Near this point, denominator is negative.
  • For $x < 2$ (with $y \approx -4$): $-2x-y \approx -4+4 = 0^-$? Need more careful analysis.
• Alternative: Use second derivative implicitly. $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-2x-y}{x+2y}\right)$ Using quotient rule and substituting $\frac{dy}{dx}$: At $(2, -4)$: $\frac{dy}{dx} = 0$, $x+2y = -6$. $\frac{d^2y}{dx^2} = \frac{(-2-\frac{dy}{dx})(x+2y) - (-2x-y)(1+2\frac{dy}{dx})}{(x+2y)^2}$ At $(2, -4)$: $= \frac{(-2)(-6) - (0)(1)}{36} = \frac{12}{36} = \frac{1}{3} > 0$, so minimum.
• At $(-2, 4)$: $x+2y = -2+8 = 6$. $\frac{d^2y}{dx^2} = \frac{(-2)(6) - (0)(1)}{36} = \frac{-12}{36} = -\frac{1}{3} < 0$, so maximum.
• $(2, -4)$ is a minimum point; $(-2, 4)$ is a maximum point. [3 marks]

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Question 8 (9 marks)

(a) Find Maclaurin series for $f(x) = e^x \cos x$ up to $x^3$. [5 marks]

Solution:

• $f(x) = e^x \cos x$
• $f(0) = 1 \cdot 1 = 1$
• $f'(x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x)$, $f'(0) = 1(1-0) = 1$
• $f''(x) = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = e^x(-2\sin x)$, $f''(0) = 0$
• $f'''(x) = e^x(-2\sin x) + e^x(-2\cos x) = -2e^x(\sin x + \cos x)$, $f'''(0) = -2(0+1) = -2$
• Maclaurin series: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ...$
• $f(x) = 1 + x + 0 \cdot x^2 - \frac{2}{6}x^3 + ... = 1 + x - \frac{1}{3}x^3 + ...$ [5 marks]

(b) Approximate $e^{0.2} \cos 0.2$ to 4 d.p. [2 marks]

Solution:

• Using series with $x = 0.2$:
• $f(0.2) \approx 1 + 0.2 - \frac{1}{3}(0.2)^3 = 1 + 0.2 - \frac{1}{3}(0.008) = 1.2 - 0.002666... = 1.1973$ (to 4 d.p.) [2 marks]

(c) Estimate $\lim_{x \to 0} \frac{e^x \cos x - 1}{x}$. [2 marks]

Solution:

• Using Maclaurin series: $e^x \cos x = 1 + x - \frac{1}{3}x^3 + ...$
• $\frac{e^x \cos x - 1}{x} = \frac{x - \frac{1}{3}x^3 + ...}{x} = 1 - \frac{1}{3}x^2 + ...$
• As $x \to 0$, limit $= 1$
• Alternatively, using small angle approximations: $\cos x \approx 1 - x^2/2$, $e^x \approx 1 + x + x^2/2$, product $\approx 1 + x$, so limit $= 1$. [2 marks]

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Question 9 (10 marks)

(a) Show that $\frac{dx}{dt} = 1 - \frac{x}{20}$. [3 marks]

Solution:

• Rate of salt entering = concentration × flow rate = $0.2 \times 5 = 1$ kg/min
• Rate of salt leaving = $\frac{x}{100} \times 5 = \frac{x}{20}$ kg/min (since concentration in tank = $x/100$ kg/L)
• Net rate: $\frac{dx}{dt} = 1 - \frac{x}{20}$ [3 marks]

(b) Solve the differential equation. [4 marks]

Solution:

• $\frac{dx}{dt} = 1 - \frac{x}{20} = \frac{20-x}{20}$
• Separate variables: $\frac{dx}{20-x} = \frac{1}{20} dt$
• Integrate: $-\ln|20-x| = \frac{t}{20} + C$
• $\ln|20-x| = -\frac{t}{20} - C$
• $20-x = Ae^{-t/20}$ where $A = e^{-C}$
• $x = 20 - Ae^{-t/20}$
• Initial condition: $x(0) = 0 \implies 0 = 20 - A \implies A = 20$
• $x = 20(1 - e^{-t/20})$ [4 marks]

(c) Amount of salt after a long time. [1 mark]

Solution:

• As $t \to \infty$, $e^{-t/20} \to 0$, so $x \to 20$ kg. [1 mark]

(d) Time to reach 15 kg. [2 marks]

Solution:

• $15 = 20(1 - e^{-t/20}) \implies 1 - e^{-t/20} = 0.75 \implies e^{-t/20} = 0.25$
• $-\frac{t}{20} = \ln 0.25 \implies t = -20 \ln 0.25 = 20 \ln 4 \approx 27.73$ minutes. [2 marks]

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Question 10 (9 marks)

(a) Find $a$, $b$, and $c$. [5 marks]

Solution:

• $f(x) = \frac{ax^2 + bx + c}{x-1}$
• Perform polynomial division: $ax^2 + bx + c = (x-1)(ax + (a+b)) + (a+b+c)$
• So $f(x) = ax + (a+b) + \frac{a+b+c}{x-1}$
• Oblique asymptote is $y = ax + (a+b)$. Given $y = 2x + 3$, so $a = 2$ and $a+b = 3 \implies b = 1$.
• Vertical asymptote at $x = 1$ is consistent with denominator.
• Curve passes through $(2, 10)$: $f(2) = \frac{a(4) + b(2) + c}{2-1} = 4a + 2b + c = 10$
• Substitute $a=2, b=1$: $8 + 2 + c = 10 \implies c = 0$
• Therefore $a = 2$, $b = 1$, $c = 0$. [5 marks]

(b) Sketch the curve $y = f(x)$. [4 marks]

Solution:

• $f(x) = \frac{2x^2 + x}{x-1} = 2x + 3 + \frac{3}{x-1}$
• Vertical asymptote: $x = 1$
• Oblique asymptote: $y = 2x + 3$
• $y$-intercept: $x = 0$, $f(0) = 0$
• $x$-intercepts: $2x^2 + x = 0 \implies x(2x+1) = 0 \implies x = 0, -\frac{1}{2}$
• As $x \to 1^+$, $f(x) \to +\infty$; as $x \to 1^-$, $f(x) \to -\infty$
• Sketch shows curve with vertical asymptote at $x=1$, oblique asymptote $y=2x+3$, crossing axes at $(0,0)$ and $(-0.5, 0)$. [4 marks]

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Question 11 (9 marks)

(a) Show that $\frac{dy}{dx} = -\tan \theta$. [3 marks]

Solution:

• $x = \cos^3 \theta$, $\frac{dx}{d\theta} = 3\cos^2 \theta (-\sin \theta) = -3\cos^2 \theta \sin \theta$
• $y = \sin^3 \theta$, $\frac{dy}{d\theta} = 3\sin^2 \theta \cos \theta$
• $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\sin^2 \theta \cos \theta}{-3\cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$ [3 marks]

(b) Find equation of tangent at $\theta = \pi/4$. [3 marks]

Solution:

• At $\theta = \pi/4$: $x = \cos^3(\pi/4) = (\frac{\sqrt{2}}{2})^3 = \frac{\sqrt{2}}{4}$, $y = \sin^3(\pi/4) = \frac{\sqrt{2}}{4}$
• Gradient: $\frac{dy}{dx} = -\tan(\pi/4) = -1$
• Tangent equation: $y - \frac{\sqrt{2}}{4} = -1(x - \frac{\sqrt{2}}{4}) \implies y = -x + \frac{\sqrt{2}}{2}$ [3 marks]

(c) Find exact area bounded by $C$ and axes. [3 marks]

Solution:

• Area $= \int_0^1 y \, dx$ (since $x$ goes from 1 to 0 as $\theta$ goes from 0 to $\pi/2$)
• $dx = -3\cos^2 \theta \sin \theta \, d\theta$
• When $\theta = 0$, $x = 1$; when $\theta = \pi/2$, $x = 0$
• Area $= \int_{\theta=\pi/2}^0 \sin^3 \theta \cdot (-3\cos^2 \theta \sin \theta) \, d\theta = \int_0^{\pi/2} 3\sin^4 \theta \cos^2 \theta \, d\theta$
• $= 3\int_0^{\pi/2} \sin^4 \theta (1-\sin^2 \theta) \, d\theta = 3\int_0^{\pi/2} (\sin^4 \theta - \sin^6 \theta) \, d\theta$
• Using reduction formula: $\int_0^{\pi/2} \sin^n \theta \, d\theta = \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \frac{1}{2} \cdot \frac{\pi}{2}$ (for even $n$)
• $\int_0^{\pi/2} \sin^4 \theta \, d\theta = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$
• $\int_0^{\pi/2} \sin^6 \theta \, d\theta = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{5\pi}{32}$
• Area $= 3(\frac{3\pi}{16} - \frac{5\pi}{32}) = 3(\frac{6\pi}{32} - \frac{5\pi}{32}) = 3 \cdot \frac{\pi}{32} = \frac{3\pi}{32}$ [3 marks]

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END OF ANSWER KEY