A Level H2 Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Maths H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics (H2) Level: A-Level Paper: Practice Paper 1 (Pure Mathematics) Version: 1 of 5 Duration: 3 hours Total Marks: 100

Name: _________________________ Class: _________________________ Date: _________________________

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Instructions to Candidates

1. This paper contains 11 questions of varying lengths.
2. Answer ALL questions.
3. The use of an approved graphing calculator (GC) is expected, except where unsupported answers are required.
4. Show all necessary working. Marks are awarded for method, not just final answers.
5. Where unsupported answers are required, show your working clearly.
6. The total mark for this paper is 100.
7. Begin each question on a fresh sheet of paper.

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Section A: Pure Mathematics (100 marks)

Question 1 (8 marks)

The functions $f$ and $g$ are defined by:

$f: x \mapsto \frac{2x+1}{x-3}, \quad x \in \mathbb{R}, \ x \neq 3$

$g: x \mapsto \sqrt{x+4}, \quad x \geq -4$

(a) Show that the composite function $gf$ exists and find $gf(x)$ in its simplest form. [4 marks]

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

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

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Question 2 (9 marks)

The curve $C$ has parametric equations:

$x = t^2 - 2t, \quad y = t^2 + 2t, \quad \text{for } t \in \mathbb{R}$

(a) Find the cartesian equation of $C$, giving your answer in the form $(y-x)^2 = k(x+y)$ where $k$ is a constant to be determined. [4 marks]

(b) Sketch the curve $C$, indicating clearly any points where the curve crosses the axes. [3 marks]

(c) The region bounded by $C$ and the $x$-axis is rotated through $2\pi$ radians about the $x$-axis. Find the exact volume of the solid formed. [2 marks]

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Question 3 (8 marks)

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

(b) Hence, or otherwise, solve the inequality $\frac{|x|^2 - 5|x| + 6}{|x| + 1} \leq 0$. [4 marks]

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Question 4 (10 marks)

A geometric progression has first term $a$ and common ratio $r$, where $a > 0$ and $0 < r < 1$. The sum to infinity of the progression is 24. The sum of the first two terms is 18.

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

(b) Find the least value of $n$ such that the sum of the first $n$ terms exceeds 23.5. [3 marks]

(c) Another geometric progression has the same first term $a$ but common ratio $2r$. State, with a reason, whether the sum to infinity of this progression exists. [3 marks]

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Question 5 (9 marks)

The line $l$ has equation $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

The plane $\Pi$ has equation $2x - y + 3z = 7$.

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

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

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

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Question 6 (9 marks)

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

(b) Hence, or otherwise, find the three cube roots of $8i$ in cartesian form $x + iy$, showing your working clearly. [5 marks]

(c) On a single Argand diagram, sketch the three cube roots found in part (b). [2 marks]

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Question 7 (10 marks)

The curve $C$ has equation $x^2 + xy + y^2 = 12$.

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

(b) Find the coordinates of the stationary points on $C$. [4 marks]

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

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Question 8 (9 marks)

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

(b) Hence find an approximation for $e^{0.2} \cos 0.2$, giving your answer to 4 decimal places. [2 marks]

(c) Use the small angle approximations to estimate $\lim_{x \to 0} \frac{e^x \cos x - 1}{x}$. [2 marks]

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Question 9 (10 marks)

A tank initially contains 100 litres of pure water. A salt solution of concentration 0.2 kg per litre flows into the tank at a rate of 5 litres per minute. The mixture is kept uniform by stirring and flows out at the same rate of 5 litres per minute. Let $x$ kg be the amount of salt in the tank at time $t$ minutes.

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

(b) Solve this differential equation to express $x$ in terms of $t$. [4 marks]

(c) Find the amount of salt in the tank after a long time. [1 mark]

(d) How long does it take for the amount of salt to reach 15 kg? [2 marks]

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Question 10 (9 marks)

The function $f$ is defined by $f(x) = \frac{ax^2 + bx + c}{x - 1}$, where $a$, $b$, and $c$ are constants. The curve $y = f(x)$ has a vertical asymptote at $x = 1$ and an oblique asymptote $y = 2x + 3$. The curve passes through the point $(2, 10)$.

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

(b) Sketch the curve $y = f(x)$, showing clearly the asymptotes and the coordinates of any points where the curve crosses the axes. [4 marks]

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Question 11 (9 marks)

A curve $C$ is defined by the parametric equations:

$x = \cos^3 \theta, \quad y = \sin^3 \theta, \quad \text{for } 0 \leq \theta \leq \frac{\pi}{2}$

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

(b) Find the equation of the tangent to $C$ at the point where $\theta = \frac{\pi}{4}$. [3 marks]

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

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END OF PAPER

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Check your work carefully. Ensure all answers are clearly presented and all working is shown.

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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