A Level H2 Mathematics Practice Paper 4

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TuitionGoWhere Practice Paper - Maths H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics H2
Level: A-Level
Paper: Pure Mathematics (Practice Paper 1)
Version: 4 of 5
Duration: 3 Hours
Total Marks: 100
Name: ____________________ Class: __________ Date: __________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided.
2. Answer ALL questions.
3. You may use an approved Graphing Calculator (GC) without CAS.
4. Show all necessary working. Mathematical notation must be used; calculator commands are not acceptable.
5. Give your answers to 3 significant figures unless otherwise stated.

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Section A: Pure Mathematics

Question 1 (a) Given $f(x) = \frac{2x+1}{x-3}$ for $x \neq 3$. (i) Find the inverse function $f^{-1}(x)$ and state its domain. [3] (ii) Sketch the graph of $y = f(x)$, clearly labeling the asymptotes and intercepts. [3] (b) Let $g(x) = \sqrt{x-2}$ for $x \ge 2$. Determine if the composite function $fg$ exists. If so, find an expression for $fg(x)$ and state its range. [4] [Answer Space]

Question 2 (a) A curve $C$ is defined by the parametric equations $x = 2\cos t$ and $y = 3\sin t$ for $0 \le t \le 2\pi$. (i) Find the Cartesian equation of $C$. [2] (ii) Find the coordinates of the points where $C$ meets the $x$-axis. [2] (b) The region bounded by $C$ and the $x$-axis in the first quadrant is rotated through $\pi$ radians about the $x$-axis. Find the exact volume of the solid formed. [5] [Answer Space]

Question 3 (a) Solve the inequality $\frac{x^2 - 5x + 6}{x-1} \le 0$. [4] (b) Find the set of values of $x$ for which $|2x - 5| < |x + 1|$. [4] [Answer Space]

Question 4 (a) Given the implicit equation $x^2 + 3xy + y^2 = 10$. (i) Show that the gradient function $\frac{dy}{dx}$ can be expressed as $\frac{dy}{dx} = \frac{-2x - 3y}{3x + 2y}$. [3] (ii) Find the equation of the tangent to the curve at the point $(1, 2)$. [3] (b) Determine the coordinates of the points where the tangent to the curve is horizontal. [4] [Answer Space]

Question 5 (a) Use the Maclaurin series for $e^x$ and $\sin x$ to find the first three non-zero terms of the series expansion for $f(x) = e^{2x} \sin x$ about $x=0$. [5] (b) State the range of convergence for the series found in (a). [1] (c) Use the approximation from (a) to estimate the value of $f(0.1)$. [3] [Answer Space]

Question 6 (a) A sequence is defined by $u_1 = 2$ and $u_{n+1} = 3u_n - 4$ for $n \in \mathbb{Z}^+$. (i) Find $u_2$ and $u_3$. [2] (ii) Show that the general term is given by $u_n = 2 \cdot 3^{n-1} - 2(3^{n-1}-1)/2$ (or similar closed form). [4] (b) Determine if the series $\sum_{n=1}^{\infty} \frac{1}{u_n}$ converges. Justify your answer. [4] [Answer Space]

Question 7 (a) The complex number $z$ satisfies $|z - (2+i)| = 3$ and $\text{arg}(z - (2+i)) = \frac{\pi}{3}$. (i) Find $z$ in Cartesian form $x+iy$. [3] (ii) On an Argand diagram, sketch the locus of $w$ such that $|w - 2| = |w - (4+2i)|$. [3] (b) Solve the equation $z^3 = -8i$, giving your answers in Cartesian form. [6] [Answer Space]

Question 8 (a) A water tank in the shape of an inverted cone (vertex down) has a height of 10m and a base radius of 4m. Water is leaking from the vertex at a constant rate of $0.5\text{ m}^3/\text{min}$. (i) Find the rate of change of the water level $h$ when $h = 5\text{m}$. [5] (ii) Find the rate of change of the surface area of the water at the same instant. [4] [Answer Space]

Question 9 (a) Solve the differential equation $\frac{dy}{dx} = \frac{y^2}{x}$ given that $y(1) = 2$. [5] (b) A population of bacteria $P$ grows at a rate proportional to the square root of $P$. If $P(0) = 100$ and $P(2) = 144$, find the expression for $P(t)$. [7] [Answer Space]

Question 10 (a) Find the volume of the solid formed when the region bounded by $y = \ln x$, the $x$-axis, and $x=e$ is rotated about the $y$-axis. [7] (b) Use integration by parts to evaluate $\int x^2 \cos x \, dx$. [5] [Answer Space]

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TuitionGoWhere Practice Paper - Maths H2 A-Level (Answers)

Version 4

Question 1 (a)(i) $y = \frac{2x+1}{x-3} \implies x(y-3) = 2x+1 \implies xy - 3y = 2x+1 \implies x(y-2) = 3y+1 \implies f^{-1}(x) = \frac{3x+1}{x-2}$. Domain: $x \neq 2$. [3] (a)(ii) Vertical asymptote $x=3$, Horizontal asymptote $y=2$. $x$-int: $(-0.5, 0)$, $y$-int: $(0, -1/3)$. [3] (b) Range of $g(x)$ is $[0, \infty)$. Domain of $f(x)$ is $x \neq 3$. Since $3 \in [0, \infty)$, we must restrict $g(x) \neq 3 \implies \sqrt{x-2} \neq 3 \implies x \neq 11$. However, for $fg$ to exist as a function on its domain, we check if Range($g$) $\subseteq$ Domain($f$). It is not, unless we exclude $x=11$. $fg(x) = \frac{2\sqrt{x-2}+1}{\sqrt{x-2}-3}$. Range: $f$ maps $[0, \infty) \setminus \{3\}$ to $(-\infty, 2) \cup (2, \infty)$. [4]

Question 2 (a)(i) $\cos t = x/2, \sin t = y/3$. $(x/2)^2 + (y/3)^2 = 1 \implies \frac{x^2}{4} + \frac{y^2}{9} = 1$. [2] (a)(ii) Set $y=0 \implies \frac{x^2}{4} = 1 \implies x = \pm 2$. Points: $(2, 0)$ and $(-2, 0)$. [2] (b) $V = \pi \int_0^2 y^2 \, dx = \pi \int_0^2 9(1 - x^2/4) \, dx = 9\pi [x - \frac{x^3}{12}]_0^2 = 9\pi (2 - 8/12) = 9\pi (4/3) = 12\pi$. [5]

Question 3 (a) $\frac{(x-2)(x-3)}{x-1} \le 0$. Critical points: $1, 2, 3$. Test intervals: $(-\infty, 1) \to (-)$, $(1, 2) \to (+)$, $(2, 3) \to (-)$, $(3, \infty) \to (+)$. Solution: $x < 1$ or $2 \le x \le 3$. [4] (b) $|2x-5|^2 < |x+1|^2 \implies 4x^2 - 20x + 25 < x^2 + 2x + 1 \implies 3x^2 - 22x + 24 < 0$. $(3x-4)(x-6) < 0$. Solution: $4/3 < x < 6$. [4]

Question 4 (a)(i) $2x + 3y + 3x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx}(3x+2y) = -2x-3y \implies \frac{dy}{dx} = \frac{-2x-3y}{3x+2y}$. [3] (a)(ii) At $(1, 2)$, $\frac{dy}{dx} = \frac{-2-6}{3+4} = -8/7$. Eq: $y-2 = -8/7(x-1) \implies 8x+7y=28$. [3] (b) $\frac{dy}{dx} = 0 \implies 2x+3y=0 \implies x = -1.5y$. Substitute into $x^2+3xy+y^2=10$: $(-1.5y)^2 + 3(-1.5y)y + y^2 = 10 \implies 2.25y^2 - 4.5y^2 + y^2 = 10 \implies -1.25y^2 = 10$ (No real solutions). [4]

Question 5 (a) $e^{2x} \approx 1 + 2x + 2x^2$, $\sin x \approx x - x^3/6$. $f(x) = (1 + 2x + 2x^2)(x - x^3/6) = x + 2x^2 + 2x^3 - x^3/6 = x + 2x^2 + \frac{11}{6}x^3$. [5] (b) $x \in \mathbb{R}$. [1] (c) $f(0.1) \approx 0.1 + 2(0.01) + \frac{11}{6}(0.001) \approx 0.1 + 0.02 + 0.001833 = 0.1218$. [3]

Question 6 (a)(i) $u_1=2, u_2=3(2)-4=2, u_3=3(2)-4=2$. [2] (a)(ii) Since $u_1=2$ and $u_2=2$, the sequence is constant $u_n = 2$. [4] (b) $\sum \frac{1}{2}$ diverges as the terms do not approach 0. [4]

Question 7 (a)(i) $z = (2+i) + 3(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2+i + 3(0.5 + i\frac{\sqrt{3}}{2}) = 3.5 + i(1 + 1.5\sqrt{3})$. [3] (a)(ii) Perpendicular bisector of $(2,0)$ and $(4,2)$. Midpoint $(3,1)$, gradient of line is 1, so locus gradient is -1. $y-1 = -1(x-3) \implies y = -x+4$. [3] (b) $z^3 = 8e^{i(3\pi/2 + 2\pi k)}$. $z = 2e^{i(\pi/2 + 2\pi k/3)}$. $k=0: 2e^{i\pi/2} = 2i$. $k=1: 2e^{i(7\pi/6)} = 2(-\frac{\sqrt{3}}{2} - \frac{1}{2}i) = -\sqrt{3}-i$. $k=2: 2e^{i(11\pi/6)} = 2(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \sqrt{3}-i$. [6]

Question 8 (a)(i) $V = \frac{1}{3}\pi r^2 h$. Since $r/h = 4/10 = 0.4$, $V = \frac{1}{3}\pi (0.4h)^2 h = \frac{0.16}{3}\pi h^3$. $\frac{dV}{dt} = 0.16\pi h^2 \frac{dh}{dt}$. $-0.5 = 0.16\pi (25) \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-0.5}{4\pi} = -0.0398\text{ m/min}$. [5] (a)(ii) $A = \pi r^2 = \pi (0.4h)^2 = 0.16\pi h^2$. $\frac{dA}{dt} = 0.32\pi h \frac{dh}{dt} = 0.32\pi (5)(-0.0398) = -0.200\text{ m}^2/\text{min}$. [4]

Question 9 (a) $\int y^{-2} dy = \int x^{-1} dx \implies -y^{-1} = \ln x + C$. $y(1)=2 \implies -1/2 = 0 + C \implies C = -1/2$. $-1/y = \ln x - 1/2 \implies y = \frac{1}{0.5 - \ln x}$. [5] (b) $\frac{dP}{dt} = k\sqrt{P} \implies \int P^{-1/2} dP = \int k dt \implies 2\sqrt{P} = kt + C$. $t=0, P=100 \implies 2(10) = C \implies C=20$. $t=2, P=144 \implies 2(12) = 2k + 20 \implies 2k = 4 \implies k=2$. $2\sqrt{P} = 2t + 20 \implies \sqrt{P} = t+10 \implies P(t) = (t+10)^2$. [7]

Question 10 (a) $V = \pi \int_0^1 x^2 dy$ (where $y = \ln x \implies x = e^y$). $V = \pi \int_0^1 (e^y)^2 dy = \pi \int_0^1 e^{2y} dy = \pi [\frac{1}{2}e^{2y}]_0^1 = \frac{\pi}{2}(e^2 - 1)$. [7] (b) $u=x^2, dv=\cos x dx \implies du=2xdx, v=\sin x$. $\int x^2 \cos x dx = x^2 \sin x - \int 2x \sin x dx$. For $\int 2x \sin x dx$: $u=2x, dv=\sin x dx \implies du=2dx, v=-\cos x$. $\int 2x \sin x dx = -2x \cos x - \int -2 \cos x dx = -2x \cos x + 2 \sin x$. Final: $x^2 \sin x + 2x \cos x - 2 \sin x + C$. [5]