A Level H2 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key - Version 3

Question 1 (a) $y = \frac{2x+1}{x-3} \implies xy - 3y = 2x + 1 \implies x(y-2) = 3y+1 \implies x = \frac{3y+1}{y-2}$. $f^{-1}(x) = \frac{3x+1}{x-2}$, Domain: $x \neq 2$. [4] (b) Critical points: $x = -5, 1, 2$. Testing intervals: $x \in (-5, 1) \cup (2, \infty)$ gives positive. $x \in (-\infty, -5] \cup [1, 2]$ gives negative or zero. Wait, check $x=2$: $(0)(7)/1 = 0$ (Included). Check $x=1$: Undefined. Solution: $x \in (-\infty, -5] \cup (1, 2]$. [4]

Question 2 (a) (i) Graph is a hyperbola-like shape starting at $(1,0)$ and $(3,0)$ opening upwards. [3] (ii) Range: $[0, \infty)$. [1] (b) For $hg$ to exist, Range($g$) $\subseteq$ Domain($h$). Range($g$) = $[0, \infty)$. Domain($h$) = $(-2, \infty)$. Since $[0, \infty) \subseteq (-2, \infty)$, $hg$ exists for all $x$ in Domain($g$). $x \in (-\infty, 1] \cup [3, \infty)$. [4]

Question 3 (a) (i) $\cos t = x/2, \sin t = y/3 \implies (x/2)^2 + (y/3)^2 = 1 \implies \frac{x^2}{4} + \frac{y^2}{9} = 1$. [3] (ii) $y=0 \implies x^2/4 = 1 \implies x = \pm 2$. Points: $(2,0), (-2,0)$. [2] (b) $V = \pi \int_{-2}^2 y^2 dx = \pi \int_{-2}^2 9(1 - x^2/4) dx = 9\pi [x - x^3/12]_{-2}^2 = 9\pi [(2 - 8/12) - (-2 + 8/12)] = 9\pi [4 - 4/3] = 9\pi(8/3) = 24\pi$. [5]

Question 4 (a) (i) $e^{2x} - 5e^x + 6 = 0 \implies (e^x - 2)(e^x - 3) = 0 \implies x = \ln 2, \ln 3$. [3] (ii) $f'(x) = 2e^{2x} - 5e^x = e^x(2e^x - 5)$. Stationary point: $e^x = 2.5 \implies x = \ln 2.5$. $y = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25$. Point: $(\ln 2.5, -0.25)$. $f''(x) = 4e^{2x} - 5e^x$. At $x = \ln 2.5$, $f'' = 4(6.25) - 5(2.5) = 25 - 12.5 = 12.5 > 0 \implies$ Minimum. [4] (b) Sketch with intercepts $(\ln 2, 0), (\ln 3, 0)$, y-intercept $(0, 2)$, and min $(\ln 2.5, -0.25)$. [3]

Question 5 (a) Using Gaussian elimination or Cramer's rule: $x=1, y=1, z=4$. [6] (b) $|2x-5| < k \implies 5-k < 2x < 5+k \implies \frac{5-k}{2} < x < \frac{5+k}{2}$. For solutions to be in $(1, 4)$, the interval $(\frac{5-k}{2}, \frac{5+k}{2})$ must overlap with $(1, 4)$. Also, for the solution set to be contained or intersect? Usually "has solutions in" means intersection is non-empty. $\frac{5-k}{2} < 4$ and $\frac{5+k}{2} > 1 \implies k > -3$ and $k > -3$. However, $k$ must be positive for the modulus inequality to have any solution. $k > 0$. [4]

Question 6 (a) (i) $f'(x) = -1/(x+1)^2$. Since $f'(x) < 0$ for all $x > -1$, $f$ is strictly decreasing, thus one-to-one. [3] (ii) $y = 1/(x+1) \implies x+1 = 1/y \implies x = 1/y - 1$. $f^{-1}(x) = \frac{1}{x} - 1$. Domain: $x > 0$. [3] (b) $f(x) \to f(x-2)$ (Translation 2 units right) $\to 3f(x-2)$ (Stretch vertical scale factor 3) $\to 3f(x-2)+1$ (Translation 1 unit up). [4]

Question 7 (a) (i) $2x + 3(x + y\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0 \implies 3y\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - 3y \implies \frac{dy}{dx}(3y+2y) = -(2x+3y) \implies \frac{dy}{dx} = -\frac{2x+3y}{3x+2y}$. [4] (ii) At $(1, 2)$, $\frac{dy}{dx} = -\frac{2(1)+3(2)}{3(1)+2(2)} = -\frac{8}{7}$. Eq: $y - 2 = -\frac{8}{7}(x - 1) \implies 8x + 7y = 22$. [3] (b) Horizontal $\implies \frac{dy}{dx} = 0 \implies 2x + 3y = 0 \implies x = -1.5y$. 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. No points with horizontal tangents. [3]

Question 8 (a) (i) $u_2 = 0.5(2)+3 = 4$; $u_3 = 0.5(4)+3 = 5$. [2] (ii) Let $v_n = u_n - 6$. $v_{n+1} = u_{n+1} - 6 = (0.5u_n + 3) - 6 = 0.5u_n - 3 = 0.5(u_n - 6) = 0.5v_n$. Since $v_{n+1}/v_n = 0.5$, it is a GP. [4] (iii) $v_1 = 2 - 6 = -4$. $v_n = -4(0.5)^{n-1}$. $u_n = 6 - 4(0.5)^{n-1}$. [3] (b) As $n \to \infty, (0.5)^{n-1} \to 0$, so $u_n \to 6$. [1]

Question 9 (a) (i) $f(x) = \ln(x^2-1)$. Range of $f$: $(-\infty, \infty)$. Domain of $f^{-1}$ = Range of $f = \mathbb{R}$. Range of $f^{-1}$ = Domain of $f = (1, \infty)$. [4] (ii) $f(x)$ is increasing from $x=1$ (asymptote $x=1$) passing through $( \sqrt{2}, 0)$. $f^{-1}(x)$ is the reflection across $y=x$. [4] (b) $\ln(x^2-1) = 0 \implies x^2-1 = 1 \implies x^2 = 2 \implies x = \sqrt{2}$ (since $x>1$). [2]

Question 10 (a) (i) $V = \frac{1}{3}\pi r^2 h$. By similar triangles, $r/h = 5/10 = 1/2 \implies r = h/2$. $V = \frac{1}{3}\pi (h/2)^2 h = \frac{\pi h^3}{12}$. $\frac{dV}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt} \implies 2 = \frac{\pi (4)^2}{4} \frac{dh}{dt} \implies 2 = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{2\pi}\text{ m/min}$. [6] (ii) $S = \pi r^2 = \pi (h/2)^2 = \frac{\pi h^2}{4}$. $\frac{dS}{dt} = \frac{\pi h}{2} \frac{dh}{dt} = \frac{\pi (4)}{2} (\frac{1}{2\pi}) = 1\text{ m}^2/\text{min}$. [4] (b) $V = \frac{\pi (8)^3}{12} = \frac{512\pi}{12} = \frac{128\pi}{3}$. Time $t = V / (dV/dt) = \frac{128\pi/3}{2} = \frac{64\pi}{3} \approx 67.02\text{ min}$. [2]