A Level H2 Mathematics Algebra Functions Quiz

A-Level Maths H2 Quiz - Algebra Functions (Answer Key)

Total Marks: 45

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Section A: Functions and Composite Functions [20 marks]

1.(a) Show that the composite function $fg$ exists. [2]

Answer: For $fg$ to exist, the range of $g$ must be a subset of the domain of $f$. Range of $g(x) = x^2 - 4$: $g(x) \geq -4$, so range is $[-4, +\infty)$ Domain of $f$: $\mathbb{R} \setminus \{3\}$ Since $[-4, +\infty) \subset \mathbb{R} \setminus \{3\}$, $fg$ exists.

Marking: 1 mark for identifying range of $g$, 1 mark for correct conclusion

1.(b) Find an expression for $fg(x)$ and state its domain. [3]

Answer: $fg(x) = f(g(x)) = f(x^2 - 4) = \frac{2(x^2 - 4) + 1}{(x^2 - 4) - 3} = \frac{2x^2 - 7}{x^2 - 7}$

Domain: $x^2 - 7 \neq 0$, so $x \neq \pm\sqrt{7}$ Domain is $\mathbb{R} \setminus \{\pm\sqrt{7}\}$

Marking: 2 marks for correct expression, 1 mark for domain

2.(a) Find $hk(x)$. [2]

Answer: $hk(x) = h(k(x)) = h\left(\frac{1}{x+1}\right) = 3\left(\frac{1}{x+1}\right) - 2 = \frac{3}{x+1} - 2 = \frac{3 - 2(x+1)}{x+1} = \frac{1-2x}{x+1}$

Marking: 2 marks for correct simplification

2.(b) Solve the equation $hk(x) = 1$. [2]

Answer: $\frac{1-2x}{x+1} = 1$ $1-2x = x+1$ $-2x - x = 1 - 1$ $-3x = 0$ $x = 0$

Marking: 1 mark for setting up equation, 1 mark for correct solution

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Section B: Parametric Equations and Curves [15 marks]

3.(a) Find the cartesian equation of $C$. [3]

Answer: From $x = 2\cos t$: $\cos t = \frac{x}{2}$ From $y = 3\sin t$: $\sin t = \frac{y}{3}$ Using $\cos^2 t + \sin^2 t = 1$: $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$ $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Marking: 1 mark for eliminating parameter, 2 marks for correct final equation

3.(b) Sketch the curve $C$. [2]

Answer: Ellipse centered at origin with $x$-intercepts at $(\pm 2, 0)$ and $y$-intercepts at $(0, \pm 3)$

Marking: 1 mark for ellipse shape, 1 mark for correct intercepts

4. Find the exact volume of the solid formed. [4]

Answer: From parametric equations: $y = 2t$, $x = t^2$ So $t = \frac{y}{2}$ and $x = \frac{y^2}{4}$, giving $y = 2\sqrt{x}$ When $t = 0$: $x = 0$; when $t = 2$: $x = 4$ Volume = $\pi \int_0^4 y^2 \, dx = \pi \int_0^4 (2\sqrt{x})^2 \, dx = \pi \int_0^4 4x \, dx = 4\pi \left[\frac{x^2}{2}\right]_0^4 = 4\pi \cdot 8 = 32\pi$

Marking: 1 mark for setup, 2 marks for integration, 1 mark for final answer

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Section C: Differential Equations and Applications [10 marks]

5.(a) Write down a differential equation. [1]

Answer: $\frac{dP}{dt} = kP$ where $k > 0$

Marking: 1 mark for correct form

5.(b) Find the population after 8 hours. [3]

Answer: General solution: $P = Ae^{kt}$ Initial condition: $P(0) = 500$, so $A = 500$ Doubling condition: $P(3) = 1000 = 500e^{3k}$ $e^{3k} = 2$, so $k = \frac{\ln 2}{3}$ Therefore: $P(t) = 500e^{\frac{t\ln 2}{3}} = 500 \cdot 2^{t/3}$ $P(8) = 500 \cdot 2^{8/3} = 500 \cdot 2^{2.667} \approx 3175$

Marking: 1 mark for general solution, 1 mark for finding $k$, 1 mark for final answer

6.(a) Show that $\frac{dy}{dx} = -\frac{x + y}{x + 2y}$ [2]

Answer: Differentiating $x^2 + 2xy + y^2 = 9$ implicitly: $2x + 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ $2x + 2y + (2x + 2y)\frac{dy}{dx} = 0$ $(2x + 2y)\frac{dy}{dx} = -(2x + 2y)$ $\frac{dy}{dx} = -\frac{2x + 2y}{2x + 4y} = -\frac{x + y}{x + 2y}$

Marking: 1 mark for implicit differentiation, 1 mark for correct simplification

6.(b) Find the gradient at $(1, 2)$. [1]

Answer: $\frac{dy}{dx}\bigg|_{(1,2)} = -\frac{1 + 2}{1 + 2(2)} = -\frac{3}{5}$

Marking: 1 mark for correct substitution and answer