A Level H2 Mathematics Practice Paper 2

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

1. Domain of $f(x) = \frac{2x+1}{x-3}$ $x \neq 3$ or $x \in \mathbb{R}, x \neq 3$. (1 mark)

2. $g(x) = \sqrt{4-x^2}$ Domain: $4-x^2 \ge 0 \implies x^2 \le 4 \implies -2 \le x \le 2$. Range: Since $x^2 \in [0, 4]$, $4-x^2 \in [0, 4]$, so $\sqrt{4-x^2} \in [0, 2]$. (2 marks)

3. $|2x - 5| \le 3$ $-3 \le 2x - 5 \le 3$ $2 \le 2x \le 8$ $1 \le x \le 4$. (2 marks)

4. $f(x) = x^2 - 4x + 7$ for $x \in [0, 5]$ Complete square: $f(x) = (x-2)^2 + 3$. Vertex at $(2, 3)$. Endpoints: $f(0) = 7$, $f(5) = 25 - 20 + 7 = 12$. Minimum value is 3, maximum is 12. Range: $[3, 12]$. (3 marks)

5. $k(x) = x^3 - x$ Not one-to-one. Justification: $k(x) = x(x-1)(x+1)$. $k(0) = k(1) = k(-1) = 0$. Multiple $x$ values map to the same $y$. (2 marks)

6. $fg(x)$ $f(g(x)) = 3(x^2) + 2 = 3x^2 + 2$. (2 marks)

7. $gf(x)$ $g(f(x)) = (3x+2)^2 = 9x^2 + 12x + 4$. (2 marks)

8. $f^{-1}(x)$ for $f(x) = \frac{x+1}{x-2}$ $y = \frac{x+1}{x-2} \implies xy - 2y = x + 1 \implies x(y-1) = 2y + 1 \implies x = \frac{2y+1}{y-1}$. $f^{-1}(x) = \frac{2x+1}{x-1}$. Domain: $x \neq 1$. (3 marks)

9. Existence of $fg$ $g(x) = \ln(x-1)$ for $x > 1$. Range of $g$ is $\mathbb{R}$. $f(x) = e^{2x}$ for $x \in \mathbb{R}$. Domain of $f$ is $\mathbb{R}$. Since Range($g$) $\subseteq$ Domain($f$), $fg$ exists. (3 marks)

10. $fg(x)$ and Range $fg(x) = e^{2\ln(x-1)} = e^{\ln(x-1)^2} = (x-1)^2$. Since $x > 1$, $(x-1)^2 > 0$. Range: $(0, \infty)$. (3 marks)

11. Domain of $h^{-1}(x)$ Domain of $h^{-1}$ = Range of $h$. $h(x) = \sqrt{x-3}$ for $x \ge 3$. Range of $h$ is $[0, \infty)$. Domain of $h^{-1}$ is $x \ge 0$. (2 marks)

12. $fg(x) = 0$ $fg(x) = 2(\frac{1}{x+2}) - 1

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

**1. Domain of $f(x) = \frac{2x+1}{x-3}$**
$x \neq 3$ or $x \in \mathbb{R}, x \neq 3$.
(1 mark)

**2. $g(x) = \sqrt{4-x^2}$**
Domain: $4-x^2 \ge 0 \implies x^2 \le 4 \implies -2 \le x \le 2$.
Range: Since $x^2 \in [0, 4]$, $4-x^2 \in [0, 4]$, so $\sqrt{4-x^2} \in [0, 2]$.
(2 marks)

**3. $|2x - 5| \le 3$**
$-3 \le 2x - 5 \le 3$
$2 \le 2x \le 8$
$1 \le x \le 4$.
(2 marks)

**4. $f(x) = x^2 - 4x + 7$ for $x \in [0, 5]$**
Complete square: $f(x) = (x-2)^2 + 3$.
Vertex at $(2, 3)$.
Endpoints: $f(0) = 7$, $f(5) = 25 - 20 + 7 = 12$.
Minimum value is 3, maximum is 12.
Range: $[3, 12]$.
(3 marks)

**5. $k(x) = x^3 - x$**
Not one-to-one.
Justification: $k(x) = x(x-1)(x+1)$. $k(0) = k(1) = k(-1) = 0$. Multiple $x$ values map to the same $y$.
(2 marks)

**6. $fg(x)$**
$f(g(x)) = 3(x^2) + 2 = 3x^2 + 2$.
(2 marks)

**7. $gf(x)$**
$g(f(x)) = (3x+2)^2 = 9x^2 + 12x + 4$.
(2 marks)

**8. $f^{-1}(x)$ for $f(x) = \frac{x+1}{x-2}$**
$y = \frac{x+1}{x-2} \implies xy - 2y = x + 1 \implies x(y-1) = 2y + 1 \implies x = \frac{2y+1}{y-1}$.
$f^{-1}(x) = \frac{2x+1}{x-1}$.
Domain: $x \neq 1$.
(3 marks)

**9. Existence of $fg$**
$g(x) = \ln(x-1)$ for $x > 1$. Range of $g$ is $\mathbb{R}$.
$f(x) = e^{2x}$ for $x \in \mathbb{R}$. Domain of $f$ is $\mathbb{R}$.
Since Range($g$) $\subseteq$ Domain($f$), $fg$ exists.
(3 marks)

**10. $fg(x)$ and Range**
$fg(x) = e^{2\ln(x-1)} = e^{\ln(x-1)^2} = (x-1)^2$.
Since $x > 1$, $(x-1)^2 > 0$.
Range: $(0, \infty)$.
(3 marks)

**11. Domain of $h^{-1}(x)$**
Domain of $h^{-1}$ = Range of $h$.
$h(x) = \sqrt{x-3}$ for $x \ge 3$.
Range of $h$ is $[0, \infty)$.
Domain of $h^{-1}$ is $x \ge 0$.
(2 marks)

**12. $fg(x) = 0$**
$fg(x) = 2(\frac{1}{x+2}) - 1 = 0$
$\frac{2}{x+2} = 1 \implies x+2 = 2 \implies x = 0$.
(3 marks)

**13. Sketch of $y = \frac{1}{x-2} + 3$**
Vertical asymptote: $x=2$. Horizontal asymptote: $y=3$.
y-intercept: $x=0 \implies y = \frac{1}{-2} + 3 = 2.5$.
x-intercept: $y=0 \implies \frac{1}{x-2} = -3 \implies x-2 = -\frac{1}{3} \implies x = \frac{5}{3}$.
(3 marks)

**14. Transformations of $y = f(2x - 4)$**
$f(2(x-2))$.
1. Horizontal stretch by factor $\frac{1}{2}$ towards y-axis.
2. Horizontal translation by 2 units to the right.
(2 marks)

**15. Sketch of $y = |x^2 - 2x|$**
$f(x) = x(x-2)$. Roots at 0 and 2. Vertex at $(1, -1)$.
For $|f(x)|$, the part of the graph between $x=0$ and $x=2$ is reflected above the x-axis.
Vertex becomes $(1, 1)$.
(3 marks)

**16. Transformations of $y = e^x$ to $y = 3e^{x+1} - 2$**
1. Translation by 1 unit to the left.
2. Vertical stretch by factor 3 from x-axis.
3. Translation by 2 units downwards.
(3 marks)

**17. Cartesian equation of $C$**
$y = 2t \implies t = \frac{y}{2}$.
$x = (\frac{y}{2})^2 \implies x = \frac{y^2}{4}$ or $y^2 = 4x$.
(3 marks)

**18. $\frac{x-1}{x+3} \le 0$**
Critical values: $x=1, x=-3$.
Testing intervals:
$x < -3$: $(-)/(-) = (+)$
$-3 < x \le 1$: $(-)/(+) = (-)$
$x > 1$: $(+)/(+) = (+)$
Solution: $-3 < x \le 1$.
(3 marks)

**19. $|2x - 1| = |x + 4|$**
Case 1: $2x - 1 = x + 4 \implies x = 5$.
Case 2: $2x - 1 = -(x + 4) \implies 3x = -3 \implies x = -1$.
Solutions: $x = 5, x = -1$.
(3 marks)

**20. $\frac{x^2 - 5x + 6}{x-1} > 0$**
$\frac{(x-2)(x-3)}{x-1} > 0$.
Critical values: $x=1, 2, 3$.
Intervals:
$(-\infty, 1)$: $(-)(-)/(-) = (-)$
$(1, 2)$: $(-)(-)/(+) = (+)$
$(2, 3)$: $(+)(-)/(+) = (-)$
$(3, \infty)$: $(+)(+)/(+) = (+)$
Solution: $1 < x < 2$ or $x > 3$.
(4 marks)