A Level H2 Mathematics Practice Paper 4

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

1.
$f(x) = 2 \implies \frac{2x-1}{x+3} = 2$
$2x - 1 = 2(x+3)$
$2x - 1 = 2x + 6$
$-1 = 6$ (Contradiction)
Alternatively, find $f^{-1}(x)$:
$y = \frac{2x-1}{x+3} \implies y(x+3) = 2x-1 \implies xy+3y = 2x-1 \implies x(y-2) = -1-3y \implies x = \frac{3y+1}{2-y}$.
$f^{-1}(2)$ involves division by zero.
Answer: Undefined / Does not exist.
(Note: If the question implies finding $x$ such that $f(x)=2$, there is no solution. If asking for $f^{-1}(2)$, it is undefined as 2 is the horizontal asymptote.)
[1]

2.
$fg(x) = f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 + 1 = x - 1 + 1 = x$.
Answer: $x$
[1]

3.
$|2x - 4| \le 6$
$-6 \le 2x - 4 \le 6$
$-2 \le 2x \le 10$
$-1 \le x \le 5$
Answer: $-1 \le x \le 5$
[2]

4.
$e^{2x} > 0$ for all real $x$.
$e^{2x} - 3 > -3$.
Answer: $y > -3$ or $(-3, \infty)$
[1]

5.
Critical values: $x=1, x=-2$.
Test intervals:
$x > 1$: $(+) / (+) > 0$ (True)
$-2 < x < 1$: $(-) / (+) < 0$ (False)
$x < -2$: $(-) / (-) > 0$ (True)
Answer: $x < -2$ or $x > 1$
[2]

6.
(a) Range of $g(x) = x^2+1$ is $[1, \infty)$. Domain of $f(x)$ is $\mathbb{R} \setminus \{2\}$.
Since $2 \in [1, \infty)$, the range of $g$ is not a subset of the domain of $f$. Specifically, $g(x)=2$ when $x=1$, and $f(2)$ is undefined.
Answer: Range of $g$ is not contained in Domain of $f$ (or $g(1)=2$ which is not in domain of $f$).
[1]

(b) We must exclude $x$ such that $g(x) = 2$.
$x^2 + 1 = 2 \implies x^2 = 1 \implies x = \pm 1$.
Largest domain is $\mathbb{R} \setminus \{-1, 1\}$.
Answer: $\{x \in \mathbb{R} : x \neq 1, x \neq -1\}$
[2]

7.
(a) $y = \frac{3x+1}{x-2} \implies y(x-2) = 3x+1 \implies xy - 2y = 3x + 1 \implies xy - 3x = 2y + 1 \implies x(y-3) = 2y+1 \implies x = \frac{2y+1}{y-3}$.
$f^{-1}(x) = \frac{2x+1}{x-3}$.
[2]

(b) Domain of $f^{-1}$ is Range of $f$.
As $x > 2$, $f(x)$ has vertical asymptote $x=2$ and horizontal asymptote $y=3$.
Check monotonicity: $f'(x) = \frac{3(x-2) - (3x+1)(1)}{(x-2)^2} = \frac{-7}{(x-2)^2} < 0$. Decreasing.
Limit $x \to 2^+ \implies f(x) \to \infty$.
Limit $x \to \infty \implies f(x) \to 3$.
Range is $(3, \infty)$.
Answer: $x > 3$
[1]

8.
(a) Vertex at $(1, 3)$. $y$-intercept: $g(0)$ is undefined in domain $x \ge 1$, but if extended, $(0-1)^2+3=4$. However, domain is $x \ge 1$, so starting point is $(1,3)$. Graph is half-parabola opening upwards.
[2]

(b) $y = (x-1)^2 + 3 \implies y-3 = (x-1)^2 \implies x-1 = \sqrt{y-3}$ (since $x \ge 1$).
$x = 1 + \sqrt{y-3}$.
$g^{-1}(x) = 1 + \sqrt{x-3}$.
Domain: $x \ge 3$.
[2]

9.
(a) $gf(x) = g(\ln(x+1)) = e^{\ln(x+1)} - 1 = (x+1) - 1 = x$.
[1]

(b) $f(g(x)) = 2$. Since $f$ and $g$ are inverses (shown in a), $f(g(x)) = x$.
Therefore, $x = 2$.
Check validity: $g(2) = e^2-1 > -1$, so valid for $f$.
Answer: $x = 2$
[2]

10.
$y = \frac{ax+b}{cx+d} \implies y(cx+d) = ax+b \implies cxy + dy = ax+b \implies x(cy-a) = b-dy \implies x = \frac{-dy+b}{cy-a}$.
$f^{-1}(x) = \frac{-dx+b}{cx-a}$.
Given $f^{-1}(x) = f(x) = \frac{ax+b}{cx+d}$.
Comparing coefficients: $\frac{-d}{c} = \frac{a}{c} \implies -d = a \implies a+d=0$.
[2]

11.
(a) Reflect negative parts of $f(x)$ in x-axis. Points A and C remain on axis. B(0,3) remains. Asymptote $x=1$ remains. Curve stays above x-axis.
[2]

(b) $y=f(|x|)$ is even symmetry. Keep right side ($x \ge 0$) and reflect it to left side.
Right side has asymptote $x=1$. Left side will have asymptote $x=-1$.
Passes through $(2,0)$ and $(-2,0)$. Y-intercept $(0,3)$.
[2]

12.

1. Translation by vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (1 unit right).
2. Stretch parallel to y-axis, scale factor 2.
3. Translation by vector $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ (3 units up).
   (Order of stretch and translations can vary if specified correctly, e.g., stretch then translate).
   [2]

13.
V-shape graph. Vertex at $(0.5, 0)$.
y-intercept: $|2(0)-1| = 1 \implies (0,1)$.
x-intercept: $(0.5, 0)$.
[2]

14.
(a) $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$.
[1]

(b) Straight line $y=x+2$ with a "hole" (open circle) at $x=2$.
Coordinate of hole: $(2, 4)$.
[2]

15.
(a) $x^2 - 4x + 5 = (x-2)^2 - 4 + 5 = (x-2)^2 + 1$.
[1]

(b) Since $(x-2)^2 + 1 \ge 1$, the function is always positive.
$|f(x)| = f(x)$. Graph is standard parabola vertex $(2,1)$.
[2]

16.
(a) Max value $\sqrt{4-0}=2$. Min value $\sqrt{4-4}=0$.
Answer: $0 \le y \le 2$
[1]

(b) The function is not one-to-one (fails horizontal line test, e.g., $f(1)=f(-1)$).
[1]

17.
(a) Plot $y$ against $\frac{1}{x}$.
[1]

(b) Equation: $y = A(\frac{1}{x}) + B$.
Gradient $A = -2$.
Intercept $B = 5$.
Answer: $A = -2, B = 5$
[2]

18.
$|x^2 - 5| < 4 \implies -4 < x^2 - 5 < 4$.
Add 5: $1 < x^2 < 9$.
$x^2 > 1 \implies x > 1$ or $x < -1$.
$x^2 < 9 \implies -3 < x < 3$.
Intersection: $(-3, -1) \cup (1, 3)$.
Answer: $-3 < x < -1$ or $1 < x < 3$
[2]

19.
$f(f(x)) = f(1/x) = \frac{1}{1/x} = x$.
This holds for all $x$ in the domain of $f(f(x))$.
Domain of $f$: $x \neq 0$.
Domain of $f(f(x))$: $x \neq 0$ and $f(x) \neq 0$.
$f(x) = 1/x$ is never 0.
So valid for all $x \neq 0$.
Answer: $\{x \in \mathbb{R} : x \neq 0\}$
[1]

20.
$fg(x) = f(x^2) = 2x^2 + 1$.
$gf(x) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1$.
$2x^2 + 1 = 4x^2 + 4x + 1$.
$2x^2 + 4x = 0$.
$2x(x+2) = 0$.
$x = 0$ or $x = -2$.
Answer: $x = 0, -2$
[2]