A Level H2 Mathematics Algebra Functions Quiz

A-Level Maths H2 Quiz - Algebra Functions

Answer Key and Marking Scheme

---

Question 1 [5 marks]

(a) [2]
The composite $fg(x) = f(g(x))$ exists if the range of $g$ is a subset of the domain of $f$.

• Range of $g$: Since $g(x) = \sqrt{x+3}$ for $x \geq -3$, we have $g(x) \geq 0$. So range of $g$ is $[0, \infty)$.
• Domain of $f$: $x \geq 0$, i.e., $[0, \infty)$.
• Since range of $g$ = $[0, \infty)$ ⊆ domain of $f$ = $[0, \infty)$, the composite $fg$ exists. ✓

Marking: 1 mark for finding range of $g$, 1 mark for comparing with domain of $f$ and concluding.

(b) [3]
$fg(x) = f(g(x)) = f(\sqrt{x+3}) = (\sqrt{x+3})^2 - 4 = x + 3 - 4 = x - 1$.

• Domain of $fg$: Since $g$ requires $x \geq -3$, the domain of $fg$ is $x \geq -3$, i.e., $[-3, \infty)$.
• Range of $fg$: Since $fg(x) = x - 1$ for $x \geq -3$, the range is $[-3 - 1, \infty) = [-4, \infty)$.

Marking: 1 mark for correct expression, 1 mark for domain, 1 mark for range.

Common mistake: Students may write $fg(x) = f(x) \cdot g(x)$. Emphasise that $fg(x) = f(g(x))$, i.e., function composition, not multiplication.

---

Question 2 [6 marks]

(a) [3]
Let $y = f(x) = \dfrac{2x+1}{x-3}$. To find the inverse, swap $x$ and $y$ and solve for $y$:

$x = \dfrac{2y + 1}{y - 3}$ $x(y - 3) = 2y + 1$ $xy - 3x = 2y + 1$ $xy - 2y = 3x + 1$ $y(x - 2) = 3x + 1$ $y = \dfrac{3x + 1}{x - 2}$

So $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$.

Domain of $f^{-1}$: Since the denominator cannot be zero, $x \neq 2$. Domain is $\mathbb{R} \setminus \{2\}$.

Marking: 1 mark for setting up the swap, 1 mark for correct algebra to isolate $y$, 1 mark for correct domain.

(b) [3]
Set $f(x) = f^{-1}(x)$:

$\dfrac{2x + 1}{x - 3} = \dfrac{3x + 1}{x - 2}$

Cross-multiplying: $(2x + 1)(x - 2) = (3x + 1)(x - 3)$ $2x^2 - 4x + x - 2 = 3x^2 - 9x + x - 3$ $2x^2 - 3x - 2 = 3x^2 - 8x - 3$ $0 = x^2 - 5x - 1$

Using the quadratic formula: $x = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2}$

Both values are valid (neither equals 3 or 2).

Marking: 1 mark for setting up the equation, 1 mark for correct expansion and simplification, 1 mark for correct final answers.

Teaching note: When $f(x) = f^{-1}(x)$, the solutions lie on the line $y = x$ (or symmetrically placed). Students can also solve $f(x) = x$ as an alternative approach since any intersection of $f$ and $f^{-1}$ on $y=x$ satisfies $f(x) = x$.

---

Question 3 [5 marks]

(a) [2]
The composite $gf(x) = g(f(x))$ exists if the range of $f$ is a subset of the domain of $g$.

• $f(x) = e^{2x} - 1$. Since $e^{2x} > 0$ for all $x$, we have $f(x) > -1$. Range of $f$ is $(-1, \infty)$.
• Domain of $g$: $x > -4$, i.e., $(-4, \infty)$.
• Since $(-1, \infty) \subset (-4, \infty)$, the range of $f$ is contained in the domain of $g$. Therefore $gf$ exists. ✓

Marking: 1 mark for finding range of $f$, 1 mark for comparison and conclusion.

(b) [3]
$gf(x) = g(f(x)) = g(e^{2x} - 1) = \ln((e^{2x} - 1) + 4) = \ln(e^{2x} + 3)$.

• Range of $gf$: Since $e^{2x} > 0$, we have $e^{2x} + 3 > 3$, so $\ln(e^{2x} + 3) > \ln 3$. As $x \to -\infty$, $e^{2x} \to 0$, so $gf(x) \to \ln 3$. As $x \to \infty$, $gf(x) \to \infty$.
• Range of $gf$ is $(\ln 3, \infty)$.

Marking: 1 mark for correct composite expression, 1 mark for correct inequality reasoning, 1 mark for correct range.

---

Question 4 [7 marks]

(a) [4]
Let $y = f(x) = 3 - \sqrt{x + 1}$.

Swap $x$ and $y$: $x = 3 - \sqrt{y + 1}$ $\sqrt{y + 1} = 3 - x$ $y + 1 = (3 - x)^2$ $y = (3 - x)^2 - 1 = 9 - 6x + x^2 - 1 = x^2 - 6x + 8$

So $f^{-1}(x) = x^2 - 6x + 8 = (x - 3)^2 - 1$.

• Domain of $f^{-1}$: Since $\sqrt{y+1} \geq 0$, we need $3 - x \geq 0$, so $x \leq 3$. Domain is $(-\infty, 3]$.
• Range of $f^{-1}$: This equals the domain of $f$, which is $[-1, \infty)$.

Marking: 1 mark for swapping and solving, 1 mark for correct expression, 1 mark for domain, 1 mark for range.

(b) [3]
The graph of $y = f(x) = 3 - \sqrt{x+1}$:

• Starts at $(-1, 3)$ and decreases, passing through $(0, 2)$, $(3, 1)$, $(8, -1)$.
• Shape: reflected square root curve (decreasing, concave up).

The graph of $y = f^{-1}(x) = (x-3)^2 - 1$:

• A parabola with vertex at $(3, -1)$, but only the left branch ($x \leq 3$) is drawn.
• Passes through $(3, -1)$, $(2, 0)$, $(1, 3)$, $(-1, 8)$.

The two graphs are reflections of each other in the line $y = x$.

Marking: 1 mark for correct shape of $f$, 1 mark for correct shape of $f^{-1}$, 1 mark for showing reflection symmetry about $y = x$.

Expected visual features for Q4-fig1: The graph should show $y = f(x)$ as a decreasing curve starting at $(-1,3)$ going through $(0,2)$, $(3,1)$, $(8,-1)$. The inverse $y = f^{-1}(x)$ should be the left half of an upward parabola with vertex at $(3,-1)$, passing through $(2,0)$, $(1,3)$, $(-1,8)$. The line $y=x$ should be shown as a dashed line. Both curves should be mirror images across $y=x$.

---

Question 5 [5 marks]

(a) [1]
$f(x) = x^2 - 6x + 5$ for $x \geq 3$. The derivative $f'(x) = 2x - 6 \geq 0$ for $x \geq 3$, with $f'(3) = 0$ and $f'(x) > 0$ for $x > 3$. Thus $f$ is strictly increasing on $[3, \infty)$, which means $f$ is one-one. Therefore $f$ has an inverse function.

Marking: 1 mark for correct reasoning (strictly increasing / one-one on the given domain).

(b) [3]
Let $y = x^2 - 6x + 5$. Complete the square: $y = (x - 3)^2 - 4$.

Swap: $x = (y - 3)^2 - 4$, so $(y-3)^2 = x + 4$, giving $y - 3 = \sqrt{x + 4}$ (taking positive root since $y \geq 3$).

So $y = 3 + \sqrt{x + 4}$, and $f^{-1}(x) = 3 + \sqrt{x + 4}$.

• Domain of $f^{-1}$: Need $x + 4 \geq 0$, so $x \geq -4$. Domain is $[-4, \infty)$.

Marking: 1 mark for completing the square, 1 mark for correct inverse expression, 1 mark for domain.

(c) [1]
The range of $f^{-1}$ equals the domain of $f$, which is $[3, \infty)$.

Marking: 1 mark for correct range.

---

Question 6 [6 marks]

(a) [2]
$f(x) = \dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2} = x + 2$ for $x \neq 2$.

At $x = 2$, the original function is undefined (division by zero). The graph of $y = f(x)$ is the line $y = x + 2$ with a hole (removable discontinuity) at the point $(2, 4)$.

Marking: 1 mark for correct simplification, 1 mark for identifying the hole/discontinuity at $x = 2$.

(b) [2]
Since $f(x) = x + 2$ for all $x \neq 2$, the output can take any real value except $2 + 2 = 4$.

Range of $f$ is $\mathbb{R} \setminus \{4\}$.

Marking: 1 mark for reasoning, 1 mark for correct range.

(c) [2]
The student's claim is incorrect because $f(x) = x + 2$ for $x \neq 2$, so $f(x) \neq 4$ for any value of $x$ in the domain. The value 4 is excluded from the range because $x = 2$ is not in the domain of $f$. The graph has a hole at $(2, 4)$, so the output 4 is never achieved.

Marking: 1 mark for identifying that $f(x) \neq 4$, 1 mark for linking to the domain restriction.

---

Question 7 [6 marks]

(a) [1]
Since $|2x - 3| \geq 0$ for all real $x$, and $|2x - 3| = 0$ when $x = \dfrac{3}{2}$, the range of $f$ is $[0, \infty)$.

Marking: 1 mark for correct range.

(b) [2]
$f$ is not one-one. For example, $f(0) = |0 - 3| = 3$ and $f(3) = |6 - 3| = 3$. Since $f(0) = f(3)$ but $0 \neq 3$, the function fails the horizontal line test and is not one-one.

Alternatively: The graph of $y = |2x - 3|$ is V-shaped with a minimum at $x = \dfrac{3}{2}$, so it is not monotonic.

Marking: 1 mark for stating "not one-one", 1 mark for valid justification (counter-explanation or graphical reasoning).

(c) [3]
For $x \geq \dfrac{3}{2}$, we have $2x - 3 \geq 0$, so $g(x) = 2x - 3$.

Let $y = 2x - 3$. Then $x = \dfrac{y + 3}{2}$, so $g^{-1}(x) = \dfrac{x + 3}{2}$.

The domain of $g^{-1}$ is the range of $g$. Since $g(x) = 2x - 3$ for $x \geq \dfrac{3}{2}$, the range is $[0, \infty)$.

So $g^{-1}(x) = \dfrac{x + 3}{2}$ for $x \geq 0$.

Marking: 1 mark for simplifying $g(x)$ on the restricted domain, 1 mark for correct inverse, 1 mark for correct domain.

---

Question 8 [5 marks]

(a) [3]
Let $y = \dfrac{ax + b}{cx + d}$. Swap $x$ and $y$:

$x = \dfrac{ay + b}{cy + d}$ $x(cy + d) = ay + b$ $cxy + dx = ay + b$ $cxy - ay = b - dx$ $y(cx - a) = b - dx$ $y = \dfrac{b - dx}{cx - a} = \dfrac{-dx + b}{cx - a}$

So $f^{-1}(x) = \dfrac{b - dx}{cx - a}$.

Note: Since $ad \neq bc$, we have $f^{-1}$ is well-defined (the determinant condition ensures $f$ is not a constant function).

Marking: 1 mark for setting up the swap, 1 mark for correct algebraic manipulation, 1 mark for correct final expression.

(b) [2]
$f(f^{-1}(x)) = f\!\left(\dfrac{b - dx}{cx - a}\right) = \dfrac{a\left(\frac{b-dx}{cx-a}\right) + b}{c\left(\frac{b-dx}{cx-a}\right) + d}$

Numerator: $\dfrac{a(b-dx) + b(cx-a)}{cx-a} = \dfrac{ab - adx + bcx - ab}{cx-a} = \dfrac{x(bc - ad)}{cx-a}$

Denominator: $\dfrac{c(b-dx) + d(cx-a)}{cx-a} = \dfrac{bc - cdx + dcx - ad}{cx-a} = \dfrac{bc - ad}{cx-a}$

Therefore: $f(f^{-1}(x)) = \dfrac{x(bc-ad)}{cx-a} \cdot \dfrac{cx-a}{bc-ad} = x$. ✓

Marking: 1 mark for correct substitution, 1 mark for correct simplification to $x$.

---

Question 9 [8 marks]

(a) [2]

• Domain of $f$: $[-3, 3]$ (given).
• Range of $f$: Since $f(x) = \sqrt{9 - x^2}$, the maximum value is $\sqrt{9} = 3$ (at $x = 0$) and the minimum is $\sqrt{0} = 0$ (at $x = \pm 3$). Range is $[0, 3]$.

Marking: 1 mark for domain, 1 mark for range.

(b) [2]
$f$ is not one-one on $[-3, 3]$. For example, $f(-1) = \sqrt{8} = f(1)$ but $-1 \neq 1$. The graph is the upper semicircle of $x^2 + y^2 = 9$, which fails the horizontal line test. Therefore $f$ does not have an inverse.

Marking: 1 mark for stating not one-one, 1 mark for valid justification.

(c) [4]
For $h(x) = \sqrt{9 - x^2}$ with domain $0 \leq x \leq 3$:

• $h$ is strictly decreasing on $[0, 3]$ (since $h'(x) = \frac{-x}{\sqrt{9-x^2}} < 0$ for $x > 0$), so $h$ is one-one and has an inverse.

Let $y = \sqrt{9 - x^2}$. Then $y^2 = 9 - x^2$, so $x^2 = 9 - y^2$, and $x = \sqrt{9 - y^2}$ (taking positive root since $x \geq 0$).

So $h^{-1}(x) = \sqrt{9 - x^2}$.

• Domain of $h^{-1}$: The range of $h$ is $[0, 3]$ (since $h(0) = 3$ and $h(3) = 0$). So domain of $h^{-1}$ is $[0, 3]$.
• Range of $h^{-1}$: The domain of $h$ is $[0, 3]$. So range of $h^{-1}$ is $[0, 3]$.

Marking: 1 mark for showing $h$ is one-one, 1 mark for correct inverse expression, 1 mark for domain, 1 mark for range.

Teaching note: Interestingly, $h^{-1}(x) = \sqrt{9-x^2} = h(x)$ — the function is its own inverse on this restricted domain. This is because the upper-right quarter circle is symmetric about $y = x$.

---

Question 10 [7 marks]

(a) [2]
$f(x) = x^2 + 2x + 3 = (x+1)^2 + 2$.

Since $(x+1)^2 \geq 0$, the minimum value of $f$ is $2$ (at $x = -1$).

Range of $f$ is $[2, \infty)$.

Marking: 1 mark for completing the square, 1 mark for correct range.

(b) [3]

• Range of $g$: Since $g(x) = x - 1$ for all real $x$, range of $g$ is $\mathbb{R}$.
• Domain of $f$: $\mathbb{R}$.
• Since range of $g$ = $\mathbb{R}$ ⊆ domain of $f$ = $\mathbb{R}$, the composite $fg$ exists.

$fg(x) = f(g(x)) = f(x - 1) = (x-1)^2 + 2(x-1) + 3 = x^2 - 2x + 1 + 2x - 2 + 3 = x^2 + 2$.

Marking: 1 mark for showing the composite exists, 1 mark for correct substitution, 1 mark for correct simplification.

(c) [2]
$fg(x) = 3$: $x^2 + 2 = 3$, so $x^2 = 1$, giving $x = \pm 1$.

Marking: 1 mark for setting up equation, 1 mark for both solutions.

---

Question 11 [5 marks]

(a) [1]
From the graph, the maximum value of $f$ is $4$ and the parabola opens downward. The range is $(-\infty, 4]$.

Marking: 1 mark for correct range.

(b)(i) [2]
$y = f(x + 2)$ represents a translation of $y = f(x)$ by 2 units in the negative $x$-direction.

• Vertex moves from $(0, 4)$ to $(-2, 4)$.
• $x$-intercepts move from $(-2, 0)$ and $(3, 0)$ to $(-4, 0)$ and $(1, 0)$.
• $y$-intercept: $f(0 + 2) = f(2)$. From the original, $f(2) = -2(2) + 4$ ... using the fact that the parabola passes through $(3,0)$ and $(-2,0)$ with vertex $(0,4)$: $f(x) = -\frac{4}{5}(x+2)(x-3)$ ... actually, let's find the equation. With roots at $-2$ and $3$: $f(x) = a(x+2)(x-3)$. Vertex at $x = \frac{-2+3}{2} = 0.5$ — but the vertex is given as $(0,4)$. So: $f(x) = a(x+2)(x-3)$, and $f(0) = a(2)(-3) = -6a = 4$, so $a = -\frac{2}{3}$. Thus $f(x) = -\frac{2}{3}(x+2)(x-3)$.

For $y = f(x+2)$: vertex at $(-2, 4)$, $x$-intercepts at $(-4, 0)$ and $(1, 0)$, $y$-intercept at $f(2) = -\frac{2}{3}(4)(-1) = \frac{8}{3}$.

Marking: 1 mark for correct vertex and intercepts, 1 mark for correct shape/diagram.

(b)(ii) [2]
$y = 2f(x)$ represents a stretch parallel to the $y$-axis by scale factor 2.

• Vertex moves from $(0, 4)$ to $(0, 8)$.
• $x$-intercepts remain at $(-2, 0)$ and $(3, 0)$ (unchanged since $y = 0$ is invariant under vertical stretch).
• $y$-intercept: $2f(0) = 2(4) = 8$.

Marking: 1 mark for correct vertex and intercepts, 1 mark for correct shape/diagram.

Expected visual features for Q11-fig1: A downward-opening parabola with vertex at (0,4), passing through (-2,0) and (3,0). The y-intercept is at (0,4). Axes should be labelled with appropriate scale.

---

Question 12 [8 marks]

(a) [3]
Starting from $y = \dfrac{1}{x}$:

Step 1: Replace $x$ with $x - 3$ to get $y = \dfrac{1}{x-3}$. This is a translation of 3 units in the positive $x$-direction.

Step 2: Multiply by 2 to get $y = \dfrac{2}{x-3}$. This is a stretch parallel to the $y$-axis by scale factor 2.

Step 3: Add 4 to get $y = \dfrac{2}{x-3} + 4$. This is a translation of 4 units in the positive $y$-direction.

Marking: 1 mark per correct transformation (order must be correct).

(b) [2]

• Vertical asymptote: $x = 3$ (denominator zero).
• Horizontal asymptote: $y = 4$ (as $x \to \pm\infty$, $\dfrac{2}{x-3} \to 0$).

Marking: 1 mark each.

(c) [3]
Set $\dfrac{2}{x-3} + 4 = x$:

$\dfrac{2}{x-3} = x - 4$ $2 = (x-4)(x-3) = x^2 - 7x + 12$ $x^2 - 7x + 10 = 0$ $(x-2)(x-5) = 0$ $x = 2 \text{ or } x = 5$

When $x = 2$: $y = 2$. When $x = 5$: $y = 5$.

The intersection points are $(2, 2)$ and $(5, 5)$.

Marking: 1 mark for setting up equation, 1 mark for solving quadratic, 1 mark for both points.

---

Question 13 [4 marks]

We work backwards from the transformed function $y = \dfrac{6x + 15}{x + 2}$.

Undo translation (shift 2 units in positive $x$-direction, i.e., replace $x$ with $x - 2$):

$y = \dfrac{6(x-2) + 15}{(x-2) + 2} = \dfrac{6x - 12 + 15}{x} = \dfrac{6x + 3}{x}$

Undo stretch (divide $y$ by 3, i.e., compress parallel to $y$-axis by factor 3):

$y = \frac{1}{3} \cdot \dfrac{6x + 3}{x} = \dfrac{2x + 1}{x}$

So $f(x) = \dfrac{2x + 1}{x} = 2 + \dfrac{1}{x}$.

Marking: 1 mark for undoing translation correctly, 1 mark for undoing stretch correctly, 1 mark for correct final expression, 1 mark for showing working clearly.

---

Question 14 [7 marks]

(a) [2]
$f(3) = f(1 + 1 + 1) = f(1) + f(1) + f(1) = 3f(1) = 3 \times 5 = 15$.

Alternatively: $f(2) = f(1+1) = f(1) + f(1) = 10$, then $f(3) = f(2+1) = f(2) + f(1) = 10 + 5 = 15$.

Marking: 1 mark for using the functional equation, 1 mark for correct answer.

(b) [1]
$f(0) = f(0 + 0) = f(0) + f(0) = 2f(0)$, so $f(0) = 0$.

Marking: 1 mark for correct answer with reasoning.

(c) [2]
We use induction. Base case: $n = 1$, $f(1) = 5 = 5(1)$. ✓

Inductive step: Assume $f(k) = 5k$ for some positive integer $k$. Then: $f(k+1) = f(k) + f(1) = 5k + 5 = 5(k+1)$. ✓

By induction, $f(n) = 5n$ for all positive integers $n$.

Marking: 1 mark for base case, 1 mark for inductive step.

(d) [2]
Since $f$ is continuous and satisfies Cauchy's functional equation $f(x+y) = f(x) + f(y)$, and $f(1) = 5$, the only continuous solution is $f(x) = 5x$ for all $x \in \mathbb{R}$.

Marking: 1 mark for identifying this as Cauchy's equation, 1 mark for stating $f(x) = 5x$.

---

Question 15 [6 marks]

(a) [2]
$f(x) = x^2 - 4x + 7 = (x - 2)^2 + 3$.

So $a = 2$ and $b = 3$.

Marking: 1 mark for correct $a$, 1 mark for correct $b$.

(b) [2]
Since $(x-2)^2 \geq 0$, the minimum value is $3$, occurring at $x = 2$.

Marking: 1 mark for minimum value, 1 mark for $x$-value.

(c) [2]
$g(x) = f(x + k) = (x + k - 2)^2 + 3$.

The minimum occurs when $x + k - 2 = 0$, i.e., $x = 2 - k$.

Given that the minimum is at $x = -1$: $2 - k = -1$, so $k = 3$.

Marking: 1 mark for finding the minimum point of $g$ in terms of $k$, 1 mark for correct value of $k$.

---

Question 16 [7 marks]

(a) [2]
$fg(x) = f(g(x)) = f(\log_2 x) = 2^{\log_2 x} = x$ for all $x > 0$. ✓

Marking: 1 mark for correct substitution, 1 mark for using $2^{\log_2 x} = x$.

(b) [2]
$gf(x) = g(f(x)) = g(2^x) = \log_2(2^x) = x$ for all $x \in \mathbb{R}$. ✓

Marking: 1 mark for correct substitution, 1 mark for using $\log_2(2^x) = x$.

(c) [3]
$f(2x + 1) = 2^{2x+1}$ and $g(x - 3) = \log_2(x - 3)$.

Equation: $2^{2x+1} = \log_2(x - 3)$.

This requires $x > 3$ (for the logarithm to be defined).

Using a graphing calculator to solve $2^{2x+1} = \log_2(x - 3)$:

Let $h(x) = 2^{2x+1} - \log_2(x - 3)$. For $x > 3$, $2^{2x+1}$ grows very rapidly while $\log_2(x-3)$ grows slowly. At $x = 3.01$: $2^{7.02} \approx 129.7$ and $\log_2(0.01) \approx -6.64$. So $h(x) > 0$ for $x$ just above 3. Since $2^{2x+1}$ is always positive and increasing, and $\log_2(x-3)$ is negative for $3 < x < 4$ and positive but small for larger $x$, there is no solution to this equation.

More rigorously: for $x > 3$, $2^{2x+1} > 2^7 = 128$ while $\log_2(x-3) < 128$ for $x - 3 < 2^{128}$. But we need to check if they can be equal. At $x = 3.1$: LHS $= 2^{7.2} \approx 147.0$, RHS $= \log_2(0.1) \approx -3.32$. LHS > RHS. At $x = 4$: LHS $= 2^9 = 512$, RHS $= \log_2(1) = 0$. LHS > RHS. Since LHS is always much larger than RHS for $x > 3$, there is no solution.

Answer: No solution.

Marking: 1 mark for noting $x > 3$, 1 mark for attempting to solve, 1 mark for correct conclusion of no solution.

Teaching note: This question tests whether students can reason about the relative growth rates of exponential and logarithmic functions rather than just attempting algebraic manipulation.

---

Question 17 [9 marks]

(a) [3]
Let $y = \dfrac{x-2}{x+1}$. Swap:

$x = \dfrac{y-2}{y+1}$ $x(y+1) = y - 2$ $xy + x = y - 2$ $xy - y = -2 - x$ $y(x - 1) = -(x + 2)$ $y = \frac{x+2}{1-x}$

So $f^{-1}(x) = \dfrac{x + 2}{1 - x}$.

Marking: 1 mark for swapping, 1 mark for correct algebra, 1 mark for correct final expression.

(b) [3]
$f(f(x)) = f\!\left(\dfrac{x-2}{x+1}\right) = \dfrac{\frac{x-2}{x+1} - 2}{\frac{x-2}{x+1} + 1}$

Numerator: $\dfrac{x-2 - 2(x+1)}{x+1} = \dfrac{x-2-2x-2}{x+1} = \dfrac{-x-4}{x+1}$

Denominator: $\dfrac{x-2 + (x+1)}{x+1} = \dfrac{2x-1}{x+1}$

So $f(f(x)) = \dfrac{-x-4}{2x-1}$.

Wait — let me recheck. The question states $f(f(x)) = -\dfrac{1}{x}$. Let me recompute carefully.

$f(f(x)) = \dfrac{\frac{x-2}{x+1} - 2}{\frac{x-2}{x+1} + 1}$

Numerator: $\frac{x-2}{x+1} - 2 = \frac{x-2 - 2(x+1)}{x+1} = \frac{x-2-2x-2}{x+1} = \frac{-x-4}{x+1}$

Denominator: $\frac{x-2}{x+1} + 1 = \frac{x-2+x+1}{x+1} = \frac{2x-1}{x+1}$

$f(f(x)) = \frac{-x-4}{2x-1}$.

This doesn't equal $-\frac{1}{x}$. Let me re-examine the question. The function is $f(x) = \frac{x-2}{x+1}$. Let me verify the claim $f(f(x)) = -\frac{1}{x}$:

$\frac{-x-4}{2x-1} = -\frac{1}{x}$ would mean $x(x+4) = 2x-1$, i.e., $x^2 + 4x = 2x - 1$, i.e., $x^2 + 2x + 1 = 0$, $(x+1)^2 = 0$. This only holds at $x = -1$, which is excluded. So the claim is false for this function.

Let me reconsider. Perhaps the function should be $f(x) = \frac{x-1}{x+1}$ or similar. Actually, for $f(x) = \frac{1}{x}$, $f(f(x)) = x$. For $f(x) = 1 - \frac{1}{x}$, $f(f(x)) = \frac{1}{1-x}$ and $f(f(f(f(x)))) = x$.

Let me try $f(x) = \frac{x-2}{x+1}$ and compute $f(f(f(f(x))))$ directly instead.

Actually, let me just proceed with the computation as stated. The answer key should reflect the actual mathematics of the function as given.

$f(f(x)) = \frac{-x-4}{2x-1}$

$f(f(f(x))) = f\!\left(\frac{-x-4}{2x-1}\right) = \frac{\frac{-x-4}{2x-1} - 2}{\frac{-x-4}{2x-1} + 1} = \frac{-x-4 - 2(2x-1)}{-x-4 + 2x-1} = \frac{-x-4-4x+2}{x-5} = \frac{-5x-2}{x-5}$

$f(f(f(f(x)))) = f\!\left(\frac{-5x-2}{x-5}\right) = \frac{\frac{-5x-2}{x-5} - 2}{\frac{-5x-2}{x-5} + 1} = \frac{-5x-2-2(x-5)}{-5x-2+x-5} = \frac{-5x-2-2x+10}{-4x-7} = \frac{-7x+8}{-4x-7} = \frac{7x-8}{4x+7}$

This is getting complicated. Let me reconsider the question design. The question asks to show $f(f(x)) = -\frac{1}{x}$, which is a nice result. Let me check what function would give this.

If $f(x) = \frac{ax+b}{cx+d}$ and $f(f(x)) = -\frac{1}{x}$, then $f$ has order 4 (applying it 4 times gives $x$). This is a known property of certain Möbius transformations.

For $f(x) = \frac{1}{1-x}$: $f(f(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{1}{\frac{1-x-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} = 1 - \frac{1}{x}$.

$f(f(f(x))) = f(1-\frac{1}{x}) = \frac{1}{1-(1-\frac{1}{x})} = \frac{1}{\frac{1}{x}} = x$. So this has order 3, not 4.

For $f(x) = \frac{x-1}{x+1}$: $f(f(x)) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{x-1-x-1}{x-1+x+1} = \frac{-2}{2x} = -\frac{1}{x}$. ✓

So the function should be $f(x) = \frac{x-1}{x+1}$, not $\frac{x-2}{x+1}$. Let me redo this question with the corrected function.

REVISED Q17: The function $f$ is defined by $f(x) = \dfrac{x - 1}{x + 1}$ for $x \in \mathbb{R}$, $x \neq -1$.

(a) [3]
Let $y = \dfrac{x-1}{x+1}$. Swap:

$x = \dfrac{y-1}{y+1}$

$x(y+1) = y - 1$

$xy + x = y - 1$

$xy - y = -1 - x$

$y(x-1) = -(x+1)$

$y = \frac{x+1}{1-x}$

So $f^{-1}(x) = \dfrac{x+1}{1-x}$.

Marking: 1 mark for swapping, 1 mark for correct algebra, 1 mark for correct final expression.

(b) [3]
$f(f(x)) = f\!\left(\dfrac{x-1}{x+1}\right) = \dfrac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1}$

Numerator: $\dfrac{x-1-(x+1)}{x+1} = \dfrac{-2}{x+1}$

Denominator: $\dfrac{x-1+x+1}{x+1} = \dfrac{2x}{x+1}$

$f(f(x)) = \dfrac{-2}{2x} = -\dfrac{1}{x}$. ✓

Valid for $x \neq 0$ and $x \neq -1$.

Marking: 1 mark for correct substitution, 1 mark for correct simplification of numerator and denominator, 1 mark for final result.

(c) [3]
$f(f(f(f(x)))) = f(f(f(f(x)))) = f(f(-\frac{1}{x}))$ since $f(f(x)) = -\frac{1}{x}$.

$f(f(-\frac{1}{x})) = -\frac{1}{(-\frac{1}{x})} = x$.

So $f(f(f(f(x)))) = x$.

Domain: We need $x \neq -1$ (for $f(x)$ defined), $x \neq 0$ (for $f(f(x)) = -1/x$ defined), and $f(x) \neq -1$ (for $f(f(x))$ defined). $f(x) = -1$ means $\frac{x-1}{x+1} = -1$, so $x-1 = -x-1$, $2x = 0$, $x = 0$ (already excluded). Also $f(f(x)) \neq -1$ for $f(f(f(x)))$ to be defined: $-\frac{1}{x} = -1$ means $x = 1$. And $f(f(f(x))) \neq -1$ for $f(f(f(f(x))))$ to be defined: $f(f(f(x))) = f(-1/x) = \frac{-1/x - 1}{-1/x + 1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}$. Setting this $\neq -1$: $\frac{x+1}{1-x} = -1$ means $x+1 = x-1$, impossible. So no additional restriction.

Domain: $x \neq -1, 0, 1$.

Marking: 1 mark for using the result from (b), 1 mark for correct answer $x$, 1 mark for correct domain.

---

Question 18 [8 marks]

(a) [3]
Let $y = \dfrac{3x-2}{x+4}$. Swap:

$x = \dfrac{3y-2}{y+4}$

$x(y+4) = 3y - 2$

$xy + 4x = 3y - 2$

$xy - 3y = -2 - 4x$

$y(x - 3) = -4x - 2$

$y = \frac{4x+2}{3-x}$

So $f^{-1}(x) = \dfrac{4x + 2}{3 - x}$.

Marking: 1 mark for swapping, 1 mark for correct algebra, 1 mark for correct final expression.

(b) [2]

• Domain of $f^{-1}$: $x \neq 3$ (since denominator $3 - x \neq 0$). Domain is $\mathbb{R} \setminus \{3\}$.
• Range of $f^{-1}$: This equals the domain of $f$, which is $\mathbb{R} \setminus \{-4\}$.

Marking: 1 mark for domain, 1 mark for range.

(c) [3]
Set $f(x) = f^{-1}(x)$:

$\dfrac{3x-2}{x+4} = \dfrac{4x+2}{3-x}$

Cross-multiplying:

$(3x-2)(3-x) = (4x+2)(x+4)$

$9x - 3x^2 - 6 + 2x = 4x^2 + 16x + 2x + 8$

$11x - 3x^2 - 6 = 4x^2 + 18x + 8$

$0 = 7x^2 + 7x + 14$

$x^2 + x + 2 = 0$

Discriminant: $1 - 8 = -7 < 0$.

There are no real solutions.

Marking: 1 mark for setting up equation, 1 mark for correct expansion, 1 mark for correct conclusion.

Teaching note: The graphs of $f$ and $f^{-1}$ are reflections of each other across $y = x$. If they don't intersect, it means the curve and its reflection have no common points. This can happen when the function doesn't intersect the line $y = x$.

---

Question 19 [9 marks]

(a) [2]
$f(-x) = \sqrt{(-x)^2 + 4} = \sqrt{x^2 + 4} = f(x)$.

Since $f(-x) = f(x)$ for all $x \in \mathbb{R}$, $f$ is an even function.

Marking: 1 mark for computing $f(-x)$, 1 mark for correct classification.

(b) [2]
Since $x^2 \geq 0$, we have $x^2 + 4 \geq 4$, so $\sqrt{x^2 + 4} \geq \sqrt{4} = 2$.

The minimum value is $2$ (at $x = 0$), and as $|x| \to \infty$, $f(x) \to \infty$.

Range of $f$ is $[2, \infty)$.

Marking: 1 mark for minimum value, 1 mark for correct range.

(c) [2]
$f$ is not one-one because it is an even function: $f(-x) = f(x)$ for all $x$. For example, $f(1) = \sqrt{5} = f(-1)$ but $1 \neq -1$. Therefore $f$ does not have an inverse function.

Marking: 1 mark for stating not one-one, 1 mark for justification using even function property.

(d) [3]
For $h(x) = \sqrt{x^2 + 4}$ with domain $x \geq 0$:

$h$ is strictly increasing on $[0, \infty)$ (since $h'(x) = \frac{x}{\sqrt{x^2+4}} > 0$ for $x > 0$), so $h$ is one-one.

Let $y = \sqrt{x^2 + 4}$. Then $y^2 = x^2 + 4$, so $x^2 = y^2 - 4$, and $x = \sqrt{y^2 - 4}$ (taking positive root since $x \geq 0$).

So $h^{-1}(x) = \sqrt{x^2 - 4}$.

• Domain of $h^{-1}$: Range of $h$ is $[2, \infty)$, so domain is $[2, \infty)$.
• Range of $h^{-1}$: Domain of $h$ is $[0, \infty)$, so range is $[0, \infty)$.

Marking: 1 mark for correct inverse, 1 mark for domain, 1 mark for range.

---

Question 20 [10 marks]

(a) [1]
$P(0) = \dfrac{1200}{1 + 3e^0} = \dfrac{1200}{1 + 3} = \dfrac{1200}{4} = 300$.

The initial population is 300.

Marking: 1 mark for correct answer.

(b) [2]
As $t \to \infty$, $e^{-0.4t} \to 0$, so $P(t) \to \dfrac{1200}{1 + 0} = 1200$.

The long-term population approaches 1200. This is the carrying capacity of the environment in this logistic model.

Marking: 1 mark for limit calculation, 1 mark for interpretation as carrying capacity.

(c) [3]
Set $P(t) = 600$:

$\dfrac{1200}{1 + 3e^{-0.4t}} = 600$

$1 + 3e^{-0.4t} = 2$

$3e^{-0.4t} = 1$

$e^{-0.4t} = \dfrac{1}{3}$

$-0.4t = \ln\!\left(\dfrac{1}{3}\right) = -\ln 3$

$t = \dfrac{\ln 3}{0.4} = \dfrac{5\ln 3}{2}$

Marking: 1 mark for setting up equation, 1 mark for correct algebra, 1 mark for exact answer.

(d) [1]
$Q(t) = P(t - c)$ represents a translation of $c$ units in the positive $t$-direction (a horizontal shift to the right by $c$ units). This means the population curve is delayed by $c$ years — the same growth pattern occurs but starts later.

Marking: 1 mark for correct description of transformation.

(e) [3]
Set $Q(t) = 900$, i.e., $P(t - c) = 900$:

$\dfrac{1200}{1 + 3e^{-0.4(t-c)}} = 900$

$1 + 3e^{-0.4(t-c)} = \dfrac{4}{3}$

$3e^{-0.4(t-c)} = \dfrac{1}{3}$

$e^{-0.4(t-c)} = \dfrac{1}{9}$

$-0.4(t-c) = \ln\!\left(\dfrac{1}{9}\right) = -\ln 9 = -2\ln 3$

$t - c = \dfrac{2\ln 3}{0.4} = 5\ln 3$

$t = c + 5\ln 3$

Wait — the question doesn't give a value for $c$. Let me re-read. The question says "the model is adjusted so that the population function becomes $Q(t) = P(t - c)$ for some constant $c > 0$." But no value of $c$ is given. This means the answer should be expressed in terms of $c$, or perhaps I need to provide a value. Let me adjust: I'll state that $c = 2$ for concreteness, or I'll leave the answer in terms of $c$.

Actually, looking at the question again, part (e) asks for a numerical answer to 3 s.f., so $c$ must be determined from context. Since it isn't given, let me revise the question to specify $c = 2$.

REVISED Q20(d) and (e):

(d) The model is adjusted with $c = 2$, so that $Q(t) = P(t - 2)$. Describe the effect of this transformation on the graph of $P$. [1]

(e) For the adjusted model $Q$, find the time at which the population reaches 900, giving your answer correct to 3 significant figures. [3]

(d) [1]
$Q(t) = P(t - 2)$ represents a translation of 2 units in the positive $t$-direction. The entire population curve is shifted 2 years to the right, meaning the growth pattern is identical but delayed by 2 years.

(e) [3]
Set $Q(t) = 900$: $P(t-2) = 900$.

From the working above: $t - 2 = 5\ln 3$, so $t = 2 + 5\ln 3$.

$t = 2 + 5 \times 1.0986... = 2 + 5.493... = 7.493...$

$t \approx 7.49$ years (to 3 s.f.).

Marking: 1 mark for setting up equation, 1 mark for correct algebra, 1 mark for correct numerical answer.

---

Mark Summary

| Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 | Q11 | Q12 | Q13 | Q14 | Q15 | Q16 | Q17 | Q18 | Q19 | Q20 | Total | |----|----|----|----|----|----|----|----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-------| | 5 | 6 | 5 | 7 | 5 | 6 | 6 | 5 | 8 | 7 | 5 | 8 | 4 | 7 | 6 | 7 | 9 | 8 | 9 | 10 | 133 |

Wait — the total should be 60 marks. Let me recount and adjust.

The marks per question add up to: 5+6+5+7+5+6+6+5+8+7+5+8+4+7+6+7+9+8+9+10 = 133. This is way over 60. I need to reduce the marks significantly.

Let me redesign the mark allocation to total exactly 60 marks:

Q	Marks
1	3
2	4
3	3
4	4
5	3
6	3
7	3
8	3
9	4
10	3
11	3
12	4
13	2
14	3
15	3
16	3
17	4
18	4
19	4
20	6
Total	69

Still over. Let me adjust further:

Q	Marks
1	3
2	3
3	3
4	4
5	3
6	3
7	3
8	2
9	4
10	3
11	3
12	4
13	2
14	3
15	3
16	3
17	4
18	3
19	3
20	4
Total	60 ✓

Now I need to adjust the mark allocations in the questions and answer key accordingly. Let me revise the quiz to reflect these mark allocations properly.

Actually, the simplest approach is to update the bracketed marks in the questions and adjust the answer key marking schemes. Let me redo the quiz with correct mark allocations.

Given the complexity of this revision, let me provide the corrected quiz with proper mark allocations totalling exactly 60 marks.

---

REVISED QUIZ (corrected mark allocations)

I need to redo the entire quiz with corrected marks. Let me do this carefully.

Q1 [3]: (a) Show fg exists [1], (b) Find fg(x), state domain and range [2] Q2 [3]: (a) Find f⁻¹(x) and state domain [2], (b) Find x where f(x) = f⁻¹(x) [1] — too little for part (b). Let me restructure.

Let me take a different approach. I'll keep the questions as they are but adjust the bracketed marks to total 60.

Revised mark scheme:

• Q1: (a)[1] (b)[2] = 3
• Q2: (a)[2] (b)[1] = 3 — but part (b) deserves more. Let me restructure Q2.

OK, I think the cleanest approach is to just adjust the marks in the questions. Here is the final corrected version:

---

A-Level Maths H2 Quiz - Algebra Functions (CORRECTED)

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 60

Duration: 90 minutes
Total Marks: 60

Instructions:

• Answer ALL questions.
• Show all working clearly. Unsupported answers may not receive full credit.
• An approved graphing calculator (without CAS) may be used where indicated.
• Give exact answers where possible; otherwise, correct to 3 significant figures unless stated otherwise.
• The number of marks available is shown in brackets [ ] at the end of each question or part-question.

---

Section A: Composite Functions and Inverses (Questions 1–5)

1. Two functions are defined as $f(x) = x^2 - 4$ for $x \in \mathbb{R}$, $x \geq 0$, and $g(x) = \sqrt{x + 3}$ for $x \geq -3$.

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

(b) Find an expression for $fg(x)$ and state its domain and range. [2]

2. The function $f$ is defined by $f(x) = \dfrac{2x + 1}{x - 3}$ for $x \in \mathbb{R}$, $x \neq 3$.

(a) Find $f^{-1}(x)$ and state its domain. [2]

(b) Find the value of $x$ for which $f(x) = f^{-1}(x)$, giving your answer in exact form. [2]

3. Functions $f$ and $g$ are defined by $f: x \mapsto e^{2x} - 1$ for $x \in \mathbb{R}$, and $g: x \mapsto \ln(x + 4)$ for $x > -4$.

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

(b) Find an expression for $gf(x)$ and state the range of $gf$. [2]

4. The function $f$ is defined by $f(x) = 3 - \sqrt{x + 1}$ for $x \geq -1$.

(a) Find $f^{-1}(x)$ and state its domain and range. [3]

(b) Sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$ on the same set of axes. [2]

Image pending generation: graph for Q4.

5. The function $f$ is defined by $f(x) = x^2 - 6x + 5$ for $x \in \mathbb{R}$, $x \geq 3$.

(a) Explain why $f$ has an inverse function. [1]

(b) Find $f^{-1}(x)$ and state its domain. [2]

---

Section B: Domain, Range, and Function Properties (Questions 6–10)

6. The function $f$ is defined by $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \in \mathbb{R}$, $x \neq 2$.

(a) Simplify $f(x)$ and explain what happens at $x = 2$. [1]

(b) State the range of $f$ and justify your answer. [2]

7. The function $f$ is defined by $f(x) = |2x - 3|$ for $x \in \mathbb{R}$.

(a) Find the range of $f$. [1]

(b) Determine whether $f$ is one-one. Justify your answer. [1]

(c) The function $g$ is defined by $g(x) = |2x - 3|$ for $x \in \mathbb{R}$, $x \geq \dfrac{3}{2}$. Find $g^{-1}(x)$ and state its domain. [2]

8. A function $f$ is defined by $f(x) = \dfrac{ax + b}{cx + d}$ where $a, b, c, d \in \mathbb{R}$, $c \neq 0$, and $ad \neq bc$.

Find $f^{-1}(x)$ in terms of $a, b, c, d$. [2]

9. The function $f$ is defined by $f(x) = \sqrt{9 - x^2}$ for $-3 \leq x \leq 3$.

(a) State the domain and range of $f$. [1]

(b) Explain why $f$ does not have an inverse function. [1]

(c) By restricting the domain of $f$ to $0 \leq x \leq 3$, a new function $h$ is defined. Find $h^{-1}(x)$ and state its domain and range. [3]

10. Two functions are defined as $f(x) = x^2 + 2x + 3$ for $x \in \mathbb{R}$, and $g(x) = x - 1$ for $x \in \mathbb{R}$.

(a) Find the range of $f$. [1]

(b) Show that the composite function $fg$ exists and find an expression for $fg(x)$. [2]

(c) Solve the equation $fg(x) = 3$. [1]

---

Section C: Graphical Transformations and Applications (Questions 11–15)

11. The graph of $y = f(x)$ is a downward-opening parabola passing through the points $(-2, 0)$, $(0, 4)$, and $(3, 0)$, with a maximum at $(0, 4)$.

Image pending generation: graph for Q11.

Wait, I realize the points (-2,0), (0,4), (3,0) with maximum at (0,4) are geometrically inconsistent for a parabola. A parabola with x-intercepts at -2 and 3 has its axis of symmetry at x = 0.5, not x = 0. Let me fix this.

Let me use: x-intercepts at (-2, 0) and (2, 0), with maximum at (0, 4). Then f(x) = 4 - x². This is consistent.

Actually, let me just use a consistent set. Let me say the parabola has x-intercepts at $(-2, 0)$ and $(2, 0)$ with a maximum at $(0, 4)$. Then $f(x) = 4 - x^2$.

Let me redo Q11 with consistent data.

11. The graph of $y = f(x)$ is a downward-opening parabola with x-intercepts at $(-2, 0)$ and $(2, 0)$, and a maximum at $(0, 4)$.

Image pending generation: graph for Q11.

(a) State the range of $f$. [1]

(b) On separate diagrams, sketch the graphs of:

    (i) $y = f(x + 2)$ [1]

    (ii) $y = 2f(x)$ [1]

In each case, state the coordinates of the vertex and any intercepts with the axes.

12. The function $f$ is defined by $f(x) = \dfrac{1}{x}$ for $x \in \mathbb{R}$, $x \neq 0$.

(a) The graph of $y = f(x)$ is transformed to give the graph of $y = \dfrac{2}{x - 3} + 4$. Describe the transformations involved. [2]

(b) State the equations of any asymptotes of the transformed function. [1]

(c) Find the coordinates of the point where the transformed graph intersects the line $y = x$. [2]

13. The graph of $y = f(x)$ undergoes the following transformations in succession:

• A stretch parallel to the $y$-axis by scale factor 3
• A translation of 2 units in the negative $x$-direction

The resulting equation is $y = \dfrac{6x + 15}{x + 2}$.

Find an expression for $f(x)$. [2]

14. A function $f$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, and $f(1) = 5$.

(a) Find $f(3)$. [1]

(b) Find $f(0)$. [1]

(c) Show that $f(n) = 5n$ for all positive integers $n$. [2]

15. The function $f$ is defined by $f(x) = x^2 - 4x + 7$ for $x \in \mathbb{R}$.

(a) Express $f(x)$ in the form $(x - a)^2 + b$, stating the values of $a$ and $b$. [1]

(b) Hence state the minimum value of $f(x)$ and the value of $x$ at which it occurs. [1]

(c) The function $g$ is defined by $g(x) = f(x + k)$ where $k$ is a positive constant. Given that the minimum point of $g$ is at $x = -1$, find the value of $k$. [2]

---

Section D: Advanced Function Problems (Questions 16–20)

16. The functions $f$ and $g$ are defined by $f(x) = 2^x$ for $x \in \mathbb{R}$, and $g(x) = \log_2(x)$ for $x > 0$.

(a) Show that $fg(x) = x$ for all $x > 0$. [1]

(b) Show that $gf(x) = x$ for all $x \in \mathbb{R}$. [1]

(c) Explain why the equation $f(2x + 1) = g(x - 3)$ has no real solution. [2]

17. The function $f$ is defined by $f(x) = \dfrac{x - 1}{x + 1}$ for $x \in \mathbb{R}$, $x \neq -1$.

(a) Find $f^{-1}(x)$. [2]

(b) Show that $f(f(x)) = -\dfrac{1}{x}$ for $x \neq 0, x \neq -1$. [2]

(c) Hence find $f(f(f(f(x))))$ and state its domain. [2]

18. A one-one function $f$ is defined by $f(x) = \dfrac{3x - 2}{x + 4}$ for $x \in \mathbb{R}$, $x \neq -4$.

(a) Find $f^{-1}(x)$. [2]

(b) State the domain and range of $f^{-1}$. [1]

(c) Show that the equation $f(x) = f^{-1}(x)$ has no real solution. [2]

19. The function $f$ is defined by $f(x) = \sqrt{x^2 + 4}$ for $x \in \mathbb{R}$.

(a) State whether $f$ is even, odd, or neither. Justify your answer. [1]

(b) Find the range of $f$. [1]

(c) Explain why $f$ does not have an inverse function. [1]

(d) A new function $h$ is defined by $h(x) = \sqrt{x^2 + 4}$ for $x \geq 0$. Find $h^{-1}(x)$ and state its domain and range. [3]

20. In a biological model, the population $P$ of a species at time $t$ years is modelled by the function

$P(t) = \dfrac{1200}{1 + 3e^{-0.4t}}, \quad t \geq 0.$

(a) Find the initial population. [1]

(b) State the long-term population predicted by this model. Justify your answer. [1]

(c) Find the exact time at which the population reaches 600. [2]

(d) The model is adjusted with a horizontal shift of 2 units in the positive $t$-direction, giving $Q(t) = P(t - 2)$. Find the time at which $Q$ reaches a population of 900, giving your answer correct to 3 significant figures. [2]

---

End of Quiz

---

Mark Total Verification

Q	Marks
1	1+2 = 3
2	2+2 = 4
3	1+2 = 3
4	3+2 = 5
5	1+2 = 3
6	1+2 = 3
7	1+1+2 = 4
8	2
9	1+1+3 = 5
10	1+2+1 = 4
11	1+1+1 = 3
12	2+1+2 = 5
13	2
14	1+1+2 = 4
15	1+1+2 = 4
16	1+1+2 = 4
17	2+2+2 = 6
18	2+1+2 = 5
19	1+1+1+3 = 6
20	1+1+2+2 = 6
Total	78

Still over 60. I need to be more aggressive. Let me cut marks further.

Q	Target
1	3
2	3
3	3
4	4
5	3
6	3
7	3
8	2
9	4
10	3
11	3
12	3
13	2
14	3
15	3
16	3
17	4
18	3
19	4
20	4
Total	60 ✓

Now let me write the final clean version with these exact marks.