Secondary 3 Elementary Mathematics Algebra Functions Quiz

Secondary 3 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

Total Marks: 40

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Section A (Questions 1–10, 2 marks each = 20 marks)

1. Given $f(x) = 2x^2 - 5x + 3$, find $f(-2)$.

Working: $f(-2) = 2(-2)^2 - 5(-2) + 3$ $= 2(4) + 10 + 3$ $= 8 + 10 + 3$ $= 21$

Answer: 21 [2]

Marking notes: 1 mark for correct substitution, 1 mark for correct evaluation. Common error: $(-2)^2 = -4$ instead of $4$.

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2. $g(x) = \frac{4}{x-1}$, $x \neq 1$. Find $x$ when $g(x) = 2$.

Working: $\frac{4}{x-1} = 2$ $4 = 2(x-1)$ $4 = 2x - 2$ $2x = 6$ $x = 3$

Check: $x = 3 \neq 1$, valid.

Answer: 3 [2]

Marking notes: 1 mark for setting up equation correctly, 1 mark for solving. Must state $x \neq 1$ or check validity.

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3. $h(x) = 3x - 7$. Find $h^{-1}(x)$.

Working: Let $y = 3x - 7$. Swap $x$ and $y$: $x = 3y - 7$. $3y = x + 7$ $y = \frac{x+7}{3}$

So $h^{-1}(x) = \frac{x+7}{3}$.

Answer: $h^{-1}(x) = \frac{x+7}{3}$ [2]

Marking notes: 1 mark for correct method (swap and solve), 1 mark for correct expression. Alternative: $h^{-1}(x) = \frac{1}{3}x + \frac{7}{3}$.

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4. Graph of $y = kx^2$ passes through $(2, 12)$. Find $k$.

Working: Substitute $(2, 12)$ into $y = kx^2$: $12 = k(2)^2$ $12 = 4k$ $k = 3$

Answer: 3 [2]

Marking notes: 1 mark for substitution, 1 mark for solving. The graph passes through origin, confirming $y = kx^2$ form.

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5. $f(x) = x^2 - 4x + 5$. Find minimum value.

Method 1 (Complete the square): $f(x) = (x^2 - 4x + 4) + 1$ $= (x-2)^2 + 1$

Minimum value is $1$ when $x = 2$.

Method 2 (Vertex formula): For $ax^2 + bx + c$, vertex at $x = -\frac{b}{2a} = \frac{4}{2} = 2$. $f(2) = 4 - 8 + 5 = 1$.

Answer: 1 [2]

Marking notes: 1 mark for correct method (completing square or vertex), 1 mark for correct minimum value. Must give the $y$-value (minimum value), not the $x$-coordinate.

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

Working: $fg(2) = f(g(2))$ $g(2) = 2^2 - 3 = 4 - 3 = 1$ $f(1) = 2(1) + 1 = 3$

Answer: 3 [2]

Marking notes: 1 mark for finding $g(2) = 1$, 1 mark for finding $f(1) = 3$. Common error: computing $gf(2)$ instead of $fg(2)$.

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7. $f(x) = \frac{2x+3}{x-2}$, $x \neq 2$. Find $f^{-1}(5)$.

Method 1 (Find inverse first): $y = \frac{2x+3}{x-2}$ $y(x-2) = 2x+3$ $xy - 2y = 2x + 3$ $xy - 2x = 2y + 3$ $x(y-2) = 2y + 3$ $x = \frac{2y+3}{y-2}$ $f^{-1}(x) = \frac{2x+3}{x-2}$

$f^{-1}(5) = \frac{2(5)+3}{5-2} = \frac{13}{3}$

Method 2 (Direct): Solve $f(x) = 5$: $\frac{2x+3}{x-2} = 5$ $2x+3 = 5(x-2)$ $2x+3 = 5x-10$ $3x = 13$ $x = \frac{13}{3}$

Answer: $\frac{13}{3}$ [2]

Marking notes: 1 mark for correct method, 1 mark for correct answer. Method 2 is faster for single value.

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8. Quadratic $y = ax^2 + bx + c$, vertex $(-1, 4)$, passes through $(0, 2)$. Find $a, b, c$.

Working: Vertex form: $y = a(x+1)^2 + 4$. Substitute $(0, 2)$: $2 = a(1)^2 + 4 \Rightarrow a = -2$. $y = -2(x+1)^2 + 4 = -2(x^2 + 2x + 1) + 4 = -2x^2 - 4x - 2 + 4 = -2x^2 - 4x + 2$.

So $a = -2$, $b = -4$, $c = 2$.

Answer: $a = -2$, $b = -4$, $c = 2$ [2]

Marking notes: 1 mark for finding $a = -2$, 1 mark for correct $b$ and $c$. Can also use vertex formula $-\frac{b}{2a} = -1$ and $c = 2$ from y-intercept.

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9. $y = \frac{6}{x}$, $x > 0$. Gradient of tangent at $x = 2$.

Working: $y = 6x^{-1}$ $\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}$ At $x = 2$: $\frac{dy}{dx} = -\frac{6}{4} = -1.5$

Answer: $-1.5$ (or $-\frac{3}{2}$) [2]

Marking notes: 1 mark for differentiation (or gradient formula for reciprocal), 1 mark for evaluation. In Sec 3 E-Math, gradient of tangent to $y = \frac{k}{x}$ is $-\frac{k}{x^2}$ (can be quoted or derived).

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

Working: $fg(x) = f(g(x)) = f(x^2 + 1) = 3 - 2(x^2 + 1) = 3 - 2x^2 - 2 = 1 - 2x^2$. Set $1 - 2x^2 = 1$: $-2x^2 = 0$ $x^2 = 0$ $x = 0$

Answer: $x = 0$ [2]

Marking notes: 1 mark for correct composite function $fg(x) = 1 - 2x^2$, 1 mark for solving. Check: $g(0) = 1$, $f(1) = 1$. ✓

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Section B (Questions 11–15, 3 marks each = 15 marks)

11. $f(x) = 2x^2 - 8x + 7$.

(a) Express in form $a(x-h)^2 + k$.

Working: $f(x) = 2(x^2 - 4x) + 7$ $= 2[(x-2)^2 - 4] + 7$ $= 2(x-2)^2 - 8 + 7$ $= 2(x-2)^2 - 1$

Answer (a): $2(x-2)^2 - 1$ [1]

(b) Vertex coordinates.

From (a), vertex is $(2, -1)$.

Answer (b): $(2, -1)$ [1]

(c) Range of $f$.

Since $a = 2 > 0$, parabola opens upwards. Minimum value is $-1$. Range: $f(x) \ge -1$ or $[-1, \infty)$.

Answer (c): $f(x) \ge -1$ [1]

Marking notes: (a) 1 mark for correct completed square form. (b) 1 mark for reading vertex from (a). (c) 1 mark for correct range notation. Follow-through from (a) allowed.

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12. Graph of $y = f(x)$ (cubic) given.

(a) Values of $x$ for which $f(x) = 0$.

From graph: x-intercepts at $x = -2, 0, 2$.

Answer (a): $x = -2, 0, 2$ [1]

(b) Estimate gradient at $x = 1$ by drawing tangent.

At $x = 1$, the curve is decreasing. Tangent drawn at $x = 1$ passes approximately through $(1, -1.5)$ and $(2, -2.5)$. Gradient $\approx \frac{-2.5 - (-1.5)}{2 - 1} = \frac{-1}{1} = -1$. (Acceptable range: $-1.2$ to $-0.8$)

Answer (b): $-1$ (accept $-1.2$ to $-0.8$) [1]

(c) Sketch $y = f(x) + 2$.

Translation of original graph 2 units upwards. New y-intercept at $(0, 0)$, x-intercepts shift accordingly (solve $f(x) = -2$), turning points at $(-1.2, 3.5)$ and $(1.2, 0.5)$.

Answer (c): Graph shifted up by 2 units [1]

Marking notes: (a) 1 mark for all three intercepts. (b) 1 mark for reasonable tangent and gradient estimate. (c) 1 mark for correct vertical translation shown.

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13. $f(x) = \frac{5x-2}{x+3}$, $x \neq -3$.

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

Working: $y = \frac{5x-2}{x+3}$ $y(x+3) = 5x-2$ $xy + 3y = 5x - 2$ $xy - 5x = -3y - 2$ $x(y-5) = -3y - 2$ $x = \frac{-3y-2}{y-5} = \frac{3y+2}{5-y}$

$f^{-1}(x) = \frac{3x+2}{5-x}$, $x \neq 5$.

Answer (a): $f^{-1}(x) = \frac{3x+2}{5-x}$ [1]

(b) Value excluded from domain of $f^{-1}$.

Denominator $5-x \neq 0 \Rightarrow x \neq 5$. (Alternatively, range of $f$ is $y \neq 5$, so domain of $f^{-1}$ excludes 5.)

Answer (b): $x = 5$ [1]

(c) Solve $f(x) = f^{-1}(x)$.

When $f(x) = f^{-1}(x)$, the graphs intersect on line $y = x$ (for monotonic functions). Solve $f(x) = x$: $\frac{5x-2}{x+3} = x$ $5x-2 = x(x+3)$ $5x-2 = x^2 + 3x$ $x^2 - 2x + 2 = 0$ Discriminant: $(-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$. No real solutions.

Answer (c): No real solutions [1]

Marking notes: (a) 1 mark for correct inverse. (b) 1 mark for $x=5$. (c) 1 mark for correct equation and conclusion. Alternative: solve $\frac{5x-2}{x+3} = \frac{3x+2}{5-x}$ directly, leads to same quadratic.

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14. $f(x) = x^2 - 6x + 10$, $x \ge 3$.

(a) Explain why $f^{-1}$ exists.

$f(x) = (x-3)^2 + 1$. For $x \ge 3$, $x-3 \ge 0$, so $f$ is strictly increasing (derivative $2x-6 \ge 0$). A strictly increasing function is one-to-one, so inverse exists.

Answer (a): $f$ is one-to-one (strictly increasing) on $x \ge 3$ [1]

(b) Find $f^{-1}(x)$.

$y = (x-3)^2 + 1$, $x \ge 3$ $(x-3)^2 = y-1$ $x-3 = \sqrt{y-1}$ (positive root since $x \ge 3$) $x = 3 + \sqrt{y-1}$

$f^{-1}(x) = 3 + \sqrt{x-1}$, $x \ge 1$.

Answer (b): $f^{-1}(x) = 3 + \sqrt{x-1}$ [1]

(c) Domain and range of $f^{-1}$.

Domain of $f^{-1}$ = Range of $f$ = $[1, \infty)$ (since min at $x=3$ gives $f(3)=1$). Range of $f^{-1}$ = Domain of $f$ = $[3, \infty)$.

Answer (c): Domain: $x \ge 1$, Range: $f^{-1}(x) \ge 3$ [1]

Marking notes: (a) 1 mark for one-to-one reasoning. (b) 1 mark for correct expression with positive root. (c) 1 mark for both domain and range correct.

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15. Graph of $y = a^x$, $a > 1$, passes through $(2, 9)$.

(a) Find $a$.

$9 = a^2 \Rightarrow a = 3$ (since $a > 1$).

Answer (a): $3$ [1]

(b) Find $y$ when $x = -1$.

$y = 3^{-1} = \frac{1}{3}$.

Answer (b): $\frac{1}{3}$ [1]

(c) Graph reflected in $y$-axis. Equation of new graph.

Reflection in $y$-axis: replace $x$ with $-x$. New equation: $y = 3^{-x}$ or $y = \left(\frac{1}{3}\right)^x$.

Answer (c): $y = 3^{-x}$ [1]

Marking notes: (a) 1 mark for $a=3$. (b) 1 mark for $\frac{1}{3}$. (c) 1 mark for correct transformation.

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Section C (Questions 16–20, 1 mark each = 5 marks)

16. $f(x) = 4x - 5$. Find $f^{-1}(11)$.

Working: $f^{-1}(x) = \frac{x+5}{4}$ $f^{-1}(11) = \frac{11+5}{4} = 4$

Or solve $4x-5=11 \Rightarrow 4x=16 \Rightarrow x=4$.

Answer: 4 [1]

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17. $g(x) = \frac{1}{x+2}$, $x \neq -2$. Range of $g$.

$g(x)$ can take any real value except $0$ (numerator is constant 1, never zero). Range: $y \in \mathbb{R}, y \neq 0$ or $(-\infty, 0) \cup (0, \infty)$.

Answer: $y \neq 0$ [1]

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18. $y = x^2 - 4x$ translated 3 units right, 2 units up.

Original: $y = (x-2)^2 - 4$. Translate right 3: replace $x$ with $x-3$: $y = (x-5)^2 - 4$. Translate up 2: $y = (x-5)^2 - 2$. Expand: $y = x^2 - 10x + 25 - 2 = x^2 - 10x + 23$.

Answer: $y = (x-5)^2 - 2$ or $y = x^2 - 10x + 23$ [1]

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19. $h(x) = 2^x$. Find $x$ when $h(x) = 32$.

$2^x = 32 = 2^5 \Rightarrow x = 5$.

Answer: 5 [1]

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20. $f(x) = \sqrt{x-1}$, $x \ge 1$. Find $ff(5)$.

$f(5) = \sqrt{5-1} = \sqrt{4} = 2$. $ff(5) = f(2) = \sqrt{2-1} = \sqrt{1} = 1$.

Answer: 1 [1]

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End of Answer Key