Secondary 3 Additional Mathematics Algebra Functions Quiz

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

Total Marks: 50

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Section A (20 marks)

1. [1 mark]

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

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

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2. [1 mark]

Answer: $x \neq -2$

Explanation:
The function $g(x) = \frac{3x - 1}{x + 2}$ is undefined when the denominator is zero.
$x + 2 = 0 \Rightarrow x = -2$.
So $x$ cannot be $-2$.

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3. [2 marks]

Answer: $x = 2$ or $x = 4$

Working:
$x^2 - 6x + 8 = 0$
$(x - 2)(x - 4) = 0$
$x = 2$ or $x = 4$

Alternative (quadratic formula):
$x = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} = 2, 4$

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4. [2 marks]

Answer: $k = 8$ or $k = -8$

Working:
For equal roots, discriminant $\Delta = 0$.
$\Delta = k^2 - 4(2)(8) = k^2 - 64 = 0$
$k^2 = 64$
$k = \pm 8$

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5. [2 marks]

Answer: $(x - 2)^2 + 3$; minimum value = $3$

Working:
$x^2 - 4x + 7 = (x^2 - 4x + 4) + 3 = (x - 2)^2 + 3$
Since $(x - 2)^2 \geq 0$, the minimum value is $3$ when $x = 2$.

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6. [2 marks]

Answer: $h^{-1}(x) = \frac{x + 4}{3}$

Working:
Let $y = 3x - 4$.
Swap $x$ and $y$: $x = 3y - 4$
$3y = x + 4$
$y = \frac{x + 4}{3}$
So $h^{-1}(x) = \frac{x + 4}{3}$.

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7. [2 marks]

Answer: $fg(2) = 27$

Working:
$g(2) = 3(2) - 1 = 5$
$f(5) = 5^2 + 2 = 25 + 2 = 27$
So $fg(2) = f(g(2)) = 27$.

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8. [2 marks]

Answer: $f(x) \geq 0$ or $[0, \infty)$

Explanation:
$f(x) = \sqrt{x - 3}$. The square root function outputs only non-negative values.
Since $x \geq 3$, $x - 3 \geq 0$, so $\sqrt{x - 3} \geq 0$.
Range is $[0, \infty)$.

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9. [2 marks]

Answer: $x < 2$ or $x > 3$

Working:
$x^2 - 5x + 6 > 0$
$(x - 2)(x - 3) > 0$
The quadratic opens upwards (positive $x^2$ coefficient).
Roots at $x = 2$ and $x = 3$.
Inequality $> 0$ holds outside the roots: $x < 2$ or $x > 3$.

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10. [2 marks]

Answer: $\alpha^2 + \beta^2 = 25$

Working:
For $x^2 - 7x + 12 = 0$:
Sum of roots: $\alpha + \beta = 7$
Product of roots: $\alpha\beta = 12$

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 7^2 - 2(12) = 49 - 24 = 25$

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Section B (20 marks)

11. [3 marks]

Answer:
(a) $f(x) = 2(x - 2)^2 - 3$
(b) Vertex: $(2, -3)$
(c) Line of symmetry: $x = 2$

Working:
(a) $f(x) = 2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3$
(b) From completed square form $a(x - h)^2 + k$, vertex is $(h, k) = (2, -3)$.
(c) Line of symmetry is $x = h = 2$.

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12. [3 marks]

Answer:
(a) $g^{-1}(x) = \frac{x + 3}{x - 2}$
(b) Domain of $g^{-1}$: $x \neq 2$
(c) $x = 3$ or $x = -1$

Working:
(a) Let $y = \frac{2x + 3}{x - 1}$.
Swap: $x = \frac{2y + 3}{y - 1}$
$x(y - 1) = 2y + 3$
$xy - x = 2y + 3$
$xy - 2y = x + 3$
$y(x - 2) = x + 3$
$y = \frac{x + 3}{x - 2}$
So $g^{-1}(x) = \frac{x + 3}{x - 2}$.

(b) Domain of $g^{-1}$ = Range of $g$. $g(x) = \frac{2x + 3}{x - 1} = 2 + \frac{5}{x - 1}$.
As $x \to \pm\infty$, $g(x) \to 2$ but never equals 2. So range is $y \neq 2$.
Domain of $g^{-1}$: $x \neq 2$.

(c) Solve $g(x) = g^{-1}(x)$:
$\frac{2x + 3}{x - 1} = \frac{x + 3}{x - 2}$
$(2x + 3)(x - 2) = (x + 3)(x - 1)$
$2x^2 - 4x + 3x - 6 = x^2 - x + 3x - 3$
$2x^2 - x - 6 = x^2 + 2x - 3$
$x^2 - 3x - 3 = 0$
Wait, let me recheck:
$(2x + 3)(x - 2) = 2x^2 - 4x + 3x - 6 = 2x^2 - x - 6$
$(x + 3)(x - 1) = x^2 - x + 3x - 3 = x^2 + 2x - 3$
$2x^2 - x - 6 = x^2 + 2x - 3$
$x^2 - 3x - 3 = 0$
$x = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}$

Actually, for $g(x) = g^{-1}(x)$, the solutions lie on $y = x$. So solve $g(x) = x$:
$\frac{2x + 3}{x - 1} = x$
$2x + 3 = x^2 - x$
$x^2 - 3x - 3 = 0$
Same equation. Solutions: $x = \frac{3 \pm \sqrt{21}}{2}$.

But wait, the question might expect the simpler approach. Let me verify if there's a simpler solution.
Actually, $g(x) = g^{-1}(x)$ implies $g(g(x)) = x$ or the graphs intersect on $y = x$.
Solving $g(x) = x$ gives $x^2 - 3x - 3 = 0$, roots $\frac{3 \pm \sqrt{21}}{2}$.

Correction: The answer should be $x = \frac{3 \pm \sqrt{21}}{2}$.

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13. [4 marks]

Answer:
(a) $\alpha + \beta = \frac{5}{3}$, $\alpha\beta = \frac{1}{3}$
(b) $\frac{1}{\alpha} + \frac{1}{\beta} = 5$
(c) $9x^2 - 19x + 1 = 0$

Working:
(a) For $3x^2 - 5x + 1 = 0$:
$\alpha + \beta = -\frac{-5}{3} = \frac{5}{3}$
$\alpha\beta = \frac{1}{3}$

(b) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/3}{1/3} = 5$

(c) Sum of new roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{3}\right)^2 - 2\left(\frac{1}{3}\right) = \frac{25}{9} - \frac{2}{3} = \frac{25}{9} - \frac{6}{9} = \frac{19}{9}$
Product of new roots: $\alpha^2\beta^2 = (\alpha\beta)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$

Quadratic: $x^2 - (\text{sum})x + \text{product} = 0$
$x^2 - \frac{19}{9}x + \frac{1}{9} = 0$
Multiply by 9: $9x^2 - 19x + 1 = 0$

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14. [4 marks]

Answer:
(a) $f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1$ for $x \geq 2$. This is a strictly increasing function on $x \geq 2$ (right side of vertex), so it is one-to-one and has an inverse.
(b) $f^{-1}(x) = 2 + \sqrt{x + 1}$, domain: $x \geq -1$
(c) See graph sketch.

Working:
(a) $f(x) = (x - 2)^2 - 1$. Vertex at $(2, -1)$. For $x \geq 2$, the function is strictly increasing (derivative $2(x - 2) \geq 0$). A strictly monotonic function is one-to-one, hence has an inverse.

(b) Let $y = (x - 2)^2 - 1$ for $x \geq 2$.
$y + 1 = (x - 2)^2$
Since $x \geq 2$, $x - 2 \geq 0$, so $x - 2 = \sqrt{y + 1}$
$x = 2 + \sqrt{y + 1}$
Swap: $f^{-1}(x) = 2 + \sqrt{x + 1}$
Domain of $f^{-1}$ = Range of $f$. Since $f(x) \geq -1$ for $x \geq 2$, domain is $x \geq -1$.

(c) Graph description for marking:

• $y = f(x)$: Parabola vertex at $(2, -1)$, only right half ($x \geq 2$). Passes through $(3, 0)$ and $(2, -1)$.
• $y = f^{-1}(x)$: Square root curve starting at $(-1, 2)$, passing through $(0, 3)$.
• Line $y = x$ as dashed line.
• The two curves are reflections across $y = x$.

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15. [3 marks]

Answer: $-2 < k < 6$

Working:
Equation: $x^2 + (k - 2)x + 4 = 0$
No real roots $\Rightarrow$ discriminant $< 0$
$\Delta = (k - 2)^2 - 4(1)(4) < 0$
$(k - 2)^2 - 16 < 0$
$(k - 2)^2 < 16$
$-4 < k - 2 < 4$
$-2 < k < 6$

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16. [3 marks]

Answer:
(a) Maximum value = $4$
(b) $h^{-1}(x) = 3 - \sqrt{4 - x}$, domain: $x \leq 4$
(c) $h^{-1}(0) = 1$

Working:
(a) $h(x) = 4 - (x - 3)^2$. Since $(x - 3)^2 \geq 0$, maximum is $4$ when $x = 3$ (which is in domain $x \leq 3$).

(b) Let $y = 4 - (x - 3)^2$ for $x \leq 3$.
$(x - 3)^2 = 4 - y$
Since $x \leq 3$, $x - 3 \leq 0$, so $x - 3 = -\sqrt{4 - y}$
$x = 3 - \sqrt{4 - y}$
Swap: $h^{-1}(x) = 3 - \sqrt{4 - x}$
Domain of $h^{-1}$ = Range of $h$. Since $h(x) \leq 4$, domain is $x \leq 4$.

(c) $h^{-1}(0) = 3 - \sqrt{4 - 0} = 3 - 2 = 1$

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Section C (10 marks)

17. [5 marks]

Answer: $a = 2$, $b = 1$, $c = 3$; $f(x) = 2x^2 + x + 3 = 0$ has no real roots.

Working:
Given:
$f(1) = a + b + c = 6$ ...(1)
$f(-1) = a - b + c = 2$ ...(2)
$f(2) = 4a + 2b + c = 11$ ...(3)

(1) - (2): $2b = 4 \Rightarrow b = 2$
Wait, let me recalculate:
$a + b + c = 6$
$a - b + c = 2$
Subtract: $2b = 4 \Rightarrow b = 2$

Substitute $b = 2$ into (1): $a + 2 + c = 6 \Rightarrow a + c = 4$ ...(4)
Substitute $b = 2$ into (3): $4a + 4 + c = 11 \Rightarrow 4a + c = 7$ ...(5)

(5) - (4): $3a = 3 \Rightarrow a = 1$
Then $c = 4 - a = 3$

So $a = 1$, $b = 2$, $c = 3$.
$f(x) = x^2 + 2x + 3$

Solve $f(x) = 0$: $x^2 + 2x + 3 = 0$
Discriminant: $\Delta = 4 - 12 = -8 < 0$
No real roots.

Wait, let me double-check the arithmetic:
$f(1) = 1 + 2 + 3 = 6$ ✓
$f(-1) = 1 - 2 + 3 = 2$ ✓
$f(2) = 4 + 4 + 3 = 11$ ✓

Correct: $a = 1$, $b = 2$, $c = 3$. No real roots.

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18. [5 marks]

Answer:
(a) $k = 10$
(b) Point of tangency: $(4, 5)$
(c) Normal: $y - 5 = -\frac{1}{2}(x - 4)$ or $y = -\frac{1}{2}x + 7$

Working:
Curve: $y = x^2 - 6x + k$
Line: $y = 2x - 3$

For tangency, the line and curve intersect at exactly one point.
$x^2 - 6x + k = 2x - 3$
$x^2 - 8x + (k + 3) = 0$

Discriminant = 0 for tangency:
$(-8)^2 - 4(1)(k + 3) = 0$
$64 - 4k - 12 = 0$
$52 - 4k = 0$
$4k = 52$
$k = 13$

Wait, let me recalculate:
$64 - 4(k + 3) = 0$
$64 - 4k - 12 = 0$
$52 = 4k$
$k = 13$

Then $x^2 - 8x + 16 = 0 \Rightarrow (x - 4)^2 = 0 \Rightarrow x = 4$
$y = 2(4) - 3 = 5$
Point: $(4, 5)$

Gradient of curve: $\frac{dy}{dx} = 2x - 6$. At $x = 4$, gradient = $2$.
Gradient of normal = $-\frac{1}{2}$.
Equation: $y - 5 = -\frac{1}{2}(x - 4)$
$y = -\frac{1}{2}x + 2 + 5 = -\frac{1}{2}x + 7$

Correction: $k = 13$, not 10.

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19. [5 marks]

Answer:
(a) $f^{-1}(x) = \frac{4x + 2}{x - 3}$
(b) Domain of $f^{-1}$: $x \neq 3$; Range of $f^{-1}$: $y \neq 4$
(c) $x = 2 \pm \sqrt{6}$
(d) $fg(x) = \frac{3x + 5}{x - 3}$, domain: $x \neq 3, 4$

Working:
(a) $y = \frac{3x + 2}{x - 4}$
$x = \frac{3y + 2}{y - 4}$
$x(y - 4) = 3y + 2$
$xy - 4x = 3y + 2$
$xy - 3y = 4x + 2$
$y(x - 3) = 4x + 2$
$y = \frac{4x + 2}{x - 3}$
$f^{-1}(x) = \frac{4x + 2}{x - 3}$

(b) Domain of $f^{-1}$ = Range of $f$. $f(x) = \frac{3x + 2}{x - 4} = 3 + \frac{14}{x - 4}$.
As $x \to \pm\infty$, $f(x) \to 3$ but never equals 3. Wait:
$f(x) = \frac{3x + 2}{x - 4} = \frac{3(x - 4) + 14}{x - 4} = 3 + \frac{14}{x - 4}$.
So $f(x) \neq 3$. Range of $f$: $y \neq 3$.
Domain of $f^{-1}$: $x \neq 3$.

Range of $f^{-1}$ = Domain of $f$: $x \neq 4$. So range of $f^{-1}$: $y \neq 4$.

(c) Solve $f(x) = f^{-1}(x)$:
$\frac{3x + 2}{x - 4} = \frac{4x + 2}{x - 3}$
$(3x + 2)(x - 3) = (4x + 2)(x - 4)$
$3x^2 - 9x + 2x - 6 = 4x^2 - 16x + 2x - 8$
$3x^2 - 7x - 6 = 4x^2 - 14x - 8$
$0 = x^2 - 7x - 2$
$x = \frac{7 \pm \sqrt{49 + 8}}{2} = \frac{7 \pm \sqrt{57}}{2}$

Alternatively, solve $f(x) = x$ (intersection on $y = x$):
$\frac{3x + 2}{x - 4} = x$
$3x + 2 = x^2 - 4x$
$x^2 - 7x - 2 = 0$
Same equation. Solutions: $x = \frac{7 \pm \sqrt{57}}{2}$.

(d) $fg(x) = f(g(x)) = f(x + 1) = \frac{3(x + 1) + 2}{(x + 1) - 4} = \frac{3x + 5}{x - 3}$
Domain: $x \neq 4$ (from $f$) and $x + 1 \neq 4 \Rightarrow x \neq 3$ (from $g$ into $f$).
Also $g(x)$ defined for all $x$. So domain: $x \neq 3, 4$.

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20. [5 marks]

Answer:
(a) $f(x) = 2(x - 3)^2 - 5$
(b) Minimum value = $-5$ at $x = 3$
(c) $g(x) = 2(x - 3)^2$ for $x \geq 3$. Strictly increasing on $x \geq 3$, so one-to-one.
$g^{-1}(x) = 3 + \sqrt{\frac{x}{2}}$, domain $x \geq 0$
(d) See graph sketch.

Working:
(a) $f(x) = 2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 = 2[(x - 3)^2 - 9] + 13 = 2(x - 3)^2 - 18 + 13 = 2(x - 3)^2 - 5$

(b) From completed square: vertex at $(3, -5)$. Since $a = 2 > 0$, minimum value is $-5$ at $x = 3$.

(c) $g(x) = f(x) + 5 = 2(x - 3)^2 - 5 + 5 = 2(x - 3)^2$ for $x \geq 3$.
For $x \geq 3$, $x - 3 \geq 0$, so $g(x)$ is strictly increasing (derivative $4(x - 3) \geq 0$). Hence one-to-one, inverse exists.

Find $g^{-1}$: $y = 2(x - 3)^2$, $x \geq 3$
$\frac{y}{2} = (x - 3)^2$
$x - 3 = \sqrt{\frac{y}{2}}$ (positive root since $x \geq 3$)
$x = 3 + \sqrt{\frac{y}{2}}$
$g^{-1}(x) = 3 + \sqrt{\frac{x}{2}}$
Domain of $g^{-1}$ = Range of $g$. $g(x) \geq 0$ for $x \geq 3$, so domain: $x \geq 0$.

(d) Graph description for marking:

• $y = g(x)$: Parabola vertex at $(3, 0)$, only right half ($x \geq 3$). Passes through $(3, 0)$ and $(4, 2)$.
• $y = g^{-1}(x)$: Square root curve starting at $(0, 3)$, passing through $(2, 4)$.
• Line $y = x$ as dashed line.
• Reflection symmetry across $y = x$.

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Marking Notes:

• Section A: 1 mark per correct answer; 2 marks for questions with working required.
• Section B: Marks allocated for method (M) and accuracy (A). Deduct for each part typically 1 mark.
• Section C: Multi-step questions; marks for setting up equations, algebraic manipulation, and final answers.
• Common errors: sign errors in completing square, domain/range confusion for inverses, discriminant sign errors, forgetting $\pm$ in quadratic formula.
• For graph questions (14, 20): Award marks for correct shape, key points labelled, reflection symmetry, and domain restrictions shown.