Secondary 4 Additional Mathematics Algebra Functions Quiz

Secondary 4 Additional Mathematics Quiz - Algebra Functions

Answer Key

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Section A: Short Answer Questions

1. $f(x) = 2x^2 - 8x + 5$

(a) Completing the square:

$f(x) = 2(x^2 - 4x) + 5$

$= 2[(x - 2)^2 - 4] + 5$

$= 2(x - 2)^2 - 8 + 5$

$= 2(x - 2)^2 - 3$

Answer: $a = 2$, $h = 2$, $k = 3$ → $f(x) = 2(x - 2)^2 - 3$ [2 marks]

(b) Since $a = 2 > 0$, the parabola opens upwards.

Minimum value is $-3$, occurring at $x = 2$. [2 marks]

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2. $g(x) = x^2 - 6x + 7$, domain $x \geq 3$

(a) Let $y = x^2 - 6x + 7$

Complete the square: $y = (x - 3)^2 - 2$

So $(x - 3)^2 = y + 2$

$x - 3 = \sqrt{y + 2}$ (taking positive root since $x \geq 3$)

$x = 3 + \sqrt{y + 2}$

Therefore $g^{-1}(x) = 3 + \sqrt{x + 2}$

Domain of $g^{-1}$: Since the range of $g$ is $[-2, \infty)$, the domain of $g^{-1}$ is $x \geq -2$. [4 marks]

(b) Sketch: $y = g(x)$ is a parabola with vertex at $(3, -2)$, restricted to $x \geq 3$. $y = g^{-1}(x)$ is the reflection across $y = x$, starting at $(-2, 3)$ and curving upwards. [4 marks]

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

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

$y(x + 2) = 3x - 1$

$yx + 2y = 3x - 1$

$2y + 1 = 3x - yx = x(3 - y)$

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

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

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

Answer: $5$ [4 marks]

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4. $f(x) = ax^2 + bx + c$, minimum value $-7$ at $x = 1$, and $f(0) = -4$

From the vertex form: $f(x) = a(x - 1)^2 - 7$

$f(0) = a(0 - 1)^2 - 7 = a - 7 = -4$

So $a = 3$

$f(x) = 3(x - 1)^2 - 7 = 3(x^2 - 2x + 1) - 7 = 3x^2 - 6x + 3 - 7 = 3x^2 - 6x - 4$

Answer: $a = 3$, $b = -6$, $c = -4$ [4 marks]

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5. $h(x) = \sqrt{x + 3}$, $x \geq -3$

(a) Let $y = \sqrt{x + 3}$

$y^2 = x + 3$

$x = y^2 - 3$

$h^{-1}(x) = x^2 - 3$ [2 marks]

(b) Domain of $h^{-1}$: Since range of $h$ is $[0, \infty)$, domain of $h^{-1}$ is $x \geq 0$.

Range of $h^{-1}$: Since domain of $h$ is $[-3, \infty)$, range of $h^{-1}$ is $[-3, \infty)$. [2 marks]

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Section B: Structured Questions

6. $f(x) = x^2 - 4x + 3$

(a) Complete the square: $f(x) = (x - 2)^2 - 1$

Since $(x - 2)^2 \geq 0$, the minimum value is $-1$.

Range of $f(x)$ is $[-1, \infty)$. [2 marks]

(b) $f$ is a quadratic function, which is a parabola. It fails the horizontal line test — for any $y > -1$, there are two distinct $x$-values giving the same $y$-value. Therefore $f$ is not one-one and does not have an inverse. [2 marks]

(c) The least value of $k$ is $2$ (the $x$-coordinate of the vertex), so that $g$ is one-one on $[2, \infty)$.

$g(x) = (x - 2)^2 - 1$, domain $x \geq 2$

Let $y = (x - 2)^2 - 1$

$(x - 2)^2 = y + 1$

$x - 2 = \sqrt{y + 1}$ (positive root since $x \geq 2$)

$x = 2 + \sqrt{y + 1}$

$g^{-1}(x) = 2 + \sqrt{x + 1}$ [4 marks]

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7. $f(x) = 2x + 1$, $g(x) = \dfrac{x^2 - 4}{x - 2}$, $x \neq 2$

(a) $g(x) = \dfrac{(x - 2)(x + 2)}{x - 2} = x + 2$, for $x \neq 2$

$x = 2$ is excluded because the original expression has denominator zero at $x = 2$, making it undefined. [2 marks]

(b) $fg(x) = f(g(x)) = f(x + 2) = 2(x + 2) + 1 = 2x + 5$ [2 marks]

(c) $fg(x) = 13$

$2x + 5 = 13$

$2x = 8$

$x = 4$ (valid since $x \neq 2$) [4 marks]

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8. $f(x) = \dfrac{2x + 3}{x - 1}$, $x \neq 1$

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

$y(x - 1) = 2x + 3$

$yx - y = 2x + 3$

$yx - 2x = y + 3$

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

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

$f^{-1}(x) = \dfrac{x + 3}{x - 2}$, $x \neq 2$ [3 marks]

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

$= \dfrac{2\left(\dfrac{2x + 3}{x - 1}\right) + 3}{\left(\dfrac{2x + 3}{x - 1}\right) - 1}$

Numerator: $\dfrac{2(2x + 3) + 3(x - 1)}{x - 1} = \dfrac{4x + 6 + 3x - 3}{x - 1} = \dfrac{7x + 3}{x - 1}$

Denominator: $\dfrac{2x + 3 - (x - 1)}{x - 1} = \dfrac{x + 4}{x - 1}$

$f(f(x)) = \dfrac{7x + 3}{x + 4}$

Wait — let me recheck. Actually, let me verify if $f(f(x)) = x$:

$f(f(x)) = \dfrac{7x + 3}{x + 4}$

For this to equal $x$: $7x + 3 = x(x + 4) = x^2 + 4x$

$x^2 - 3x - 3 = 0$ — this is not an identity.

Let me recalculate more carefully:

$f(f(x)) = \dfrac{2 \cdot \frac{2x+3}{x-1} + 3}{\frac{2x+3}{x-1} - 1}$

Numerator: $\dfrac{2(2x+3)}{x-1} + 3 = \dfrac{4x+6+3(x-1)}{x-1} = \dfrac{4x+6+3x-3}{x-1} = \dfrac{7x+3}{x-1}$

Denominator: $\dfrac{2x+3}{x-1} - 1 = \dfrac{2x+3-(x-1)}{x-1} = \dfrac{x+4}{x-1}$

$f(f(x)) = \dfrac{7x+3}{x+4}$

This does not simplify to $x$. The question asks to "show that $f(f(x)) = x$" — this appears to be incorrect for this function. Let me adjust the function to make this work.

Actually, for $f(f(x)) = x$ to hold, we need $f = f^{-1}$, which requires $f(x) = f^{-1}(x)$.

From part (a): $f^{-1}(x) = \dfrac{x+3}{x-2}$

For $f = f^{-1}$: $\dfrac{2x+3}{x-1} = \dfrac{x+3}{x-2}$

$(2x+3)(x-2) = (x+3)(x-1)$

$2x^2 - 4x + 3x - 6 = x^2 - x + 3x - 3$

$2x^2 - x - 6 = x^2 + 2x - 3$

$x^2 - 3x - 3 = 0$ — not an identity.

The function as given does not satisfy $f(f(x)) = x$. This is an error in the question design. For a self-inverse function, we could use $f(x) = \dfrac{x+3}{x-1}$ or similar.

Let me provide the answer based on the calculation:

$f(f(x)) = \dfrac{7x + 3}{x + 4}$ [3 marks for working]

(c) $f(4) = \dfrac{2(4) + 3}{4 - 1} = \dfrac{11}{3}$

$f(f(4)) = f\left(\dfrac{11}{3}\right) = \dfrac{2 \cdot \frac{11}{3} + 3}{\frac{11}{3} - 1} = \dfrac{\frac{22}{3} + 3}{\frac{8}{3}} = \dfrac{\frac{31}{3}}{\frac{8}{3}} = \dfrac{31}{8}$

$f(f(f(4))) = f\left(\dfrac{31}{8}\right) = \dfrac{2 \cdot \frac{31}{8} + 3}{\frac{31}{8} - 1} = \dfrac{\frac{62}{8} + 3}{\frac{23}{8}} = \dfrac{\frac{86}{8}}{\frac{23}{8}} = \dfrac{86}{23}$

Answer: $\dfrac{86}{23}$ [2 marks]

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Section C: Application and Problem Solving

9. Rectangular garden, three sides fenced, 40 m of fencing.

(a) Let $x$ = length perpendicular to wall, $y$ = length parallel to wall.

Fencing: $2x + y = 40$, so $y = 40 - 2x$

Area: $A = xy = x(40 - 2x) = 40x - 2x^2$ ✓ [2 marks]

(b) $A = 40x - 2x^2 = -2x^2 + 40x$

$= -2(x^2 - 20x)$

$= -2[(x - 10)^2 - 100]$

$= -2(x - 10)^2 + 200$

Maximum area is $200$ m² when $x = 10$. [3 marks]

(c) When $x = 10$: $y = 40 - 2(10) = 20$

Dimensions: 10 m perpendicular to wall, 20 m parallel to wall. [3 marks]

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10. $f(x) = ax^2 + bx + 12$, always positive, $f(2) = 32$

(a) For $f(x)$ to be always positive for all real $x$:

• $a > 0$ (parabola opens upwards)
• Discriminant $b^2 - 4ac < 0$

$b^2 - 4a(12) < 0$

$b^2 < 48a$ ✓ [2 marks]

(b) $f(2) = a(4) + b(2) + 12 = 32$

$4a + 2b + 12 = 32$

$4a + 2b = 20$

$2a + b = 10$

$b = 10 - 2a$

Substituting into $b^2 < 48a$:

$(10 - 2a)^2 < 48a$

$100 - 40a + 4a^2 < 48a$

$4a^2 - 88a + 100 < 0$

$a^2 - 22a + 25 < 0$

Using the quadratic formula: $a = \dfrac{22 \pm \sqrt{484 - 100}}{2} = \dfrac{22 \pm \sqrt{384}}{2} = \dfrac{22 \pm 8\sqrt{6}}{2} = 11 \pm 4\sqrt{6}$

So $11 - 4\sqrt{6} < a < 11 + 4\sqrt{6}$

Approximately: $11 - 9.8 < a < 11 + 9.8$, so $1.2 < a < 20.8$

Since we need specific values, and the problem states "find the values" (implying unique values), there may be additional constraints. If we assume integer values and the simplest case:

If $a = 4$: $b = 10 - 8 = 2$, check $b^2 = 4 < 48(4) = 192$ ✓

If $a = 3$: $b = 10 - 6 = 4$, check $b^2 = 16 < 48(3) = 144$ ✓

The problem as stated has infinitely many solutions. For a unique answer, we need an additional constraint. Assuming the simplest integer solution:

Answer: $a = 4$, $b = 2$ (or other valid pairs) [3 marks]

(c) With $a = 4$, $b = 2$: $f(x) = 4x^2 + 2x + 12$

$f(x) = 4(x^2 + \frac{1}{2}x) + 12 = 4[(x + \frac{1}{4})^2 - \frac{1}{16}] + 12 = 4(x + \frac{1}{4})^2 - \frac{1}{4} + 12 = 4(x + \frac{1}{4})^2 + \frac{47}{4}$

Minimum value is $\dfrac{47}{4} = 11.75$ [3 marks]

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Section D: Further Practice

11. $f(x) = 3x - 5$, find $f^{-1}(7)$

Let $y = 3x - 5$, then $x = \dfrac{y + 5}{3}$

$f^{-1}(x) = \dfrac{x + 5}{3}$

$f^{-1}(7) = \dfrac{7 + 5}{3} = 4$

Answer: $4$

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

Complete the square: $f(x) = (x + 1)^2 - 9$

Minimum value is $-9$.

Answer: Range is $[-9, \infty)$

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13. $g(x) = \dfrac{1}{x - 3}$, $x \neq 3$

Let $y = \dfrac{1}{x - 3}$

$x - 3 = \dfrac{1}{y}$

$x = \dfrac{1}{y} + 3$

$g^{-1}(x) = \dfrac{1}{x} + 3$, $x \neq 0$

Answer: $g^{-1}(x) = \dfrac{1}{x} + 3$

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14. $f(x) = x^2 - 6x + 10$

$= (x - 3)^2 - 9 + 10$

$= (x - 3)^2 + 1$

Minimum value is $1$ at $x = 3$.

Answer: $1$

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15. $f(x) = \sqrt{2x + 5}$, find $f^{-1}(3)$

Let $y = \sqrt{2x + 5}$

$y^2 = 2x + 5$

$x = \dfrac{y^2 - 5}{2}$

$f^{-1}(x) = \dfrac{x^2 - 5}{2}$

$f^{-1}(3) = \dfrac{9 - 5}{2} = 2$

Answer: $2$

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16. $f(x) = 4x - 1$, $g(x) = x^2 + 2$, find $fg(3)$

$g(3) = 9 + 2 = 11$

$f(11) = 4(11) - 1 = 43$

Answer: $43$

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17. $f(x) = x^2 + px + q$, minimum $-9$ at $x = 2$

$f(x) = (x - 2)^2 - 9 = x^2 - 4x + 4 - 9 = x^2 - 4x - 5$

Answer: $p = -4$, $q = -5$

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18. $f(x) = \dfrac{x + 4}{x - 2}$, find $f^{-1}(5)$

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

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

$yx - 2y = x + 4$

$yx - x = 2y + 4$

$x(y - 1) = 2y + 4$

$x = \dfrac{2y + 4}{y - 1}$

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

$f^{-1}(5) = \dfrac{10 + 4}{5 - 1} = \dfrac{14}{4} = \dfrac{7}{2}$

Answer: $\dfrac{7}{2}$

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19. For $ax^2 + bx + c$ to be always positive for all real $x$:

• $a > 0$
• Discriminant $b^2 - 4ac < 0$

Answer: $a > 0$ and $b^2 - 4ac < 0$

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20. $f(x) = 2x^2 - 12x + 19$

$= 2(x^2 - 6x) + 19$

$= 2[(x - 3)^2 - 9] + 19$

$= 2(x - 3)^2 - 18 + 19$

$= 2(x - 3)^2 + 1$

Minimum point: $(3, 1)$

Answer: $f(x) = 2(x - 3)^2 + 1$, minimum at $(3, 1)$

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