Secondary 4 Elementary Mathematics Algebra Functions Quiz

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Secondary 4 Elementary Mathematics Quiz - Algebra Functions

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all necessary working clearly. No marks will be given for correct answers without working.
4. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless otherwise specified.
5. The use of an approved scientific calculator is expected.

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Section A: Short Questions (10 Marks)

Answer questions 1 to 5. Each question carries 2 marks.

1. Given that $f(x) = 3x - 5$ and $g(x) = x^2 + 2$, find the value of $fg(2)$.

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2. The function $h$ is defined by $h(x) = \frac{4}{x-3}$, for $x \neq 3$. Find the inverse function $h^{-1}(x)$.

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3. Solve the equation $2^{3x-1} = 16$.

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4. The graph of $y = x^2 - 6x + 11$ can be written in the form $y = (x-a)^2 + b$. Find the value of $a$.

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5. Given that $y$ varies inversely as the square root of $x$, and $y=10$ when $x=4$, find the constant of variation $k$.

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Section B: Structured Questions (15 Marks)

Answer questions 6 to 10. Marks are indicated at the end of each question or part question.

6. The functions $p$ and $q$ are defined as: $p(x) = 2x + 1$ $q(x) = x^2 - 3$

(a) Find an expression for $qp(x)$ in its simplest form. [2]

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(b) Solve the equation $qp(x) = 13$. [2]

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7. Consider the quadratic function $f(x) = -2x^2 + 8x - 5$.

(a) Express $f(x)$ in the form $a(x-h)^2 + k$ by completing the square. [2]

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(b) State the coordinates of the maximum point of the graph of $y = f(x)$. [1]

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8. The diagram below shows the graph of $y = f(x)$ for $-2 \le x \le 4$. (Note: Imagine a standard parabola opening upwards with vertex at $(1, -4)$ and passing through $(3,0)$ and $(-1,0)$).

(a) Write down the roots of the equation $f(x) = 0$. [1]

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(b) On the same axes, sketch the graph of $y = |f(x)|$. Clearly indicate the coordinates of the turning point. [2]

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9. A population of bacteria grows exponentially according to the formula $N = N_0 e^{kt}$, where $N$ is the number of bacteria at time $t$ hours, and $N_0$ is the initial population. Initially, there are 500 bacteria. After 3 hours, there are 1200 bacteria.

(a) Find the value of $k$, correct to 3 decimal places. [2]

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(b) Calculate the time taken for the population to reach 5000 bacteria. [2]

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10. Given that $f(x) = \frac{2x+1}{x-4}$, for $x \neq 4$.

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

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

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Section C: Application Questions (15 Marks)

Answer questions 11 to 15. These questions require detailed reasoning and working.

11. The cost $C$ (in dollars) of producing $x$ items is given by the function $C(x) = 0.5x^2 - 20x + 400$. The revenue $R$ (in dollars) from selling $x$ items is given by $R(x) = 30x$.

(a) Find the profit function $P(x)$, where Profit = Revenue - Cost. [2]

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(b) Determine the number of items $x$ that must be sold to maximize the profit. [2]

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12. Let $f(x) = x^2 - 4x + 3$ and $g(x) = 2x - 1$.

(a) Find the values of $x$ for which $f(x) = g(x)$. [3]

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(b) Hence, find the coordinates of the points of intersection of the graphs $y = f(x)$ and $y = g(x)$. [2]

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13. The function $f$ is defined by $f(x) = \sqrt{x+2}$ for $x \ge -2$. The function $g$ is defined by $g(x) = x^2 - 1$ for $x \ge 0$.

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

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(b) Find the composite function $fg(x)$ and state its domain. [3]

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14. The height $h$ meters of a ball thrown vertically upwards is given by $h(t) = 20t - 5t^2$, where $t$ is the time in seconds.

(a) Calculate the maximum height reached by the ball. [2]

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(b) Find the total time the ball is in the air before hitting the ground. [2]

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15. Consider the function $y = \frac{1}{x-2} + 3$.

(a) State the equation of the vertical asymptote. [1]

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(b) State the equation of the horizontal asymptote. [1]

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(c) Find the coordinates of the x-intercept. [2]

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Section D: Advanced Concepts (10 Marks)

Answer questions 16 to 20. Marks are indicated at the end of each question.

16. The function $f$ is defined by $f(x) = \frac{3x}{x+1}$ for $x \neq -1$. Find the value of $x$ such that $f(x) = f^{-1}(x)$. [2]

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17. Given that $y = kx^n$, and that $y=4$ when $x=2$, and $y=32$ when $x=4$. Find the values of $k$ and $n$. [2]

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18. The function $g(x) = 2^{x-1} - 3$. (a) State the equation of the horizontal asymptote. [1]

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(b) Find the x-intercept of the graph. [1]

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19. Solve the inequality $x^2 - 5x + 6 < 0$. [2]

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20. The functions $f$ and $g$ are defined by $f(x) = x+2$ and $g(x) = x^2$. Find the value of $x$ for which $fg(x) = gf(x)$. [2]

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End of Quiz

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Secondary 4 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

1. $g(2) = 2^2 + 2 = 6$ $f(6) = 3(6) - 5 = 18 - 5 = 13$ Answer: 13 [2]

2. Let $y = \frac{4}{x-3}$ Swap $x$ and $y$: $x = \frac{4}{y-3}$ $x(y-3) = 4$ $y-3 = \frac{4}{x}$ $y = \frac{4}{x} + 3$ Answer: $h^{-1}(x) = \frac{4}{x} + 3$ [2]

3. $16 = 2^4$ $2^{3x-1} = 2^4$ $3x - 1 = 4$ $3x = 5$ $x = \frac{5}{3}$ Answer: $x = \frac{5}{3}$ [2]

4. $y = x^2 - 6x + 11$ Complete the square: $(x-3)^2 - 3^2 + 11$ $= (x-3)^2 - 9 + 11$ $= (x-3)^2 + 2$ Comparing to $(x-a)^2 + b$: Answer: $a = 3$ [2]

5. $y = \frac{k}{\sqrt{x}}$ $10 = \frac{k}{\sqrt{4}} \Rightarrow 10 = \frac{k}{2} \Rightarrow k = 20$ Answer: $k = 20$ [2]

6. (a) $qp(x) = q(p(x)) = q(2x+1)$ $= (2x+1)^2 - 3$ $= 4x^2 + 4x + 1 - 3$ $= 4x^2 + 4x - 2$ Answer: $4x^2 + 4x - 2$ [2]

(b) $4x^2 + 4x - 2 = 13$ $4x^2 + 4x - 15 = 0$ $(2x+5)(2x-3) = 0$ $x = -\frac{5}{2}$ or $x = \frac{3}{2}$ Answer: $x = -2.5, 1.5$ [2]

7. (a) $f(x) = -2(x^2 - 4x) - 5$ $= -2[(x-2)^2 - 4] - 5$ $= -2(x-2)^2 + 8 - 5$ $= -2(x-2)^2 + 3$ Answer: $-2(x-2)^2 + 3$ [2]

(b) Vertex is at $(2, 3)$. Since coefficient of $x^2$ is negative, it is a maximum. Answer: $(2, 3)$ [1]

8. (a) Roots are x-intercepts. From description: $x = -1$ and $x = 3$. Answer: $x = -1, 3$ [1]

(b) Sketch:

• The part of the graph below the x-axis (between -1 and 3) is reflected above the x-axis.
• Turning point was $(1, -4)$, becomes $(1, 4)$. Answer: Correct sketch with turning point $(1,4)$. [2]

9. (a) $N = 500 e^{kt}$ $1200 = 500 e^{3k}$ $2.4 = e^{3k}$ $\ln(2.4) = 3k$ $k = \frac{\ln(2.4)}{3} \approx 0.29178$ Answer: $k = 0.292$ [2]

(b) $5000 = 500 e^{0.29178t}$ $10 = e^{0.29178t}$ $\ln(10) = 0.29178t$ $t = \frac{\ln(10)}{0.29178} \approx 7.896$ Answer: 7.90 hours [2]

10. (a) $y = \frac{2x+1}{x-4}$ $x = \frac{2y+1}{y-4}$ $x(y-4) = 2y+1$ $xy - 4x = 2y + 1$ $xy - 2y = 4x + 1$ $y(x-2) = 4x + 1$ $y = \frac{4x+1}{x-2}$ Answer: $f^{-1}(x) = \frac{4x+1}{x-2}$ [2]

(b) Denominator cannot be zero. Answer: $x \neq 2$ [1]

11. (a) $P(x) = R(x) - C(x)$ $P(x) = 30x - (0.5x^2 - 20x + 400)$ $P(x) = -0.5x^2 + 50x - 400$ Answer: $P(x) = -0.5x^2 + 50x - 400$ [2]

(b) Max occurs at vertex $x = \frac{-b}{2a}$ $x = \frac{-50}{2(-0.5)} = \frac{-50}{-1} = 50$ Answer: 50 items [2]

12. (a) $x^2 - 4x + 3 = 2x - 1$ $x^2 - 6x + 4 = 0$ $x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$ Answer: $x = 3 + \sqrt{5}, 3 - \sqrt{5}$ [3]

(b) If $x = 3 + \sqrt{5}$, $y = 2(3+\sqrt{5}) - 1 = 5 + 2\sqrt{5}$ If $x = 3 - \sqrt{5}$, $y = 2(3-\sqrt{5}) - 1 = 5 - 2\sqrt{5}$ Answer: $(3+\sqrt{5}, 5+2\sqrt{5})$ and $(3-\sqrt{5}, 5-2\sqrt{5})$ [2]

13. (a) $f(x) = \sqrt{x+2}$. Since $\sqrt{} \ge 0$, range is $f(x) \ge 0$. Answer: $f(x) \ge 0$ [1]

(b) $fg(x) = f(g(x)) = f(x^2-1) = \sqrt{(x^2-1)+2} = \sqrt{x^2+1}$ Domain: Inner function $g(x)$ requires $x \ge 0$. Outer function $f(u)$ requires $u \ge -2$. Here $u = x^2-1$. $x^2-1 \ge -2 \Rightarrow x^2 \ge -1$, which is always true for real x. So domain is determined by $g(x)$'s domain. Answer: $fg(x) = \sqrt{x^2+1}$, Domain: $x \ge 0$ [3]

14. (a) $h(t) = 20t - 5t^2$. Vertex at $t = \frac{-20}{2(-5)} = 2$. Max height $h(2) = 20(2) - 5(2)^2 = 40 - 20 = 20$. Answer: 20 m [2]

(b) Hits ground when $h(t) = 0$. $20t - 5t^2 = 0$ $5t(4 - t) = 0$ $t = 0$ (start) or $t = 4$. Answer: 4 seconds [2]

15. (a) Vertical asymptote where denominator is zero. Answer: $x = 2$ [1]

(b) As $x \to \infty$, $\frac{1}{x-2} \to 0$, so $y \to 3$. Answer: $y = 3$ [1]

(c) x-intercept when $y=0$. $0 = \frac{1}{x-2} + 3$ $-3 = \frac{1}{x-2}$ $-3(x-2) = 1$ $-3x + 6 = 1$ $-3x = -5$ $x = \frac{5}{3}$ Answer: $(\frac{5}{3}, 0)$ [2]

16. For $f(x) = f^{-1}(x)$, we solve $f(x) = x$ (provided the function is increasing and intersects $y=x$). $\frac{3x}{x+1} = x$ $3x = x(x+1)$ $3x = x^2 + x$ $x^2 - 2x = 0$ $x(x-2) = 0$ $x = 0$ or $x = 2$. Check: If $x=0, f(0)=0, f^{-1}(0)=0$. Valid. If $x=2, f(2)=6/3=2, f^{-1}(2)=2$. Valid. Answer: $x = 0, 2$ [2]

17. $y = kx^n$

1. $4 = k(2)^n$
2. $32 = k(4)^n$ Divide (2) by (1): $\frac{32}{4} = \frac{k(4)^n}{k(2)^n}$ $8 = (\frac{4}{2})^n = 2^n$ $2^3 = 2^n \Rightarrow n = 3$ Substitute $n=3$ into (1): $4 = k(2)^3 = 8k$ $k = 0.5$ Answer: $k = 0.5, n = 3$ [2]

18. (a) As $x \to -\infty$, $2^{x-1} \to 0$, so $y \to -3$. Answer: $y = -3$ [1]

(b) x-intercept when $y=0$. $0 = 2^{x-1} - 3$ $3 = 2^{x-1}$ $\log_2(3) = x - 1$ $x = 1 + \log_2(3)$ Answer: $x = 1 + \log_2(3)$ (or approx 2.58) [1]

19. $x^2 - 5x + 6 < 0$ $(x-2)(x-3) < 0$ Critical values: $x=2, x=3$. Since parabola opens upward, values are negative between roots. Answer: $2 < x < 3$ [2]

20. $fg(x) = f(x^2) = x^2 + 2$ $gf(x) = g(x+2) = (x+2)^2 = x^2 + 4x + 4$ Set $fg(x) = gf(x)$: $x^2 + 2 = x^2 + 4x + 4$ $2 = 4x + 4$ $4x = -2$ $x = -0.5$ Answer: $x = -0.5$ [2]