Secondary 4 Elementary Mathematics Algebra Functions Quiz

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

Total Marks: 40

Section A (10 marks)

1. $f(-2) = 2(-2)^2 - 5(-2) + 3 = 2(4) + 10 + 3 = 8 + 10 + 3 = 21$
Answer: 21 [1]

2. For $g(x) = \frac{4}{x-2}$, the denominator cannot be zero.
$x - 2 \neq 0 \Rightarrow x \neq 2$
Answer: $x \neq 2$ [1]

3. $h(x) = 3x - 7$. Let $y = 3x - 7$. Then $x = \frac{y+7}{3}$. So $h^{-1}(x) = \frac{x+7}{3}$.
Answer: $h^{-1}(x) = \frac{x+7}{3}$ [1]

4. From the graph, the vertex (minimum point) is at $(1, -4)$.
Answer: $(1, -4)$ [1]

5. $fg(3) = f(g(3)) = f(3-1) = f(2) = 2^2 + 2(2) = 4 + 4 = 8$
Answer: 8 [1]

6. $f(x) = 5 - 2x$. For inverse: $y = 5 - 2x \Rightarrow 2x = 5 - y \Rightarrow x = \frac{5-y}{2}$. So $f^{-1}(x) = \frac{5-x}{2}$.
Solve $f(x) = f^{-1}(x)$: $5 - 2x = \frac{5-x}{2} \Rightarrow 10 - 4x = 5 - x \Rightarrow 5 = 3x \Rightarrow x = \frac{5}{3}$.
Answer: $\frac{5}{3}$ [1]

7. $f(4) = \frac{2(4)+1}{4-3} = \frac{8+1}{1} = 9$
Answer: 9 [1]

8. $x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3) = 0 \Rightarrow x = 1$ or $x = 3$.
Points: $A(1, 0)$ and $B(3, 0)$ (or vice versa).
Answer: $A(1, 0)$ and $B(3, 0)$ [1]

9. $gf(1) = g(f(1)) = g(3(1)^2 - 2) = g(1) = \sqrt{1+5} = \sqrt{6}$
Answer: $\sqrt{6}$ [1]

10. $gf(x) = g(f(x)) = g(2x+5) = (2x+5)^2 - 1 = 4x^2 + 20x + 25 - 1 = 4x^2 + 20x + 24$
Answer: $4x^2 + 20x + 24$ [1]

Section B (18 marks)

11. (a) $f(x) = 2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2[(x-2)^2 - 4] + 5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3$
Answer: $2(x-2)^2 - 3$ [2]
Marking: 1 mark for factorising 2, 1 mark for completing square correctly.

(b) Vertex is at $(h, k) = (2, -3)$.
Answer: $(2, -3)$ [1]

(c) Since $a = 2 > 0$, minimum value is $-3$. Range: $f(x) \ge -3$ or $[-3, \infty)$.
Answer: $f(x) \ge -3$ [1]

(d) $g(x) = f(x) + 3 = 2(x-2)^2 - 3 + 3 = 2(x-2)^2$. Minimum is $0$. Range: $g(x) \ge 0$ or $[0, \infty)$.
Answer: $g(x) \ge 0$ [1]

12. (a) y = \begin: y = \frac{3x-2}{x+1} \Rightarrow y(x+1) = 3x-2 \Rightarrow yx + y = 3x - 2 \Rightarrow yx - 3x = -2 - y \Rightarrow x(y-3) = -(y+2) \Rightarrow x = \frac{y+2}{3-y}
So $f^{-1}(x) = \frac{x+2}{3-x}$ for $x \neq 3$.
Answer: $f^{-1}(x) = \frac{x+2}{3-x}$ [3]
Marking: 1 mark for cross-multiplying, 1 mark for collecting x terms, 1 mark for final expression with domain.

(b) Denominator of $f^{-1}(x)$ cannot be zero: $3 - x \neq 0 \Rightarrow x \neq 3$.
Answer: $x \neq 3$ [1]

(c) Solve $f(x) = f^{-1}(x)$: $\frac{3x-2}{x+1} = \frac{x+2}{3-x}$
Cross-multiply: $(3x-2)(3-x) = (x+2)(x+1)$
$9x - 3x^2 - 6 + 2x = x^2 + x + 2x + 2$
$-3x^2 + 11x - 6 = x^2 + 3x + 2$
$0 = 4x^2 - 8x + 8 = 4(x^2 - 2x + 2)$
Discriminant: $(-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$. No real solutions.
Answer: No real solutions [2]
Marking: 1 mark for correct equation setup, 1 mark for correct conclusion.

13. (a) $fg(x) = f(g(x)) = f(x^2 + 2) = 4(x^2 + 2) - 3 = 4x^2 + 8 - 3 = 4x^2 + 5$
Answer: $4x^2 + 5$ [2]
Marking: 1 mark for substitution, 1 mark for simplification.

(b) $gf(x) = g(f(x)) = g(4x - 3) = (4x - 3)^2 + 2 = 16x^2 - 24x + 9 + 2 = 16x^2 - 24x + 11$
Answer: $16x^2 - 24x + 11$ [2]
Marking: 1 mark for substitution, 1 mark for expansion and simplification.

(c) $fg(x) = gf(x) \Rightarrow 4x^2 + 5 = 16x^2 - 24x + 11$
$0 = 12x^2 - 24x + 6 = 6(2x^2 - 4x + 1)$
$2x^2 - 4x + 1 = 0$
$x = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$
Answer: $x = 1 + \frac{\sqrt{2}}{2}$ or $x = 1 - \frac{\sqrt{2}}{2}$ [2]
Marking: 1 mark for correct quadratic, 1 mark for correct solutions.

14. (a) Graph passes through $(2, 6)$: $6 = \frac{k}{2} \Rightarrow k = 12$.
Answer: $k = 12$ [1]

(b) Graph of $y = \frac{12}{x} + 2$: Same shape as $y = \frac{12}{x}$ but shifted up by 2 units. Asymptotes: $x = 0$ (vertical), $y = 2$ (horizontal). Passes through $(2, 8)$.
Answer: Sketch showing reciprocal curve shifted up 2 units, with correct asymptotes and point (2, 8). [2]
Marking: 1 mark for correct shape and asymptotes, 1 mark for correct shift and key point.

(c) Vertical asymptote: $x = 0$. Horizontal asymptote: $y = 2$.
Answer: $x = 0$ and $y = 2$ [2]
Marking: 1 mark each.

15. (a) For $x \ge 3$, $f(x) = x^2 - 6x + 10 = (x-3)^2 + 1$. This is strictly increasing on $x \ge 3$ (derivative $2x-6 \ge 0$), so it is one-to-one. Hence inverse exists.
Answer: $f$ is strictly increasing on $x \ge 3$, so it is one-to-one. [1]

(b) $y = (x-3)^2 + 1 \Rightarrow (x-3)^2 = y - 1 \Rightarrow x - 3 = \sqrt{y-1}$ (positive root since $x \ge 3$)
$x = 3 + \sqrt{y-1}$. So $f^{-1}(x) = 3 + \sqrt{x-1}$. Domain of $f^{-1}$ = Range of $f = [1, \infty)$.
Answer: $f^{-1}(x) = 3 + \sqrt{x-1}$, domain: $x \ge 1$ [3]
Marking: 1 mark for completing square, 1 mark for correct inverse with positive root, 1 mark for domain.

(c) Sketch: $y = f(x)$ is parabola with vertex $(3, 1)$, opening upwards, for $x \ge 3$. $y = f^{-1}(x)$ is its reflection in $y = x$, starting at $(1, 3)$ and increasing. Both graphs intersect on $y = x$ at $(3, 3)$? Wait: $f(3) = 1$, so intersection of $f$ and $f^{-1}$ is not at $(3,3)$. Actually $f(x) = x \Rightarrow (x-3)^2 + 1 = x \Rightarrow x^2 - 7x + 10 = 0 \Rightarrow (x-2)(x-5)=0 \Rightarrow x=5$ (since $x \ge 3$). So they intersect at $(5, 5)$.
Answer: Sketch showing both curves, reflection in $y=x$, intersection at $(5,5)$, $f$ starts at $(3,1)$, $f^{-1}$ starts at $(1,3)$. [2]
Marking: 1 mark for correct shapes and reflection, 1 mark for correct key points.

Section C (12 marks)

16. (a) $f(x) = 3x^2 - 12x + 11 = 3(x^2 - 4x) + 11 = 3[(x-2)^2 - 4] + 11 = 3(x-2)^2 - 12 + 11 = 3(x-2)^2 - 1$
Answer: $3(x-2)^2 - 1$ [2]
Marking: 1 mark for factorising 3, 1 mark for completing square.

(b) $g(x) = f(x) - 2 = 3(x-2)^2 - 3$. Vertex: $(2, -3)$.
Answer: $(2, -3)$ [1]

(c) $h(x) = f(x+1) = 3(x+1)^2 - 12(x+1) + 11 = 3(x^2 + 2x + 1) - 12x - 12 + 11 = 3x^2 + 6x + 3 - 12x - 1 = 3x^2 - 6x + 2$
Answer: $3x^2 - 6x + 2$ [2]
Marking: 1 mark for substitution, 1 mark for expansion and simplification.

(d) $f(x) = h(x) \Rightarrow 3x^2 - 12x + 11 = 3x^2 - 6x + 2 \Rightarrow -12x + 11 = -6x + 2 \Rightarrow -6x = -9 \Rightarrow x = \frac{3}{2}$
Answer: $x = \frac{3}{2}$ [2]
Marking: 1 mark for equating and simplifying, 1 mark for solution.

17. (a) $gf(x) = g(f(x)) = g(2x+1) = \frac{(2x+1)-3}{2} = \frac{2x-2}{2} = x - 1$.
Answer: Shown. [2]
Marking: 1 mark for substitution, 1 mark for simplification to $x-1$.

(b) $fg(x) = f(g(x)) = f\left(\frac{x-3}{2}\right) = 2\left(\frac{x-3}{2}\right) + 1 = x - 3 + 1 = x - 2$
Answer: $x - 2$ [1]

(c) $h(x) = fg(x) - gf(x) = (x-2) - (x-1) = -1$.
$h(x) = -1$ is a constant function. It is not one-to-one, so $h^{-1}$ does not exist.
Answer: $h^{-1}$ does not exist because $h(x) = -1$ is not one-to-one. [3]
Marking: 1 mark for finding $h(x) = -1$, 1 mark for stating inverse does not exist, 1 mark for correct reason.

18. (a) $f(x) = ax^2 + bx + c$.
$f(0) = 4 \Rightarrow c = 4$.
$f(1) = 1 \Rightarrow a + b + 4 = 1 \Rightarrow a + b = -3$.
$f(3) = 4 \Rightarrow 9a + 3b + 4 = 4 \Rightarrow 9a + 3b = 0 \Rightarrow 3a + b = 0$.
Subtract: $(3a + b) - (a + b) = 0 - (-3) \Rightarrow 2a = 3 \Rightarrow a = \frac{3}{2}$.
Then $b = -3a = -\frac{9}{2}$.
Answer: $a = \frac{3}{2}, b = -\frac{9}{2}, c = 4$ [3]
Marking: 1 mark for $c=4$, 1 mark for solving system, 1 mark for correct $a$ and $b$.

(b) Line of symmetry: $x = -\frac{b}{2a} = -\frac{-9/2}{2(3/2)} = \frac{9/2}{3} = \frac{3}{2}$. Or from vertex $(1,1)$? Wait, vertex is at $(1,1)$ from graph. But $x = -\frac{b}{2a} = \frac{9/2}{3} = 1.5$. There's inconsistency. Let's recheck: $f(1) = 1$, $f(3) = 4$, $f(0) = 4$. The axis of symmetry is midway between $x=0$ and $x=3$? No, $f(0)=f(3)=4$, so axis is $x = 1.5$. But graph says vertex at $(1,1)$. This is a contradiction in the problem setup. Let's use the calculated values: $a=1.5, b=-4.5, c=4$. Vertex at $x = -b/2a = 4.5/3 = 1.5$. $f(1.5) = 1.5(2.25) - 4.5(1.5) + 4 = 3.375 - 6.75 + 4 = 0.625$. So vertex is $(1.5, 0.625)$. The graph description said vertex at $(1,1)$ but points $(0,4), (1,1), (3,4)$ don't form a symmetric parabola with vertex at $(1,1)$. Actually $f(0)=4, f(1)=1, f(3)=4$. The axis of symmetry is $x=1.5$ (midpoint of 0 and 3). So vertex is at $x=1.5$. The point $(1,1)$ is not the vertex. The graph description was slightly off. We'll use the calculated values.
Answer: $x = \frac{3}{2}$ [1]

(c) $g(x) = f(x) + k = \frac{3}{2}x^2 - \frac{9}{2}x + 4 + k$. Touches $x$-axis $\Rightarrow$ discriminant $= 0$.
$b^2 - 4ac = 0 \Rightarrow \left(-\frac{9}{2}\right)^2 - 4\left(\frac{3}{2}\right)(4+k) = 0$
$\frac{81}{4} - 6(4+k) = 0 \Rightarrow \frac{81}{4} - 24 - 6k = 0 \Rightarrow \frac{81}{4} - \frac{96}{4} = 6k \Rightarrow -\frac{15}{4} = 6k \Rightarrow k = -\frac{15}{24} = -\frac{5}{8}$.
Answer: $k = -\frac{5}{8}$ [2]
Marking: 1 mark for discriminant condition, 1 mark for correct $k$.

19. (a) $y = \frac{2x+5}{x-1} \Rightarrow y(x-1) = 2x+5 \Rightarrow yx - y = 2x + 5 \Rightarrow yx - 2x = y + 5 \Rightarrow x(y-2) = y+5 \Rightarrow x = \frac{y+5}{y-2}$.
$f^{-1}(x) = \frac{x+5}{x-2}$ for $x \neq 2$.
Answer: $f^{-1}(x) = \frac{x+5}{x-2}$ [3]
Marking: 1 mark for cross-multiplying, 1 mark for collecting x terms, 1 mark for final expression.

(b) $fg(x) = f(g(x)) = f(x^2 - 4) = \frac{2(x^2-4)+5}{(x^2-4)-1} = \frac{2x^2 - 8 + 5}{x^2 - 5} = \frac{2x^2 - 3}{x^2 - 5}$.
Solve $fg(x) = 3$: $\frac{2x^2 - 3}{x^2 - 5} = 3 \Rightarrow 2x^2 - 3 = 3x^2 - 15 \Rightarrow x^2 = 12 \Rightarrow x = \pm 2\sqrt{3}$.
Check: $x^2 - 5 \neq 0 \Rightarrow x \neq \pm \sqrt{5}$. Solutions valid.
Answer: $x = 2\sqrt{3}$ or $x = -2\sqrt{3}$ [3]
Marking: 1 mark for $fg(x)$ expression, 1 mark for equation setup, 1 mark for solutions.

20. (a) $f(x) = x^3 - 3x^2 + 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$.
Stationary points when $f'(x) = 0 \Rightarrow x = 0$ or $x = 2$.
$f(0) = 2$, $f(2) = 8 - 12 + 2 = -2$.
Coordinates: $(0, 2)$ and $(2, -2)$.
Answer: $(0, 2)$ and $(2, -2)$ [3]
Marking: 1 mark for derivative, 1 mark for $x$ values, 1 mark for coordinates.

(b) $f''(x) = 6x - 6$.
At $x = 0$: $f''(0) = -6 < 0 \Rightarrow$ local maximum at $(0, 2)$.
At $x = 2$: $f''(2) = 6 > 0 \Rightarrow$ local minimum at $(2, -2)$.
Answer: $(0, 2)$ is a local maximum; $(2, -2)$ is a local minimum. [2]
Marking: 1 mark for second derivative test, 1 mark for correct nature.

(c) Sketch: Cubic with positive leading coefficient. Passes through $(0, 2)$ (max), $(2, -2)$ (min). $y$-intercept: $(0, 2)$. $x$-intercepts: solve $x^3 - 3x^2 + 2 = 0$. Try $x=1$: $1-3+2=0$. Factor: $(x-1)(x^2-2x-2)=0$. Other roots: $x = 1 \pm \sqrt{3}$. So $x \approx -0.732, 1, 2.732$. For $-1 \le x \le 3$, show shape from $(-1, -2)$ to $(3, 2)$.
Answer: Sketch showing cubic shape with max at $(0,2)$, min at $(2,-2)$, $x$-intercepts at $1-\sqrt{3}$, $1$, $1+\sqrt{3}$, $y$-intercept at $(0,2)$. [2]
Marking: 1 mark for correct shape and stationary points, 1 mark for correct intercepts and endpoints.

End of Answer Key