O Level Additional Mathematics Algebra Functions Quiz

O-Level Additional Mathematics Quiz - Algebra Functions (Answer Key)

1.
(a) Let $y = 2x - 3$.
Swap $x$ and $y$: $x = 2y - 3$.
$2y = x + 3 \implies y = \frac{x+3}{2}$.
$f^{-1}(x) = \frac{x+3}{2}$.
[2]

(b) $\frac{x+3}{2} = 2x - 3$.
$x + 3 = 4x - 6$.
$3x = 9 \implies x = 3$.
[2]

2.
(a) Since $x > 2$, $x-2 > 0$. As $x \to \infty$, $g(x) \to 0$. As $x \to 2^+$, $g(x) \to \infty$.
Range: $g(x) > 0$ (or $(0, \infty)$).
[1]

(b) $gg(x) = g(g(x)) = \frac{1}{g(x) - 2} = \frac{1}{\frac{1}{x-2} - 2}$.
Multiply numerator and denominator by $(x-2)$:
$\frac{x-2}{1 - 2(x-2)} = \frac{x-2}{1 - 2x + 4} = \frac{x-2}{5-2x}$.
[2]

(c) For $g(x)$ to be defined, $x > 2$. For $g(g(x))$ to be defined, $g(x) \neq 2$.
$\frac{1}{x-2} \neq 2 \implies 1 \neq 2x - 4 \implies 2x \neq 5 \implies x \neq 2.5$.
Domain: $x > 2, x \neq 2.5$.
[1]

3.
$h(x) = x^2 + 4x = (x+2)^2 - 4$.
Vertex at $x = -2$.
For $h$ to be one-to-one (have an inverse), the domain must be restricted to one side of the vertex.
Since $x \ge k$, we need $k \ge -2$.
Smallest value $k = -2$.
[2]

4.
Let $y = \sqrt{x-1}$.
$y^2 = x - 1 \implies x = y^2 + 1$.
$p^{-1}(x) = x^2 + 1$.
Since range of $p(x)$ is $y \ge 0$, domain of $p^{-1}(x)$ is $x \ge 0$.
[2]

5.
$fg(x) = f(x^2) = 3(x^2) + 1 = 3x^2 + 1$.
$gf(x) = g(3x+1) = (3x+1)^2 = 9x^2 + 6x + 1$.
$3x^2 + 1 = 9x^2 + 6x + 1$.
$6x^2 + 6x = 0$.
$6x(x+1) = 0$.
$x = 0$ or $x = -1$.
[3]

6.
$2x^2 - 8x + 5 = 2(x^2 - 4x) + 5$.
Complete square inside: $2[(x-2)^2 - 4] + 5$.
$= 2(x-2)^2 - 8 + 5$.
$= 2(x-2)^2 - 3$.
[3]

7.
From part (6), minimum value is $-3$.
Occurs when $x - 2 = 0 \implies x = 2$.
[2]

8.
For two distinct real roots, discriminant $\Delta > 0$.
$\Delta = b^2 - 4ac = k^2 - 4(3)(k+3)$.
$k^2 - 12(k+3) > 0$.
$k^2 - 12k - 36 > 0$.
Roots of $k^2 - 12k - 36 = 0$:
$k = \frac{12 \pm \sqrt{144 - 4(1)(-36)}}{2} = \frac{12 \pm \sqrt{144 + 144}}{2} = \frac{12 \pm \sqrt{288}}{2}$.
$\sqrt{288} = 12\sqrt{2}$.
$k = 6 \pm 6\sqrt{2}$.
Since inequality is $>0$ (outside roots):
$k < 6 - 6\sqrt{2}$ or $k > 6 + 6\sqrt{2}$.
[4]

9.
Intersection: $x^2 - 4x + 7 = 2x + c$.
$x^2 - 6x + (7-c) = 0$.
Tangent condition: $\Delta = 0$.
$(-6)^2 - 4(1)(7-c) = 0$.
$36 - 28 + 4c = 0$.
$8 + 4c = 0 \implies 4c = -8 \implies c = -2$.
[3]

10.
$f(x) = x^2 - 6x + 10$.
Discriminant $\Delta = (-6)^2 - 4(1)(10) = 36 - 40 = -4$.
Since $\Delta < 0$, there are no real roots.
Alternatively, $f(x) = (x-3)^2 + 1$. Since $(x-3)^2 \ge 0$, $f(x) \ge 1$. Thus $f(x)$ can never be 0.
[3]

11.
$P(1) = 0 \implies 2(1)^3 - 5(1)^2 + a(1) + b = 0$.
$2 - 5 + a + b = 0 \implies a + b = 3$ --- (1).
$P(-2) = -20 \implies 2(-8) - 5(4) + a(-2) + b = -20$.
$-16 - 20 - 2a + b = -20$.
$-36 - 2a + b = -20 \implies -2a + b = 16$ --- (2).
Subtract (2) from (1): $(a+b) - (-2a+b) = 3 - 16$.
$3a = -13 \implies a = -\frac{13}{3}$.
Substitute into (1): $-\frac{13}{3} + b = 3 \implies b = 3 + \frac{13}{3} = \frac{22}{3}$.
$a = -\frac{13}{3}, b = \frac{22}{3}$.
[4]

12.
$P(x) = 2x^3 - 5x^2 - \frac{13}{3}x + \frac{22}{3}$.
Since $(x-1)$ is a factor, divide $P(x)$ by $(x-1)$.
Using synthetic division or long division with coefficients $2, -5, -13/3, 22/3$:
Quotient is $2x^2 - 3x - \frac{22}{3}$.
To factorise completely, we can write $P(x) = \frac{1}{3}(x-1)(6x^2 - 9x - 22)$.
Check discriminant of quadratic: $81 - 4(6)(-22) = 81 + 528 = 609$ (not a perfect square).
So, $P(x) = \frac{1}{3}(x-1)(6x^2 - 9x - 22)$.
Note: If integer coefficients were intended in question design, values might differ, but based on calculated a,b:
Factors: $(x-1)$ and $(6x^2 - 9x - 22)$.
[2]

13.
$\frac{3x^2 + 5x - 2}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$.
$3x^2 + 5x - 2 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$.
Let $x = 1$: $3+5-2 = A(3)^2 \implies 6 = 9A \implies A = \frac{2}{3}$.
Let $x = -2$: $3(4) - 10 - 2 = C(-3) \implies 12 - 12 = -3C \implies 0 = -3C \implies C = 0$.
Compare coeff of $x^2$: $3 = A + B \implies 3 = \frac{2}{3} + B \implies B = \frac{7}{3}$.
Answer: $\frac{2}{3(x-1)} + \frac{7}{3(x+2)}$.
[5]

14.
$\int_{2}^{3} \left( \frac{2}{3(x-1)} + \frac{7}{3(x+2)} \right) dx$.
$= \left[ \frac{2}{3} \ln|x-1| + \frac{7}{3} \ln|x+2| \right]_{2}^{3}$.
Upper limit ($x=3$): $\frac{2}{3} \ln 2 + \frac{7}{3} \ln 5$.
Lower limit ($x=2$): $\frac{2}{3} \ln 1 + \frac{7}{3} \ln 4 = 0 + \frac{7}{3} \ln(2^2) = \frac{14}{3} \ln 2$.
Value: $\frac{2}{3} \ln 2 + \frac{7}{3} \ln 5 - \frac{14}{3} \ln 2$.
$= \frac{7}{3} \ln 5 - \frac{12}{3} \ln 2 = \frac{7}{3} \ln 5 - 4 \ln 2$.
[4]

15.
Since $(x-1)$ is a factor, divide $x^3 - 2x^2 - 5x + 6$ by $(x-1)$.
$(x^3 - 2x^2 - 5x + 6) \div (x-1) = x^2 - x - 6$.
Factorise $x^2 - x - 6 = (x-3)(x+2)$.
Other factors are $(x-3)$ and $(x+2)$.
[3]

16.
Let $u = 3^x$. Equation becomes $u^2 - 10u + 9 = 0$.
$(u-9)(u-1) = 0$.
$u = 9$ or $u = 1$.
$3^x = 9 \implies x = 2$.
$3^x = 1 \implies x = 0$.
Solutions: $x = 0, 2$.
[3]

17.
$\log_a 20 = \log_a (4 \times 5) = \log_a (2^2 \times 5)$.
$= \log_a (2^2) + \log_a 5 = 2 \log_a 2 + \log_a 5$.
$= 2p + q$.
[2]

18.
(a) $t = 2030 - 2020 = 10$.
$P = 500 e^{0.03(10)} = 500 e^{0.3}$.
$P \approx 500(1.34986) \approx 674.93$.
Population is 674,930 (or 675 thousand).
[2]

(b) $800 = 500 e^{0.03t}$.
$1.6 = e^{0.03t}$.
$\ln 1.6 = 0.03t$.
$t = \frac{\ln 1.6}{0.03} \approx \frac{0.4700}{0.03} \approx 15.67$.
Year: $2020 + 15.67 \approx 2035.67$.
First exceed in year 2036.
[3]

19.
$\log_2(x(x-2)) = 3$.
$x(x-2) = 2^3 = 8$.
$x^2 - 2x - 8 = 0$.
$(x-4)(x+2) = 0$.
$x = 4$ or $x = -2$.
Since $\log_2(x)$ requires $x > 0$, reject $x = -2$.
Solution: $x = 4$.
[3]

20.
Use product rule: $u = e^{2x}, v = \ln(x)$.
$u' = 2e^{2x}, v' = \frac{1}{x}$.
$\frac{dy}{dx} = u'v + uv' = 2e^{2x}\ln(x) + e^{2x}\left(\frac{1}{x}\right)$.
$\frac{dy}{dx} = e^{2x}\left(2\ln(x) + \frac{1}{x}\right)$.
[2]