O Level Additional Mathematics Algebra Functions Quiz

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

1. (a) $f(4) = 2(4) - 3 = 8 - 3 = 5$ [1] (b) Let $y = 2x - 3$. Swap $x$ and $y$: $x = 2y - 3 \Rightarrow 2y = x + 3 \Rightarrow y = \frac{x+3}{2}$. $f^{-1}(x) = \frac{x+3}{2}$ [2]

2. (a) As $x > 2$, $x-2 > 0$, so $\frac{1}{x-2} > 0$. Range is $g(x) > 0$ or $(0, \infty)$. [1] (b) $g$ is strictly decreasing for $x > 2$ (or one-one), so it has an inverse. [1]

3. $x^2 + 1 = 10 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. Since domain is $x \ge 0$, $x = 3$. [2]

4. (a) $fg(x) = f(x^2) = 3(x^2) + 1 = 3x^2 + 1$. [2] (b) $gf(x) = g(3x+1) = (3x+1)^2 = 9x^2 + 6x + 1$. [2]

5. (a) $4-x \ge 0 \Rightarrow x \le 4$. Domain: $x \le 4$ or $(-\infty, 4]$. [2] (b) $\sqrt{4-x} \ge 0$. Range: $k(x) \ge 0$ or $[0, \infty)$. [1]

6. $pq(x) = p(\frac{10}{x}) = 2(\frac{10}{x}) + 5 = \frac{20}{x} + 5$. $\frac{20}{x} + 5 = 15 \Rightarrow \frac{20}{x} = 10 \Rightarrow 10x = 20 \Rightarrow x = 2$. [3]

7. (a) $y = \frac{2x+1}{x-3} \Rightarrow y(x-3) = 2x+1 \Rightarrow xy - 3y = 2x + 1$. $xy - 2x = 3y + 1 \Rightarrow x(y-2) = 3y + 1 \Rightarrow x = \frac{3y+1}{y-2}$. $f^{-1}(x) = \frac{3x+1}{x-2}$. [3] (b) Domain of $f^{-1}$ is Range of $f$. As $x > 3$, asymptote is $y=2$. Since $x>3$, $f(x) > 2$. Domain: $x > 2$. [1]

8. (a) $y = (x-2)^2 + 1 \Rightarrow y-1 = (x-2)^2$. Since $x \ge 2$, $x-2 = \sqrt{y-1} \Rightarrow x = 2 + \sqrt{y-1}$. $g^{-1}(x) = 2 + \sqrt{x-1}$. [3] (b) Graph $g$: Vertex $(2,1)$, passes through $(3,2)$. Graph $g^{-1}$: Vertex $(1,2)$, passes through $(2,3)$. Intersection on $y=x$ at approx $(2.6, 2.6)$ (exact: $x^2-5x+5=0 \Rightarrow x = \frac{5+\sqrt{5}}{2}$). [3]

9. (a) For $0 \le x \le 2$, the graph is the left half of the parabola with vertex at $x=2$. It is strictly decreasing, hence one-one. [1] (b) $y = x^2 - 4x + 7 = (x-2)^2 + 3$. $y-3 = (x-2)^2$. Since $x \le 2$, $x-2 = -\sqrt{y-3} \Rightarrow x = 2 - \sqrt{y-3}$. $f^{-1}(x) = 2 - \sqrt{x-3}$. [3]

10. $h^{-1}(x) = \frac{dx-b}{-cx+a}$. For self-inverse, $h(x) = h^{-1}(x)$. $\frac{ax+b}{cx+d} = \frac{dx-b}{-cx+a}$. Comparing coefficients or using condition $a = -d$ and $b,c$ arbitrary? No, standard condition for $\frac{ax+b}{cx+d}$ to be self-inverse is $a = -d$. Given $a=3, d=-3$, this holds. The relationship between $b$ and $c$ is not restricted by the self-inverse property alone provided $a=-d$, but usually questions imply specific forms. Wait, if $a=-d$, then $h(h(x)) = x$ for any $b,c$. However, often "relationship" implies checking if $b$ or $c$ must be 0? No. Let's check: $h(h(x)) = \frac{a(\frac{ax+b}{cx+d})+b}{c(\frac{ax+b}{cx+d})+d} = \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} = \frac{(a^2+bc)x + b(a+d)}{c(a+d)x + (bc+d^2)}$. If $a=-d$, numerator: $(a^2+bc)x$, denominator: $(bc+a^2)$. Result $x$. So any $b,c$ work. Correction based on typical exam pattern: Often asks for $a+d=0$. Here it is given. If the question implies a specific constraint like $h(x)=x$ has no solution or similar, it might differ. But strictly, if $a=-d$, it is self-inverse for all $b,c$. Alternative interpretation: Maybe the question meant $h(x) = \frac{ax+b}{cx-a}$. Let's assume the question asks for the condition $a+d=0$. Since $3+(-3)=0$, it is satisfied. If forced to find a relationship between $b$ and $c$ for a specific type of self-inverse (e.g. symmetric about $y=x$), there isn't one unless $b=c=0$ (identity) or similar. Re-reading standard templates: Usually, if $a \neq -d$, it's not self-inverse. If $a=-d$, it is. Perhaps the question implies $h(x) = h^{-1}(x)$ leads to $b=c$? No. Let's provide the standard condition: $a = -d$. Since this is given, the relationship is that $b$ and $c$ can be any real numbers. However, in many O-Level contexts, if asked for a relationship, it might be a trick or I should check if $b=c$ makes it symmetric? No. Let's stick to: The condition is $a = -d$. Since this is met, there is no specific constraint linking $b$ and $c$ other than they are real. Self-Correction: If the question meant $h(x) = \frac{3x+b}{cx-3}$, then $h^{-1}(x) = \frac{3x+b}{cx-3}$? Inverse of $\frac{3x+b}{cx-3}$ is $\frac{3x+b}{-cx+3}$? No. $h^{-1}(x) = \frac{-3x-b}{-cx+3} = \frac{3x+b}{cx-3}$. Yes. So any $b,c$ works. Marking Note: Award marks for stating $a = -d$ is the condition. [3]

11. (a) $fg(x) = f(\ln(x+1)) = e^{2\ln(x+1)} = e^{\ln((x+1)^2)} = (x+1)^2$. $(x+1)^2 = 5 \Rightarrow x+1 = \pm\sqrt{5}$. Domain of $g$ is $x > -1$, so $x+1 > 0$. $x = \sqrt{5} - 1$. [3] (b) Domain of $g$ is $x > -1$. $f(x) = e^{2x}$ is always $>0$. $gf(x) = \ln(e^{2x}+1)$. Since $e^{2x} > 0$, $e^{2x}+1 > 1$, so $\ln$ is defined. Domain is $x \in \mathbb{R}$. [2]

12. (a) Vertex of $x^2-4x+7$ is at $x = -(-4)/2 = 2$. For one-one, domain must be $x \ge 2$ or $x \le 2$. Given $x \ge k$, smallest $k=2$. [2] (b) Min value at $x=2$ is $4-8+7=3$. Range is $f(x) \ge 3$. [2]

13. (a) V-shape graph. Vertex at $(3,0)$. Y-intercept at $(0,6)$. [2] (b) $|2x-6|=4 \Rightarrow 2x-6=4$ or $2x-6=-4$. $2x=10 \Rightarrow x=5$. $2x=2 \Rightarrow x=1$. $x=1, 5$. [2]

14. (a) $gf(x) = \log_3(3^x) = x \log_3 3 = x(1) = x$. [2] (b) Let $u = 3^x$. $u^2 - 4u + 3 = 0 \Rightarrow (u-3)(u-1)=0$. $u=3 \Rightarrow 3^x=3 \Rightarrow x=1$. $u=1 \Rightarrow 3^x=1 \Rightarrow x=0$. $x=0, 1$. [4]

15. $f(x) = f^{-1}(x)$ intersects on $y=x$. $\frac{x+1}{2x-1} = x \Rightarrow x+1 = 2x^2 - x \Rightarrow 2x^2 - 2x - 1 = 0$. $x = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}$. Both are valid ($x \neq 0.5$). [4]

16. (a) $fg(x) = \sqrt{x^2-2+2} = \sqrt{x^2} = |x|$. Domain: Inside square root $\ge 0 \Rightarrow x^2 \ge 0$ (always true). But $g(x)$ output must be in domain of $f$ ($x \ge -2$). $x^2-2 \ge -2 \Rightarrow x^2 \ge 0$. So Domain is $\mathbb{R}$. [3] (b) $gf(x) = (\sqrt{x+2})^2 - 2 = x+2-2 = x$. Domain: $x+2 \ge 0 \Rightarrow x \ge -2$. [3]

17. (a) As $x > 1$, $x-1 > 0$, so $\frac{2}{x-1} > 0$. Thus $h(x) > 3$. Range $(3, \infty)$. [2] (b) $y = \frac{2}{x-1} + 3 \Rightarrow y-3 = \frac{2}{x-1} \Rightarrow x-1 = \frac{2}{y-3} \Rightarrow x = \frac{2}{y-3} + 1$. $h^{-1}(x) = \frac{2}{x-3} + 1$. [3]

18. $f(f(x)) = a(ax+b)+b = a^2x + ab + b$. $a^2x + (ab+b) = 4x + 9$. $a^2 = 4 \Rightarrow a = 2$ or $a = -2$. Case 1: $a=2$. $2b+b=9 \Rightarrow 3b=9 \Rightarrow b=3$. Case 2: $a=-2$. $-2b+b=9 \Rightarrow -b=9 \Rightarrow b=-9$. Solutions: $a=2, b=3$ or $a=-2, b=-9$. [4]

19. (a) $g(x) = g(-x)$, so it is many-one (fails horizontal line test). [1] (b) $y = \frac{1}{x^2} \Rightarrow x^2 = \frac{1}{y} \Rightarrow x = \frac{1}{\sqrt{y}}$ (since $x>0$). $g^{-1}(x) = \frac{1}{\sqrt{x}}$. [2]

20. (a) $x > -2 \Rightarrow x+2 > 0 \Rightarrow \frac{4}{x+2} > 0$. Range $g(x) > 0$. [2] (b) $g(f(x)) = \frac{4}{(2x-1)+2} = \frac{4}{2x+1}$. $\frac{4}{2x+1} = 1 \Rightarrow 2x+1=4 \Rightarrow 2x=3 \Rightarrow x=1.5$. [3] (c) $g^{-1}(x)$: $y = \frac{4}{x+2} \Rightarrow x+2 = \frac{4}{y} \Rightarrow x = \frac{4}{y}-2$. $g^{-1}(x) = \frac{4}{x}-2$. $2x-1 = \frac{4}{x}-2 \Rightarrow 2x+1 = \frac{4}{x} \Rightarrow 2x^2+x-4=0$. $x = \frac{-1 \pm \sqrt{1 - 4(2)(-4)}}{4} = \frac{-1 \pm \sqrt{33}}{4}$. Check domain of $g^{-1}$ ($x \neq 0$) and range of $g$ ($x>0$ for input to $g^{-1}$? No, domain of $g^{-1}$ is range of $g$, which is $x>0$). So we need $x > 0$. $\sqrt{33} \approx 5.7$. $-1+5.7 > 0$. $-1-5.7 < 0$. Only $x = \frac{-1+\sqrt{33}}{4}$ is valid. [4]