# Chapter 4 : Functions - Multiple Choice Questions

မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။

1. If $A=\{a, b, c\}$, then $n(A \times A)=$
Explanation
\begin{aligned} A &=\{a, b, c\}\\\\ A \times A &=\{(a, a),(a, b),(a, c),\\\\ &\quad\ (b, a),(b, b),(b, c),\\\\ &\quad\ (c, a),(c, b),(c, c)\}\\\\ \therefore\quad n(A \times A)&=9\\\\ \text{Note that}\ & n(A \times A)=[n(A)]^{2} \end{aligned}
2. Given that $A=\{a\}$, then $A \times A$=
Explanation
$\begin{array}{l} A=\{a\} \\\\ A \times A=\{a\} \times\{a\} \\\\ \hspace{1.3cm}=\{(a, a)\} \end{array}$

product sets ၏ အစု၀င်များကို orderd pair $(x,y)$ ပုံစံဖြင့်ရေးရသည်။
3. Given that $f(x)=x^{2}+3 x+1 .$ If $f(a)=\dfrac{31}{4}$ where $a>0$, then what is the value of $a$ ?
Explanation
$\begin{array}{l} f(x)=x^{2}+3 x+1 \\\\ f(a)=\dfrac{31}{4} \\\\ a^{2}+2 a+1=\dfrac{31}{4} \\\\ 4 a^{2}+12 a-27=0\\\\ (2 a+9)(2 a-3)=0\\\\ a=-\dfrac{9}{2}\ \text{(or)}\ a=\dfrac{3}{2}\\\\ \end{array}$
Since $a>0$, the correct solution is $a=\dfrac{3}{2} .$
4. What is the domain of $f(x)=\dfrac{1}{x^{2}-4}$.
Explanation
$f(x)$ is not defined when
$x^{2}-4=0$
$x^{2}=4$
$x=\pm 2$
$\therefore$ dom $(f)=\mathbb{R} \smallsetminus\{-2,2\}$.
5. Given that $A=\{x \mid x>0, x \in \mathbb{R}\}$ and function $f: A \rightarrow \mathbb{R}$ and $g: A \rightarrow \mathbb{R}$ ane defined as $f(x)=x-2$ and $g(x)=\dfrac{x^{2}-4}{x+2}$. Which of the following is(are) true?
I. $f(2)=g(2)\quad$ II. $f=g\quad$ III. $f \ne g$
Explanation
$\operatorname{dom}(f)=\operatorname{dom}(g)=A=\{x \mid x>0, x \in \mathbb{R}\}$
$\therefore \operatorname{dom}(f)$ and $\operatorname{dom}(g)$ are the set of positive real nembers.
$f(x)=x-2$ and $g(x)=\dfrac{x^{2}-4}{x+2}=\dfrac{(x-2)(x+2)}{x+2}=x-2$ when $x \ne-2$
Since $-2 \notin A$, we can say $f(x)=g(x)$ for all $x \in A$
6. The graph of the function $y=a x^{2}+b x+c$ when $a=0$ is
Explanation
Generally $y=a x^{2}+b x+c$ is a quadratic function.
But when $a=0, y=b x+c$ is a linear function and the graph is a straight line.
7. What is the equation of horizontal asymptote of the curve $y=\dfrac{3}{x-1}+2 .$
Explanation
We have known that the graph $y=\dfrac{k}{x-p}+q$ has horizontal asymptote $y=q$ and vertical asymptote $x=p$.
$\therefore$ The horizontal arympatote of $y=\dfrac{3}{x-1}+2$ is $y=2$.
8. The furction $f(x)=\dfrac{3}{x-1}+2$ is not defined when
Explanation
A rational function is not defined when its denominator is equal to zero.
$\therefore f(x)=\dfrac{3}{x-1}+2$ is not defined when
$x-1=0 \text { or } x=1$,
9. The vertical asymptote of the graph of function $y=\dfrac{-3 x+4}{x-2}$ is
Explanation
We have known that the graph $y=\dfrac{k}{x-p}+q$ has horizontal asymptote $y=q$ and vertical asymptote $x=p$.
$y=-\dfrac{3 x+4}{x-2}=-\dfrac{2}{x-2}-3$
$\therefore$ The vertical asymptote of $y=-\dfrac{3 x+4}{x-2}$ is $x=2$.
10. Which of the following is one to one?
Explanation
See: Definition of one to one furction Chapter (4), Section $(4.3 .2)$
11. If $f^{-1}(x)=\dfrac{x-3}{2}$, then $f(x)=\ldots$
Explanation
$f^{-1}(x)=\dfrac{x-3}{2}$
Let $f(x)=y$, then
$f^{-1}(y)=x$
$\dfrac{y-3}{2}=x$
$y=2 x+3$
$\therefore f(x)=2 x+3$
12. Given that $f(x)=\dfrac{4}{2-3 x}$, then the domain of $f^{-1}$ is
Explanation
$f(x)=\dfrac{4}{2-3 x}$
If $f^{-1}(x)=y$ then
$f(y)=x$
$\dfrac{4}{2-3 y}=x$
$2-3 y=\dfrac{4}{x}$
$3 y=-\dfrac{4}{x}+2$
$y=\dfrac{2 x-4}{x}$
$f^{-1}(x)=\dfrac{2 x-4}{3 x}$
$\therefore f^{-1}$ exists when $x \ne 0 .$
13. If $f(x)=\dfrac{3 x-1}{2 x+1}$, $f^{-1}(1)=\ldots$
Explanation
$f(x)=\dfrac{3 x-1}{2 x+1}$
$f^{-1}(1)=a$
$f(a)=1$
$\dfrac{3 a-1}{2 a+1}=1$
$3a-1=2 a+1$
$a=2$
$\therefore\ f^{-1}(1)=2$
14. The function $f$ is given by $f(x)=10^{x}-2$, then $f^{-1}(2)=\cdots$
Explanation
$f(x)=10^{x}-2$
Let $f^{-1}(2)=a$
$f(a)=2$
$10^{a}-2=2$
$10^{a}=4$
$a=\log _{10} 4$
$\therefore\ f^{-1}(2)=\log _{10} 4$
15. Given that $f(x)=x^{2}$, what is the domain of $f$ for which $f^{-1}$ exists?
Explanation
$f^{-1}$ exists if and only if $f$ is a one to one function.
$f(x)$ is one to one only when $x \ge 0$.
$\therefore \operatorname{dom}(f)=\{x \mid x \ge 0, x \in \mathbb{R}\}$.
16. If $f(x)=x^{2}$ and $g(x)=2 x$,$(f \circ g)\left(-\dfrac{1}{2}\right)=\ldots$
Explanation
$f(x)=x^{2}$
$g(x)=2 x$
$(f \circ g)\left(-\dfrac{1}{2}\right)$
$=f\left(g\left(-\dfrac{1}{2}\right)\right)$
$=f\left(2\left(-\dfrac{1}{2}\right)\right)$
$=f(-1)$
$=(-1)^{2}$
$=1$
17. If $f(x)=x^{2}$ and $g(x)=\dfrac{2 x+1}{x-4}$, what is the domain of $g\circ f$ ?
Explanation
\begin{aligned} f(x)&=x^{2} \\\\ g(x)&=\dfrac{2 x+1}{x-4} \\\\ (g \circ f)(x)&=g(f(x)) \\\\ &=g\left(x^{2}\right)\\\\ &=\dfrac{2 x^{2}+1}{x^{2}-4} \\\\ (g \circ f)(x) &\text { exists when } \\\\ x^{2}-4 &\neq 0 \\\\ x^{2} &\neq 4 \\\\ x &\neq \pm 2 \\\\ \therefore\ \operatorname{dom}(g \circ f)&=\mathbb{R} \smallsetminus\{\pm 2\} \end{aligned}
18. If $g(x)=\dfrac{x+2}{2 x-1}$ and $h(x)=2 x$, what is the range of $g \circ h ?$
Explanation
\begin{aligned} g(x)&=\dfrac{x+2}{2 x-1} \\\\ h(x)&=2 x \\\\ (g \circ h)(x) &=g(h(x))\\\\ &=g(2 x) \\\\ &=\dfrac{2 x+2}{4 x-1} \\\\ &=\dfrac{5 / 2}{4 x-1}+\dfrac{1}{2} \\\\ \therefore \operatorname{ran}(g \cdot f)&=\left\{y \mid y \neq \dfrac{1}{2}, y \in \mathbb{R}\right\} \end{aligned}
19. The function $f: A \rightarrow B$ is onto function then the range of $f$ is
Explanation
A function $f$ is onto function when range of $f=$ codomain.
20. If $f$ is a function an a set $A=\{1,2,3,4,5\}$ such that $f=\{(1,2),(2,3),(3,4),(4, x),(5,5)\}$ is a one to function, then $x=\ldots$
Explanation
To be one to ove furction $f$, $A$ must be related with $1$.