Chapter 4 : Functions - Multiple Choice Questions August 04, 2021 TargetMathematics chapter-4, functions, grade 10, mcq, multiple choice, new syllabus, သင်ရိုးသစ် No comments Edit မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။ If $A=\{a, b, c\}$, then $n(A \times A)=$ A. $3$ B. $6$ C. $9$ D. $12$ Explanation $\begin{aligned} A &=\{a, b, c\}\\\\ A \times A &=\{(a, a),(a, b),(a, c),\\\\ &\quad\ (b, a),(b, b),(b, c),\\\\ &\quad\ (c, a),(c, b),(c, c)\}\\\\ \therefore\quad n(A \times A)&=9\\\\ \text{Note that}\ & n(A \times A)=[n(A)]^{2} \end{aligned}$ Given that $A=\{a\}$, then $A \times A$= A. $\{a\}$ B. $\{(a, a)\}$ C.$\{a, a\}$ D. $a^{2}$ 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။ Given that $f(x)=x^{2}+3 x+1 .$ If $f(a)=\dfrac{31}{4}$ where $a>0$, then what is the value of $a$ ? A. $\dfrac{3}{2}$ or $\dfrac{9}{2}$ B. $-\dfrac{3}{2}$ or $\dfrac{9}{2}$ C.$-\dfrac{9}{2}$ or $\dfrac{3}{2}$ D. $\dfrac{3}{2}$ 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 $a=\dfrac{3}{2} .$ What is the domain of $f(x)=\dfrac{1}{x^{2}-4}$. A. $\mathbb{R} \smallsetminus\{-2,2\}$ B. $\mathbb{R} \smallsetminus\{4\}$ C. $\mathbb{R}$ D. $\varnothing$ Explanation $f(x)$ is not defined when $x^{2}-4=0$ $x^{2}=4$ $x=\pm 2$ $\therefore$ dom $(f)=\mathbb{R} \smallsetminus\{-2,2\}$. 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. $f \ne g$ A. I only B. II only C.I and II only D. I and III only 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 $x \in A$ The graph of the function $y=a x^{2}+b x+c$ when $a=0$ is A. straight line B. parabola C. circle D. ellipse 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. What is the equation of horizontal asymptote of the curve $y=\dfrac{3}{x-1}+2 .$ A. $y=2$ B. $y=3$ C. $x=1$ D. $x=-1$ 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 $y=2$. The furction $f(x)=\dfrac{3}{x-1}+2$ is not defined when A. $y=2$ B. $y=3$ C. $x=1$ D. $x=-1$ 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 $x-1=0 \text { or } x=1$, The vertical asymptote of the graph of function $y=\dfrac{-3 x+4}{x-2}$ is A. $x=2$ B. $x=-2$ C. $x=\dfrac{4}{3}$ D. $y=-3$ 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 $x=2$. Which of the following is one to one? A. $ f(x)=x^{2}$ B. $f(x)=|x|$ C.$f(x)=x^{4}-1$ D. $f(x)=x^{3}+3$ Explanation See: Definition of one to one furction Chapter (4), Section $(4.3 .2)$ If $f^{-1}(x)=\dfrac{x-3}{2}$, then $f(x)=\ldots$ A. $x-3$ B. $2 x-3$ C. $2 x+3$ D. $\dfrac{x-2}{3}$ Explanation $f^{-1}(x)=\dfrac{x-3}{2}$ Let $f(x)=y$, then $f^{-1}(y)=x$ $\dfrac{y-3}{2}=x$ $y=2 x+3$ $\therefore f(x)=2 x+3$ Given that $f(x)=\dfrac{4}{2-3 x}$, then the domain of $f^{-1}$ is A. $\left\{n \mid x \ne \dfrac{3}{2}, x \in \mathbb{R}\right\}$ B. $\left\{x \mid x \ne \dfrac{2}{3}, x \in \mathbb{R}\right\}$ C. $\left\{x \mid x \ne-\dfrac{2}{3}, x \in \mathbb{R}\right\}$ D. $\{x \mid x \neq 0, x \in \mathbb{R}\}$ 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 $x \ne 0 .$ If $f(x)=\dfrac{3 x-1}{2 x+1}$, $f^{-1}(1)=\ldots$ A. $0$ B. $1$ C. $2$ D. $3$ Explanation $f(x)=\dfrac{3 x-1}{2 x+1}$ $f^{-1}(1)=a$ $f(a)=1$ $\dfrac{3 a-1}{2 a+1}=1$ $3a-1=2 a+1$ $a=2$ $\therefore\ f^{-1}(1)=2$ The function $f$ is given by $f(x)=10^{x}-2$, then $f^{-1}(2)=\cdots$ A. $ 88 $ B. $ 0.4 $ C. $\ln 4$ D. $\log _{10} 4$ Explanation $f(x)=10^{x}-2$ Let $f^{-1}(2)=a$ $f(a)=2$ $10^{a}-2=2$ $10^{a}=4$ $a=\log _{10} 4$ $\therefore\ f^{-1}(2)=\log _{10} 4$ Given that $f(x)=x^{2}$, what is the domain of $f$ for which $f^{-1}$ exists? A. $\{x \mid x \neq 0, x \in \mathbb{R}\}$ B. $\{x \mid x>0, x \in \mathbb{R}\}$ C.$\{x \mid x \ge 0, x \in \mathbb{R}\}$ D. $\{x \mid x <0, x \in \mathbb{R}\}$ Explanation $f^{-1}$ exists if and only if $f$ is a one to one function. $f(x)$ is one to one only when $x \ge 0$. $\therefore \operatorname{dom}(f)=\{x \mid x \ge 0, x \in \mathbb{R}\}$. If $f(x)=x^{2}$ and $g(x)=2 x,(f \circ g)\left(-\dfrac{1}{2}\right)=\ldots$ A. $1$ B. $0$ C. $-1$ D. $\dfrac{1}{4}$ Explanation $f(x)=x^{2}$ $g(x)=2 x$ $(f \circ g)\left(-\dfrac{1}{2}\right)$ $=f\left(g\left(-\dfrac{1}{2}\right)\right)$ $=f\left(2\left(-\dfrac{1}{2}\right)\right)$ $=f(-1)$ $=(-1)^{2}$ $=1$ If $f(x)=x^{2}$ and $g(x)=\dfrac{2 x+1}{x-4}$, what is the domain of $g\circ f$ ? A. $\{-2,2\}$ B. $\mathbb{R} \smallsetminus\{4\}$ C. $\mathbb{R} \smallsetminus\{\pm 2\}$ D. $\mathbb{R}$ Explanation $\begin{aligned} f(x)&=x^{2} \\\\ g(x)&=\dfrac{2 x+1}{x-4} \\\\ (g \circ f)(x)&=g(f(x)) \\\\ &=g\left(x^{2}\right)\\\\ &=\dfrac{2 x^{2}+1}{x^{2}-4} \\\\ (g \circ f)(x) &\text { exists when } \\\\ x^{2}-4 &\neq 0 \\\\ x^{2} &\neq 4 \\\\ x &\neq \pm 2 \\\\ \therefore\ \operatorname{dom}(g \circ f)&=\mathbb{R} \smallsetminus\{\pm 2\} \end{aligned}$ If $g(x)=\dfrac{x+2}{2 x-1}$ and $h(x)=2 x$, what is the range of $g \circ h ?$ A. $\mathbb{R}$ B. $\varnothing$ C.$\mathbb{R} \smallsetminus\{2\}$ D. $\mathbb{R} \smallsetminus\left\{\dfrac{1}{2}\right\}$ Explanation $\begin{aligned} g(x)&=\dfrac{x+2}{2 x-1} \\\\ h(x)&=2 x \\\\ (g \circ h)(x) &=g(h(x))\\\\ &=g(2 x) \\\\ &=\dfrac{2 x+2}{4 x-1} \\\\ &=\dfrac{5 / 2}{4 x-1}+\dfrac{1}{2} \\\\ \therefore \operatorname{ran}(g \cdot f)&=\left\{y \mid y \neq \dfrac{1}{2}, y \in \mathbb{R}\right\} \end{aligned}$ The function $f: A \rightarrow B$ is onto function then the range of $f$ is A. $A$ B. $B$ C. subset of $A$ D. subset of $B$ Explanation A function $f$ is onto function when range of $f=$ codomain. 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 $x=\ldots$ A. $5$ B. $3$ C. $2$ D. $1$ Explanation To be one to ove furction $f$, $A$ must be related with $1$. Submit answers Your Score: Email ThisBlogThis!Share to TwitterShare to Facebook Share This: Facebook Twitter Google+ Stumble Digg
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