MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 12

$P(S)$ denotes the set of all subsets of a given set S, then the number of one-to-one functions from the set

$S=\{1,2,3\}$ to he set

$P(S)$

0%
$24$
0%
$8$
0%
$336$
0%
$320$

$\begin{aligned} \text { If } A & = \{ x | x / 2 \in Z , 0 \leq x \leq 10 \} \\ B & = \{ x | x \text { is one digit prime } \} \\ C & = \{ x | x / 3 \in N , x \leq 12 \} \end{aligned}$ ,

Then

$A \cap ( B \cup C )$ is equal to-

0%
$\{ 2,6 \}$
0%
$\{3,6,12 \}$
0%
$\{ 2,6,12 \}$
0%
$\{ 6,8 \}$

The function

$f:N\rightarrow N$ defined by

$f\left( x \right) =x-5\left[ \dfrac { x }{ 5 } \right]$ , where

$N$ is the set of natural numbers and

$[x]$ denotes the greatest integer less then or equal to

$x$ is

0%
One-one and onto
0%
One-one but not onto
0%
Onto but not one-one
0%
Neigher one-one nor onto

Let

$f:R\rightarrow R$ is defined as

$f(x)=\begin{cases} 2x+{ \alpha }^{ 2 },\ x \ge 2 \\ \frac { \alpha x }{ 2 } +10,\ x <2 \end{cases}$ . If

$f(x)$ is onto function then set of values of

$\alpha$ is

0%
$[1,4]$
0%
$[-2,3]$
0%
$(0,3]$
0%
$[2,5]$

The domain of the function

$f ( x ) = \underbrace { \log _ { 2 } \log _ { 2 } \log _ { 2 } \dots \log _ { 2 } x } _ { n times }$ is

0%
$( - \infty , - 2 ) \cup [ 4 , \infty )$
0%
$( - \infty , - 2 ] \cup [ 4 , \infty )$
0%
$( - \infty , - 2 ) \cup ( 4 , \infty ]$
0%
none of these

The domain of definition of the function

$y=\dfrac { 325 }{ 197 } \left[ \dfrac { \sqrt { { x }^{ 2 }-1 } }{ \sqrt { { x-1 } } } \right]$

0%
$\left(1,\infty\right)$
0%
$\left[1,\infty\right)$
0%
Set of all real numbers
0%
$\left(-\infty,-1\right)\cup \left( 1,\infty \right)$

Explanation

$y=\dfrac{325}{197}\left[\dfrac{\sqrt{x^{2}-1}}{\sqrt{x-1}}\right] \\$

$\text { For y to be defined, } \\$

$\sqrt{x-1} \neq 0 \\$

$x \neq 1 \\$

$x \in R-\{1\}-(i) \\$

$x-1>0 \\$

$x>1 \\$

$x \in(1, \infty)-(i i) \\$

$x^{2}-1>0 \\$

$x^{2}>1 \\$

$x \in(-\infty,-1) \cup(1, \infty)-(iii) \\$

All above the conditions must satisfy

$x \in(1, \infty)$

Let

$X=\left\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\right\}$ &

$Y=\{b_{1},b_{2},b_{3}\}$ the number of function

$f$ from

$X$ to

$Y$ such that it is onto and there are exactly three elements

$x$ in

$X$ such that

$f(x)=b_{1}$ is

0%
$75$
0%
$100$
0%
$120$
0%
$90$

The domain of

$f(x)=\log_{x}\log_{2}\left(\dfrac {1}{x-1/2}\right)$ is

0%
$\left(\dfrac {1}{2},1\right) \cup \left(1,\dfrac {3}{2}\right)$
0%
$\left(\dfrac {1}{2},\dfrac {3}{2}\right)$
0%
$\left[\dfrac {1}{2},\dfrac {3}{2}\right]$
0%
$\left(\dfrac {1}{2},\dfrac {3}{2}\right]$

The number of non-bijective mappings that can be defined from

$A={1,2,7}$ to itself is

0%
$21$
0%
$27$
0%
$6$
0%
$9$

The number of linear functions which map from

$[-1, 1]$ onto

$[0, 2]$ is

0%
$0$
0%
$1$
0%
$2$
0%
$infinite$

Let

$f : R \rightarrow (-1,1)$ be defined as

$f(x)=\dfrac {e^{x}-e^{-x}}{e^{x}+e^{-x}}$ then

$f$ is

0%
One-one onto
0%
One-one into
0%
Many-one onto
0%
Many-one-into

Domain of the function

$f ( x ) = \frac { 1 } { \sqrt { 4 x - \left| x ^ { 2 } - 10 x + 9 \right| } }$ is

0%
$( 7 - \sqrt { 40 } , 7 + \sqrt { 40 } )$
0%
$( 0,7 + \sqrt { 40 } )$
0%
$( 7 - \sqrt { 40 } , \infty )$
0%
None of these

$f:Z\rightarrow Z, f(n)=\begin{cases} n+1;\quad n\quad is\quad even \\ n-3;\quad n\quad is\quad odd \end{cases}$ is

$f$ is ...........

0%
only one one
0%
only Onto
0%
one one & Onto both
0%
Neither one one nor Onto

The complete set of values of

$x$ for which the function

$f(x)=2\tan^{-1}x+\sin^{-1} \dfrac{2x}{1+x^{2}}$ behaves like a constant function with positive output is equal to

0%
$x \in [-1,1]$
0%
$[1,\infty)$
0%
$(-\infty,1]$
0%
$(-\infty, -1] \cup [1,\infty)$

The function f:

$R\rightarrow R$ defined by f(x)=x-[x],

$\forall x\epsilon R\quad is$

0%
one-one
0%
onto
0%
Both one-one and onto
0%
neither one-one nor onto

$f:R\rightarrow R\quad defined\quad by\quad f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,\quad then\quad f\quad is$

0%
one-one but not onto
0%
not one-one but onto
0%
one-one and onto
0%
neither one-one noronto

$f (x) = x^4 - 10x^3 + 35x^2 - 50x + c$ is a constant. the number of real roots of . f (x) = 0 and
f'' (x) = 0 are respectively

0%
1 , 0
0%
3, 2
0%
1 , 2
0%
3 , 0

Explanation

Given,

$f(x)=x^{4}-10 x^{3}+35 x^{2}-50 x+c$

Then

$, f^{\prime}(x)=\dfrac{d}{d x}\left(-10 x^{3}+x^{4}+35 x^{2}-50 x+c\right)$

$\Rightarrow f^{\prime}(x)=4 x^{3}-30 x^{2}+70 x-50=0$

$\therefore 2 x^{3}-15 x^{2}+35 x-25=0$

First, let us factorise

$f(x)=0$

$\Rightarrow x^{4}-10 x^{3}+35 x^{2}-50 x+c=0$

$\Rightarrow(x-1)\left(x^{3}-9 x^{2}+26 x-c\right)=0$

$\Rightarrow(x-1)\left((x-2)\left(x^{2}-7 x+\dfrac{c}{2}\right)\right)=0$

$\Rightarrow f(x)=(x-1)(x-2)(x-4)(x-3)$

Roots of

$f^{\prime}(x)=0$

$\Rightarrow 2 x^{3}-15 x^{2}+35 x-25=0$

$\Rightarrow f^{\prime}(x)$ has 3 roots which are

$x=\dfrac{5}{2}, \dfrac{5+\sqrt{5}}{2}, \dfrac{-\sqrt{5}+5}{2}$

Hence

$(A)$ is the correct option.

$y^2 = ax^2 +bx+c$ , then

$y^2 \dfrac{d^2y}{dx^2}$ is

0%
a constant function
0%
a function of x only
0%
a function of y only
0%
a function of both x and y

Explanation

$y^{2}=a x^{2}+b x+c \quad$ is

Differentiating this equ. (i),

$2 y \dfrac{d y}{d x}=2 a x+6$

Differentiating above equation again,

$2 y \dfrac{d^{2} y}{d x^{2}}+2\left(\dfrac{d y}{d x}\right)\left(\dfrac{d y}{d x}\right)=2 a$

$\dfrac{d^{2} y}{d x^{2}}=\dfrac{2 a-2\left(\dfrac{d y}{d x}\right)^{2}}{2 y}=\dfrac{a-\left(\frac{d y}{d x}\right)^{2}}{y}$

$y^{2} \dfrac{\alpha^{2} y}{d x^{2}}=y^{2}\left(\dfrac{a-\left(\dfrac{d y}{d x}\right)^{2}}{y}\right)$

$=y\left(a-\left(\dfrac{d y}{d x}\right)^{2}\right) \quad\left\{\begin{array}{l}\because 2 y \dfrac{d y}{d x}=2 a x+b \\ \frac{d y}{d x}=\dfrac{2 a x+b}{2 y}\end{array}\right\}$

$=y\left(a-\left(\dfrac{2 a x+b}{2 y}\right)^{2}\right)$

$=\dfrac{y\left(4 a y^{2}-\left(4 a^{2} x^{2}+b^{2}+4 a b x\right)\right.}{4 y^{2}}$

$=\dfrac{4 a y^{2}-4 a^{2} x^{2}-b^{2}-4 a b x}{4 y}$

$=\dfrac{4 a\left(a x^{2}+6 x+c\right)-4 a^{2} x^{2}-6^{2}-4 a b x}{4 y}$

$=\dfrac{4 a^{2} x^{2}+4 a b x+4 a c-4 a^{2} x^{2}-b^{2}-4 a b x}{4 y}$

$=\dfrac{4 a c-b^{2}}{4 y}$

$\therefore$ It is function of

$y$ only.

option (C) is correct.

The domain of the real valued function

$f(x)$ for which

$4^{f(x)+4^{1-f(x)}}= 4^{x}$ is

0%
$(-1,1)$
0%
$(-\infty ,1)$
0%
$[1,\infty)$
0%
R

Explanation

Given,

$4^{f(x)}+4^{(1-f(x)}=4^{x}$

$\Rightarrow \quad 4^{f(x)}+\frac{4}{4^{f(x)}}=4^{x}$

$\Rightarrow \frac{4^{f(x)} \cdot 4^{f(x)}+4}{4^{f(x)}}=4^{x}$

$\Rightarrow y^{2 f(x)}+4=4^{x+f(x)}$

Dividing above equation by 4 on both sides

$\begin{aligned} & \frac{4^{2 f(x)}}{4}+1=\frac{y^{x+f(x)}}{4} \\ \Rightarrow & 4^{2 f(x)-1}+1=4^{x+f(x)-1}-(i) \\ &\left.1+4^{-(1-2 f(x)}=4^{x-f(x)}-(ii\right)\end{aligned}$

$\text {both equation are same, } \\$

$\text { Then, }$

$4^{2 f(x)-1} =4^{1-2 f(x)} \\$

$2 f(x)-1 =1-2 f(x) \\$

$f(x)+3 f(x) =2 \\$

$\Rightarrow 4 f(x)=2 \\$

$7 f(x)=\frac{1}{2}$

$\quad 4^{x+f(x)-1}=4^{x-f(x)}$

$\Rightarrow x+f(x)-1=x-f(x)$

$\Rightarrow \quad 2 f(x)=1$

$\Rightarrow \quad f(x)=\frac{1}{2}$

$Therefore = f(x)$ is a constant function.

but from above equation

$4^{x}=4^{f(x)}+4^{1-f(x)}$

$\text { = Putting the value of } f(x)=\frac{1}{2} \\$

$\text { in above equation. } \\$

$\Rightarrow \quad 4^{x}=4^{\frac{1}{2}}+4^{1-\frac{1}{2}} \\$

$\Rightarrow \quad 4^{x}=4$

$x=1$

$x$ has one value in i.e.

$x=1$

So,

$x \in[1, \infty)$

Option C

The function

$f:R\rightarrow R$ defined by

$f\left(x\right)=6^ {x}+6$ is

0%
one-one and onto
0%
many-one and onto
0%
one-one and into
0%
mauny-one and into

Explanation

$f: R \rightarrow R \\$

$f(x)=6^{x}+6$

$\text { For one - one, } \\$

$\qquad f\left(x_{1}\right)=f\left(x_{2}\right) \\$

$\Rightarrow 6^{x_{1}}+6$ &=

$6^{x_{2}}+$

$\Rightarrow \quad 6^{x_{1}}=6^{x_{2}} \\$

$\Rightarrow \quad x_{1}=x_{2}$

Therefore f is one - one.

Now,

$\begin{aligned} f(x) &=y=6^{x}+6 \\ & \Rightarrow(y-6)=6^{x} \end{aligned}$

Taking log both side.

$\Rightarrow \log _{6}(y-6)=x$

Now,

$f\left(\log _{6}(y-6)\right)=6^{\log _{6}(y-6)}+6$

$=y-6+6$

$=y$

$\therefore f \text { is onto }$

Hence, f is one - one and onto.

Let n(A) = 4 and n(B) =Then the number of one - one functions from A to B is

0%
120
0%
360
0%
24
0%
none of these

Explanation

$n(A)=4$

$n(B)=6$

$\text { Number of one-one functions }(A \rightarrow B)$

$=6 c_{4} \times 4 !$

$=\frac{6 !}{2 ! 4 !} \times 4 !$

$=6 \times 5 \times 4 \times 3$

$=360$

$f(x)$ is

0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
Many-one and into

The domain of the function

$f(x) = \log_{10}[1-\log_{10}(x^{2}-5x+16)]$ is

0%
$(2,3)$
0%
$[2,3]$
0%
$(2,3]$
0%
$[2,3)$

The domain of the definition of the function

$y\left(x\right)$ given by the equation

$2^ {x}+2^ {y}$ is

0%
$0 < x \le 1$
0%
$0\le x\le 1$
0%
$-\infty < x \le 0$
0%
$-\infty < x < 1$

$fxln\left(1+\dfrac{1}{x}\right)dx=p(x)ln\left(1+\dfrac{1}{x}\right)+\dfrac{1}{2}x-\dfrac{1}{2}ln(1+x)+c$ , being arbitary costant, then

0%
$p(X)=\dfrac{1}{2}x^{2}$
0%
$p(x)=0$
0%
$p(x)=1$
0%
$none\ of\ these$

The domain of

$\frac{x+1}{\sqrt{x^{2}-5x+6}}$ is

0%
R-{2,3}
0%
$(3,\infty )$
0%
$(-\infty ,\infty )$
0%
$(-\infty, 2)\cup (3,\infty )$

Explanation

$\dfrac{x+1}{\sqrt{x^{2}-5 x+6}}$

$x^{2}-5 x+6>0$

$x^{2}-3 x-2 x+6>0$

$x(x-3)-2(x-3)>0$

$(x-3)(x-2)>0$

$x \in[-\infty, 2) \cup(3, \infty)$

Domain of

$x+1 \in R$

$\therefore$ Domain of $\dfrac{x+1}{\sqrt{x^{2}-5 x+6}}$ is $R-(2,3)$

Option $A=R-(2,3)$

Consider the function

$f\left( x \right) ={ e }^{ x }$ and

$g\left( x \right)=\sin ^{ -1 }{ x }$ , then which of the following is/are necessarily true.

0%
Domain of $gof =$ Domain of $f$
0%
Range of $gof\ \subset$ Range of $g$
0%
Domain of $gof$ is $\left( -\infty ,0 \right)$
0%
Range of $gof$ is $\left( -\dfrac{\pi}{2} ,0 \right)$

Explanation

$\text { solution: } \quad f(x)=e^{x} \quad g(x)=\sin ^{-1}(x) \\$

$\text { gof }=\sin ^{-1}\left(e^{x}\right) \\$

$\therefore-1 \leqslant e^{x} \leqslant 1 \\$

$\text { so we need } x \in(-\infty \text { o] } \\$

$\text { domain of gof is }(-\infty, 0] \\$

$\text { Range of gof is same as range of } g \\$

$\text { Ansuer. : option (B) }$

$f(x)=\frac { \alpha x }{ x+1 }$ , where

$x\neq -1$ and (fof) (x) = x, then

$\alpha =$

0%
$\sqrt { 2 }$
0%
$-\sqrt { 2 }$
0%
$1$
0%
$-1$

Let N be the set of natural numbers and two functions f and g be defined as
and g(n)=

$n-{ \left( -1 \right) }^{ n }$ then fog is:

0%
one-one but not onto
0%
onto but not one-one
0%
neither one-one nor into
0%
both one-one and onto

let

$f:R\rightarrow R$ be a function defined by

$f(x)=\frac { { x }^{ 2 }-3x+4 }{ { x }^{ 2 }+3x+4 }$ then f is

0%
one-one but not onto
0%
onto but not one
0%
onto as well as one-one
0%
neither onto nor one-one

Let

$f:R\rightarrow R$ defined by

$f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ -x^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,$ then

0%
f(x) is one-one but not onto
0%
f(x) is neither one-one nor onto
0%
f(x) is many one but onto
0%
f(x) is one-one and onto

Explanation

$\text { solution: } \quad f(x)=\frac{e^{x^{2}}-e^{-x^{2}}}{e^{x^{2}}+e^{-x^{2}}}=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1} \\$

$f^{\prime}(x)=\frac{\left\{\left(e^{x^{2}}\right)^{2}+1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}-\left\{\left(e^{x^{2}}\right)^{2}-1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}}{\left(\left(e^{x^{2}}\right)^{2}+1\right)^{2}} \\$

$\text { Since } f^{\prime}(x) \text { is nither positive, nor negative } \\$

$\text { for all } x \text { so } f(x) \text { is not one - one function } \\$

$\qquad f(x)=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1}$

Range of

$f(x)$ is

$[0,1 )$

since Range of

$f(x)$ is not equal to co-domain

$f(x)$ so

$f(x)$ is not onto function.

Answer: option: (B)

The function

$f:R\rightarrow R$ defined by

$f\left( x \right) =\frac { { e }^{ \left| x \right| }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } }$ is

0%
One-One and onto
0%
One-one but not onto
0%
Not one-one but onto
0%
Neither one-one nor onto

$f:N\rightarrow N\quad where\quad f\left( x \right) =x-{ (-1) }^{ x }$ , then 'f' is

0%
one-one and into
0%
many- one and into
0%
one-one and onto
0%
many-one and onto

Let f(x+y)=f f(x) f(y) and f(x) =1+x g(x) G(x), where

$\underset { x\rightarrow 0 }{ lim } g\left( x \right) =a and \underset { x\rightarrow 0 }{ lim } G\left( x \right) =b,$ then f' (x) is equal to

0%
1+ab
0%
ab
0%
f(x)
0%
ab f(x)

Let (X) be a function satisfying f' (X) = f (X) with f (0) = 1 and g (X) be a function that satisfies f (X) + g (x) =

${ x }^{ 2 },$ Then the value of the integral

$\int _{ 0 }^{ 1 }{ f } (x)\quad g\quad (x)\quad dx,\quad is$

0%
$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 5 }{ 2 }$
0%
$e+\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 }$
0%
$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 }$
0%
$e+\frac { { e }^{ 2 } }{ 2 } +\frac { 5 }{ 2 }$

l qt f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g be the function satisfying f(x)+g(x)=

$X^{2}$ , the value of the integral

$\int _{ 0 }^{ 1 }{ f(x)g(x)\quad dx\quad is }$

0%
$\frac { 1 }{ 4 } (e-7)$
0%
$\frac { 1 }{ 4 } (e-2)$
0%
$\frac { 1 }{ 4 } (e-3)$
0%
none of the above

$f:A\rightarrow B$ will be an into function if

0%
$f\left(A\right) \subset B$
0%
$f\left(A\right)=B$
0%
$B\subset f\left(A\right)$
0%
$f\left(B\right) \subset A$

If f (x) = cosx and g (x) = x

$^2$ then (gof) (x) is ....

0%
cos $^2$ x
0%
cosx $^2$
0%
both (a)&(b)
0%
x $^2$ cosx

Suppose that

$g(x)=1+\sqrt { x } and\quad f(g(x))=3+2\sqrt { x } +x\quad then\quad f(x)\quad is$

0%
$\quad 1+2{ x }^{ 2 }$
0%
$\quad 2+{ x }^{ 2 }$
0%
$1+x$
0%
$2+x$

Which one of the following is one-one?

0%
$f:R\rightarrow R$ given by $f(x) =\left| x-1 \right|$ for all $x\in R$
0%
$g:\left[ -\pi /2,\pi /2 \right] \rightarrow given by$ $g(x)=\left| sinx \right|$
0%
$h:\left[ -\pi /2,\pi /2 \right] \in R$ given by $h(x) =sin x$ for all $x\in \left[ -\pi /2,\pi /2 \right]$
0%
$\phi :R\rightarrow R$ given by $f(x)={ x }^{ 2 }-4$ for all $x\in R$

Let

$f:R\rightarrow R$ , be defined as

$f(x)={ e }^{ x^{ 2 } }+cosx$ then f is

0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
many-one and into

$f : R \rightarrow R$ defined by

$f ( x ) = \dfrac { x } { x ^ { 2 } + 1 } , \forall x \in R$ is

0%
one-one
0%
onto
0%
bijective
0%
neither one one nor onto

Explanation

$\begin{aligned} f(x) &=\dfrac{x}{1+x^{2}} \\-f(4)=& \dfrac{4}{1+16}=\dfrac{4}{17} \\ f\left(\dfrac{1}{4}\right) &=\frac{\dfrac{1}{4}}{1+\dfrac{1}{16}}=\dfrac{4}{17} \end{aligned}$

Hence two different input has Same output

$\cdot$ so not one-one

$f(x)=y$

$y=\dfrac{x}{1+x^{2}}=1 \quad y+y x^{2}-x=0$

$x=\dfrac{1 \pm \sqrt{1-4 y^{2}}}{2 y}$

$1-4 y^{2} \geq 0$

$(1+2 y)(1-x y) \geq 0$

$\dfrac{1}{-3} \leq y<\dfrac{1}{2}$

Range of

$f(x)$ is

$\left[\dfrac{-1}{2}, \dfrac{1}{2}\right]$

Range of

$f(x) \neq$ codomain

$f(x)$ Hence f(x) is neither one-one nor onto option D is correct.

State which of the following defines a mapping from A to B, if

$A={a,b,c,}$ and

$B={x,y,z}.$

0%
0%
0%
0%
None of these.

Choose correct answer (s) from given choice
If f(x) = x + 4, g (x) = 5x and h(x) = 12/x. Find the value of

${ f }^{ -1 }(g(h(6)))$

0%
10
0%
14
0%
6
0%
0

$f ( x ) = \sqrt { x ^ { 2 } + 1 } , g ( x ) = \dfrac { x + 1 } { x ^ { 2 } + 1 }$ and

$h ( x ) = 2 x - 3 ,$ then

$f ^ { \prime } \left( h ^ { \prime } \left( g ^ { \prime } ( x ) \right) =\right.$

0%
$0$
0%
$\dfrac { 1 } { \sqrt { x ^ { 2 } + 1 } }$
0%
$\dfrac { 2 } { \sqrt { 5 } }$
0%
$\dfrac { x } { \sqrt { x ^ { 2 } + 1 } }$

${ f }:{ R }\rightarrow { R }$ where

$f ( x ) = \dfrac { x ^ { 2 } + a x + 1 } { x ^ { 2 } + x + 1 }.$ Complete set of values of

$'a'$ such that

$f ( x )$ is onto to is :

0%
$( - \infty , \infty )$
0%
$( - \infty , 0 )$
0%
$( 0 , \infty )$
0%
None

Explanation

$f(x)=\dfrac{x^{2}+a x+1}{x^{2}+x+1}$

$f: R \rightarrow R$

domain co-domain

$D\left(x^{2}+x+1\right)=-3<0$

Hence

$x^{2}+x+1$ is always positive

$x^{2}+a x+1$ coefficient of

$x^{2}=1>0$

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow \infty} \dfrac{x^{2}+a x+1}{x^{2}+x+1} \dfrac{\infty}{\infty}$ form

$=\lim _{x \rightarrow 0} \dfrac{x^{2}\left(1+\dfrac{a}{x}+\dfrac{1}{x^{2}}\right)}{x^{2}\left(1+\frac{1}{x}+\dfrac{1}{x^{2}}\right)}=\dfrac{1+0+0}{1+0+0}=1$

similarly

$lim_{x\rightarrow-\infty} f(x)=1$

Hence

$f(x)$ is bounded

Range

$\neq R$

$\neq$ co-domain

$f(x)$ cannot be onto for any value of

${a}$

option

$D$ is correct.

The domain of

$f(x) = \sin^{-1} log_2 (x^2/2)$ is

0%
$(0, \infty)$
0%
$(0, \surd{2})$
0%
$(-1, 0) \cup (0, 1)$
0%
$[-2, -1] \cup [1, 2]$

The domain of real valued function

$f(x)$ for which

$4^ {f(x)}+4^ {1-f(x)}=4^ {x}$ is

0%
$(-1,1)$
0%
$(1,\infty)$
0%
$(\infty ,1)$
0%
$(-\infty ,-1)$

Which of the following function(s) have the same domain and range?

0%
$f\left( x \right) =\sqrt { 1-{ x }^{ 2 } }$
0%
$g\left( x \right) =\dfrac { 1 }{ x }$
0%
$h\left( x \right) =\sqrt { x }$
0%
$l\left( x \right) =\sqrt { 4-x }$

Domain of the function

$y=\sqrt{\dfrac{x-2}{x+2}}+\sqrt{\dfrac{1-x}{1+x}}$ is

0%
$(-\infty, 0)$
0%
$R$
0%
$(-\infty, 0]$
0%
$\phi$

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 12 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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