If $$f(x) = \sqrt {x^{2} - 4}$$, identify its domain.

All x such that $$x \leq -2$$ or $$x \geq 2$$

All x such that $$-2\leq x\leq 2$$

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 5

$p(x) = \dfrac{x}{x-2}$ and

$q(x) = \sqrt{9-x}$ , find the value of

$(p\circ q)(5)$

0%
$0$
0%
$\dfrac{8}{7}$
0%
$2$
0%
Undefined

Explanation

Given,

$p = \dfrac{x}{x-2}$

$q=\sqrt{9-x}$
For

$(p\circ q)$ , substitute the value

$p$ in

$q$
ie.,

$\dfrac{\sqrt{9-x}}{\sqrt{9-x}-2}(5)$ is a undefined.

Consider the functions

$\displaystyle f\left( x \right) =\sqrt { x }$ and

$\displaystyle g\left( x \right) =7x+b$ . Find the value of

$b$ , if the composite function,

$\displaystyle y=f\left( g\left( x \right) \right)$ passes through

$(4, 6)$ .

0%
$8$
0%
$-8$
0%
$-25$
0%
$-26$
0%
$\displaystyle 4-7\sqrt { 6 }$

Explanation

Given,

$f(x)=\sqrt{x}$ and

$g(x)=7x+b$

Therefore,

$f(g(x))=\sqrt{7x+b}$

Now it passes through

$(4,6)$ .

Hence,

$y_{4,6}=\sqrt{7x+b}_{4,6}$

$\Rightarrow 6=\sqrt{28+b}$

$\Rightarrow 36=28+b$

$\Rightarrow b=8$

In real number system, find the domain of the function

$\displaystyle f\left( x \right) =\frac { \sqrt { x-3 } }{ x-3 }$ .

0%
$\displaystyle x\ge 3$
0%
$\displaystyle x>3$
0%
$\displaystyle x<-3$
0%
$x>-3$
0%
$\displaystyle x<-3$ or $\displaystyle x>3$

Explanation

The domain of a function

$f(x)$ is the set of all values for which the function is defined.

A rational function is a function of the form

$f(x)=\dfrac { p(x) }{ q(x) }$ , where

$p(x)$ and

$q(x)$ are polynomials and

$q(x)≠0$ .

The domain of a rational function consists of all the real numbers

$x$ except those for which the denominator is

$0$ .

To find these

$x$ values to be excluded from the domain of a rational function, equate the denominator to zero and solve for

$x$ .

To find the excluded value in the domain of the function

$f(x)=\dfrac { \sqrt { x-3 } }{ x-3 }$ , equate the denominator to zero and solve for

$x$ .

$x-3=0⇒x=3$

So, the domain of the function is

$x>3$ .

$f(x) =$

$\sqrt{x}$ and

$g(x) =$

$\sqrt{x^2+4}$ , calculate the value of

$f(g(2))$ .

0%
$0$
0%
$1.41$
0%
$1.68$
0%
$2.45$
0%
$2.83$

Explanation

Given functions are

$f(x)=\sqrt {x}$ and

$g(x)=\sqrt {x^2+4}$ .

Now

$g(2)=\sqrt{2^2+4}=2\sqrt2$

Then

$f(g(2))=\sqrt { g(2) } = \sqrt { 2\sqrt { 2 } } ={ 2 }^{ 3/4 }=1.681$

$f(x) = 2x$ and

$f(f(x)) = x + 1$ , then the value of

$x$ is

0%
$\dfrac {1}{3}$
0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

Given

$f(x)=2x$ and

$f(f(x))=x+1$

Then

$f(f(x))=x+1$

$\Rightarrow 2[f(x)]=x+1$

$\Rightarrow 2(2x)=x+1$

$\Rightarrow 4x=x+1$

$\Rightarrow x=\dfrac {1}{3}$

$f(x) = \sqrt {x^{2} - 4}$ , identify its domain.

0%
All real numbers
0%
All x such that $x \geq 2$
0%
All x such that $x \leq 2$
0%
All x such that $-2\leq x\leq 2$
0%
All x such that $x \leq -2$ or $x \geq 2$

Explanation

Domain of given function is

${ x }^{ 2 }-4>0$

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

Here both

$x+2$ and

$x-2$ should be greater than zero or both should be less than zero.

$\Rightarrow x>2 , x>-2$ or

$x<2 , x<-2$

$\Rightarrow x>2$ and

$x<-2$

Which of the following is the domain of the function

$f(x) = \dfrac {3 - x}{\sqrt {x^{2} - 9}}$ ?

0%
$-3 > x > 3$
0%
$-x\leq x \leq 3$
0%
$-3\leq x < 3$
0%
$x < -3$ or $x > 3$
0%
$x \leq -3$ or $x \geq 3$

Explanation

For

$f(x)$ to exist

$x^2-9\neq0$ and

$x^2-9\geq0$

Combining we get

$x^2-9>0$

$\Rightarrow |x|>3$

$\Rightarrow -3>x$ or

$x>3$

$f(x) =$

$x^2$ and

$g(x) = 2x$ , calculate the value of

$f(g(-3))-g(f(-3))$ .

0%
54
0%
18
0%
0
0%
-18
0%
-54

Explanation

Given

$f(x)=x^2$ and

$g(x)=2x$ .

Then

$f(-3)={(-3)}^2=9$ and

$g(-3)=2(-3)=-6$

So,

$f(g(-3))={[g(-3)]^2}={[-6]}^2=36$

and

$g(f(-3))=2*f(-3)=2*9=18$

The value of

$f(g(-3))-g(f(-3))$ is

$36-18=18$

Find

$g(x)$ , if

$f(x) = 7x + 12$ and

$f(g(x) = 21x^{2} + 40$

0%
$21x^{2} + 28$
0%
$21x^{2}$
0%
$7x^{2} + 4$
0%
$3x^{2} + 28$
0%
$3x^{2} + 4$

Explanation

Given,

$f(x) = 7x+12, f(g(x))=21x^2+40$

$\therefore f(g(x)) = 7g(x)+12 = 21 {x}^{2} + 40$

$\therefore 7 g(x) = 21 {x}^{2}+28$

$\therefore g(x) = 3 {x}^{2} + 4$

$\left\{ \left( 7,11 \right) ,\left( 5,a \right) \right\}$ represents a constant function, then the value of '

$a$ ' is :

0%
$7$
0%
$11$
0%
$5$
0%
$9$

Explanation

Since we know that in constant function image of every element is same.
Here, image of 7 is 11, so image of any element will be 11.
Therefore, image of 5 is too 11.
Hence,

$a=11$

Given a function

$f(x) = \dfrac {1}{2}x - 4$ and the composite function

$f(g(x)) = g(f(x))$ , determine which among the following can be

$g(x)$ :
I.

$2x - \dfrac {1}{4}$
II.

$2x + 8$
III.

$\dfrac {1}{2}x - 4$

0%
I only
0%
II only
0%
III only
0%
II and III only
0%
I, II, and III

Explanation

$g(x)=2x-\dfrac14$

$g(f(x))=x-8-\dfrac14$

$f(g(x))=x-\dfrac18-4$

Case II

$g(f(x))=2(\dfrac{x}{2}-4)+8=x$

$f(g(x))=\dfrac12(2x+8)-4=x+4-4=x$

Case III

$g(f(x))=\dfrac12(\dfrac{x-8}{2})-4=\dfrac x4-6$

$f(g(x))=\dfrac12(\dfrac{x-8}{2})-4=\dfrac x4-6$

Option D is correct

Find the values of

$x$ for which

$\dfrac {1}{\sqrt {x + 1}}$ is undefined

0%
$-1$ only
0%
$1$ only
0%
All real numbers greater than $-1$
0%
All real numbers less than $-1$
0%
All real numbers less than or equal to $-1$

Explanation

$\dfrac { 1 }{ \sqrt { x+1 } }$ , for it to exist, the expression under square root must be positive.
Therefore for it not to exist,

$x+1\le 0$ , which implies

$x\le -1$ .

Find the value of

$g(f(2))$ , if

$f(x) = e^{x}$ and

$g(x) = \dfrac {x}{2}$

0%
$2.7$
0%
$3.7$
0%
$4.2$
0%
$5.4$
0%
$6.1$

Explanation

Given,

$f(x)=e^x, g(x)=\dfrac {x}{2}$

$\Rightarrow f(2) = { e }^{ 2 }=7.389$

$\Rightarrow g(f(2)) = g(7.389) =\dfrac { 7.389}{2} = 3.7$

$f(x) = 4x - 3$ and

$g(x) = x - 4$ , determine which of the following composite function has a value of

$-11$ .

0%
$f(g(2))$
0%
$g(f(2))$
0%
$g(f(3))$
0%
$f(g(3))$
0%
$f(g(4))$

Explanation

Given,

$f(x)= 4x-3, g(x)=x-4$

$\Rightarrow f(g(2)) = f(-2) = -11$

$\Rightarrow g(f(2))=g(5) = 1$

$\Rightarrow g(f(3)) = g(9) = 5$

$\Rightarrow f(g(3)) = f(-1) = -7$

$\Rightarrow f(g(4)) = f(0) = -3$

$f(x) = x^{2} - 10$ and

$g(x) = 4x + 3$ , calculate the value of

$f(g(2))$ .

0%
$-24$
0%
$-21$
0%
$12$
0%
$27$
0%
$111$

Explanation

Given,

$f(x)=x^2-10$ and

$g(x)=4x+3$

$\therefore g(2) = 4 \times 2 + 3 = 8+3 = 11$
Then,

$f(g(2)) = f(11) = { 11 }^{ 2 }-10=121-10=111$

Find the domain of the function

$f(x) = \sqrt {x^{2} + 3}$ .

0%
$-1.73 \leq x \leq 1.73$
0%
$-1.32 \leq x \leq 1.32$
0%
$x > 1.32$
0%
$x > 1.73$
0%
All real numbers

Explanation

$x^2+3 > 0$
substract

$3$ from both sides.

$x^2 > -3$

which is always true for all real numbers.

The above figure shows the graph of the function

$f(x)$ , the value of

$f(f(3))$ is:

0%
$-4$
0%
$-2$
0%
$0$
0%
$1$
0%
$3$

Explanation

From the graph shown, we need to find

$f(f(3))$

The value of the function at

$x = 3$ is given by

$f(3) = -2$

Next,

$f(f(3)) = f(-2) = 0$

$2$ does not lie in the domain of which of the following function?

0%
$f(x) = \dfrac {x^{2} - 2x}{x^{2} - 2x^{3}}$
0%
$f(x) = \dfrac {x^{-2} - \dfrac {2}{x}}{x^{2} - 2x^{2}}$
0%
$f(x) = \dfrac {2x^{2}}{x^{3} - 2x^{2}}$
0%
$f(x) = \dfrac {x^{2} - 2x}{x^{2} - x}$
0%
$f(x) = \dfrac {4x^{2}}{x^{-2} + 2x}$

Explanation

Function will not be defined if denominator is zero.

In option

$C$

$f\left( x \right) =\dfrac { 2{ x }^{ 2 } }{ { x }^{ 3 }-2{ x }^{ 2 } } \\ f\left( x \right) =\dfrac { 2{ x }^{ 2 } }{ { x }^{ 2 }(x-2) }$

Clearly at

$x=2$ , the denominator is zero, so

$2$ does not lies in its domain.

Option C is correct.

$f(x) = 3x - 5$ and

$g(x) = x^2 + 1, f [g(x)] =$

0%
$3x^2-5$
0%
$3x^2+6$
0%
$x^2-5$
0%
$3x^2-2$
0%
$3x^2+5x-2$

Explanation

Given

$f(x) = 3x-5$ and

$g(x) = {x}^{2}+1$

$\therefore f(g(x)) = f({x}^{2}+1)$

$= 3({x}^{2}+1)-5$

$= 3{x}^{2}-2$

If k is a positive constant different from 1, which of the following could be the graph of

$\displaystyle y-x=k(x+y)$ in the xy-plane?

Explanation

to predict graph of

$y-x=k(x+y)$

$\Rightarrow y-x=k(x+y)$

$\Rightarrow y-x =kx+ky$

$\Rightarrow y=\left(\dfrac{1+k}{\bot-l}\right)x$

we get the equation of form

$y=m$

where

$m=\dfrac{\bot+k}{\bot -k}$

this graph is a line that passes through the origin.

(option B ) is the only graph that fulfil the criteria.

$g(x) = \dfrac {5x - 3}{2x^{2} - 11 - 6}$ , what is the sum of all the real numbers that are not in the domain of

$g(x)$ ?

0%
$-2$
0%
$0.5$
0%
$2$
0%
$5.5$
0%
$6.5$

Explanation

The only cause for

$g(x)$ to become undefined would be its denominator turning 0.

$2x^2-11x-6=0 \Rightarrow 2x(x-6)+1(x-6)=0 \Rightarrow (2x+1)(x-6)=0$

$x=-\dfrac12$ or

$x= 6$

Now

$6-\dfrac12 =5.5$

The Set

$A$ has

$4$ elements and the Set

$B$ has

$5$ elements then the number of injective mappings that can be defined from

$A$ to

$B$ is

0%
$144$
0%
$72$
0%
$60$
0%
$120$

Explanation

$\textbf{Step 1: Number of injective function from set A to B.}$

$\text{Given,}$

$\text{Number of elements in A is 4}$

$\& \text{ Number of elements in B is 5 }$

$\because \text{For injective functions distinct elements has distinct image}$

$\text{First element of A has 5 choices in B}$

$\text{Second element of A has 4 choices in B}$

$\text{Third element of A has 3 choices in B}$

$\text{Last element of A has 2 choices in B}$

$\text{Total Injective functions = }5\times 4\times 3\times 2$

$=120$

$\textbf{Hence, Option D is correct.}$

Let

$f:R\rightarrow R$ be defined by

$f(x)=2x+6$ which is a bijective mapping then

${ f }^{ -1 }(x)\quad$ is given by

0%
$\cfrac { x }{ 2 } -3$
0%
$2x+6$
0%
$x-3$
0%
$6x+2$

Explanation

Let

$y=f(x)=2x+6$

$\Rightarrow 2x=y-6$

$\Rightarrow x =\dfrac{y}{2}-\dfrac{6}{3}$

$\Rightarrow x=\dfrac{y}{2}-3=f^{-1}(y)$

Hence

$f^{-1}(x)=\dfrac{x}{2}-3$

$f(x)=\dfrac{x+1}{x-1}$ and

$g(x)=2x-1, f[g(x)]=$

0%
$\dfrac{x-1}{x}$
0%
$\dfrac{x}{x+1}$
0%
$\dfrac{x+1}{x}$
0%
$\dfrac{x}{x-1}$
0%
$\dfrac{2x-1}{2x+1}$

Explanation

Given,

$f(x) = \dfrac { x+1 }{ x-1 }$ and

$g(x) = 2x-1$

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

$= \dfrac { 2x-1+1 }{ 2x-1-1 }$

$=\dfrac { 2x }{ 2x-2 }$

$=\dfrac { x }{ x-1 }$

The domain of

$g(x)=\cfrac { 3 }{ \sqrt { 4-{ x }^{ 2 } } }$ is:

0%
$\left[ -2,2 \right]$
0%
$\left( -2,2 \right)$
0%
$\left( 0,2 \right)$
0%
$\left( -\infty ,-2 \right)$
0%
$\left( -\infty ,2 \right)$

Explanation

For

$g\left(x\right)$ to be defined, quantity under square-root in denominator should be strictly positive, therefore

$\Rightarrow 4-{x}^{2}>0$

$\Rightarrow x^2-4<0$

$\Rightarrow (x+2)(x-2)<0$

$\Rightarrow -2<x<2$

$\Rightarrow x \in \left( -2,2 \right)$

Let f :

$N \rightarrow N$ defined by

$f(n)=\left\{\begin{matrix} \dfrac{n+1}{2} & \text{if }\, n \, \text{is odd} \\ \dfrac{n}{2} & \text{if}\, n \, \text{is even} \end{matrix}\right.$
then

$f$ is.

0%
Many-one and onto
0%
One-one and not onto
0%
Onto but not one-one
0%
Neither one-one nor onto

Explanation

By substituting values of

$n$ , the elements in the range of function are,

$\{1,2,3,4,5,6,...............\}$ , i.e. set of all natural number

So the range and co-domain are same, ergo function is onto.

Also

$f(1)=f(2)$ , so the function is many one.

Hence the given function is many-one and onto

The number of onto functions from the set

$\{1, 2, .........., 11\}$ to set

$\{1, 2, ....., 10\}$ is

0%
$5\times \underline{|11}$
0%
$\underline{|10}$
0%
$\dfrac{\underline{|11}}{2}$
0%
$10\times \underline{|11}$

Explanation

Finding no. of such functions is same as putting 11 balls in 10 boxes with each box getting at least one ball. Then the only possible way is to put 2 balls in a box which can be done in

$^{11}C_2$ ways.

$^{11}C_{2}\times ^{9}C_{1} \times ^8C_1...................^1C_1$

$=\dfrac{11\times 10}{2}\times 9\times8\times7.................................$

$11!\times 5$

$f: R\rightarrow R$ is defined by

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

$f(f(2))$

0%
$\dfrac {1}{29}$
0%
$\dfrac {10}{29}$
0%
$\dfrac {29}{10}$
0%
$29$

Explanation

We have,

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

So,

$f(2)=\dfrac{2}{2^2+1}=\dfrac{2}{5}$

Hence

$f(f(2))=f\left(\dfrac{2}{5}\right)=\dfrac{\frac{2}{5}}{\left(\frac{2}{5}\right)^2+1}=\dfrac{2\cdot 5}{5^2+2^2}=\dfrac{10}{29}$

The domain of the function

$f(x)=\sqrt{\cos x}$ is.

0%
$\left [ 0, \frac{\pi}{2} \right ]$
0%
$\left [ 0, \frac{\pi}{2} \right ]\cup \left [ \frac{3\pi}{2}, 2\pi \right ]$
0%
$\left [ \frac{3\pi}{2}, 2 \pi \right ]$
0%
$\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$

Explanation

We know that, square root is defined only for non negative number

Hence for

$f(x)=\sqrt{\cos x}$ to be defined

$\Rightarrow \cos\ge 0$

$\Rightarrow x$ should lies in 1st and 4th quadrant including ends of the quadrant

$\Rightarrow x\in [2n\pi+0,2n\pi+\frac{\pi}{2}]\cup[2n\pi+\frac{3\pi}{2},2n\pi+2\pi]$ , where

$n\in Z$ , for

$n = 0$ the option

$B$ is true

Let

$f(x)=2x-\sin x$ and

$g(x)=\sqrt[3] x$ , then

0%
Range of $gof$ is $R$
0%
$gof$ is one-one
0%
both $f$ and $g$ are one-one
0%
both $f$ and $g$ are onto

Explanation

Given :

$f\left( x \right) =2x-\sin { x }$

$g\left( x \right) =3\sqrt { x }$

For all values of

$x$ , in

$f(x) \quad and \quad g(x)$ each element of the domain is mapped to exactly one element of the domain, Then it is known as Bijective Function.

$\Rightarrow f\left( x \right) =2x-\sin { (x) }$

$\Rightarrow f^{ 1 }\left( x \right) =2-\cos { x }$ which is between 1 and 3 (inclusive for all x.Therefor

$f(x)$ is strictly increasing which implies it is bijective (one-one and onto)

$\Rightarrow g\left( x \right) =\sqrt [ 3 ]{ x } ={ (x) }^{ \dfrac { 1 }{ 3 } }$

$\Rightarrow g^{ 1 }\left( x \right) =\dfrac { 1 }{ 3 } { (x) }^{ \dfrac { 1 }{ 3 } -1 }=\dfrac { 1 }{ 3 } { (x) }^{ \dfrac { -2 }{ 3 } }$

Since for every

$x$ there is distinct values of

$g(x)$ which implies

$g(x)$ is bijective (one - one and onto)

$gof=g(x)of(x)\\$

For

$f=\sqrt [ 3 ]{ x }$ substitute x with

$g\left( x \right) =2x-\sin { x }$

$gof=\sqrt [ 3 ]{ 2x-\sin { (x) } }$

Since

$sinx$ function range is [-1,+1]

$\therefore \quad 2x>\sin { x }$

$\Rightarrow (2x-\sin { x } )>0$

$gof=\sqrt [ 3 ]{ 2x-\sin { (x) } }$ will be in the range of real numbers (R) and one-one function ,since both

$g(x) \quad and \quad f(x)$ are one-one function.

Therefore option A,B,C,D are correct.

For the function

$f(x) = \left [\dfrac {1}{[x]}\right ]$ , where

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

$x$ , which of the following statements are true?

0%
The domain is $(-\infty, \infty)$
0%
The domain is $\left \{0\right \} \cup \left \{-1\right \} \cup \left \{1\right \}$
0%
The domain is $\left (-\infty, 0\right ) \cup [1, \infty)$
0%
The domain is $\left \{0\right \}\cup \left \{1\right \}$

Explanation

$f\left( x \right) =\left[ \dfrac { 1 }{ \left[ x \right] } \right] ,\left[ x \right]$ denotes gretaest integer.

Greatest integer function defined by,

$f\left( x \right)$

$=\left[ x \right]$

$=\left\{ -1\quad if\quad -1\le x<0\\ 0\quad if\quad 0\le x<1\\ 1\quad if\quad 1\le x<2 \right\\$

so the Domain of

$f\left( x \right)$ =

$(-\infty ,0) \cup [1,\infty)$

If the function

$f : R \rightarrow R$ is defined by

$f(x) = (x^2+1)^{35} \forall \in R$ , then

$f$ is

0%
One-one but not onto
0%
Onto but not one-one
0%
Neither one-one nor onto
0%
Both one-one and onto

Explanation

$f(x)=(x^2+1)^{35}$

$f'(x)=35(x^2+1)^{34}(2x)=70(x^2+1)^{34}x$

So for

$x<0,$

$f'(x)<0$ and for

$x>0, f'(x)>0$

That means the function is increasing for

$x>0$ and decreasing for

$x>0$

So the function is many one,

Also the range of

$f(x)$ is

$[1,\infty)$ which is not same as codomain

Hence

$f$ is into function

Therefore option

$C$ is correct choice

Let

$f(x) = 2^{100}x+1$

$g(x) = 3^{100}x+1$
Then the set of real numbers x such that

$f(g(x)) = x$ is

0%
Empty
0%
A singleton
0%
A finite se with more than one element
0%
Infinite

Explanation

$f(g(x)) =2^{100}(3^{100}x+1)+1=x$

$\Rightarrow x(6^{100}-1)=-(1+2^{100})$

Clearly there is just one of value of x satisfying this equation. Option B is correct.

The number of real linear functions

$f(x)$ satisfying

$f(f(x))=x+f(x)$ is

0%
$0$
0%
$4$
0%
$5$
0%
$2$

Explanation

Since

$f(x)$ is a linear function, so take

$f(x)=ax+b$

Given

$f(f(x))=x+f(x)$

$\Rightarrow$

$a(ax+b)+b=x+(ax+b)$

Comparing the coefficients we get,

${ a }^{ 2 }=a+1$ and

$ab+b=b$

$a^{ 2 }-a-1=0$ and

$ab=0\Rightarrow a\ne 0,b=0$

$\therefore a=\cfrac { 1\pm \sqrt { 5 } }{ 2 }$ and

$b=0$
Thus

$f(x)=\left( \cfrac { 1\pm \sqrt { 5 } }{ 2 } \right) x+0$

So there are two possible such functions exists

Let

$f : R\rightarrow R$ be defined by

$f(x) = \dfrac {1}{x} \ \ \forall \ x \ \in \ R$ , then

$f$ is _____

0%
One-one
0%
Onto
0%
Bijective
0%
$f$ is not defined

Explanation

The function

$f(x)=\dfrac{1}{x}$ is not defined for

$x=0$

But in the question, domain is given

$R$

Therefore given function is not defined

Consider the function

$f(x)=\displaystyle\frac{x-1}{x+1}$ . What is

$f(f(x))$ equal to?

0%
$x$
0%
$-x$
0%
$-\displaystyle\frac{1}{x}$
0%
None of the above

Explanation

$f(x)=\dfrac{x-1}{x+1}$

$f(f(x))=\dfrac{\dfrac{x-1}{x+1} - 1}{\dfrac{x-1}{x+1}+1}$

$f(f(x)) = \dfrac{-1}{x}$

hence

$Ans$ is

$option C$

Let N denote the set of all non-negative integers and Z denote the set of all integers. The function

$f : Z \rightarrow N$ given by f(x) = |x| is :

0%
One-one but not onto
0%
Onto but not one-one
0%
Both one-one and onto
0%
Neither one-one nor onto

Explanation

$f: X\rightarrow Y$

For a function to be one one or injective, every element in the domain is the image of at most one element of it's co-domain.

In simple words, no value of

$y$ must be same for

$2$ or more different values of

$x$ .

For

$f(x)=\left| x \right|$ , we see that

$f(a)=f(-a)$ , for

$a\in Z$

Hence, the function is not one one

For a function

$f: X\rightarrow Y$ , to be surjective,

every element

$y$ in the co-domain Y must be linked with at least one element

$x$ in the domain.

Every element in the co-domain of

$f(x)=\left| x \right|$ is linked to at-least one element in domain.

Thus,

$f(x)=\left| x \right|$ is onto but not one one.

Hence, b is correct.

The curve

$a^{2}y^{2} = x^{2}(a^{2} - x^{2})$ is defined for

0%
$x\leq a$ and $x\geq -a$
0%
$x < a$ and $x > -a$
0%
$x \leq -a$ and $x\geq a$
0%
$x\leq a$ and $x > -a$

Explanation

Since we know that, square of any real number can never be negative,

And LHS of the given curve

$a^2y^2=x^2(a^2-x^2)$ is non negative

So RHS must also be non-negative

$\Rightarrow x^2(a^2-x^2)\ge 0$

$\Rightarrow a^2-x^2\ge 0$ , since

$x^2\ge 0\forall x\in R$

$\Rightarrow x^2-a^2\le 0$

$\Rightarrow (x+a)(x-a)\le 0$

$\therefore x\le a$ and

$x\ge -a$

$f(x) = \log_{e}\left (\dfrac {1 + x}{1 - x}\right ), g(x) = \dfrac {3x + x^{3}}{1 + 3x^{2}}$ and

$go f(t) = g(f(t))$ , then what is

$go f\left (\dfrac {e - 1}{e + 1}\right )$ equal to?

0%
$2$
0%
$1$
0%
$0$
0%
$\dfrac {1}{2}$

Explanation

$f\left(\dfrac{e-1}{e+1}\right)=\log_e\left\lbrace\dfrac{1+\left(\dfrac{e-1}{e+1}\right)}{1-\left(\dfrac{e-1}{e+1}\right)}\right\rbrace=\log_e\left(\dfrac{2e}{2}\right)=\log_ee=1$

$\therefore gof\left(\dfrac{e-1}{e+1}\right)=g\left(f\left(\dfrac{e-1}{e+1}\right)\right)=g(1)=\dfrac{3+1}{1+3}=1$

$g(x)=\dfrac{1}{f(x)}$ and

$f(x)=x, x\ne 0,$ then which one of the following is correct?

0%
$f(f(f(g(g(f(x))))))=g(g(f(g(f(x)))))$
0%
$f(g(f(g(g(f(g(x)))))))=g(g(f(g(f(x)))))$
0%
$f(g(f(g(g(f(g(x)))))))=f(g(f(g(f(x)))))$
0%
$f(f(f(g(g(f(x))))))=f(f(f(g(f(x)))))$

Explanation

From the given data, it is evident that

$gof(x)=\dfrac{1}{x}=fog(x)$

$f(f(f(g(g(f(x))))))=x$ and

$g(g(f(g(f(x)))))=\dfrac{1}{x}$

$f(g(f(g(g(f(g(x)))))))=x$ and

$g(g(f(g(f(x)))))=\dfrac{1}{x}$

$f(g(f(g(g(f(g(x)))))))=x$ and

$f(g(f(g(f(x)))))=x$

$f(f(f(g(g(f(x))))))=x$ and

$f(f(f(g(f(x)))))=\dfrac{1}{x}$

The domain of the function

$f(x) = \dfrac {1}{\log_{10}(1 - x)} + \sqrt {x + 2}$ is

0%
$[-3, -2.5]\cap[ - 2.5, -2]$
0%
$[-2, 0)\cup(0, 1)$
0%
$[0, 1]$
0%
None of the above

Explanation

The given function

$f(x) = \dfrac {1}{\log_{10}(1 - x)} + \sqrt {x + 2}$ is defined if

$x + 2\geq 0$ , i.e.,

$x \geq -2$ or

$-2\leq x$ and

$\log_{10}(1 - x)\neq 0$

$\Rightarrow 1 - x \neq 1 \Rightarrow x \neq 0$

Again,

$1 - x > 0$

Combining all the results for values of

$x$ , we get

$-2\leq x < 0$ and

$0 < x < 1$

i.e.

$x\in [-2,0)$ or

$x\in (0,1)$

Hence,

$[-2,0)\cup (0,1)$ is the domain for

$f(x)$ .

Let

$f(x)=\dfrac{x+1}{x-1}$ for all

$x \neq 1$ .

Let

$f^1(x)=f(x), f^2(x)=f(f(x))$ and generally

$f^n(x)=f(f^{n-1}(x))$ for

$n > 1$
Let

$P= f^1(2)f^2(3)f^3(4)f^4(5)$
Which of the following is a multiple of P ?

0%
$125$
0%
$375$
0%
$250$
0%
$147$

Explanation

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

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

$=\dfrac { 2x }{ 2 } =x$

Now,

$f^{3}(x)=f(f^{2}(x))=f(x)=f^{1}(x)$

This will repeat so,

$f^{ 3 }(x)=f^{ 1 }(x)$

$f^{ 4 }(x)=f^{ 2 }(x)$

We have,

$f^{ 1 }(2)=3,$

$f^{ 2 }(3)=3,$

$f^{ 3 }(4)=\dfrac { 5 }{ 3 },$

$f^4(5)=5$

So,

$P= 75$ and

$375$ is the multiple of

$75$ .

Hence, the answer is

$375$ .

Number of values of

$x\in [0, \pi]$ where

$f(x) = [4\sin x-7]$ is non-derivable is
[Note: [k] denotes the greatest integer less than or equal to k.]

0%
7
0%
8
0%
9
0%
10

Explanation

$f\left( x \right) =\left[ 4\sin { x } -7 \right]$

$f(x)$ is derivable, if

$f$ is defined

Now,

$f$ is defined if

$\left[ 4\sin { x } -7 \right]$ is defined

We know that,

$-1\le \sin { x } \le 1$ for

$0\le x\leq 2\pi$

$-4\le 4\sin { x } \le 4$

$-11\le 4\sin { x } -7\le -3$

$\left[ 4\sin { x } -7 \right] =-11,-10,-9,-8,-7,-6,-5,-4$
Total 8 number

Let

$M$ be the set of all

$2 \times 2$ matrices with entries from the set of real numbers

$R$ . Then the function

$f : M \rightarrow R$ defined by

$f\left( A \right) =\left| A \right|$ for every

$A\in M$ , is

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

Explanation

$f : M \rightarrow R$ defined by

$f\left( A \right) =\left| A \right|$ for every

$A\in M$

The function is the determinant of the matrix

We know that, two different matrices can have a same determinant

For example,

$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and

$B=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$

Then

$|A|=1=|B|$ , but

$A\neq B$

So, the function will not be one-one.

Now, determinants can have any real values, so range of the function will be

$R$ .

Thus, the function will be onto.

Hence, option D is correct

Consider the following statements :
Statement 1 : The function

$f:R \rightarrow R$ such that

$f(x)=x^3$ for all

$x\in R$ is one-one.
Statement 2 :

$f(a) = f(b) \Rightarrow a=b$ for all

$a, b \in R$ if the function

$f$ is one-one.
Which one of the following is correct in respect of the above statements?

0%
Both the statements are true and Statement 2 is the correct explanation of Statement 1.
0%
Both the statements are true and Statement 2 is not the correct explanation of Statement 1.
0%
Statement 1 is true but Statement 2 is false.
0%
Statement 1 is true but Statement 2 is true.

Explanation

Solution:

Statement

$1$ is correct, since

$f(x)=x^3$

For any

$x,y\in R$

$f(x)=f(y)$

$\Rightarrow x^{3}=y^3$

$\Rightarrow x=y$

$\therefore\ f$ is one-one

$\forall x\in R.$

$\because f(a)=f(b)\Longrightarrow a^3=b^3\Longrightarrow a=b$

Statement

$2$ is also correct and is the correct explanation of statement

$1.$

Hence, A is the correct option.

$f(x) = 8x^3, g(x) = x^{1/3}$ , then fog (x) is

0%
$8^3x$
0%
$(8x)^{1/3}$
0%
$8x^3$
0%
$8x$

Explanation

$f\left( x \right) =8{ x }^{ 3 }$

$g\left( x \right) ={ x }^{ { 1 }/{ 3 } }$

$fog\left( x \right) =f\left[ g\left( x \right) \right]$

$=f\left[ { x }^{ { 1 }/{ 3 } } \right]$

$=8{ \left[ { x }^{ { 1 }/{ 3 } } \right] }^{ 3 }$

$=8x$

The mid-point of the domain of the function

$f\left( x \right) =\sqrt { 4-\sqrt { 2x+5 } }$ for real

$x$ is

0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$
0%
$-\dfrac{2}{5}$

Explanation

The term inside the square root has to be non-negative.

$\Rightarrow 4 - \sqrt{2x + 5} \geq 0$

$\therefore 4 \geq \sqrt{2x + 5}$

Squaring both sides,

$16 \geq 2x + 5$

$\therefore 2x \leq 11$ or

$x \leq 5.5$

Also, since the term

$2x + 5$ is also inside a square root,

$2x + 5 \geq 0$

i.e.

$x \geq -2.5$

The domain of the function therefore becomes

$[-2.5, 5.5]$

Midpoint of the domain is

$\cfrac{-2.5 + 5.5}{2} = 1.5 = \cfrac{3}{2}$

The domain of the function

$f(x)=\sin ^{ -1 }{ \left\{ \log _{ 2 }{ \left( \cfrac { 1 }{ 2 } { x }^{ 2 } \right) } \right\} }$ is

0%
$\left[ -2,1 \right] \cup \left[ 1,2 \right]$
0%
$(-2,-1] \cup [1,2]$
0%
$\left[ -2,-1 \right] \cup \left[ 1,2 \right]$
0%
$(-2,-1)\cup (1,2)$

Explanation

$f(x)=\sin ^{ -1 }{ \left\{ \log _{ 2 }{ \left( \cfrac { 1 }{ 2 } { x }^{ 2 } \right) } \right\} }$

For

$\sin ^{ -1 }{ y }$ function

$y$ should define as,

$-1 < y < 1$

Using this fact,

$-1 \le \log _{ 2 }{ \left( \cfrac { 1 }{ 2 } { x }^{ 2 } \right) \le 1 }$

Taking antilog with base

$2$ ,

${ 2 }^{ -1 } \le \cfrac { 1 }{ 2 } { x }^{ 2 } \le { 2 }^{ 1 }$

$\Rightarrow \cfrac { 1 }{ 2 } \le \cfrac { 1 }{ 2 } { x }^{ 2 } \le { 2 }$

$\Rightarrow 1 \le { x }^{ 2 } \le 4$

$\Rightarrow 1 \le x^2$ and

$x^2\le4$

$\Rightarrow x^2-1 \ge0$ and

$x^2-4\le 0$

$\Rightarrow (x+1)(x-1)\ge 0$ and

$(x-2)(x+2) \le 0$

$\Rightarrow x\ge1\ and\ x\le-1$ and

$-2\le x \le2$

taking the intersection of the above inequalities

we get

$\therefore$ Domain is

$\left[ -2,-1 \right] \cup \left[ 1,2 \right]$

Let

$f (x) = \sqrt {2 - x - x^2}$ and g(x) = cos x. Which of the following statements are true?
(I) Domain of

$f((g(x))^2) =$ Domain of f(g(x))
(II) Domain of f(g(x)) + g(f(x)) = Domain of g(f(x))
(III) Domain of f(g(x)) = Domain of g(f(x))
(IV) Domain of

$g((f(x))^3) =$ Domain of f(g(x))

0%
Only (I)
0%
Only (I) and (II)
0%
Only (III) and (IV)
0%
Only (I) and (IV)

Explanation

Domain of

$f(g(x))$ is R

$\because f(g(x)) = 2- \cos x - \cos x^2 \geq 0$

$\Rightarrow(\cos x + 2)(\cos x - 1) \leq 0$

$\Rightarrow$

$-2$

$\leq$

$\cos x$

$\leq$

$1$

$\therefore$ x

$\in$ R

Domain of

$g(f(x))$ is [

$-2$ ,

$1$ ]

$\because$ In

$\cos \sqrt{2 - x - x^2}$

$2 - x - x^2$

$\geq$

$0$

Domain of

$f((g(x))^2)$ is R

since

$2$ -

$\cos^2 x$ -

$\cos^4 x$

$\geq$

$0$

(

$\cos^2 x$ +

$2$ )(

$\cos^2 x$ -

$1$ )

$\leq$

$0$

$-1$

$\leq$

$\cos x$

$\leq$

$1$

$\Rightarrow$ x

$\in$ R

Domain of

$g((f(x))^3)$ is same as Domain of

$g(f(x))$

i.e. [

$-2$ ,

$1$ ]

From above discussion statement

$\textrm{I}$ is true and statements

$\textrm{II, III}$ and

$\textrm{IV}$ are false.

$f:R\rightarrow R, g:R \rightarrow R$ be two functions given by

$f(x)=2x-3$ and

$g(x)=x^3+5$ , then

$(fog)^{-1}(x)$ is equal to

0%
$\begin{pmatrix}\dfrac{x+7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$
0%
$\begin{pmatrix}\dfrac{x-7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$
0%
$\begin{pmatrix}x-\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$
0%
$\begin{pmatrix}x+\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$

Explanation

Given,

$f(x)=2x-3$ and

$g(x)=x^3+5$

Now,

$(fog)(x)=f(g(x))=2(x^3+5)-3$

$=2x^3+10-3=2x^3+7$

Let,

$(fog)(x)=y,$ then,

$y=2x^3+7$

$\Longrightarrow x=\left(\cfrac{y-7}{2}\right)^{\cfrac 13}$

$\therefore fog^{-1}(x)=\left(\cfrac{x-7}{2}\right)^{\cfrac 13}$

Hence, B is the correct option.

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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