If $$f:A\rightarrow B $$ is surjective then

no two elements of $$A$$ have the same image in $$B$$

every element of $$A$$ has an image in $$B$$

$$A$$ and $$B$$ are finite non empty sets

MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 1

The domain of

$f(x)=\frac{1}{\sqrt{|x|-x}}$ is

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

A constant function

$f:A\rightarrow B$ will be onto if

0%
$n(A) = n(B)$
0%
$n (A) =1$
0%
$n (B) = 1$
0%
$n(A) > n(B)$

Explanation

Given the

$f:A\rightarrow$ B is onto & is constant.
Co domain=Range and so it should contain only one element.

$\therefore n(B)=1$

The number of non-bijective mappings possible from

$A= \{1,2,3\}$ to

$B=\{4, 5\}$ is

0%
$9$
0%
$8$
0%
$12$
0%
$6$

Explanation

From

$A\rightarrow B$ we cannot form any bijective functions because

$n(a)\neq n(b)$
So, total no of non bijective functions possible =

$n(b)^{n(a)}=2^3=8$ (nothing but total no functions possible)

A constant function

$f:A\rightarrow B$ will be one-one if

0%
$n (A) = n(B)$
0%
$n(A) = 1$
0%
$n (B) = 1$
0%
$n (A) < n (B)$

Explanation

Given f is a constant functions.

$\Rightarrow$ range of f is

$\left \{ c \right \}(say)$
Since f is one-one

$\Rightarrow$ domain of A should also contain
one element.

$\therefore n(A)=1.$

$f:\mathbb{N} \rightarrow \mathbb{N}$ and

$f(x) = x^{2}$ then the function is

0%
not one to one function
0%
one to one function
0%
into function
0%
none of these

Explanation

Given :

$f(x)=x^2$

$\because x^2>0 \implies f(x)>0$ for every

$x\in \mathbb{N}$

Let' find domain and range of

$f(x)$

$x =1\longrightarrow f(x)=1$

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

$x =3\longrightarrow f(x)=9$

$x=4\longrightarrow f(x)=16$

$x =5\longrightarrow f(x)=25$

Here we get that, no two elements of the domain has the same image and no element of co-domain is the image of more than one element in the domain.

$\therefore$

$f$ is one-one.

Function

$f:X\rightarrow Y$ is onto if, for any

$y\in {Y}$ there exist

$x\in {X}$ such that

$f(x)=y$

Let's prove that

$f(x)=x^2$ is not onto.

Let's take an example

$y=3\in \mathbb{N}$

$\implies f(x)=y$ ............ by definition of onto function

$\implies x^2=3$

$\implies x=\sqrt{3}\ or\ -\sqrt3$ which does not belong to

$\mathbb{N}$

Hence, for

$y=3\in \mathbb{N}$ there does not exist any

$x\in \mathbb{N}$ such that

$f(x)=y$ .

$\therefore$

$f$ is not onto.

Hence,

$f$ in only one-one.

$f(x)=1$ , if

$x$ is rational and

$f(x)=0$ , if

$x$ is irrational
then

$(fof) (\sqrt{5})=$

0%
$0$
0%
$1$
0%
$\sqrt{5}$
0%
$\dfrac{1}{\sqrt{5}}$

Explanation

Given,

$f(x)=\begin{cases}1 & \text {; if x is rational} \\ 0 & \text{; if x is irrational}\end{cases}$

Consider

$x=\sqrt{5}$ ,

$\text{(f o f)}(\sqrt{5})=f(f(\sqrt{5}))$

$=f(0)$ ..........

$[\because \sqrt{5}$ is irrational

$]$

$=1$ ..........

$[\because 0$ is rational

$]$

$\therefore \text{(f o f)}(\sqrt{5})=1$
Hence, option B is correct.

The domain of the function, f(x) =

$\sqrt {x-1}+\sqrt {5-x}$ is

0%
$[1, \infty )$
0%
$(\infty , 5)$
0%
(1, 5)
0%
[1, 5]

Explanation

Given,

$f(x) = \sqrt {x-1}+\sqrt {5-x}$

For given function to be defined,

$x-1\geq 0$ and

$5-x\geq 0$

$\Rightarrow x\geq 1$ and

$x\leq 5$

$\therefore x \in [1,5]$

$f(x) = 3x + 2, g(x) = x^2 + 1$ , then the value of

$(fog) (x^2 +1)$ is

0%
$3x^4 + 6x^2 + 8$
0%
$3x^4 + 3x + 4$
0%
$6x^4 + 3x^2 + 2$
0%
$3x^2 + 6x + 2$

Explanation

We have

$f(x) = 3x + 2, g(x) = x^2 + 1$

$fog(x) = f[g(x)] =3(x^2+1)+2= 3x^2 + 5$
so

$(fog) (x^2 +1) = 3(x^2 + 1)^2 + 5 =3(x^4+2x^2+1)+5= 3x^4+ 6x^2 + 8$

$f:A\rightarrow B$ is surjective then

0%
no two elements of $A$ have the same image in $B$
0%
every element of $A$ has an image in $B$
0%
every element of $B$ has at least one pre-image in $A$
0%
$A$ and $B$ are finite non empty sets

Explanation

Surjective means onto function.
co domain

$=$ Range
So every element of

$B$ has at least one pre-image in

$A$ .

$f:R\rightarrow R , g:R\rightarrow R$ and

$f(x)= \sin x$ ,

$g(x)=x^{2}$ then

$fog(x)=$

0%
$x^{2}+\sin x$
0%
$x^{2}\sin x$
0%
$\sin^{2}x$
0%
$\sin x^{2}$

Explanation

$fog\left ( x \right )=f\left(g ( x \right ))=f (x^{2})$

$=\sin x^{2}$

Find the value of

$\displaystyle \left( g\circ f \right) \left( 6 \right)$ if

$\displaystyle g\left( x \right) ={ x }^{ 2 }+\frac { 5 }{ 2 }$ and

$\displaystyle f\left( x \right) =\frac { x }{ 4 } -1$ .

0%
2.75
0%
3
0%
3.5
0%
8.625

Explanation

$f(x)=\dfrac {x}{4}-1$ and

$g(x)=x^2+\dfrac {5}{2}$

$\therefore f(6)=\dfrac{6}{4}-1=\dfrac{3}{2}-1=\dfrac{1}{2}$

$\therefore (gof)(6)=g(\dfrac{1}{2})=\left(\dfrac{1}{2}\right)^2+\dfrac{5}{2}=\dfrac{11}{4}=2.75$

$f: R \rightarrow R$ and

$g: R \rightarrow R$ are defined by

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

$g(x)=2 + 3x$ , find

$(g^{-1}\, of^{-1})(5)$ .

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

Explanation

Given,

$f(x)=3x-4$

Let

$y=3x-4$

$\Rightarrow x=\dfrac { y+4 }{ 3 } \\ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x+4 }{ 3 }$

Also given

$g(x)=2+3x$

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

Thus

$g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { \left( \dfrac { x+4 }{ 3 } \right) -2 }{ 3 }$

$\Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { x-2 }{ 9 } \\ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( 5 \right) =\dfrac { 5-2 }{ 9 } =\dfrac { 1 }{ 3 }$

So, option C is correct.

$f: R \rightarrow R$ and

$g: R \rightarrow R$ are defined by

$f(x) =2x +3, g(x)=x^2 + 7$ , what are the values of

$x$ such that

$g(f(x))=8$ ?

0%
$1, 2$
0%
$-1, 2$
0%
$-1, -2$
0%
$1, -2$

Explanation

Given,

$f(x)=2x+3$

Also given

$g(x)={ x }^{ 2 }+7$

Therefore,

$g[f(x)]={ \left( f(x) \right) }^{ 2 }+7$

$\Rightarrow g[f(x)]={ \left( 2x+3 \right) }^{ 2 }+7$

Thus

${ \left( 2x+3 \right) }^{ 2 }+7=8\\ \Rightarrow { \left( 2x+3 \right) }^{ 2 }=1\\ \Rightarrow 2x+3=\pm 1\\ \Rightarrow 2x=\pm 1-3\\ \Rightarrow 2x=-2,-4\\ \Rightarrow x=-1,-2$

Option C is correct.

Find the correct expression for

$\displaystyle f\left( g\left( x \right) \right)$ given that

$\displaystyle f\left( x \right) =4x+1$ and

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

0%
$\displaystyle -{ x }^{ 2 }+4x+1$
0%
$\displaystyle { x }^{ 2 }+4x-1$
0%
$\displaystyle 4{ x }^{ 2 }-7$
0%
$\displaystyle 4{ x }^{ 2 }-1$
0%
$\displaystyle 16{ x }^{ 2 }+8x-1$

Explanation

Given,

$g(x)=x^2-2$

$f(g(x))=f(x^2-2)$

Also given,

$f(x)=4x+1$

$\therefore f(x^2-2)=4(x^2-2)+1$

$\Rightarrow 4x^2-8+1$

$\Rightarrow 4x^2-7$

$f(x) = \sqrt {x^{2} - 3x + 6}$ and

$g(x) = \dfrac {156}{x +17}$ , find the value of the composite function

$g(f(4))$ .

0%
$5.8$
0%
$7.4$
0%
$7.7$
0%
$8.2$
0%
$10.3$

Explanation

$f(x)=\sqrt { x^{ 2 }-3x+6 } ,\quad g(x)=\frac{ 156 }{ x+17 }\\ g(f(4))=\quad ?\\ f(4)=\sqrt { 10 } \\ g(f(4))=g(\sqrt { 10 } )=\quad 3.16\\ g(f(4))=7.7$

Find the domain of following function:

$\sqrt { { x }^{ 2 }-9 }$

0%
$(-\infty,-3]\cup[3,\infty)$
0%
$(-\infty,-5)\cup(3,\infty)$
0%
$(-\infty,-3)\cup(6,\infty)$
0%
None of these

Explanation

We have,

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

We observe that

$f(x)$ is defined for all

$x$ satisfying

$x^2-9\geq0$

$\Rightarrow (x-3)(x+3)\geq0$

$\Rightarrow x\leq-3$ or

$x\geq3$

$\Rightarrow x\in (-\inf,-3]\cup[3,\inf)$

Hence,

$\text{D}$ is the correct option.

For what value of x is

$fog = gof$ if

$f(x)=x - 2$ and

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

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

Explanation

$f(x)=x-2\\ g(x)={ x }^{ 3 }+3\\ f[g(x)]=g(x)-2\\ \Rightarrow fog={ x }^{ 3 }+3-2={ x }^{ 3 }+1\\ g[f(x)]={ \{ f(x)\} }^{ 3 }+3\\ \Rightarrow gof={ (x-2) }^{ 3 }+3\\ fog=gof\\ { x }^{ 3 }+1={ (x-2) }^{ 3 }+3\\ { x }^{ 3 }-2={ (x-2) }^{ 3 }\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6x(x-2)\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6{ x }^{ 2 }+12x\\ \Rightarrow { x }^{ 2 }+1+2x=0\\ \Rightarrow { (x+1) }^{ 2 }=0\\ \Rightarrow x=-1$

So none of the above options are correct

Which of the following functions is/are constant ?

0%
$f(x)=x^{2}+2$
0%
$f(x)=x+\dfrac{1}{x}$
0%
$f(x)=7$
0%
$f(x)=6+x$

Explanation

$f(x)=7$ is constant function as its values do not depend on the variable

$x$ .

Its value is

$7$ for any value of

$x$ .

Option

$C$ is correct.

Let

$f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that for any irrational number

$r,$ and any real number

$x$ we have

$f(x)=f(x+r)$ . Then,

$f$ is

0%
an identity function
0%
a constant function
0%
a zero function
0%
onto function

Explanation

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).

So in here our input is

$x$ i f we change our input to say

$x+r$

and still output does'nt change i.e.

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

it means it is a constant function,

example:

$f(x)=5=5(x)^{0}$

$f(x+r)=5(x+r)^{0}$

$f(x+r)=5\times 1$

$f(x+r)=5=f(x)$

Let

$f$ be a function from the set of natural numbers to the set of even natural numbers given by

$f\left( x \right)=2x$ . Then

$f$ is

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

Explanation

Now, for

$x_1,x_2$ let

$f(x_1)=f(x_2)$ then

$2x_1=2x_2 \Rightarrow x_1=x_2$ .

So the given function is one-one.

Here range will be

$2,4,6,8,10....$ i.e every even natural number.

$range = co-domain$ .

Thus, the given function is onto.

Hence, C is correct.

$\{ (x, 2), (4, y) \}$ represents an identity function, then

$( x, y)$ is :

0%
(2, 4)
0%
(4, 2)
0%
(2, 2)
0%
(4, 4)

Explanation

Identity funtion returns the same value as of input.

Since input

$x$ gives output

$2$

$\therefore$

$x=2$

Similarly input of

$4$ gives output

$y$

$\therefore y=4$

$\Rightarrow (x,y)=(2,4)$

Option (a) is correct.

Find number of all such functions

$y = f(x)$ which are one-one?

0%
$0$
0%
$3^{5}$
0%
$^{5}P_{3}$
0%
$5^{3}$

Explanation

One-one functions are those in which each element in the domain has a unique image in the range of the function.

Here

$B$ is the co-domain of the function which has lesser number of elements than the domain itself. This shows that it is not possible for the function to be one-one.

The number of real linear functions

$f(x)$ satisfying

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

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

Explanation

Let

$f(x) = ax+b$

$f(f(x))=a(ax+b)+b = a^2x+ab+b$

Given,

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

$\implies a^2x+ab+b = x+ ax+b$

$\implies (a^2-a-1)x+ab=0$

Since above equation is valid for all values of x, we have

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

$ab=0$

$a^2-a-1 =0$

$a=\dfrac{ 1\pm \sqrt{1+4}}{2}$

$\implies a=\dfrac{ 1\pm \sqrt{5}}{2}$

$ab=0$

Since

$a$ is non-zero,

$\implies b=0$

$f(x) =\dfrac{ 1+ \sqrt{5}}{2}x$ and

$f(x) =\dfrac{ 1- \sqrt{5}}{2}x$ satisfy the condition.

Hence the answer is option (D)

Let

$f:N\times N\rightarrow N-\left\{ 1 \right\}$ be defined as

$f(m,n)=m+n$ , then function

$f$ is ______.

0%
One-One and onto
0%
Many- One and not onto
0%
One- One and not onto
0%
Many- One and onto

Explanation

Given

$f(m,n)=m+n$

Range of this function is set of all natural numbers except

$1$

[since least value of

$m$ and

$n$ is

$1$ ]

Therefore the range and co-domain are equal

Therefore it is onto

For every

$(m,n)$ , there exists only one

$m+n$

Therefore it is one-one also

Hence,

$f$ is one-one and onto.

Let E = {1, 2, 3, 4} and F {1, 2}. Then the number of onto functions from E to F is

0%
14
0%
16
0%
6
0%
4

Explanation

The total number of functions are

$2^{4}=16$ . From the total number of functions, subtract those functions which do not have 2 as the range =1. Subtract those functions which do not have 1 as the range.

So, the answer is

$16-2=14$

Find the domain for which the functions

$f(x)=2x^2-1$ and

$g(x)=1-3x$ are equal.

0%
{-2,2}
0%
{2, 1/2}
0%
{-2,1/2}
0%
None of these

Explanation

The values of x for which f(x) and g(x) are equal are given by

$f(x)=g(x)$

$\implies 2x^2-1 =1-3x$

$2x^2+3x-2=0$

$(x+2)(2x-1)=0$

$x=-2, 1/2$ .

Thus,

$f(x)$ and

$g(x)$ are equal on the set {-2, 1/2}.

Suppose that

$g\left( x \right) =1+\sqrt { x }$ and

$f\left( g\left( x \right) \right) =3+2\sqrt { x } +x$ , then

$f\left( x \right)$ is

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

Explanation

$g(x)=1+\sqrt{x}$

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

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

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

Find the domain of

$\sqrt{\dfrac{\sqrt{x-2}}{x-3}}$ .

0%
$(2,\infty)$
0%
$(3,\infty)$
0%
$(-\infty, 2)$
0%
$(-\infty, 3)$

Explanation

Number inside the square root must be non negative and denominator should not be equal to zero

$\Rightarrow x-2\geq 0$

$\Rightarrow x\geq 2$

And

$x-3>0$

$\Rightarrow x>3$

Combining both we get

$x\in (3,\infty)$

If domain of

$f(x)=\sqrt{{x}^{2}+bx+4}$ is

$R$ , then maximum possible integral value of

$b$ is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

If Domain of

$f(x)$ is

$R$

Then

${x^2} + bx + 4 \geqslant$

Since Coefficient of

$x>0$

$D < 0$

$\Rightarrow {b^2} - 4 \times 1 \times 4 \leqslant 0$

$\Rightarrow {b^2} - {\left( 4 \right)^2} \leqslant 0$

$\Rightarrow \left( {b - 4} \right)\left( {b + 4} \right) \leqslant 0$

$\Rightarrow b \in \left[ { - 4,4} \right]$

There fore maximum value of

$b$ is

$4$

Let A = {0, 1} and N the set of all natural numbers. Then the mapping

$f : N \rightarrow A$ defined by

$f(2n - 1) = 0, f (2n) = 1 \forall n \epsilon N$
is many-one onto.

0%
True
0%
False

Explanation

Let

$A={0,1}$ and N the set of all natural numbers. Then the mapping

$f:N \longrightarrow A$ defines by

$f(2n-1)=0,$

$f(2x)=1$

$\forall$

$n$

$\epsilon$

$N$ is many-one onto. The above statement is absolutely true.

$f:R \to R,f\left( x \right) = \dfrac {{x^2} + 2x + c}{{x^2} + 4x + c}$ is onto only if

0%
$c < 2$
0%
$c \ge 0$
0%
$c < 0$
0%
$|c| \le 1$

Explanation

$f:R\rightarrow R,f\left( x \right) =\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow y=\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow { x }^{ 2 }y+4xy+cy={ x }^{ 2 }+2x+c\\ \Rightarrow { x }^{ 2 }(y-1)+2x(2y-1)+(cy-c)=0\\ \because x\epsilon R\quad \quad \therefore 4{ (2y-1) }^{ 2 }-4c{ (y-1) }^{ 2 }\ge 0\\ { (2y-1) }^{ 2 }-c{ (y-1) }^{ 2 }\ge 0\\ 4{ y }^{ 2 }+1-4y-c{ y }^{ 2 }+2cy-1\ge 0\\ { y }^{ 2 }(4-c)-y(4-2c)\ge 0\\ \because y\epsilon R\quad \quad { (4-2c) }^{ 2 }-4(4-c)\times 0<0\\ { (4-2c) }^{ 2 }<0\\ { (2c-4) }^{ 2 }<0\\ 2c-4<0\\ c<2$

Let

$f(x) = \sqrt{log_2 \dfrac{10x - 4}{4 - x^2}-1}$ . Then sum of all integers in domain of

$f(x)$ is

0%
$-15$
0%
$-16$
0%
$-17$
0%
$-20$

Explanation

The function

$f(x) = \sqrt{log_2 \dfrac{10x - 4}{4 - x^2}-1}$ is defined for,

${\log_2 \dfrac{10x - 4}{4 - x^2}-1}\ge 0$

or,

${\log_2 \dfrac{10x - 4}{4 - x^2}}\ge 1$

or,

$\dfrac{10x-4}{4-x^2}\ge 2$

or,

$5x-2\ge4-x^2$

or,

$x^2+5x-6\le 0$

or,

$(x+6)(x-1)\le 0$

or,

$-6\le x\le 1$ .

So the domain is

$[-6,1]$

So the sum of all the integer in domain of

$f(x)$ is

$-6-5-4-3-2-1+0+1=-20$ .

Domain of function as

$f\left( x \right) = \dfrac{{^3\sqrt {{x^2} - 5x + 6} }}{{{x^2} - x - 6}}$ is

0%
$\left[ {2,3} \right]$
0%
$R - \left\{ { - 3,2} \right\}$
0%
$R - \left\{ { - 2,3} \right\}$
0%
$R$

Explanation

$f(x)=(\frac{\sqrt[3]{x^2-5x+6}}{x^2-x-6})$

$Cube\>root\>expression\>is\>fine\>for\>any\>value\>of\>the\>term\>inside\>the\>root.$

$But\>the\>denomiantor\>must\>not\>be\>equal\>to\>zero.$

$x^2-x-6\neq0$

$(x-3)(x+2)\neq0$

$x\neq3\>or\>x\neq-2$

$\therefore$ domain: R- {-2,3}

The domain of the fuction

$f(x)=\sqrt{\log_{16}x^2}$ is ?

0%
$x=0$
0%
$|x|\ge4$
0%
$|x|\ge 1$
0%
$|x|\ge 2$

Explanation

$Given$

$f\left( x \right)=\sqrt{\log _{16}^{{}}{{x}^{2}}}$

$The\text{ domain of a funcion is the set of input or argument values }$

$\text{for which the function is real and defined}$

$\log _{16}^{{}}{{x}^{2}}\ge 0$

$\Rightarrow {{x}^{2}}\ge {{16}^{0}}$

$\Rightarrow {{x}^{2}}\ge 1$

$\Rightarrow {{x}^{2}}-1\ge 0$

$\Rightarrow \left( x-1 \right)\left( x+1 \right)\ge 0$

$x\ge 1\,or\,x\le 1$

$(-\propto ,1]\cup [1,\propto )\,$

$Hence$

$\,Domain:\left| x \right|\ge 1$

Domain of

$f(x)=\dfrac {4x}{\sqrt{x^2-16}}$

0%
$|x|>4$
0%
$|x|<4$
0%
$|x|=4$
0%
$x\in R$

Explanation

The expression is

$\dfrac {4x}{\sqrt{x^2-16}}$

The function is defined only if

$x^2-16>0\\x^2>16\\|x|>4$

If the domain of the function

$f(x)= \dfrac{x^2-5x+66}{x-4}$ is

$R-\{a\}$ , then the value of a -

0%
4
0%
6
0%
8
0%
12

Explanation

The function

$f(x)=\dfrac{x^2-5x+66}{x-4}$ is defined only when

$a\neq4$

$x\in R-\{4\}\implies a=4$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
Consider the following statements:

$f[g(x)]$ is a polynomial of degree 3.

$g[g(x)]$ is a polynomial of degree 2.
Which of the above statements is/are correct ?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Neither 1 nor 2

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

(i)

$f[g(x)]$

$=f[5x+30]$

$=(5x+30)^2+2(5x+30)-5$

upon solving the above equation, we get,

$f[g(x)]=25x^2+310x+955$

degree

$=2$

(ii)

$g[g(x)]$

$=g[5x+30]$

$=5(5x+30)+30$

$=25x+150+30$

$=25x+180$

$g[g(x)]=25x+180$

degree

$=1$

Let

$f(x)=\cfrac { 1 }{ 1-x }$ . Then

$\left\{ f\circ \left( f\circ f \right) \right\} (x)$

0%
$x$ for all $x\in R$
0%
$x$ for all $x\in R-\left\{ 1 \right\}$
0%
$x$ for all $x\in R-\left\{ 0,1 \right\}$
0%
none of these

Explanation

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

$x\in R-\{1\}$

$\{fof\}(x)=\dfrac{1}{1-\dfrac{1}{1-x}}=\dfrac{1}{\dfrac{1-x-1}{1-x}}=\dfrac{x-1}{x}=1-\dfrac{1}{x}$ for

$x\in R - \{0,1\}$

$\{fofof\}(x)=\dfrac{1}{1-fof(x)}=\dfrac{1}{1-\dfrac{x-1}{x}}=\dfrac{1}{\dfrac{x-x+1}{x}}=x$ for

$x\in R-\{0,1\}$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
What are the roots of the equation

$g[f(x)] = 0$ ?

0%
$1, -1$
0%
$-1, -1$
0%
$1, 1$
0%
$0, 1$

Explanation

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

$g[f(x)]=0$

$g[x^2+2x-5]=0$

$5(x^2+2x-5)+30=0$

$5x^2+10x-25+30=0$

$5x^2+10x+5=0$

$x^2+2x+1=0$

$(x+1)^2=0$

$\therefore x=-1,-1$

Read the following information and answer the three items that follow :
Let

$f(x) = x^2 + 2x - 5$ and

$g(x) = 5x + 30$
If

$h(x) = 5f(x) - xg (x)$ , then what is the derivative of

$h(x)$ ?

0%
$-40$
0%
$-20$
0%
$-10$
0%
$0$

Explanation

Given,

$f(x)=x^2+2x-5$

$g(x)=5x+30$

$h(x)=5f(x)-xg(x)$

$=5(x^2+2x-5)-x(5x+30)$

$=5x^2+10x-25-5x^2-30x$

$h(x)=-20x-25$

$\dfrac{d}{dx}[h(x)]=\dfrac{d}{dx}(-20x-25)$

$=-20$

The number of one-one functions that can be defined from

$A=\{4,8,12,16\}$ to

$B$ is

$5040,$ then

$n(B)=$

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

Explanation

The first element

$4$ of

$A$ can be mapped to any of the

$n$ elements in

$B$
Similarly second element

$8$ of

$A$ can be mapped to any of

$(n-1)$ elements in

$B$

$\Rightarrow$ Total no of one - one functions is

$n(n-1)(n-2)(n-3)=5040$

$\Rightarrow n=10$

The number of injections that are possible from

$A$ to itself is

$720,$ then

$n (A) =$

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

first element of

$A$ in domain can be mapped to
any of the

$n$ elements of

$A$ in range.

$2^{nd}$ element of

$A$ in domain can be mapped to any of

$(n-1)$ elements of

$A$ in range.

$\therefore$ Total no of injections possible is

$n\times (n-1)\times (n-2)...\times 3\times 2\times 1=n!$

$\therefore n!=720$

$\Rightarrow n=6$

The number of one-one functions that can be defined from

$A = \left \{ 1,2,3 \right \}$ to

$B = \left \{ a,e,i,o,u \right \}$ is

0%
$3^{5}$
0%
$5^{3}$
0%
${_{}}^{5}P_{3}$
0%
$5!$

Explanation

You can correlate this to permutation and combination problems.

We have to arrange 3 people on 5 places.

So total no of permutations

$={_{}}^{5}P_{3}$

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

$A = \left \{ 1,4,9,16 \right \}$ to

$B=\left \{ 2,8,16,32,64 \right \}$ is

0%
$1024$
0%
$20$
0%
$505$
0%
$625$

Explanation

Here

$n(B)>n(A)$ , no function can be a surjective.

No of functions

$=$ No of non-surjectives

Any elements of

$A$ can be mapped to any of

$5$ elements in

$B$
So non surjective mapping from

$A$ to

$B$

$=5\times5\times5\times5$

$=625$

$A = \left \{ 11, 12, 13, 14 \right \}$ and

$B = \left \{ 6,8,9,10 \right \}$ then the number of bijections defined from

$A$ to

$B$ is

0%
$256$
0%
$24$
0%
$16$
0%
$64$

Explanation

No of bijections from

$A$ to

$B$

$=4\times3\times2\times1$

$=24$

If function

$f$ has an inverse, then which of the following conditions is necessary and sufficient

0%
$f$ is a one-one function.
0%
$f$ is an onto function.
0%
$f$ is a one-one and onto function.
0%
$f$ is an identity function.

Explanation

To have an inverse for the function

$f$ ,

$f$ should be one-one and onto.
Each point in the domain should have a unique image in range.
So, it should be one-one.
And if it is not onto, then there will be no pre-image of the extra element, so inverse will not be possible.

$f:A\rightarrow B$ is a constant function which is onto then

$B$ is

0%
a singleton set
0%
a null set
0%
an infinite set
0%
a finite set

Explanation

$f(x)$ is a constant fucntion

$\Rightarrow$ Range of

$f(x)$ is a singleton set.
For

$f$ to be an onto function, Co domain

$B$ should be equal to Range.

$\therefore B$ is a singleton set.

$f:A\rightarrow B$ is a bijection then

$f^{-1} of =$

0%
$fof^{-1}$
0%
$f$
0%
$f^{-1}$
0%
an identity

Explanation

Since

$f$ is bijective, its inverse will also be bijective.

$f(x)=y$

$x=f^{-1}(y)$

$f(x)=f^{-1}(y)$

$y=f(f)^{-1}(y)=f(x)=y$
Hence, it is an identity.

The number of bijection that can be defined from

$A = \left \{ 1,2,8,9 \right \}$ to

$B = \left \{ 3,4,5,10 \right \}$ is

0%
$4^{4}$
0%
$4^{2}$
0%
$24$
0%
$18$

Explanation

There are 4 inputs

$\{1, 2, 8, 9\}$ and 4 outputs

$\{3, 4, 5, 10\}$ . Hence function will be bijective if and only if each output is connected with only one input.
Therefore number of bijective function is

$4! = 24$

Assertion (A) :

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

$g(x) = x+2$ are equal.

Reason (R): Two functions

$f$ and

$g$ are said to be equal if their domains and ranges are equal and

$f(x)=g(x) \forall x \in$ domain.

0%
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
0%
Both $A$ and $R$ are true and $R$ is not correct explanation of $A$
0%
$A$ is true but $R$ is false
0%
$A$ is false but $R$ is true

Explanation

The domain for

$f \left ( x \right )$ is

$R- \left \{ 2 \right \}$ and domain of

$g \left ( x \right )$ is

$R$ . Hence Assertion is false as the domain is not equal.

Reason is true as two functions

$f$ and

$g$ are said to be equal if

The domain of

$f$ = domain of

$g$ .

The co-domain of

$f$ = co-domain of

$g$ .

The range of

$f$ = range of

$g$ .

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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