If $$f : IR \rightarrow IR$$ is defined by $$f(x) = 2x + 3$$, then $$f^{-1}(x)$$

Is given by $$\dfrac{1}{2x + 3}$$

Does not exist because 'f' is not injective

Does not exist because 'f' is not surjective

If $$f : R \rightarrow R$$ is defined by $$f(x) = x^{3}$$ then $$f^{-1}(8) =$$

$$\left \{2, 2\omega, 2\omega^{2}\right \}$$

$$\left \{2, 2\omega\right \}$$

If $$fog = |\sin x|$$ and $$gof = \sin^{2}\sqrt {x}$$, then $$f(x)$$ and $$g(x)$$ are

$$f(x) = \sqrt {x}, g(x) = \sin^{2}x$$

$$f(x) = \sqrt {\sin x}, g(x) = x^{2}$$

$$f(x) = \sin \sqrt {x}, g(x) = x^{2}$$

If $$f(x)=\left| x \right| ,x\in R$$, then

$$f(x)=\left( f\circ f \right) \left( x \right) $$

$$f(x)=\left( f\times f \right) \left( x \right) $$

$$f(x)=\left( f\times f \right) \left( x^2 \right) $$

MCQ Questions for Class 12 Commerce Maths Relations And Functions Quiz 6

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,

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

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

$\Rightarrow 36=28+b$

$\Rightarrow b=8$

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,

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

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

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

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

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

Extraction of a cube root of a given number is

0%
binary operation
0%
relation
0%
unary operation
0%
relation in some set

Explanation

Extraction of cube root of a given number is binary operation

Therefore the correct option is

$A$

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

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

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

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

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

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

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

$*$ is a binary operation on

$Z$ such that:

$a * b = a + b + ab$ .
The solution of

$(3* 4) *x = -1$ is

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

Explanation

$(3*4)=(3+4)+(3\times 4)=7+12=19$ ...(i)
Now

$(3*4)*x=19*x=19+x+19x=-1$ or

$20x=-20$ or

$x=-1$ .
Hence the value of x is -1.

$f(x) = 3x - 5$ 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

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

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

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

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

$a*b = |a-b|$ then

$6*8$ will be

0%
$-2$
0%
$14$
0%
$2$
0%
Cannot be determined

Explanation

$\mathbf{{\text{Step -1: Identifying the binary function definition}}{\text{.}}}$

${\text{In this * is a binary function, which performs an operation defined as}}$

${\text{If the operation is performed between two variables a and b such as }a*b \text{ then}}$

${{a*b = |a - b| }}{\text{ i.e}}{\text{. the binary operation between two variables will return the }}$

${\text{absolute difference between them}}.$

$\mathbf{{\text{Step -2: Finding the value of 6*8}}{\text{.}}}$

${\text{Working according to the definiton of the operation we can write,}}$

${{6*8 = |6 - 8| = | - 2| = 2}}$

$\mathbf{{\text{Therefore, the value of 6*8 = 2}}{\text{.}}}$

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

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

$\text{Given,}$

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

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

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

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

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

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

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

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

$=120$

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

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

$g(x) = 2x-1$

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

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

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

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

$f(x) = 2x - 6$ , then

$f^{-1}(x)$ is

0%
$6 - 2x$
0%
$\dfrac {1}{2}x - 6$
0%
$\dfrac {1}{2}x - 3$
0%
$\dfrac {1}{2}x + 3$
0%
$\dfrac {1}{2} x + 6$

Explanation

Here take

$y=f(x)$
Then

$y=2x-6$

$\Rightarrow 2x=y+6$

$\Rightarrow x=\dfrac{y}{2}+3$
Now switch

$x$ and

$y$ in the resulting equation. The resulting function of

$x$ is the inverse.

$\Rightarrow y=\dfrac{x}{2}+3$
Thus the inverse function is

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

$f(x) = 4/\sqrt [3]{x + 1}$ , what is

$f^{-1}(7)$ ?

0%
$-7$
0%
$8$
0%
- $\dfrac{279}{343}$
0%
$1.75$
0%
$7.2$

Explanation

$y^3=\dfrac {64}{x+1}$

$\Rightarrow x= \dfrac{64}{y^3}-1$

$f^{-1}(x)=\dfrac{64}{x^3}-1$

$f^{-1}(7)=\dfrac{64-343}{343}=-\dfrac{279}{343}$

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

If the operation

$\oplus$ is defined by

$a\oplus b = a^{2} + b^{2}$ for all real numbers

$'a'$ and

$'b'$ , the

$(2\oplus 3)\oplus 4 =$

0%
$120$
0%
$175$
0%
$129$
0%
$185$
0%
$312$

Explanation

We have

$(2\oplus 3)\oplus 4=(2^2+3^2)\oplus 4=13\oplus4=13^2+4^2=169+16=185$ [the given condition

$a\oplus b=a^2+b^2$ ]

$f: R\rightarrow R$ 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

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

$f : IR \rightarrow IR$ is defined by

$f(x) = 2x + 3$ , then

$f^{-1}(x)$

0%
Is given by $\dfrac{x-3}{2}$
0%
Is given by $\dfrac{1}{2x + 3}$
0%
Does not exist because 'f' is not injective
0%
Does not exist because 'f' is not surjective

Explanation

$f: IR \rightarrow IR$

$f(x) = 2x + 3$

$\Rightarrow x = \cfrac{f(x) - 3}{2}$

i.e.

$f^{-1}x = \cfrac{x - 3}{2}$

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.

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

In the set of integers under the operation

$\ast$ defined by

$a\ast b=a+b-1$ , the identity element is:

0%
$0$
0%
$1$
0%
$a$
0%
$b$

Explanation

Let

$e$ be the identity element, then

$\Rightarrow a*e=a$
or

$a+e-1=a$

$e-1=0$

$e=1$ .
Hence the identity element for the above operation is

$1$ .

Which of the following is not a binary operation on

$R$ ?

0%
$a \times b = ab$
0%
$a \times b = a - b$
0%
$a \times b = \sqrt{ab}$
0%
$a \times b = \sqrt{a^2 + b^2}$

Explanation

When you will multiply, add , subtract, any two real number the resulting

number will be real number, so option

$A$ and

$B$ are binary operations.

Also if

$a$ and

$b$ are real number then

$\sqrt{a^2+b^2}$ will be a real number, so option

$D$ is also binary operation

Check option 'C', If sign of product of

$a$ and

$b$ will be negative, then

$\sqrt{ab}$ will not be real number,

Hence it is not a binary operation

Therefore option

$C$ is correct choice

$N$ is a set of natural numbers, then under binary operation

$a\cdot b = a + b, (N, .)$ is

0%
Quasi-group
0%
Semi-group
0%
Monoid
0%
Group

Explanation

Let

$a,b$ be any element in

$N$

$\implies$

$a+b\ \epsilon N$

$\therefore a \cdot b=a+b\ \epsilon N$

$\implies\ a \cdot b\ \epsilon N$ .........

$(1)$

$(a \cdot b) \cdot c=(a\cdot b)+c=a+b+c$

Also,

$a\cdot (b\cdot c)=a+(b \cdot c)=a+b+c$

$\therefore\ a \cdot (b \cdot c)=(a \cdot b) \cdot c$ ........

$(2)$

$a\cdot b=a+b=b+a=b\cdot a$ .........

$(3)$

From

$(1), (2)$ and

$(3)$ ,

The structure

$(N, .)$ satisfies the closure property, associativity and commutativity but the identity element

$0$ does not belong to

$N$ .

Hence,

$(N,.)$ is a semi-group.

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

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

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

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

$375$ .

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.

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

$option C$

On the set

$Z$ , of all integers

$\ast$ is defined by

$a\ast b = a + b - 5$ . If

$2\ast (x\ast 3) = 5$ then

$x =$

0%
$0$
0%
$3$
0%
$5$
0%
$10$

Explanation

Given

$a*b=a+b-5$ where

$*$ is a unique operator

$2*(x*3)=2*(x+3-5)=2*(x-2)=2+x-2-5=x-5$

Its value is given as

$5$ . so

$x-5=5 \implies x=10$

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

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

Explanation

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

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

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

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

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

$=8x$

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

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

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.

$f : R \rightarrow R$ is defined by

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

$f^{-1}(8) =$

0%
$\left \{2\right \}$
0%
$\left \{2, 2\omega, 2\omega^{2}\right \}$
0%
$\left \{2, -2\right \}$
0%
$\left \{2, 2\omega\right \}$

Explanation

The inverse function of the given function is given by interchanging the

$x$ and

$y$ and finding the function (the inverse function which is now

$y$ ) in terms of

$x$ .

For

$f(x)=x^3$

Let

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

$y=x^3$

$x=y^{\tfrac { 1 }{ 3 } }$

$\therefore f^{-1}(y)=$

$y^{\tfrac { 1 }{ 3 } }$

$f^{-1}(8)=8^{\tfrac { 1 }{ 3 } }=2\times (1)^{\tfrac13}$

We know that the cube roots of unity

$(1)$ are

$1, \omega$ and

$\omega^2$

$\therefore f^{-1}(8) = \{2,2\omega, 2\omega^2\}$

Hence, option

$B$ is correct.

Let

$A=\left\{ x\in R|x\ge 0 \right\}$ . A function

$f:A\rightarrow A$ is defined by

$f(x)={ x }^{ 2 }$ . Which one of the following is correct?

0%
The function does not have inverse
0%
$f$ is its own inverse
0%
The function has an inverse but $f$ is not its own inverse
0%
None of the above

Explanation

$f\left( x \right) ={ x }^{ 2 }\\ \Longrightarrow f^{ -1 }\left( x \right) =\sqrt { x }$
Clearly,

$f^{ -1 }\left( x \right)$ exists in the domain of definition of

$f$ .
Also,

$f\left( x \right)$ is not its own inverse.
Option C is the answer.

$f:[0, \infty)\to [0,\infty)$ and

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

$f$ is

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

Explanation

Here,

$f:[0, \infty) \rightarrow [0, \infty)$ i.e, domain is

$[0, \infty)$ and codomain is

$[0, \infty)$

For one-one

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

$f'(x) = \dfrac{1}{(1+x)^2} > 0, \forall x \in [0, \infty)$

$\therefore f(x)$ is increasing in this domain. Thus

$f(x)$ is one-one in its domain

For onto (we find range)

$f(x) = \dfrac{x}{1+x}$ i.e.

$y =\dfrac{x}{1+x}$

$\Rightarrow y+yx=x$

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

$\Rightarrow x=\dfrac{y}{1-y} \ge 0$ as

$x \ge 0$

$\therefore 0 \le y \neq 1$ and

$y<1$

$\therefore$ Range

$\neq$ Codomain

$\therefore f(x)$ is one-one but not onto

$\ast$ is the operation defined by

$a\ast b={ a }^{ b }$ for

$a,b\in N$ , then

$\left( 2\ast 3 \right) \ast 2$ is equal to

0%
$81$
0%
$512$
0%
$216$
0%
$64$
0%
$243$

Explanation

Given

$a\ast b={ a }^{ b }$

$\therefore (2\ast 3)\ast 2={ (2 }^{ 3 })\ast 2=64$

The function

$f:A\rightarrow B$ given by

$f(x) = x ,x\in A$ , is one to one but not onto. Then;

0%
$B\subset A$
0%
$A=B$
0%
$A'\subset B'\quad$
0%
$A\subset B$
0%
$A'\cap B'=\phi$

Explanation

$f(x)=x$ is identity function
It is given that

$f(x)$ is one to one but not on to.
It means some element in

$B$ has no pre image in

$A$

$A\subset B$

$fog = |\sin x|$ and

$gof = \sin^{2}\sqrt {x}$ , then

$f(x)$ and

$g(x)$ are

0%
$f(x) = \sqrt {\sin x}, g(x) = x^{2}$
0%
$f(x) = |x|, g(x) = \sin x$
0%
$f(x) = \sqrt {x}, g(x) = \sin^{2}x$
0%
$f(x) = \sin \sqrt {x}, g(x) = x^{2}$

Explanation

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

Also,

$gof = g\left \{f(x)\right \} = \sin^{2}\sqrt {x}$

Obviously,

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

and

$\sin^{2} \sqrt {x} = \sin^{2} \left \{f(x)\right \}$
i.e.,

$g (x) = \sin^{2}x$

and

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

Let

$f(x) = |x - 2|$ , where

$x$ is a real number. Which one of the following is true?

0%
$f$ is periodic
0%
$f(x + y) = f(x) + f(y)$
0%
$f$ is an odd function
0%
$f$ is not one-one function
0%
$f$ is an even function

Explanation

From the graph of the function it can be observed clearly that

$f$ is not one-one function.

So option

$D$ is correct.

The value of

$\alpha (\neq 0)$ for which the function

$f(x) = 1 + \alpha x$ is the inverse of itself is

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

Explanation

We have,

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

or,

$y=1+\alpha x$

or,

$x=\cfrac{y-1}{\alpha}$

Change

$x$ with

$y$ and

$y$ with

$x$

or,

$y=\cfrac{x-1}{\alpha}$

or,

$f^{-1}(x)=\cfrac{x-1}{\alpha}$

For,

$f(x)=f^{-1}(x)$ value of

$\alpha$ should be

$-1.$

Hence, C is the correct option.

$A = \left \{ 1 , 3 , 5 , 7 \right \}$ and

$B = \left \{ 1 , 2 , 3, 4 , 5 , 6 , 7 , 8 \right \}$ then the number of one-to-one functions from

$A$ into

$B$ is

0%
1340
0%
1860
0%
1430
0%
1880
0%
1680

Explanation

Given,

$A = \{1, 3, 5, 7\}$
and

$B =\{1, 2, 3, 4, 5, 6, 7, 8\}$
Here,

$n(A) = 4$ and

$n(B) = 8$
Thus number of one-one function from

$A$ into

$B$

$= \,^8 P_4 = 8.7.6.5 = 1680$

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

$g(x)={ x }^{ 2 }-1$ , then

$\left( f\circ g \right)$

$({ x }^{ 2 }-1)$ is equal to

0%
$3{ x }^{ 4 }-3x+5$
0%
$3{ x }^{ 4 }-6{ x }^{ 2 }+5$
0%
$6{ x }^{ 4 }+3{ x }^{ 2 }+5$
0%
$6{ x }^{ 4 }-6x+5$
0%
$3{ x }^{ 2 }+6x+4$

Explanation

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

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

$\therefore f\circ g\left( { x }^{ 2 }-1 \right) =f[g\left( { x }^{ 2 }-1 \right) ]=f\left[ { \left( { x }^{ 2 }-1 \right) }^{ 2 }-1 \right]$

$=3\left( { x }^{ 4 }-2{ x }^{ 2 } \right) +5=3{ x }^{ 4 }-6{ x }^{ 2 }+5$

$\ast$ is defined by

$a\ast b = a - b^{2}$ and

$\oplus$ is defined by

$a\oplus b = a^{2} + b$ , where

$a$ and

$b$ are integers, then

$(3\oplus 4)\ast 5$ is equal to

0%
$164$
0%
$38$
0%
$-12$
0%
$-28$
0%
$144$

Explanation

Given,

$a\ast b = a - b^{2}$ .... (i)

and

$a\oplus b = a^{2} + b$ .... (ii)

where,

$a$ and

$b$ are integers

Then,

$(3\oplus 4) \ast 5 = \left \{(3)^{2} + 4\right \}\ast 5$

$= (9 + 4) \ast 5 = 13\ast 5$

$= 13 - (5)^{2} = 13 - 25 = -12$

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

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

$f(x)$ is equal to

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

Explanation

Given,

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

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

$(i)$

$\Rightarrow f(1+\sqrt{x})=3+2\sqrt{x}+x$
Put

$1+\sqrt{x}=y\Rightarrow x=(y-1)^2$

$\therefore f(y)=3+2(y-1)+(y-1)^2=3+2y-2+y^{2}-2y+1=2+y^2$

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

Let

$f(x)=\cot ^{ -1 }{ \left( \cfrac { 1-{ x }^{ 2 } }{ 2x } \right) } +\cot ^{ -1 }{ \left( \cfrac { 1-3{ x }^{ 2 } }{ 3x-{ x }^{ 3 } } \right) } -\cot ^{ -1 }{ \left( \cfrac { 1-6{ x }^{ 2 }+{ x }^{ 4 } }{ 4x-4{ x }^{ 3 } } \right) }$ , the

$F'(x)$ equals

0%
$\cfrac { -1 }{ \sqrt { 1-{ x }^{ 2 } } }$
0%
$\cfrac { -1 }{ 1+{ x }^{ 2 } }$
0%
$\cfrac { 1 }{ 1+{ x }^{ 2 } }$
0%
$\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

Explanation

$F(x)=\tan ^{ -1 }{ \left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 3x-{ x }^{ 3 } }{ 1-3{ x }^{ 2 } } \right) } -\tan ^{ -1 }{ \left( \cfrac { 4x-4{ x }^{ 3 } }{ 1-6{ x }^{ 2 }+{ x }^{ 4 } } \right) }$

$=2\tan ^{ -1 }{ x } +3\tan ^{ -1 }{ x } -4\tan ^{ -1 }{ x } \\=\tan ^{ -1 }{ x }$

$\Rightarrow F'(x)=\cfrac { 1 }{ 1+{ x }^{ 2 } }$

$f(x)=\left| x \right| ,x\in R$ , then

0%
$f(x)=\left( f\times f \right) \left( x \right)$
0%
$f(x)=x$
0%
$f(x)=\left( f\times f \right) \left( x^2 \right)$
0%
$f(x)=\left( f\circ f \right) \left( x \right)$

$g(x)={ x }^{ 2 }+x-2$ and

$\cfrac { 1 }{ 2 } (g\circ f)(x)=2{ x }^{ 2 }-5x+2$ , then

$f(x)$ is

0%
$2x-3$
0%
$2x+3$
0%
$2{ x }^{ 2 }+3x+1$
0%
$2{ x }^{ 2 }+3x-1$

Explanation

$\cfrac { 1 }{ 2 } \left( g\circ f \right) (x)=2{ x }^{ 2 }-5x+2$

$\Rightarrow \cfrac { 1 }{ 2 } g\left[ f(x) \right] =2{ x }^{ 2 }-5x+2\quad$

$\left[ { \left\{ f(x) \right\} }^{ 2 }+\left\{ f(x) \right\} -2 \right] =2\left[ 2{ x }^{ 2 }-5x+2 \right] \quad \left[ \because g(x)={ x }^{ 2 }+x-2 \right]$

$\Rightarrow f{ \left( x \right) }^{ 2 }+f(x)-\left( 4{ x }^{ 2 }-10x+6 \right) =0$ ....represents quadratic equation

We have a formula for solving quadratic equation

$ax^2+bx+c=0$ is

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\therefore f(x)=\cfrac { -1\pm \sqrt { 1+4\left( 4{ x }^{ 2 }-10x+6 \right) } }{ 2 }$

$=\cfrac { -1\pm(4x-5) }{ 2 } =2x-3$ or

$2-2x$

$(ax^2 + bx + c)y +a'x^2+b'x+c=0$ , then the condition that x may be a rational function of y is

0%
$(ac'-a'c)^2=(ab'-a'b)(bc'-b'c)$
0%
$(ab'-a'b)^2=(ab'-a' c)(bc'-b'c)$
0%
$(bc'-b'c)^2=(ab'-a'b)(ac'-a'c)$
0%
None of these

$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) }$ and

$g\left( \cfrac { 5 }{ 4 } \right) =1$ , then

$g\circ f(x)$ is equal to

0%
$0$
0%
$1$
0%
$\sin { { 1 }^{ o } }$
0%
None of these

Explanation

Given,

$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) }$

$=\sin ^{ 2 }{ x } +\left( \sin { x } \cos { \cfrac { \pi }{ 3 } } +\cos { x } \sin { \cfrac { \pi }{ 3 } } \right) +\cos { x } \left( \cos { \cfrac { \pi }{ 3 } } -\sin { x } \sin { \cfrac { \pi }{ 3 } } \right)$

$=\sin ^{ 2 }{ x } +\cfrac { \sin ^{ 2 }{ x } }{ 4 } +\cfrac { 3\cos ^{ 2 }{ x } }{ 4 } +\cfrac { 2\sqrt { 3 } }{ 2.2 } \sin { x } \cos { x } +\cfrac { \cos ^{ 2 }{ x } }{ 2 } -\cos { x } \sin { x } \cfrac { \sqrt { 3 } }{ 2 }$

$=\cfrac { 5 }{ 4 } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) =\cfrac { 5 }{ 4 } \quad$
Therefore,

$g\circ f(x)=g(f(x))=g\left( \cfrac { 5 }{ 4 } \right) =1\quad \quad$

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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