Range of the function $f\left( x \right) = \frac{x}{{1 + {x^2}}}$ is

$\left( { - \infty ,\;\infty } \right)$

$\left[ { - 1,{\rm{ }}1} \right]$

$\left[ { - \sqrt 2 ,\sqrt 2 } \right]$

If $f(x)=3 x+10, g(x)=x^2-1$, then $(f \circ g)^{-1}$ is equal to

$\left(\frac{x-3}{7}\right)^{1 / 2}$

$\left(\frac{x+3}{7}\right)^{1 / 2}$

On the set of integers Z, define $f:Z \to Z$ as $f\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{n}{2},\;n\;\;is\;\;even}\\ {0,\;\;\;n\;\;\;is\;\;odd} \end{array}} \right\}$, then $'f'$ is

Neither injective nor surjective

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 91-120 of 650 questions

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91 Medium

91 of 650

Analyzing +4 | -1

The domain of the function

f\left( x \right) = \frac{{{{\cos }^{ - 1}}x}}{{\left[ x \right]}}

View answer and explanation

Correct option: A

\left[ { - 1,\;0} \right) \cup \left\{ 1 \right\}

Option A (

\left[ { - 1,\;0} \right) \cup \left\{ 1 \right\}

) is correct because it considers the domain of the inverse cosine function and the restriction that the denominator of the fraction cannot be zero.

92 Easy

92 of 650

Analyzing +4 | -1

f: \boldsymbol{R} \rightarrow \boldsymbol{R}

given by

f(x)=a x+\sin x+a

, then

f

is one-one and onto for all

View answer and explanation

Correct option: B

a \in \mathbf{R}-[-1,1]

Option B (

a \in \mathbf{R}-[-1,1]

) is correct because for f(x) to be one-one and onto, it must be strictly monotonic, which requires that

f'(x) = a + \cos x

be either always positive or always negative, implying

|a| > 1

93 Medium

93 of 650

Analyzing +4 | -1

f\left( {x + 2y,x,x - 2y} \right) = xy

, then

f\left( {x,\;y} \right)

equals

View answer and explanation

Correct option: A

\frac{{{x^2} - {y^2}}}{8}

Option A (

\frac{{{x^2} - {y^2}}}{8}

) is correct because by substituting

x + 2y = u

and

x - 2y = v

, we can express

f(u, v)

in terms of

u

and

v

, leading to the given expression.

94 Easy

94 of 650

Understanding +4 | -1

The function

f:R \to R,

defined by

f\left( x \right) = \left[ x \right],

where

\left[ x \right]

denotes the greatest integer less than or equal to

x

, is

View answer and explanation

Correct option: D

Neither one-one nor onto

Option D (Neither one-one nor onto) is correct because the greatest integer function is neither injective (one-one) nor surjective (onto).

95 Easy

95 of 650

Analyzing +4 | -1

Range of the function

f\left( x \right) = \frac{x}{{1 + {x^2}}}

View answer and explanation

Correct option: C

\left[ { - \frac{1}{2},\frac{1}{2}} \right]

Option C

\left[ { - \frac{1}{2},\frac{1}{2}} \right]

is correct because the range is found by solving the quadratic inequality derived from the function.

96 Medium

96 of 650

Application +4 | -1

f(x)=3 x+10, g(x)=x^2-1

, then

(f \circ g)^{-1}

is equal to

View answer and explanation

Correct option: A

\left(\frac{x-7}{3}\right)^{1 / 2}

Option A (

\left( \frac{{x-7}}{{3}} \right)^{{1 / 2}}

) is correct because

(f \circ g)(x) = 3x^2 + 7

, and its inverse is found by solving for

x

in terms of

y

, resulting in

x = \sqrt{{\frac{{y-7}}{{3}}}}

97 Easy

97 of 650

Understanding +4 | -1

On the set of integers Z, define

f:Z \to Z

f\left( n \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{n}{2},\;n\;\;is\;\;even}\\ {0,\;\;\;n\;\;\;is\;\;odd} \end{array}} \right\}

, then

'f'

View answer and explanation

Correct option: D

Surjective but not injective

Option D (Surjective but not injective) is correct because the function maps multiple odd integers to 0, making it not injective, but covers all integers in the range, making it surjective.

98 Medium

98 of 650

Application +4 | -1

f:R \to R

is a function defined by

f\left( x \right) = 10\;x - 7.

g = {f^{ - 1}},

then

g\left( x \right) =

View answer and explanation

Correct option: C

\frac{{x + 7}}{{10}}

Option C (

\frac{{x + 7}}{{10}}

) is correct because it is the inverse function obtained by swapping x and y in the equation y = 10x - 7 and solving for y.

99 Easy

99 of 650

Application +4 | -1

f\left( x \right) = {\left( {a - {x^n}} \right)^{1/n}}

where

a > 0\;and\;n \in N

, then

fof\left( x \right)

is equal to

View answer and explanation

Correct option: B

x

Option B (

x

) is correct because after applying the function

f

twice, the result simplifies back to the original input

x

100 Easy

100 of 650

Analyzing +4 | -1

The function

f\left( x \right) = {\log _{10}}\left( {x + \sqrt {{x^2} + 1} } \right)

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because it satisfies the condition f(-x) = -f(x), indicating odd symmetry.

101 Easy

101 of 650

Understanding +4 | -1

[x]

denotes the integer part of

x

, then the domain of the function

f(x)=\frac{e^{-x}}{1+[x]}

View answer and explanation

Correct option: A

\mathbb{R}-[-1,0)

Option A is correct because the function is undefined when the denominator is zero, which occurs when [x] = -1, meaning -1 <= x < 0.

102 Easy

102 of 650

Application +4 | -1

Let

f

and

g

be functions defined by

f(x)=\frac{x}{x+1}, g(x)=\frac{x}{1-x}

, then

(f \circ g)(x)

View answer and explanation

Correct option: D

x

Option D (

x

) is correct because when

f

is composed with

g

, the expression simplifies to

x

103 Medium

103 of 650

Application +4 | -1

If f is a real valued function such that

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)\;and\;f\left( 1 \right) = 5

, then the value of

f\left( {100} \right)

View answer and explanation

Correct option: D

500

Option D (500) is correct because the function is linear, and

f(100) = 100 * f(1) = 500

104 Easy

104 of 650

Application +4 | -1

f(x)=x^2+1

, then the value of

(f \circ f)(x)

is equal to

View answer and explanation

Correct option: A

x^4+2 x^2+2

Option A (

x^4 + 2x^2 + 2

) is correct because when you compute

(f \circ f)(x)

with

f(x) = x^2 + 1

, the result simplifies to

x^4 + 2x^2 + 2

105 Medium

105 of 650

Application +4 | -1

The domain of the function

\cos ^{-1} \log _2\left(x^2+5 x+8\right)

is-

View answer and explanation

Correct option: B

[-3,-2]

Option B ([-3,-2]) is correct because it represents the domain for which the function is defined, considering the logarithmic and inverse cosine functions.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

106 Medium

106 of 650

Application +4 | -1

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a function satisfying the relation

f(x+1)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}

for all

x \in \mathbb{R}

. Then period of

f(x)

View answer and explanation

Correct option: A

Option A (2) is correct because after two iterations of the function, f(x+2) simplifies back to f(x), indicating a period of 2.

107 Medium

107 of 650

Application +4 | -1

Let

g\left( x \right) = \cos {x^2},f\left( x \right) = \sqrt x

and

\alpha ,\beta \left( {\alpha < \beta } \right)

be the roots of the quadratic equation

18{x^2} - 9\pi x + {\pi ^2} = 0

Then, the area (in sq units) bounded by the curve

y = \left( {gof} \right)\left( x \right)

and the lines

x = \alpha ,\,x = \beta

and

y=0

View answer and explanation

Correct option: A

\frac{1}{2}\left( {\sqrt 3 - 1} \right)

Option A (

\frac{1}{2}(\sqrt 3 - 1)

) is correct because the area under the curve is found by integrating the composite function gof(x) = cos(x) between the limits alpha and beta, which are the roots of the given quadratic equation.

108 Medium

108 of 650

Analyzing +4 | -1

f\left( x \right) = \sin x + \cos x,\;\;g\left( x \right) = {x^2} - 1

then

g\left( {f\left( x \right)} \right)

is invertible in the domain

View answer and explanation

Correct option: B

\left[ { - \frac{\pi }{4},\;\frac{\pi }{4}} \right]

Option B (

[ - \frac{\pi }{4},\\;\frac{\pi }{4}]

) is correct because the composite function simplifies to sin(2x), which is invertible in the given domain.

109 Medium

109 of 650

Application +4 | -1

f\left(2 x^2+\frac{y^2}{8}, 2 x^2-\frac{y^2}{8}\right)=x y

, then the value of

f(60,48)

+f(80,48)+f(13,5)

must be:

View answer and explanation

Correct option: A

112

The correct answer is A (112). The value is obtained by substituting appropriate values of x and y into the given function and summing the results.

110 Easy

110 of 650

Understanding +4 | -1

The period of

|\sin 2 x|

is:

View answer and explanation

Correct option: B

\frac{\pi}{2}

Option B (

\frac{\pi}{2}

) is correct because the period of the absolute value of a sine function is half the period of the original sine function.

111 Easy

111 of 650

Understanding +4 | -1

The function

f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because f(x) + f(-x) = 0, which is the condition for an odd function.

112 Easy

112 of 650

Application +4 | -1

Let

f:(2,3) \rightarrow(0,1)

be defined by

f(x)=x-[x]

then

f^{-1}(x)

equals

View answer and explanation

Correct option: D

x+2

Option D (

x+2

) is correct because in the interval (2,3),

[x] = 2

, so

f(x) = x-2

. Therefore,

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

113 Easy

113 of 650

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x,\;\;if\;\;x\;\;is\;\;rational}\\ {0,\;\;if\;\;x\;\;is\;\;irrational} \end{array}} \right.

and

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {0,\;\;if\;\;x\;\;is\;\;rational}\\ {x,\;\;if\;\;x\;\;is\;\;irrational\;} \end{array}} \right.

. Then,

f - g

View answer and explanation

Correct option: D

One-one and onto

Option D (One-one and onto) is correct because the function f-g is both injective (one-to-one) and surjective (onto).

114 Easy

114 of 650

Application +4 | -1

2 f\left(x^2\right)+3 f\left(1 / x^2\right)=x^2-1

, then

f\left(x^2\right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derived expression for

f(x^2)

does not match any of the given options.

115 Easy

115 of 650

Application +4 | -1

The domain of definition of the real function

f\left( x \right) = \sqrt {{{\log }_{12}}{x^2}}

of the real variable

x

, is

View answer and explanation

Correct option: B

\left| x \right| \ge 1

Option B (

\left| x \right| \ge 1

) is correct because the logarithm must be non-negative, meaning log_{16}(x²) >= 0, which implies x² >= 1, and thus |x| >= 1.

116 Easy

116 of 650

Application +4 | -1

The domain of the function

f\left( x \right) = {\log _{10}}\left( {\sqrt {x - 4} + \sqrt {6 - x} } \right),

View answer and explanation

Correct option: A

\left[ {4,\;6} \right]

Option A (

\left[ {4,\;6} \right]

) is correct because it represents the intersection of the domains of the square root functions within the logarithm.

117 Easy

117 of 650

Understanding +4 | -1

Let A be a set containing 10 distinct elements, then the total number of distinct function from A to A is

View answer and explanation

Correct option: A

{10^{10}}

Option A (

{10^{10}}

) is correct because each of the 10 elements in A can be mapped to any of the 10 elements in A.

118 Easy

118 of 650

Understanding +4 | -1

The transforms the real line is

f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }},\;\mathrm{is}

View answer and explanation

Correct option: C

\left( { - \infty ,0} \right)

Option C (

\left( { - \infty ,0} \right)

) is correct because the function is defined only for negative values of x.

119 Medium

119 of 650

Analyzing +4 | -1

Let

f:A \to B

be a function defined by

f\left( x \right) = \sqrt 3 \sin x + \cos x + 4.

If f is invertible, then

View answer and explanation

Correct option: A

A = \left[ { - 2\pi /3,\pi /3\left] {,B = } \right[2,\;6} \right]

Option A is correct. To be invertible, the function must be a bijection. By rewriting f(x) as

2\sin(x + \pi/6) + 4

, we find the domain and range that make it a bijection.

120 Medium

120 of 650

Application +4 | -1

The domain of the function

f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}

is (where

[x]

denotes the greatest integer less than or equal to

x

)

View answer and explanation

Correct option: B

(-\infty,-2) \cup[6, \infty)

Option B

((-\infty,-2) \cup[6, \infty))

is correct because the expression inside the square root must be positive, leading to the given domain after considering the greatest integer function.

Revision notes and key takeaways

Review FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

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