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

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

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

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

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 · Easy

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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 61-90 of 478 questions

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61 Easy

61 of 478

Understanding +4 | -1

Let

f(x)=\log _e(\sqrt{x-3}+\sqrt{5-x}), x \in R

. Then domain of

f(x)

is -

View answer and explanation

Correct option: A

[3,5]

Option A ([3,5]) is correct because the domain of the function is determined by the intersection of the domains of the square roots, ensuring non-negative values inside the square roots.

62 Easy

62 of 478

Application +4 | -1

5\{x\}=x+[x]

and

[x]-\{x\}=\frac{1}{2}

when

\{x\}

and [

x

] are fractional and integral part of

x

then

x

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because solving the given equations

[x] - \{x\} = \frac{1}{2}

and

5\{x\} = x + [x]

leads to

x = \frac{3}{2}

63 Easy

63 of 478

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

64 Easy

64 of 478

Application +4 | -1

The values of

b

and

c

for which the identity

f(x+1)-f(x)=8 x+3

is satisfied, where

f(x)=b x^2+c x+d

are

View answer and explanation

Correct option: B

b=4, c=-1

Option B (

b=4, c=-1

) is correct because substituting

f(x)=bx^2+cx+d

into

f(x+1)-f(x)=8x+3

and solving for

b

and

c

yields these values.

65 Easy

65 of 478

Application +4 | -1

f:[1, \infty) \rightarrow[2, \infty)

is given by

f(x)=x+\frac{1}{x}

, then

f^{-1}(x)

equals

View answer and explanation

Correct option: A

\frac{x+\sqrt{x^2-4}}{2}

Option A (

\frac{{x+\sqrt{{x^2-4}}}}{{2}}

) is correct because it represents the inverse function of

f(x) = x + \frac{{1}}{{x}}

with the given domain and range.

66 Easy

66 of 478

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).

67 Easy

67 of 478

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.

68 Easy

68 of 478

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.

69 Easy

69 of 478

Analyzing +4 | -1

Which of the function defined below is one-one?

View answer and explanation

Correct option: B

f:[0, \infty) \rightarrow R, f(x)=x^2+4 x-5

Option B (

f:[0, \infty) \rightarrow R, f(x)=x^2+4 x-5

) is correct because it is a quadratic function with a restricted domain that makes it one-to-one.

70 Easy

70 of 478

Application +4 | -1

If domain of

y=f(x)

x \in[-3,2]

, then domain of

y=f(|[x]|):

(where

[\cdot]

denotes greatest integer function)

View answer and explanation

Correct option: B

[-2,3)

Option B ([-2,3)) is correct because it represents the domain of the composite function after considering the greatest integer function and absolute value.

71 Easy

71 of 478

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

72 Easy

72 of 478

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.

73 Easy

73 of 478

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.

74 Easy

74 of 478

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

75 Easy

75 of 478

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

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76 Easy

76 of 478

Application +4 | -1

If the function

\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2+3 \mathrm{x}+5

satisfies LMVT at

\mathrm{x}=2

on the closed interval

[1, \mathrm{a}]

then the value of '

\mathrm{a}

' is equal to -

View answer and explanation

Correct option: A

Option A (3) is correct because it satisfies the LMVT condition at x=2 on the closed interval [1, a].

77 Easy

77 of 478

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.

78 Easy

78 of 478

Application +4 | -1

Range of the function

f(x)=\sin ^2 x-4 \sin x+5

View answer and explanation

Correct option: B

[2,10]

Option B ([2,10]) is correct because by completing the square, we find the function's minimum and maximum values within the range of sin(x).

79 Easy

79 of 478

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.

80 Easy

80 of 478

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

81 Easy

81 of 478

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).

82 Easy

82 of 478

Application +4 | -1

f(x)=\frac{4^x}{4^x+2}

, then

f(x)+f(1-x)

is equal to-

View answer and explanation

Correct option: C

Option C (1) is correct because substituting (1-x) into the function and adding it to the original function results in 1.

83 Easy

83 of 478

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.

84 Easy

84 of 478

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.

85 Easy

85 of 478

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.

86 Easy

86 of 478

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.

87 Easy

87 of 478

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.

88 Easy

88 of 478

Application +4 | -1

f(x)=\frac{x-|x|}{|x|}

, then

f(-1)=

View answer and explanation

Correct option: B

-2

Option B (-2) is correct because substituting x = -1 into the function

f(x)=\frac{{x-|x|}}{{|x|}}

results in -2.

89 Easy

89 of 478

Application +4 | -1

f(x) = \,{\log _e}\left( {\frac{{1 - x}}{{1 + x}}} \right),\,\,\left| x \right| < 1,\,

then

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

is equal to

View answer and explanation

Correct option: A

2f(x)

Option A (

2f(x)

) is correct because substituting

\frac{{2x}}{{1 + {x^2}}}

into the function and simplifying yields twice the original function.

90 Easy

90 of 478

Analyzing +4 | -1

The function

f: \mathbf{R} \rightarrow \mathbf{R}

given by

f(x)=3-2 \sin x

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the function

f(x) = 3 - 2\sin x

is neither one-to-one nor onto, as its range is [1, 5] and it's periodic.

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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