If $f:R \to R$ is defined by $f\left( x \right) = \left| x \right|$, then

${f^{ - 1}}\left( x \right) = - x$

${f^{ - 1}}\left( x \right) = \frac{1}{{\left| x \right|}}$

${f^{ - 1}}\left( x \right) = \frac{1}{x}$

The domain of the function ${\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{{1 + x}}{x}} \right)$ is

$\left( { - 1,\, - \frac{1}{2}} \right] \cup \left( {0,\,\infty } \right)$

$\left[ { - \frac{1}{2},\,0} \right) \cup \left[ {1,\,\infty } \right)$

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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 391-420 of 650 questions

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391 Medium

391 of 650

Application +4 | -1

The range of the function

f\left( x \right) = \frac{1}{{2 - \cos 3\;x}},

View answer and explanation

Correct option: C

\left[ {1/3,\;1} \right]

Option C (

\left[ {1/3,\;1} \right]

) is correct because the range is found by determining the minimum and maximum values of the function based on the cosine function's range.

392 Medium

392 of 650

Application +4 | -1

The domain of the function

f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {x/3} \right)} \right)

View answer and explanation

Correct option: A

\left[ {1,\;9} \right]

Option A ([1, 9]) is correct because the argument of the arcsine function must be between -1 and 1, leading to the inequality -1 <= log_3(x/3) <= 1, which simplifies to 1 <= x <= 9.

393 Easy

393 of 650

Analyzing +4 | -1

f:R \to R

is defined by

f\left( x \right) = x - \left[ x \right] - \frac{1}{2}\;for\;x \in R

, where

\left[ x \right]

is the greatest integer not exceeding

x

, then

\left\{ {x \in R:f\left( x \right) = \frac{1}{2}} \right\}

is equal to

View answer and explanation

Correct option: C

\phi

, the empty set

Option C (

\phi

, the empty set) is correct because the equation

f(x) = \frac{1}{2}

leads to a contradiction, meaning there are no real solutions.

394 Easy

394 of 650

Application +4 | -1

If the function

f:\left[ {1,\;\infty } \right) \to \left[ {1,\;\infty } \right)

is defined by

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

then

{f^{ - 1}}\left( x \right)

View answer and explanation

Correct option: B

\frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}y} } \right)

Option B (

\frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}y} } \right)

) is correct because it is the inverse function obtained by solving for x in terms of y and considering the domain of the original function.

395 Easy

395 of 650

Understanding +4 | -1

The minimum value of

f\left( x \right) = \left| {3 - x} \right| + 7

View answer and explanation

Correct option: C

Option B (

2\,\,\sqrt {f\,\,\left( x \right)}

) is correct because the derivative of

2\sqrt{f(x)}

\frac{f'(x)}{\sqrt{f(x)}}

396 Medium

396 of 650

Application +4 | -1

The value of the function

f\left( x \right) = 3\sin \left( {\sqrt {\frac{{{\pi ^2}}}{{16}} - {x^2}} } \right)

lies in the interval

View answer and explanation

Correct option: B

\left[ {0,\;3/\sqrt 2 } \right]

Option B ([0, 3/√(2)]) is correct because it represents the range of the function, considering the domain restriction and the sine function's properties.

397 Easy

397 of 650

Understanding +4 | -1

The number of all one-one functions from set

A=\{1,2,3\}

to itself is

View answer and explanation

Correct option: B

Option B (6) is correct because the number of one-to-one functions from a set to itself is the number of permutations, which is n!.

398 Easy

398 of 650

Analyzing +4 | -1

a,b

are two fixed positive integers such that

f\left( {a + x} \right) = b + {\left[ {{b^3} + 1 - 3{b^2}f\left( x \right) + 3b\;{{\left\{ {f\left( x \right)} \right\}}^2} - {{\left\{ {f\left( x \right)} \right\}}^3}} \right]^{1/3}}

For all

x \in R,

then

f\left( x \right)

is a periodic function with period

View answer and explanation

Correct option: B

2a

The correct answer is B (2a) because by substituting f(a+x) into the original equation twice, we find that f(x+2a) = f(x), indicating a period of 2a.

399 Easy

399 of 650

Understanding +4 | -1

f:R \to R

is defined by

f\left( x \right) = \left| x \right|

, then

View answer and explanation

Correct option: C

The function

{f^{ - 1}}\left( x \right)

does not exist

Option C (The function

{f^{ - 1}}\left( x \right)

does not exist) is correct because the absolute value function is not one-to-one, and therefore does not have an inverse.

400 Easy

400 of 650

Understanding +4 | -1

The graph of the function

y=f(x)

is symmetrical about the line

x=2

, then

View answer and explanation

Correct option: B

f(2+x)=f(2-x)

Option B (

f(2+x)=f(2-x)

) is correct because it represents the condition for symmetry about the line x=2, meaning the function value is the same for points equidistant from x=2.

401 Easy

401 of 650

Understanding +4 | -1

Let

\mathrm{A}

and

\mathrm{B}

be two finite sets having

m

and

\mathrm{n}

elements respectively. Then, the total number of mappings from

A

\mathrm{B}

is :

View answer and explanation

Correct option: D

\mathrm{n}^{\mathrm{m}}

Option D (

n^m

) is correct because each of the m elements in set A can be mapped to any of the n elements in set B, resulting in

n^m

possible mappings.

402 Medium

402 of 650

Application +4 | -1

\left| {2x - 3} \right| < \left| {x + 5} \right|,

then

x

belongs to

View answer and explanation

Correct option: C

\left( { - \frac{2}{3},8} \right)

Option C

\left( { - \frac{2}{3},8} \right)

is correct because it represents the solution set of the inequality after considering different cases based on the absolute values.

403 Easy

403 of 650

Application +4 | -1

The range of '

\mathrm{a}

' for which

f(x)=a x+\cos x

is one-one is:

View answer and explanation

Correct option: B

(-\infty,-1] \cup[1, \infty)

Option B (

(-\infty,-1] \cup[1, \infty)

) is correct because the function is one-to-one when its derivative is either always non-negative or always non-positive.

404 Easy

404 of 650

Analyzing +4 | -1

Let

f:\{x, y, z\} \rightarrow\{1,2,3\}

be a one-one function. If it is given that exactly one of the following statements is true,

Statement-1:

f(x)=1

, Statement-2:

f(y) \neq 1

, Statement-3:

f(z) \neq 2

.

then

f^{-1}

(1) is

View answer and explanation

Correct option: B

y

Option B (y) is correct because if f(x) = 1, then f(y) cannot be 1, violating the one-one property. If f(z) = 1, statements 2 and 3 are both true, which is not allowed. Therefore, f(y) = 1.

405 Easy

405 of 650

Application +4 | -1

{x^2} + 6x - 27 < 0\;

and

\;{x^2} - 3x - 4 < 0,

then

View answer and explanation

Correct option: C

3 < x < 4

Option C (

3 < x < 4

) is correct because it is the intersection of the solution sets of the two inequalities

{x^2} + 6x - 27 < 0

and

{x^2} - 3x - 4 < 0

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406 Easy

406 of 650

Understanding +4 | -1

Domain of the function

f(x)=\frac{x^{2}-3 x+2}{x^{2}+x-6}

View answer and explanation

Correct option: D

\{x: x \in R, x \neq 2, x \neq-3\}

Option D (

\\{x: x \in R, x \neq 2, x \neq-3\\}

) is correct because the denominator cannot be zero, so

x^2 + x - 6 \neq 0

, which means

(x-2)(x+3) \neq 0

, so

x \neq 2

and

x \neq -3

407 Hard

407 of 650

Application +4 | -1

The domain of the function

\mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\frac{{1 + x}}{x}} \right)

View answer and explanation

Correct option: D

\left[ { - \frac{1}{2},\,\infty } \right) - \left\{ 0 \right\}

Option D (

\left[ { - \frac{1}{2},\,\infty } \right) - \left\{ 0 \right\}

) is correct because it represents the values of x for which the argument of the inverse cosecant function is greater than or equal to 1 in absolute value.

408 Medium

408 of 650

Understanding +4 | -1

Let

A=\{1,2,3,4\}

and

R

be the relation on

A

defined by

\{(a, b): a, b \in A, a \times b

is an even number

\}

, then the range of

R

View answer and explanation

Correct option: B

\{2,4\}

Option B (

\{2,4\}

) is correct because the range of the relation R consists of the second elements of the ordered pairs (a, b) where a * b is even.

409 Medium

409 of 650

Application +4 | -1

f\left( x \right)

is defined on

\left[ {0,\;1} \right],

then the domain of definition of

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

View answer and explanation

Correct option: A

\left[ {n\;\pi ,n\;\pi + \pi /4} \right],\;n \in Z

Option A (

[n \pi ,n \pi + \pi /4],\;n \in Z

) is correct because it defines the interval where

0 \le \tan x \le 1

410 Easy

410 of 650

Application +4 | -1

f:R \to R,

defined by

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

then the values of

{f^{ - 1}}\left( {17} \right)\;and\;{f^{ - 1}}\left( { - 3} \right)

respectively are

View answer and explanation

Correct option: C

\left\{ {4, - 4} \right\},\mathrm{\;}\phi

Option C (

\{\4, -4\}, \phi

) is correct because when x=17, the inverse function yields 4 and -4, but when x=-3, the inverse function has no real solution.

411 Medium

411 of 650

Application +4 | -1

A real valued function

f(x)

satisfies the function equation

f(x-y)=f(x) f(y)-f(a-x) f(a+y)

where

a

is a given constant and

f(0)=1, f(2 a-x)

is equal to

View answer and explanation

Correct option: C

-f(x)

Option C (

-f(x)

) is correct because substituting

f(0)=1

and

f(a)=0

into the given functional equation leads to

f(2a-x) = -f(x)

412 Easy

412 of 650

Understanding +4 | -1

The composite mapping

f \circ g

of the maps

f: R \rightarrow R, f(x)=\sin x, g: R \rightarrow R, g(x)=x^2

View answer and explanation

Correct option: C

\sin x^2

Option C (

\sin x^2

) is correct because the composite mapping

f \circ g

is defined as

f(g(x))

, which means we substitute

g(x) = x^2

into

f(x) = \sin x

, resulting in

\sin(x^2)

413 Easy

413 of 650

Application +4 | -1

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

, where

a > 0

and

n

is a positive integer, then

f[f(x)]=

View answer and explanation

Correct option: C

x

Option C (

x

) is correct because substituting

f(x)

into itself simplifies to

x

due to the inverse relationship created by the exponents and the subtraction from

a

414 Easy

414 of 650

Analyzing +4 | -1

Which of the following functions is inverse of itself?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because applying the function to itself results in the original input x, meaning it's its own inverse.

415 Hard

415 of 650

Application +4 | -1

Let

f:N \to N

be a function such that

f(m + n) = f(m) + f(n)

for every

m,\,\,n \in N.

f(6) = 18,

then

f(2).f(3)

is equal to

View answer and explanation

Correct option: B

Option B (54) is correct because by using the given functional equation and the value of f(6), we can find f(2) and f(3) and then multiply them.

416 Medium

416 of 650

Application +4 | -1

f(x)=\sin \log x

, then the value of

f(x y)+f\left(\frac{x}{y}\right)-2 f(x) \cdot \cos \log y

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because the expression simplifies to 0 after applying trigonometric identities and the properties of logarithms.

417 Medium

417 of 650

Understanding +4 | -1

The domain of

F(x)=\frac{\log _2(x+3)}{x^2+3 x+2}

View answer and explanation

Correct option: D

(-3, \infty)-\{-1,-2\}

Option D (

(-3, \infty)-\{-1,-2\}

) is correct because the argument of the logarithm must be positive, and the denominator cannot be zero.

418 Easy

418 of 650

Application +4 | -1

g(x)=1+\sqrt[3]{x}

then a function

f

such that

f(g(x))=3-3(\sqrt[3]{x})+x

View answer and explanation

Correct option: A

f(x)=x^3-3 x^2+x+5

Option C is correct because substituting g(x) into f(x) yields the given expression after simplification.

419 Easy

419 of 650

Analyzing +4 | -1

\left( {x - 1} \right)\left( {{x^2} - 5x + 7} \right) < \left( {x - 1} \right),

then

x

belongs to

View answer and explanation

Correct option: C

\left( { - \infty ,1} \right) \cup \left( {2,3} \right)

Option C,

\left( { - \infty ,1} \right) \cup \left( {2,3} \right)

, is correct because it represents the solution set obtained by solving the inequality after factoring.

420 Medium

420 of 650

Application +4 | -1

The range of the function

{ }^{7-x} P_{x-3}

View answer and explanation

Correct option: A

\{1,2,3\}

Option A is correct because the range of the function is {1, 2, 3} by evaluating the permutation for valid x values.

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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