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

Neither an even nor an odd function

If $f(x+2 a)=f(x-2 a)$, then $f(x)$ is

a periodic function with period $2 \mathrm{a}$

a periodic function with indeterminate period

Let $f\left( x \right) = \frac{{\sqrt {\sin x} }}{{1 + \sqrt[3]{{\sin x}}}}.$ If $D$ is the domain of f, then $D$ contains

$\left( { - 2\;\pi , - \pi } \right)$

$\left( {3\;\pi ,4\;\pi } \right)$

$\left( {4\;\pi ,6\;\pi } \right)$

If $f(x)=\frac{2 x-1}{x+5}(x \neq-5)$, then $f^{-1}(x)$ is equal to

$\frac{x+5}{2 x-1}, x \neq \frac{1}{2}$

$\frac{x-5}{2 x+1}, x \neq \frac{1}{2}$

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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 511-540 of 650 questions

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511 Medium

511 of 650

Application +4 | -1

The period of

f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2

is:

View answer and explanation

Correct option: C

2 n(n-1)

The correct answer is C (2n(n-1)) because it is the least common multiple of the periods of the sine and cosine functions.

512 Easy

512 of 650

Understanding +4 | -1

The function

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

View answer and explanation

Correct option: C

An odd function

Option C (An odd function) is correct because f(-x) = -f(x) for all x in the domain, which is the definition of an odd function.

513 Medium

513 of 650

Analyzing +4 | -1

The number of solutions of the equation

{\log _{x - 3}}\left( {{x^3} - 3{x^2} - 4x + 8} \right) = 3,

View answer and explanation

Correct option: A

Option A (1) is correct because only one of the potential solutions satisfies the logarithmic equation's domain restrictions.

514 Easy

514 of 650

Understanding +4 | -1

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

, then

f(x)

View answer and explanation

Correct option: A

a periodic function with period

4 \mathrm{a}

Option B (Only transitive) is correct because if x < y and y < z, then x < z, which satisfies the transitive property. It is neither reflexive nor symmetric.

515 Easy

515 of 650

Application +4 | -1

The solution set of the inequation

{\log _{\sin 2\pi /3}}\left( {{x^2} - 3x + 2} \right) \ge 2,

View answer and explanation

Correct option: C

\left[ {1/2,\;1} \right) \cup \left( {2,\;5/2} \right]

Option C

\left[ {1/2,\;1} \right) \cup \left( {2,\;5/2} \right]

is correct because it is the solution set of the given logarithmic inequality considering the domain.

516 Hard

516 of 650

Application +4 | -1

[x]

be the greatest integer less than or equal to

x,

then

\sum\limits_{n = 8}^{100} {\left[ {\frac{{{{( - 1)}^n}n}}{2}} \right]}

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because the sum telescopes, leaving only the first few terms after cancellations.

517 Medium

517 of 650

Understanding +4 | -1

Let

f\left( x \right) = \frac{{\sqrt {\sin x} }}{{1 + \sqrt[3]{{\sin x}}}}.

D

is the domain of f, then

D

contains

View answer and explanation

Correct option: A

\left( {0,\;\pi } \right)

Option A (

\left( {0,\;\pi } \right)

) is correct because

\sin x \ge 0

in the interval

\left( {0,\;\pi } \right)

518 Easy

518 of 650

Application +4 | -1

f:R \to R

defined by

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

, then

{f^{ - 1}}\left( 8 \right)

is equal to

View answer and explanation

Correct option: A

\left\{ 2 \right\}

Option A (

\{ 2 \}

) is correct because the inverse function of

f(x) = x^3

evaluated at 8 is 2.

519 Easy

519 of 650

Application +4 | -1

The domain of the function

f(x)=\sin ^{-1}\left(\frac{x^2-3 x+2}{x^2+2 x+7}\right)

is :

View answer and explanation

Correct option: C

[-1, \infty)

Option C ([-1,

\infty

)) is correct because the domain is determined by the range of the arcsin function, which requires the argument to be between -1 and 1.

520 Medium

520 of 650

Application +4 | -1

Let

f\left( {x + \frac{1}{x}} \right) = {x^2} + \frac{1}{{{x^2}}},\left( {x \ne 0} \right)

then

f\left( x \right)

equals

View answer and explanation

Correct option: B

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

Option B (

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

) is correct because it correctly expresses f(x) in terms of x, considering the domain restriction.

521 Medium

521 of 650

Understanding +4 | -1

Let

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

and

h(x)=f(x) \cdot g(x)

, then

h(x)=1

View answer and explanation

Correct option: D

\forall x \in R, x \neq 0

Option D (

\forall x \in R, x \neq 0

) is correct because

h(x) = f(x) \cdot g(x) = x \cdot \frac{{1}}{{x}} = 1

only when

x

is a real number not equal to 0, as

g(x)

is undefined at

x=0

522 Hard

522 of 650

Analyzing +4 | -1

Let

x = \left[ {\frac{{a + 2b}}{{a + b}}} \right]

and

y = \frac{a}{b}

, where

a

and

b

are positive integers. If

{y^2} > 2

then

View answer and explanation

Correct option: B

{x^2} < 2

Option B (

{x^2} < 2

) is correct because given the conditions, the value of x is such that its square is less than 2.

523 Easy

523 of 650

Understanding +4 | -1

The period of the function

f(x)=\sin ^2 x

is:

View answer and explanation

Correct option: B

\pi

Option B (

\pi

) is correct because the period of

\sin^2 x

can be found by rewriting it as

\frac{{1 - \cos 2x}}{{2}}

, which has a period of

\pi

524 Medium

524 of 650

Understanding +4 | -1

Assertion : The function defined by

f(x)=\cos \left(x^2\right)

is a continuous function.

Reason : The cosine function is continuous in its domain i.e.,

x \in R

View answer and explanation

Correct option: B

Assertion is correct, reason is correct; reason is not a correct explanation for assertion

Option B is correct because both the assertion and reason are correct, but the reason doesn't directly explain the assertion.

525 Easy

525 of 650

Application +4 | -1

f:R \rightarrow R

and

g:R \rightarrow R

are defined by

f(x)=2x+3

and

g(x)=x^2+7

, then values of

x

such that

g\left\{ {f\left( x \right)} \right\} = 8

are

View answer and explanation

Correct option: C

-1,-2

Option C (-1,-2) is correct because these are the x-values that satisfy the equation g{f(x)} = 8.

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526 Easy

526 of 650

Application +4 | -1

\phi(x)=\frac{1}{1+e^{-x}}

and

S=\phi(5)+\phi(4)+\phi(3)+\ldots .+

\phi(-3)+\phi(-4)+\phi(-5)

, then the value of

S

View answer and explanation

Correct option: A

5.5

Option A (5.5) is correct because the sum

S

simplifies to 5.5 by pairing terms using the property

\phi(x) + \phi(-x) = 1

527 Medium

527 of 650

Application +4 | -1

Suppose

f: \mathrm{R} \rightarrow(0, \infty)

be a differentiable function such that

5 f(x+y)=f(x) \cdot f(y), \forall x, y \in \mathrm{R}

. If

f(3)=320

, then

\sum\limits_{{n=0}}^5 f(n)

is equal to

View answer and explanation

Correct option: A

6825

Option A (6825) is correct because it is the sum of the function values from n=0 to n=5, given the functional equation and f(3)=320.

528 Medium

528 of 650

Application +4 | -1

The solution set of the inequation

{\log _{1/3}}\left( {{x^2} + x + 1} \right) + 1 > 0,

View answer and explanation

Correct option: C

\left( { - 2,\;1} \right)

Option C

\left( { - 2,\;1} \right)

is correct because it is the solution set of the given logarithmic inequality.

529 Easy

529 of 650

Analyzing +4 | -1

Let

[x]

and

\{x\}

denote the integral part and fractional part of

x

, respectively. Then the number of solutions of the equation

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

View answer and explanation

Correct option: B

The correct answer is B (2) because there are two solutions to the equation

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

, which are

x=0

and

x=5/3

, found by analyzing the properties of the fractional and integral parts of x.

530 Easy

530 of 650

Application +4 | -1

f(x)=\frac{2 x-1}{x+5}(x \neq-5)

, then

f^{-1}(x)

is equal to

View answer and explanation

Correct option: B

\frac{5 x+1}{2-x}, x \neq 2

Option B is correct because solving for x in terms of y and swapping x and y gives

f^{-1}(x) = \frac{5x+1}{2-x}

531 Easy

531 of 650

Application +4 | -1

f(x)=2 x^{n}+\lambda

and

f(4)=133, f(5)=255

, then sum of positive integral divisors of

f(3)-f(2)

View answer and explanation

Correct option: A

Option B (72) is correct because there are 3 choices for the first digit (5, 7, or 9) and then 4! ways to arrange the remaining 4 digits.

532 Medium

532 of 650

Application +4 | -1

The set of all

x

satisfying the inequality

\frac{{4x - 1}}{{3x + 1}} \ge 1

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the set of all x that satisfy the given inequality after simplification.

533 Easy

533 of 650

Application +4 | -1

f\left( x \right) = {a^x},

which of the following equalities hold?

View answer and explanation

Correct option: A

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

Option A (

f(x + 2) - 2f(x + 1) + f(x) = (a - 1)^2f(x)

) is correct because it simplifies to an identity using the properties of exponents.

534 Medium

534 of 650

Analyzing +4 | -1

f\left( x \right) = \frac{{{{\sin }^4}x + {{\cos }^2}x}}{{{{\sin }^2}x\; + {{\cos }^4}x}}

for

x \in R,

then

f\left( {2010} \right) =

View answer and explanation

Correct option: A

Option A (1) is correct because the function simplifies to 1 for all real values of x due to trigonometric identities.

535 Easy

535 of 650

Application +4 | -1

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

, then

(f+g)\left(\frac{\pi}{3}\right)=

View answer and explanation

Correct option: C

\sqrt{3}+\frac{1}{4}

Option C (

\sqrt{3}+\frac{1}{4}

) is correct because it is the result of evaluating

2\sin(x) + \cos^2(x)

x = \frac{\pi}{3}

536 Hard

536 of 650

Analyzing +4 | -1

Set of values of

x

lying in

\left[ {0,\;\,2\pi } \right]

satisfying the inequality

|\sin x|\, > 2{\sin ^2}x

contains

View answer and explanation

Correct option: A

\left( {\;0,\,\,\frac{\pi }{6}} \right)\,\, \cup \,\,\left( {\;\pi ,\,\frac{{7\pi }}{6}} \right)

Option A (

\left( {\;0,\,\frac{\pi }{{6}}} \right)\\ \cup \\ \left( {\;\pi ,\frac{{7\pi }}{{6}}} \right)

) is correct because it represents the intervals where the absolute value of sin x is between 0 and 1/2, excluding points where sin x is 0.

537 Easy

537 of 650

Application +4 | -1

f:R \to R\;and\;g:R \to R

are defined by

f\left( x \right) = \left| x \right|\;and\;g\left( x \right) = \left[ {x - 3} \right]

for

x \in R

, then

\left\{ {g\left( {f\left( x \right)} \right): - \frac{8}{5} < x < \frac{8}{5}} \right\}

is equal to

View answer and explanation

Correct option: C

\left\{ { - 3,\mathrm{ } - 2} \right\}

Option C (

\left\{ { - 3,\; - 2} \right\}

) is correct because it represents the set of values obtained by evaluating g(f(x)) for the given range of x.

538 Medium

538 of 650

Application +4 | -1

f:R \to R\;and\;g\;:R \to R

are defined by

f\left( x \right) = 2x + 3\;and\;g\left( x \right) = {x^2} + 7,

then the values of x such that

g\left( {f\left( x \right)} \right) = 8

are

View answer and explanation

Correct option: C

- 1,\; - 2

Option C (-1, -2) is correct because these are the solutions to the equation g(f(x)) = 8, found by substituting and solving the quadratic.

539 Easy

539 of 650

Application +4 | -1

f(x)=2 x

and

g

is identity function, then

View answer and explanation

Correct option: C

(f \circ g)(x)=(g+g)(x)

Option D (

(f \circ g)(x)=(g+g)(x)

) is correct because if

f(x) = 2x

and g is the identity function, then both sides of the equation simplify to

2x

540 Medium

540 of 650

Application +4 | -1

The function

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

satisfies the equation

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it satisfies the given functional equation.

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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