The domain of ${\sin ^{ - 1}}\left( {{{\log }_3}x} \right)$ is

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

Domain of the function $ f(x)=\sqrt{2-2 x-x^{2}} $ is

$ -\sqrt{3} \leq x \leq \sqrt{3} $

$ -2+\sqrt{3} \leq x \leq-2-\sqrt{3} $

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

$\left( { - 1,{\rm{ }}0} \right)$

$\left( { - 1,{\rm{ }}1} \right)$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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Practice format and chapter quiz

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 241-270 of 650 questions

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241 Easy

241 of 650

Analyzing +4 | -1

If the function

f:R - \left\{ {1, - 1} \right\} \to A

defined by

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

, is surjective, then

A

is equal to

View answer and explanation

Correct option: C

R - \left[ { - 1,\,0} \right)

Option C (

R - [-1, 0)

) is correct because the range of the function

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

R - [-1, 0)

, and for a surjective function, the range equals the codomain.

242 Easy

242 of 650

Application +4 | -1

The domain of

{\sin ^{ - 1}}\left( {{{\log }_3}x} \right)

View answer and explanation

Correct option: D

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

Option D ([{

\frac{1}{3}

,\;3}]) is correct because it represents the values of x for which the argument of the inverse sine function, log base 3 of x, lies between -1 and 1.

243 Easy

243 of 650

Analyzing +4 | -1

Let

[t]

denote the greatest integer

\le t.

Then the equation

x,[{x^2}] + 2[x + 2] - 7 = 0

has

View answer and explanation

Correct option: A

Infinitely many solutions

Option A (Infinitely many solutions) is correct because the equation simplifies to a condition on [x], leading to intervals of solutions.

244 Medium

244 of 650

Application +4 | -1

g(x)= x^2+x-2

and

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

, then

f(x)

is equal to:

View answer and explanation

Correct option: A

2 x-3

The correct answer is A (2x-3) because it is the function f(x) that satisfies the given composition equation.

245 Easy

245 of 650

Understanding +4 | -1

Let

f(x)=\cos (p x)+\sin x

be periodic, then

p

must be:

View answer and explanation

Correct option: C

Rational

Option C (Rational) is correct because for the sum of cosine and sine functions to be periodic, the ratio of their frequencies must be rational.

246 Easy

246 of 650

Analyzing +4 | -1

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

is a function from

R\;\;\mathrm{to}\;R

, then

f\left( x \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function simplifies to either 0 or 2x, depending on the value of x, and is therefore not injective, surjective, or bijective.

247 Easy

247 of 650

Application +4 | -1

The domain of the function

f\left( x \right) = {\log _2}\left( {{{\log }_3}({{\log }_4}x} \right))

View answer and explanation

Correct option: B

\left( {4,\;\infty } \right)

The correct answer is B ((4,∞)) because the argument of each logarithm must be positive, leading to the condition x > 4.

248 Medium

248 of 650

Application +4 | -1

f:R \to S

defined by

f\left( x \right) = \sin \,\,x - \sqrt 3 \,\,\cos \,\,x + 1

, is onto, then the interval of

S

View answer and explanation

Correct option: D

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

Option D is correct because the range of the function is determined by finding the minimum and maximum values of the sine and cosine expression.

249 Easy

249 of 650

Application +4 | -1

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

, then

f\left[\frac{2 x}{1+x^{2}}\right]

is equal to

View answer and explanation

Correct option: C

2 f(x)

Option C (

2 f(x)

) is correct because substituting

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

into

f(x)

and simplifying using logarithm properties results in

2f(x)

250 Medium

250 of 650

Understanding +4 | -1

The function

f:R \to R

is defined by

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

. Observe the following statements

I. f is one-one II. f is onto

III. f is a decreasing function

Out of these, true statement are

View answer and explanation

Correct option: C

Only I, III

Option C (Only I, III) is correct because the function is one-to-one and decreasing, but not onto.

251 Easy

251 of 650

Understanding +4 | -1

The function

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

View answer and explanation

Correct option: A

Periodic with period

2\;\pi .

Option A (Periodic with period

2\\;\pi .

) is correct because the sine function has a period of

2\pi

, and the given function involves the sine function in the exponent.

252 Medium

252 of 650

Application +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: A

2 f(x)

Option A (

2 f(x)

) is correct because substituting

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

into the function and simplifying using logarithm properties results in

2f(x)

253 Easy

253 of 650

Analyzing +4 | -1

The function

f\left( x \right) = \frac{{{{\sin }^4}x + \;\mathrm{co}{\mathrm{s}^4}\;x}}{{{x^3} + {x^4}\tan x}}

View answer and explanation

Correct option: B

Odd

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

254 Easy

254 of 650

Analyzing +4 | -1

If a function

f\;:\left[ {2,\;\infty } \right) \to B

defined by

f\left( x \right) = {x^2} - 4\;x + 5

is a bijection, then

B =

View answer and explanation

Correct option: B

\left[ {1,\;\infty } \right)

Option B is correct. Since f is a bijection, B must be the range of f. Completing the square,

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

. For

x \in [2, \infty)

, the range is

[1, \infty)

255 Easy

255 of 650

Application +4 | -1

A real valued function

f(x)

satisfies the functional equation

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

where a is a given constant and

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

is equal to

View answer and explanation

Correct option: A

-f(x)

Option A (

-f(x)

) is correct because substituting specific values into the functional equation and using the given condition f(0) = 0 leads to f(2a-x) = -f(x).

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256 Easy

256 of 650

Analyzing +4 | -1

The number of solutions of

\left[ {\sin x + \cos x} \right] = 3 + \left[ { - \sin x} \right] + \left[ { - \cos x} \right]

in the internal

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

is (where [.] denotes the greatest integer function)

View answer and explanation

Correct option: C

Infinite

Option C (Infinite) is correct because the two expressions are equivalent, leading to infinite solutions within the given interval.

257 Easy

257 of 650

Application +4 | -1

Let

f:R \to R

be a function which satisfies

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)\forall x,\,\,\,y \in R.\,\,

f\left( 1 \right) = 2

and

g\left( n \right) = \sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)},

n \in N,

then the value of

n,

for which

g\left( n \right) = 20

View answer and explanation

Correct option: A

The correct answer is A (5) because it is the value of n that satisfies the given functional equation and summation condition.

258 Easy

258 of 650

Understanding +4 | -1

The domain of definition of the function

f\left( x \right) = \sqrt[3]{{\frac{{2x + 1}}{{{x^2} - 10x - 11}}}},\;\mathrm{is}

View answer and explanation

Correct option: C

R - \left\{ { - 1,\;11} \right\}

Option C (

R - \left\{ { - 1,\\;11} \right\}

) is correct because the function is defined for all real numbers except where the denominator is zero.

259 Easy

259 of 650

Application +4 | -1

Let

A = \left\{ {x \in R:x \le 1} \right\}\;and\,f:A \to A

be defined as

f\left( x \right) = x\left( {2 - x} \right).

Then,

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

View answer and explanation

Correct option: B

1 - \sqrt {1 - x}

Option B (

1 - \sqrt {1 - x}

) is correct because by setting f(x) = y and solving for x, considering the domain restriction x <= 1, we find the inverse function.

260 Easy

260 of 650

Analyzing +4 | -1

The domain of definition of

f\left( x \right) = \sqrt {\frac{{{{\log }_{0.3}}\left| {x - 2} \right|}}{{\left| x \right|}}} ,

View answer and explanation

Correct option: A

\left[ {1,\;2} \right) \cup \left( {2,\;3} \right]

Option A (

[ {1,\\;2} ) \cup \left( {2,\\;3} ]

) is correct because it satisfies the conditions for the logarithm and the square root to be defined.

261 Easy

261 of 650

Application +4 | -1

Let

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

, then

f\left( x \right)

is equal to

View answer and explanation

Correct option: C

{x^2} - 2

Option C (

x^2 - 2

) is correct because substituting

x + \frac{1}{x}

into the function results in

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

262 Easy

262 of 650

Analyzing +4 | -1

Which of the four statements given below is different from others?

View answer and explanation

Correct option: B

f: x \rightarrow f(x)

Option B (

f: x \rightarrow f(x)

) is correct because it represents the mapping of an element to its image, while others describe the function itself.

263 Easy

263 of 650

Application +4 | -1

Let

R

be a relation in

N

defined by

R=\{(x, y): x+2 y=8\}

. The range of

R

View answer and explanation

Correct option: B

\{1,2,3\}

Option B (

\left\{1,2,3\right\}

) is correct because it represents the set of all possible

y

values (range) for which

x + 2y = 8

and

x, y

are natural numbers.

264 Easy

264 of 650

Understanding +4 | -1

If the mappings

f: A \rightarrow B

and

g: B \rightarrow C

are both bijective, then the mapping

g o f: A \rightarrow C

View answer and explanation

Correct option: B

One-one and onto

Option B (One-one and onto) is correct because the composition of two bijective functions is also bijective.

265 Easy

265 of 650

Application +4 | -1

Domain of the function

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

View answer and explanation

Correct option: B

-1-\sqrt{3} \leq x \leq-1+\sqrt{3}

Option B (-1-√3 ≤ x ≤ -1+√3) is correct because the domain requires the expression inside the square root to be non-negative.

266 Easy

266 of 650

Application +4 | -1

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

, then

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

will be equal to

View answer and explanation

Correct option: D

2f\left( x \right)

Option D (

2f\left( x \right)

) is correct because substituting

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

into

f(x)

and simplifying results in

2f(x)

267 Medium

267 of 650

Analyzing +4 | -1

Range of the function

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

View answer and explanation

Correct option: C

\left[ {0,\mathrm{ }1} \right)

Option C (

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

) is correct because the function is always non-negative and less than 1, approaching 1 as x approaches infinity.

268 Easy

268 of 650

Understanding +4 | -1

The function

f: R \rightarrow R

defined by

f(x)=e^{x}

View answer and explanation

Correct option: C

One-one and into

Option C (One-one and into) is correct because the exponential function

e^x

is one-to-one over the real numbers and its range is a subset of the codomain R.

269 Medium

269 of 650

Understanding +4 | -1

The domain of the function

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

View answer and explanation

Correct option: C

R

Option C (

R

) is correct because the denominator

x^2-3x+3

has no real roots, so it is never zero, and the numerator is defined for all real numbers.

270 Easy

270 of 650

Application +4 | -1

f\left( x \right)

satisfies the relation

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

for all real x, then

f\left( x \right)

View answer and explanation

Correct option: B

\frac{{{x^2} + 2x - 1}}{3}

The correct answer is B (

\frac{{{x^2} + 2x - 1}}{3}

) because it is the solution obtained by substituting (1-x) into the original equation and solving the resulting system of equations.

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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