The domain of $f\left( x \right) = \log \left| {{{\log }_e}x} \right|,$ is

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

The domain of the real function $f(x)=\frac{1}{\sqrt{4-x^2}}$ is

the set of all positive real numbers

If $|x+2| \leq 9$, then

$x \in(-\infty,-7) \cup(11, \infty)$

$x \in(-\infty,-7) \cup[11, \infty)$

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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 271-300 of 650 questions

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271 Easy

271 of 650

Analyzing +4 | -1

The domain of

f\left( x \right) = \log \left| {{{\log }_e}x} \right|,

View answer and explanation

Correct option: C

\left( {0,1} \right) \cup \left( {1,\infty } \right)

Option C (

\left( {0,1} \right) \cup \left( {1,\infty } \right)

) is correct because the argument of the outer logarithm must be positive, and the argument of the inner logarithm must also be positive and not equal to 1.

272 Easy

272 of 650

Understanding +4 | -1

Which one of the following functions is not invertible?

View answer and explanation

Correct option: B

f: \mathbb{R} \rightarrow[0, \infty), f(x)=x^2

Option B (

f: \mathbb{{R}} \rightarrow[0, \infty), f(x)=x^2

) is correct because this function is not one-to-one over the entire real numbers since

f(x) = f(-x)

, and therefore not invertible.

273 Easy

273 of 650

Understanding +4 | -1

A = \left\{ {1,\;2,\;3,\;4} \right\},\;B\left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}

are two sets, and function

f:A \to B

is defined by

f\left( x \right) = x + 2\;\forall x \in Ab

, then the function f is

View answer and explanation

Correct option: C

One-one

Option C (One-one) is correct because each element in set A maps to a unique element in set B, and the function is into.

274 Hard

274 of 650

Analyzing +4 | -1

\;{5^x} + {\left( {2\sqrt 3 } \right)^{2x}} \ge {13^x},\;

then the solution set for

x

View answer and explanation

Correct option: C

( - \infty ,2]

Option C,

( - \infty ,2]

, is correct because the inequality holds for x less than or equal to 2, based on the properties of trigonometric functions and exponents.

275 Easy

275 of 650

Application +4 | -1

R = \left\{ {\left( {x,y} \right):x,y \in Z,{x^2} + 3{y^2} \le 8} \right\}

is a relation on the set of integers Z, then the domain of

{R^{ - 1}}

View answer and explanation

Correct option: A

\left\{ { - 1,0,1} \right\}

The correct answer is A (

\left\{ { - 1,0,1} \right\}

) because it represents the domain of the inverse relation, which is the range of the original relation R.

276 Easy

276 of 650

Application +4 | -1

g(f(x))=|\sin x|

and

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

, then

f

and

g

may be given by

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it satisfies both given functional equations:

g(f(x))=|\sin x|

and

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

277 Easy

277 of 650

Application +4 | -1

Let

A=\{1,0,1,2\}, B=\{4,2,0,2\}

and

f, g: A \rightarrow B

be functions defined by

f(x)=x^2-x

and

g(x)=2\left|x-\frac{1}{2}\right|-1

. Then,

View answer and explanation

Correct option: A

f=g

Option A (

f=g

) is correct because the functions

f(x)

and

g(x)

yield the same output for all elements in set A.

278 Medium

278 of 650

Application +4 | -1

Let

f

and

g

be two real functions defined by

f(x)=\frac{1}{x+4}, x \neq-4

and

g(x)=(x+4)^3

, then

f / g

is equal to

View answer and explanation

Correct option: B

\frac{1}{(x+4)^4}

Option B (

\frac{1}{(x+4)^4}

) is correct because it represents the division of f(x) by g(x).

279 Medium

279 of 650

Understanding +4 | -1

The domain and range of the function

f

given by

f(x)=2-|x-5|

View answer and explanation

Correct option: B

Domain

=R

, Range

=(-\infty, 2]

Option B (Domain

=R

, Range

=(-\infty, 2]

) is correct because the absolute value function is defined for all real numbers, and the maximum value of the function is 2.

280 Easy

280 of 650

Analyzing +4 | -1

The domain of the function

f(x)=\log _{10} \frac{x-5}{x^2-10 x+24}-3 \sqrt{x+5}

View answer and explanation

Correct option: D

(4,5) \cup(6, \infty)

Option D ((4,5) ∪ (6, ∞)) is correct because it satisfies both conditions: the argument of the logarithm must be positive, and the expression inside the square root must be non-negative.

281 Easy

281 of 650

Application +4 | -1

The inverse of the function

f:R \to R

given by

f\left( x \right) = {\log _a}\left( {x + \sqrt {{x^2} + 1} } \right)\left( {a > 0,a \ne 1} \right),

View answer and explanation

Correct option: B

\frac{1}{2}\left( {{a^x} - {a^{ - x}}} \right)

Option B (

\frac{1}{2}\left( {{a^x} - {a^{ - x}}} \right)

) is correct because it represents the inverse function of the given logarithmic function.

282 Easy

282 of 650

Analyzing +4 | -1

Let

X=Y=\mathbf{R} \sim\{1\}

. The function

f: X \rightarrow Y

defined by

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

View answer and explanation

Correct option: D

one-one and onto

Option D (one-one and onto) is correct because the function

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

is both injective (one-to-one) and surjective (onto) when

X = Y = \mathbf{{R}} \setminus \{1\}

283 Easy

283 of 650

Understanding +4 | -1

The domain of the real function

f(x)=\frac{1}{\sqrt{4-x^2}}

View answer and explanation

Correct option: C

(-2,2)

Option C ((-2,2)) is correct because the function is defined only when the expression inside the square root is positive, i.e.,

4 - x^2 > 0

, which gives

-2 < x < 2

284 Easy

284 of 650

Application +4 | -1

If D_30 is the set of the divisors of 30,

x,y \in {D_{30}}

we define

x + y = \mathrm{LCM}\left( {x,y} \right),x.y = \gcd \left( {x,y} \right),x' = \frac{{30}}{x}\;\;and\;\;f\left( {x,y,z} \right) = \left( {x + y} \right).\left( {y' + z} \right),

then

f\left( {2,5,15} \right)

is equal to

View answer and explanation

Correct option: C

Option C (10) is correct because it follows from the given definitions of +, ., and ' within the set D_30.

285 Easy

285 of 650

Analyzing +4 | -1

Let

f(x) = |x - 2|\mathrm{ }\,\,\mathrm{and}\,\,\mathrm{ }g(x) = f(f(x)),x \in [0,4].

Then

\int_0^3 {(g(x) - f(x))\,dx}

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because evaluating the integral of the difference between g(x) and f(x) over the given interval yields 1.

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286 Medium

286 of 650

Understanding +4 | -1

f:R \to R

is defined by

f\left( x \right) = \sin x\;and\,g:\left( {1,\;\infty } \right) \to R

is defined by

g\left( x \right) = \sqrt {{x^2} - 1}

, then

gof\left( x \right)

View answer and explanation

Correct option: D

Not defined

Option D (Not defined) is correct because the range of f(x) is not a subset of the domain of g(x), so the composite function gof(x) is not defined.

287 Medium

287 of 650

Application +4 | -1

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

and

(g \circ f)(x)=4 x^{2}-10 x+5

, then

f\left(\frac{5}{4}\right)

is equal to:

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (

-\frac{1}{2}

) is correct because it is the solution to the quadratic equation obtained by substituting the given value into the composite function.

288 Easy

288 of 650

Analyzing +4 | -1

Let

f: X \rightarrow[1,27]

be a function by

f(x)=5 \sin x

+12 \cos x+14

. The set

X

so that

f

is one-one and onto is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is not one-to-one due to its periodic nature, even though its range is within [1, 27].

289 Easy

289 of 650

Application +4 | -1

|x+2| \leq 9

, then

View answer and explanation

Correct option: B

x \in[-11,7]

Option B (x ∈ [-11,7]) is correct because the absolute value inequality |x+2| ≤ 9 implies that -9 ≤ x+2 ≤ 9.

290 Easy

290 of 650

Understanding +4 | -1

The natural domain of the real valued function defined by

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

View answer and explanation

Correct option: D

(-\infty, \infty)-(-1,1)

Option D (

(-\infty, \infty)-(-1,1)

) is correct because

x^2 - 1

must be non-negative for the square root to be real, so

x^2 \ge 1

, which means

x \le -1

x \ge 1

291 Easy

291 of 650

Understanding +4 | -1

The function which map

[-1,1]

[0,2]

are

View answer and explanation

Correct option: B

Two linear function

Option B (Two linear function) is correct because two linear functions are needed to map the interval [-1, 1] to [0, 2] due to the change in direction.

292 Easy

292 of 650

Understanding +4 | -1

x < 5

, then

View answer and explanation

Correct option: C

-\mathrm{x} > -5

Option C (-x > -5) is correct because multiplying an inequality by a negative number reverses the inequality sign.

293 Easy

293 of 650

Analyzing +4 | -1

Let

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

, where

[x]

denotes the greatest integer less than or equal to

x

, then the range of

f

View answer and explanation

Correct option: C

[0,1 / 2)

Option C ([0, 1/2)) is correct because when x is an integer f(x) = 0, and when x is not an integer, 0 <= f(x) < 1/2.

294 Easy

294 of 650

Application +4 | -1

f\left( x \right) = \sin \left( {\log x} \right),

then the value of

f\left( {xy} \right) + f\left( {x/y} \right) - 2f\left( x \right)\cos \left( {\log y} \right),

View answer and explanation

Correct option: B

0

Option B (0) is correct because the expression simplifies to 0 using trigonometric identities and logarithmic properties.

295 Medium

295 of 650

Application +4 | -1

f(x)=\cos (\log x)

then

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

is equal to:

View answer and explanation

Correct option: A

Option A (0) is correct because the expression simplifies to 0 using trigonometric identities and the given function definition.

296 Medium

296 of 650

Understanding +4 | -1

The domain of the function

f(x)=\frac{|x+3|}{x+3}

View answer and explanation

Correct option: B

\mathrm{R}-\{-3\}

Option B is correct because the function is defined for all real numbers except x = -3, where the denominator becomes zero.

297 Easy

297 of 650

Analyzing +4 | -1

The interval in which the function

y = \frac{{x - 1}}{{{x^2} - 3\;x + 3}}

transforms the real line is

View answer and explanation

Correct option: D

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

The correct answer is D ([-1/3, 1]) because this interval represents the range of the given function.

298 Easy

298 of 650

Application +4 | -1

Let

f:R \to R

be define by

f\left( x \right) = 3\;x - 4.

Then,

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

View answer and explanation

Correct option: A

\frac{{x + 4}}{3}

Option A (

\frac{{x + 4}}{3}

) is correct because by setting f(x) = y and solving for x, we find the inverse function to be (x+4)/3.

299 Easy

299 of 650

Application +4 | -1

P(S)

denotes the set of all subsets of a given set

S

, then the number of one-to-one functions from the set

S=\{1,2,3\}

to the set

P(S)

View answer and explanation

Correct option: C

336

The correct answer is C (336) because it correctly calculates the number of one-to-one functions from S to P(S) using permutations.

300 Medium

300 of 650

Application +4 | -1

\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}

are defined by

\mathrm{f}(\mathrm{x})=5 \mathrm{x}-3, \mathrm{~g}(\mathrm{x})=x^2+3

, then

\left(\mathrm{go} f^{-1}\right)(3)=

View answer and explanation

Correct option: B

\frac{111}{25}

Option B (

\frac{111}{25}

) is correct because it is the result of evaluating

g(f^{-1}(3))

after finding the inverse of f(x).

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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