Solution set of inequality ${\log _e}\frac{{x - 2}}{{x - 3}}$ is

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

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

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

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

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

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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 271-300 of 478 questions

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271 Easy

271 of 478

Understanding +4 | -1

If R denotes the set of all real numbers, then the function

f:R \to R

defined by

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

View answer and explanation

Correct option: D

Neither one-one nor onto

Option D is correct because the absolute value function is neither one-to-one (since f(x) = f(-x)) nor onto (since the range is [0, ∞)).

272 Easy

272 of 478

Understanding +4 | -1

Let

f:(-1,1) \rightarrow B

, be a function defined by

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

, then

f

is both one- one and onto when

B

is the interval

View answer and explanation

Correct option: B

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Option B

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

is correct because the range of

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

for

x \in (-1, 1)

(-\frac{\pi}{2}, \frac{\pi}{2})

273 Easy

273 of 478

Analyzing +4 | -1

If sides of

\triangle A B C

are

a, b

and

c

such that

2 b=a+c

, then exhaustive range of

\frac{b}{c}

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the given condition

2b = a+c

and triangle inequalities lead to the range of

\frac{{b}}{{c}}

being between

\frac{{2}}{{3}}

and 2.

274 Easy

274 of 478

Application +4 | -1

Find the number of integers not in the domain of

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

View answer and explanation

Correct option: A

Option A (2) is correct because there are two integers that make the argument of the inverse cosine function outside the range of [-1, 1].

275 Easy

275 of 478

Analyzing +4 | -1

Solution set of inequality

{\log _e}\frac{{x - 2}}{{x - 3}}

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the inequality is satisfied when the argument of the logarithm is between 0 and 1.

276 Easy

276 of 478

Analyzing +4 | -1

A mapping from

N

N

is defined as follows:

f: N \rightarrow N, f(n)=(n+5)^2, n \in N

(

N

is the set of natural numbers). Then

View answer and explanation

Correct option: D

f

is one to one but not onto

Option D (

f

is one to one but not onto) is correct because each natural number maps to a unique square, but not all natural numbers are perfect squares.

277 Easy

277 of 478

Analyzing +4 | -1

Let

A=\{-3,-2,-1,0,1,2,3\}

, then number of functions from

A

A

which are odd and

f(i) \neq i

(\forall \, {i}=-3,-2,-1,1,2,3)

is/are

View answer and explanation

Correct option: B

216

The correct answer is [0,1) because the function f(x) = x²/(1+x²) always produces values between 0 (inclusive) and 1 (exclusive) as x varies over all real numbers.

278 Easy

278 of 478

Analyzing +4 | -1

The set of all real

x

satisfying the inequality

\frac{{3 - \left| x \right|}}{{4 - \left| x \right|}} > 0

View answer and explanation

Correct option: A

\left[ { - 3,\;3} \right] \cup \left( { - \infty ,\; - 4} \right) \cup \left( {4,\;\infty } \right)

The correct answer is A. By analyzing the inequality and considering the cases where the numerator and denominator have the same sign, we find the solution set to be (-∞, -4) U [-3, 3] U (4, ∞).

279 Easy

279 of 478

Application +4 | -1

Let

\mathrm{f}(\mathrm{x})=[x]

, where

[x]

denotes the greatest integer less than or equal to

x

. If

a=\sqrt{2011^{2}+2012}

, then the value of

\mathrm{f}(\mathrm{a})

is equal to

View answer and explanation

Correct option: B

2011

Option B (2011) is correct because the greatest integer less than or equal to

a = \sqrt{2011^2 + 2012}

is 2011.

280 Easy

280 of 478

Application +4 | -1

The domain of

\sin ^{-1}\left[\log _{3}\left(\frac{x}{3}\right)\right]

View answer and explanation

Correct option: A

[1,9]

Option A ([1,9]) is correct because the domain is found by solving the inequality -1 <= log3(x/3) <= 1.

281 Easy

281 of 478

Understanding +4 | -1

f(x)=a \cos (b x+c)+d

, then range of

f(x)

View answer and explanation

Correct option: D

[d-a, d+a]

Option D (

[d-a, d+a]

) is correct because the range of

\cos(bx+c)

is [-1, 1], so the range of

a\cos(bx+c)

is [-a, a], and adding d shifts the range to [d-a, d+a].

282 Easy

282 of 478

Application +4 | -1

f(x)

is an odd function,

f(1)=3

, and

f(x+2)

=f(x)+f(2)

, then the value of

f(3)

View answer and explanation

Correct option: A

Option A (9) is correct because by using the properties of odd functions and the given functional equation, we can deduce that f(3) = 9.

283 Easy

283 of 478

Application +4 | -1

If from

R \rightarrow R, f(x)=(x+1)^2, g(x)=x^2+1

, then

(

fog

)(-3)

equals

View answer and explanation

Correct option: A

121

Option A (121) is correct because when you evaluate the composite function

f(g(x))

x = -3

, the result is 121.

284 Easy

284 of 478

Understanding +4 | -1

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

are defined by

f\left( x \right) = x - \left[ x \right]\;and\;g\left( x \right) = \left[ x \right]

for

x \in R

, where

\left[ x \right]

is the greatest integer not exceeding x, then for every

x \in R,\;f\left( {g\left( x \right)} \right)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because

f(g(x)) = f([x]) = [x] - [[x]] = [x] - [x] = 0

285 Easy

285 of 478

Application +4 | -1

Let

f

be a function satisfying

2 f(x)-3 f(1 / x)=x^2

for any

x \neq 0

. Then the value of

f(2)

View answer and explanation

Correct option: B

-7 / 4

Option B (-7/4) is correct because it is the solution obtained by solving the system of equations derived from the given functional equation.

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286 Easy

286 of 478

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.

287 Easy

287 of 478

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.

288 Easy

288 of 478

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)}}

289 Easy

289 of 478

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!.

290 Easy

290 of 478

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.

291 Easy

291 of 478

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.

292 Easy

292 of 478

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.

293 Easy

293 of 478

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.

294 Easy

294 of 478

Application +4 | -1

The domain of the function

f(x)=\sin ^{-1}\left(\log _2\left(\frac{1}{2} x^2\right)\right)

is :

View answer and explanation

Correct option: C

[-2,-1] \cup[1,2]

The correct answer is C ([-2,-1]

\cup

[1,2]) because the argument of the arcsine function must be between -1 and 1, leading to the inequality

1 \le x^2 \le 4

295 Easy

295 of 478

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.

296 Easy

296 of 478

Analyzing +4 | -1

The domain of the function

f(x)=\sqrt{3-2^x-2^{1-x}}+\sqrt{\sin ^{-1} x}

View answer and explanation

Correct option: C

[0,1]

Option C ([0,1]) is correct because it satisfies both the domain restrictions imposed by the square roots and the inverse sine function.

297 Easy

297 of 478

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.

298 Easy

298 of 478

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

299 Easy

299 of 478

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

300 Easy

300 of 478

Analyzing +4 | -1

The domain of the function

f(x)=\frac{1}{\sqrt{|x|-x}}

View answer and explanation

Correct option: B

(-\infty, 0)

Option B

((-\infty, 0))

is correct because the expression inside the square root,

|x|-x

, must be greater than zero, which is only true when x is negative.

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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