Domain of function $ f(x)=\sin ^{-1} 5 x $ is

$ \left(0, \frac{1}{5}\right) $

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

$ - \sqrt 3 \le x \le + \sqrt 3 $

$ - 2 + \sqrt 3 \le x \le - 2 - \sqrt 3 $

If ${\sin ^x}\alpha + {\cos ^x}\alpha \ge 1,0 < \alpha < \frac{\pi }{2},\;$ then

$x \in \left( {2,\infty } \right)$

$x \in \left[ { - 1,1} \right]$

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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 211-240 of 478 questions

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211 Easy

211 of 478

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.

212 Easy

212 of 478

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.

213 Easy

213 of 478

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.

214 Easy

214 of 478

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.

215 Easy

215 of 478

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.

216 Easy

216 of 478

Analyzing +4 | -1

Let

f(x)=[x]^2+[x+1]-3

where

[x]=

the greatest integer function. Then

View answer and explanation

Correct option: D

Both (a) and (b)

Option D (Both (a) and (b)) is correct because the function is many-one and into, and it equals zero for an infinite number of x values.

217 Easy

217 of 478

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.

218 Easy

218 of 478

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.

219 Easy

219 of 478

Application +4 | -1

\phi(x)=a^{x}

, then

\{\phi(p)\}^{3}

is equal to

View answer and explanation

Correct option: A

\phi(3 p)

Option A (

\phi(3 p)

) is correct because given

\phi(x)=a^{x}

, then

{\phi(p)}^{3} = (a^p)^3 = a^{3p} = \phi(3p)

220 Easy

220 of 478

Understanding +4 | -1

Domain of function

f(x)=\sin ^{-1} 5 x

View answer and explanation

Correct option: B

\left[-\frac{1}{5}, \frac{1}{5}\right]

Option B (

\left[-\frac{{1}}{{5}}, \frac{{1}}{{5}}\right]

) is correct because the domain of

\sin^{-1}(x)

is [-1, 1], so

-1 \le 5x \le 1

221 Easy

221 of 478

Application +4 | -1

f(x+3 y, x-3 y)=12 x y

, then

f(x, y)

View answer and explanation

Correct option: C

x^2-y^2

Option C (

x^2-y^2

) is correct because substituting

x+3y

and

x-3y

into

x^2 - y^2

yields the given expression

12xy

222 Easy

222 of 478

Application +4 | -1

Let

f(x)=x^2+3 x+1

and

g(x)=x+1

. If

f(x)+\lambda g(x)>-10 \, \forall \, x \in R

then find the sum of all possible positive integral values of

\lambda

View answer and explanation

Correct option: A

Option A (10) is correct because it's derived from finding the range of λ that satisfies the given inequality for all real x, and then summing the positive integers within that range.

223 Easy

223 of 478

Application +4 | -1

Domain of the function

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

View answer and explanation

Correct option: B

- 1 - \sqrt 3 \le x \le - 1 + \sqrt 3

Option B (

- 1 - \sqrt 3 \le x \le - 1 + \sqrt 3

) is correct because it represents the interval where the quadratic expression inside the square root is non-negative.

224 Easy

224 of 478

Application +4 | -1

For

\theta > \frac{\pi}{3}

, the value of

f(\theta)=\sec ^{2} \theta+\cos ^{2} \theta

always lies in the interval

View answer and explanation

Correct option: D

[2, \infty)

Option D (

[2, \infty)

) is correct because the minimum value of the function occurs when sec²θ = cos²θ, which happens when cos²θ = 1 or cos²θ = 0, but since θ > π/3, the minimum value is 2.

225 Easy

225 of 478

Understanding +4 | -1

Let

f\left( x \right) = \left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|\;and\;g\left( x \right) = x + 1.

Then,

View answer and explanation

Correct option: C

g\left( x \right)

is neither even nor odd

Option C (

g(x)

is neither even nor odd) is correct because

g(-x) \ne g(x)

and

g(-x) \ne -g(x)

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226 Easy

226 of 478

Application +4 | -1

Let

f: Q-\{0,1\} \rightarrow Q

be a function such that

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

for all rational numbers

x \neq 0,1

. Then the value of

f(2)

View answer and explanation

Correct option: B

\frac{7}{4}

Option B (7/4) is correct because by substituting different values of x, we can create a system of equations to solve for f(2).

227 Easy

227 of 478

Application +4 | -1

{f}({x})={ax}+{b}

, where

{a}

and

{b}

are integers,

{f}(-1)=-5

and

f(3)=3

, then the value of '

a

' is

View answer and explanation

Correct option: A

Option A (2) is correct because solving the system of equations derived from the given function values yields a = 2.

228 Easy

228 of 478

Analyzing +4 | -1

Let

f(x) = \frac{{x - 1}}{{x + 1}},\,\,\,x \in {\text{R}} - \left\{ {0, - 1,1} \right\}.

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

for all

n \in N,

then

{f^6}\left( 6 \right) + {f^7}\left( 7 \right)

is equal to:

View answer and explanation

Correct option: B

- \frac{3}{2}

Option B (

- \frac{3}{2}

) is correct because after finding the pattern of the function composition, we can simplify the expression

{f^6}(6) + {f^7}(7)

{f^2}(6) + {f^3}(7)

, which evaluates to

- \frac{3}{2}

229 Easy

229 of 478

Analyzing +4 | -1

f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0

and

{S}=\{{x} \in {R}: {f}({x})={f}(-{x})\}

; then

{S}

View answer and explanation

Correct option: A

Contains exactly two elements.

Option A (Contains exactly two elements) is correct because solving the functional equation leads to f(x) = 2/x - x, and f(x) = f(-x) gives two solutions: x =

\sqrt{2}

and x = -

\sqrt{2}

230 Easy

230 of 478

Application +4 | -1

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

, then

x=

View answer and explanation

Correct option: A

f(y)

Option A is correct because solving for x in terms of y gives

x = f(y)

231 Easy

231 of 478

Analyzing +4 | -1

Let

f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}

be functions defined by

f(a)=\alpha

, where

\alpha

is the maximum of the powers of those primes

p

such that

p^\alpha

divides

a

, and

g(a)=a

+1

, for all

a \in \mathbb{N} - \{ 1\}

. Then, the function

f+g

View answer and explanation

Correct option: D

Neither one-one nor onto

Option D (Neither one-one nor onto) is correct because the function f+g does not satisfy the conditions for either injectivity (one-to-one) or surjectivity (onto).

232 Easy

232 of 478

Application +4 | -1

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

where

a > 0

and

n \in N,

then

fof\left( x \right)

is equal to

View answer and explanation

Correct option: B

x

Option B (

x

) is correct because after applying the function

f

twice, the expression simplifies back to

x

233 Easy

233 of 478

Application +4 | -1

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

, then

f^{-1}(x)

is equal to

View answer and explanation

Correct option: D

\frac{x}{(1-x)}

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

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

234 Easy

234 of 478

Application +4 | -1

n(A)=3

and

n(B)=5

then number of one-one functions that can be defined from

A

B

View answer and explanation

Correct option: D

Option D (60) is correct because it represents the number of ways to choose 3 elements from 5 and arrange them, which is 5P3 = 60.

235 Easy

235 of 478

Application +4 | -1

f(x)=x^2+1

, then

f \circ f(x)

is equal to

View answer and explanation

Correct option: C

x^4+2 x^2+2

Option C (

x^4 + 2x^2 + 2

) is correct because substituting

f(x)

into itself results in

(x^2 + 1)^2 + 1 = x^4 + 2x^2 + 2

236 Easy

236 of 478

Application +4 | -1

f:\left[ { - 2,\;2} \right] \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} { - 1,\;\;for - 2 \le x \le 0}\\ {x - 1,\;\;for\;0 \le x \le 2} \end{array}} \right.

Then

\left\{ {x \in \left[ { - 2,2} \right]:x \le 0\;\;\mathrm{and}\;\mathrm{\;}f\left( {\left| x \right|} \right) = x} \right\} =

View answer and explanation

Correct option: C

\left\{ { - 1/2} \right\}

Option C ({-1/2}) is correct because by substituting |x| into the function and solving for x, we find x = -1/2.

237 Easy

237 of 478

Analyzing +4 | -1

{\sin ^x}\alpha + {\cos ^x}\alpha \ge 1,0 < \alpha < \frac{\pi }{2},\;

then

View answer and explanation

Correct option: B

x \in \left( { - \infty ,2} \right]

Option B

x \in \left( { - \infty ,2} \right]

is correct because the inequality holds true for x less than or equal to 2.

238 Easy

238 of 478

Application +4 | -1

{x^2} - 3\left| x \right| + 2 < 0,\;

then

x

belongs to

View answer and explanation

Correct option: C

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

Option C is correct because solving the inequality with absolute values leads to the solution set (-2, -1) ∪ (1, 2).

239 Easy

239 of 478

Application +4 | -1

Let

f(x)=\frac{\alpha x^{2}}{x+1}, x \neq-1

. The value of

\alpha

for which

f(a)=a,(a \neq 0)

View answer and explanation

Correct option: C

1+\frac{1}{a}

Option C (

{ }^8 {C}_4 \cdot 3

) is correct because it selects 4 people out of 8 and then considers the 3 possible pairings within that group of 4.

240 Easy

240 of 478

Application +4 | -1

f(x)

and

g(x)

are periodic functions with periods 3 and 4 respectively. Then the period of

F(x)=f\left(\frac{x}{6}\right)+g\left(\frac{x}{3}\right)

View answer and explanation

Correct option: D

Option D (36) is correct because it is the least common multiple (LCM) of the periods of the individual functions after adjusting for the scaling factors.

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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