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 · 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 301-330 of 650 questions

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301 Easy

301 of 650

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)

302 Easy

302 of 650

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

303 Easy

303 of 650

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

304 Easy

304 of 650

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.

305 Easy

305 of 650

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.

306 Easy

306 of 650

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.

307 Medium

307 of 650

Understanding +4 | -1

Which of the following functions are periodic?

View answer and explanation

Correct option: C

f(x)=x-[x], x \in R

Option C is correct because f(x) = x - [x] has a period of 1, meaning its values repeat every interval of length 1.

308 Easy

308 of 650

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)

309 Medium

309 of 650

Analyzing +4 | -1

The domain of the function

f\left( x \right) = \sqrt {{{\log }_{10}}\left( {\frac{1}{{\left| {\sin x} \right|}}} \right)} ,

View answer and explanation

Correct option: B

R - \left\{ {n\;\pi \mathrm{|}n \in Z} \right\}

The correct answer is B (

R - \\{{n\\;\pi \mathrm{|}n \in Z\\}}

) because the function is defined when the argument of the logarithm is greater than 0 and the expression inside the square root is non-negative, which leads to

x \ne n\pi

310 Easy

310 of 650

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}

311 Easy

311 of 650

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)

312 Easy

312 of 650

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).

313 Easy

313 of 650

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

314 Hard

314 of 650

Analyzing +4 | -1

The function

f\left( x \right) = \max \left\{ {\left( {1 - x} \right),\left( {1 + x} \right),2} \right\},x \in \left( { - \infty ,\infty } \right)

is equivalent to

View answer and explanation

Correct option: A

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

Option A is correct because it accurately represents the piecewise definition of the function based on the maximum value among (1-x), (1+x), and 2.

315 Easy

315 of 650

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)}

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

316 Medium

316 of 650

Analyzing +4 | -1

f\left( x \right) = {\left( {\frac{x}{{1 - \left| x \right|}}} \right)^{1/2002}}

then

{D_f}

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the expression inside the square root must be positive, leading to the specified domain.

317 Easy

317 of 650

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.

318 Medium

318 of 650

Analyzing +4 | -1

Let

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} a&{ - 1}&0 \\ {ax}&a&{ - 1} \\ {a{x^2}}&{ax}&a \end{array}} \right|,\,a \in R.

Then the sum of the square of all the values of

a,

for which

2f'\left( {10} \right) - f'\left( 5 \right) + 100 = 0,

View answer and explanation

Correct option: C

125

The correct answer is C (125) because it is the sum of the squares of all values of a that satisfy the given condition involving the derivative of the determinant.

319 Medium

319 of 650

Application +4 | -1

a, b, c, k

are real constants and

\alpha, \beta, \gamma

are real variables and

a \tan \alpha+b \tan \beta+c \tan \gamma=k

then the minimum value of

\tan ^2 \alpha+\tan ^2 \beta+\tan ^2 \gamma

View answer and explanation

Correct option: D

\frac{k^2}{a^2+b^2+c^2}

Option D (

\frac{{k^2}}{{a^2+b^2+c^2}}

) is correct because it is derived from Cauchy-Schwarz inequality applied to the given equation.

320 Medium

320 of 650

Application +4 | -1

f: R \rightarrow S

defined by

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

is onto, then the interval of

S

View answer and explanation

Correct option: A

[-1,3]

Option A ([-1,3]) is correct because the range of the function f(x) = sin(x) - √3cos(x) + 1 is [-1, 3] when it's onto.

321 Easy

321 of 650

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

322 Medium

322 of 650

Application +4 | -1

f\left( {x + y,\;x - y} \right) = xy

, then the arithemetic mean of

f\left( {x,\;y} \right)\;and\;f\left( {y,\;x} \right)

View answer and explanation

Correct option: C

Option C (0) is correct because the arithmetic mean simplifies to zero after substituting and simplifying.

323 Easy

323 of 650

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.

324 Easy

324 of 650

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.

325 Easy

325 of 650

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).

326 Medium

326 of 650

Remembering +4 | -1

A point

\mathrm{c}

in the domain of a function

\mathrm{f}

is called a critical point of

\mathrm{f}

if

I.

\mathrm{f}^{\prime}(\mathrm{c})=0

II. fis not differentiable at

\mathrm{c}

.

Choose the correct option

View answer and explanation

Correct option: A

Either I or II are true

Option A (Either I or II are true) is correct because a critical point occurs where the derivative is zero or the function is not differentiable.

327 Medium

327 of 650

Application +4 | -1

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

, then

f\left\{f\left(\frac{1}{x}\right)\right\}

View answer and explanation

Correct option: C

\frac{x}{x-1}

Option C (

\frac{{x}}{{x-1}}

) is correct because after evaluating the composite function

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

, the expression simplifies to

\frac{{x}}{{x-1}}

328 Medium

328 of 650

Understanding +4 | -1

The function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\;\;\;x \in Q}\\ {0,\;\;\;x \notin Q} \end{array}} \right.

View answer and explanation

Correct option: D

Periodic with indeterminate period

Option D (Periodic with indeterminate period) is correct because the function repeats for every rational number, but there's no smallest positive rational number.

329 Easy

329 of 650

Application +4 | -1

The domain of the function

\sqrt{\log \left(x^{2}-6 x+6\right)}

View answer and explanation

Correct option: C

(-\infty, 1] \cup[5, \infty)

Option C ((-∞, 1] ∪[5, ∞)) is correct because the domain requires the argument of the logarithm to be greater than 0 and the logarithm itself to be non-negative.

330 Easy

330 of 650

Application +4 | -1

If the function

f:[1, \infty) \rightarrow[1, \infty)

is defined by

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

then

f^{-1}(x)

View answer and explanation

Correct option: B

\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)

Option B is correct because it is the inverse function derived by solving for x in terms of y and considering the domain of the inverse function.

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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