If $f(x)=3 x-5$, then $f^{-1}(x)$

does not exist because $f$ is not one-one

does not exist because $f$ is not onto

The function $ f: R \rightarrow R, f(x)=x^{2}, \forall x \in R $ is

Injection as well as surjection

If $\sin ^{-1} x=y$, then

$-\frac{\pi}{2} < y < \frac{\pi}{2}$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Easy

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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 451-478 of 478 questions

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451 Easy

451 of 478

Application +4 | -1

Given that

f\left( x \right)

is a differentiable function of

x

and that

f\left( x \right).\mathrm{ }f\left( y \right) = f\left( x \right) + f\left( y \right) + f\left( {xy} \right) - 2

and that

f\left( 2 \right) = 5

. Then,

f\left( 3 \right)

is equal to

View answer and explanation

Correct option: A

Option A (10) is correct because by substituting y = 1/x into the given functional equation, we can determine that f(x) = x² + 1, and therefore f(3) = 10.

452 Easy

452 of 478

Understanding +4 | -1

Let

f:R \to R

be a function defined by

f\left( x \right) = \cos \left( {5x + 2} \right).

Then,

f

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is neither injective nor surjective, as it's a periodic function with a limited range.

453 Easy

453 of 478

Analyzing +4 | -1

f\;:R \to R

is defined by

f\left( x \right) = x - \left[ x \right] - \frac{1}{2}

for all

x \in R,

, where

\left[ x \right]

denotes the greatest integer function, 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 correct answer is C (

\phi

) because the equation f(x) = 1/2 leads to a contradiction, implying there are no real solutions for x.

454 Easy

454 of 478

Application +4 | -1

The function

f(x)=(1 / 4) x^3-\sin \pi x+3

[-2,2]

takes the value

View answer and explanation

Correct option: A

Option A (1) is correct because substituting x=-2 into the function gives f(-2) = 1.

455 Easy

455 of 478

Analyzing +4 | -1

The function

f\left( x \right) = {\log _{2x - 5}}\left( {{x^2} - 3x - 10} \right)

is defined for all

x

belonging to

View answer and explanation

Correct option: B

\left( {5,\;\infty } \right)

Option B (

\left( {5,\\;\infty } \right)

) is correct because the argument of the logarithm must be positive, the base must be positive, and the base cannot be equal to 1.

456 Easy

456 of 478

Application +4 | -1

The domain of the function

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

is:

View answer and explanation

Correct option: C

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

The correct answer is C ((−∞,1] ∪ [5,∞)) because it represents the values of x for which the logarithmic expression is greater than or equal to 1.

457 Easy

457 of 478

Application +4 | -1

A polynomial function

f\left( x \right)

satisfies the condition

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

f\left( {10} \right) = 1001,

then

f\left( {20} \right) =

View answer and explanation

Correct option: C

8001

Option C (8001) is correct because by substituting f(x) = x^n + 1 into the given condition and using f(10) = 1001, we find n = 3, thus f(20) = 20³ + 1 = 8001.

458 Easy

458 of 478

Understanding +4 | -1

Which of the following is one-one function?

View answer and explanation

Correct option: A

e^x

Option A (

e^x

) is correct because it is a strictly increasing function, meaning each x-value maps to a unique y-value.

459 Easy

459 of 478

Analyzing +4 | -1

Let

f\left( x \right) = x\;and\;g\left( x \right) = \left| x \right|

for all

x \in R.

Then, the function

\phi \left( x \right)

satisfying

{\left[ {\phi \left( x \right) - f\left( x \right)} \right]^2} + {\left[ {\phi \left( x \right) - g\left( x \right)} \right]^2} = 0

View answer and explanation

Correct option: A

\phi \left( x \right) = x,x \in \left[ {0,\;\infty } \right)

Option A is correct. The given condition implies

\phi(x) = f(x) = g(x)

. Since

f(x) = x

and

g(x) = |x|

, this holds only for

x \ge 0

, thus

\phi(x) = x

for

x \in [0, \infty)

460 Easy

460 of 478

Application +4 | -1

f(x)=3 x-5

, then

f^{-1}(x)

View answer and explanation

Correct option: B

is given by

\frac{x+5}{3}

Option B (is given by

\frac{x+5}{3}

) is correct because it represents the inverse function of f(x) = 3x - 5.

461 Easy

461 of 478

Application +4 | -1

Let

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

f(x)

can be written as sum of an even function,

g(x)

and an odd function,

h(x)

then the value of

h(4)

View answer and explanation

Correct option: A

Option A (2) is correct because the odd function h(x) is found to be x/2, and h(4) = 4/2 = 2.

462 Easy

462 of 478

Application +4 | -1

A quadratic polynomial maps from

[-2,3]

onto

[0,3]

and touches

X

-axis at

x=3

, then the polynomial is

View answer and explanation

Correct option: B

\frac{3}{25}\left(x^2-6 x+9\right)

Option B (

\frac{3}{25}(x^2-6x+9)

) is correct because the quadratic touches the x-axis at x=3 and maps [-2,3] onto [0,3].

463 Easy

463 of 478

Application +4 | -1

f(\ln (1+|x|))=(1-\ln (1+|x|))^{\frac{1}{7}}

, then

f(f(\cos x))

is equal to:

View answer and explanation

Correct option: C

\cos x

Option C (

\cos x

) is correct because

f(f(x)) = x

, therefore

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

464 Easy

464 of 478

Application +4 | -1

For

x \in \mathbb{R}

, two real valued functions

f(x)

and

g(x)

are such that,

g(x)=\sqrt{x}+1

and

fog(x)=x+3-\sqrt{x}

. Then

f(0)

is equal to

View answer and explanation

Correct option: B

Option B (5) is correct because by substituting

t = \sqrt{x} + 1

into the given functional equation, we can find

f(t)

and then evaluate

f(0)

465 Easy

465 of 478

Analyzing +4 | -1

f\left( x \right) = \frac{{{4^x}}}{{{4^x} + 2}}

, then

f\left( {\frac{1}{{97}}} \right) + f\left( {\frac{2}{{97}}} \right) + . \ldots + f\left( {\frac{{96}}{{97}}} \right)

is equal to

View answer and explanation

Correct option: B

Option B (48) is correct because the sum can be paired such that each pair sums to 1.

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466 Easy

466 of 478

Understanding +4 | -1

The function

f: R \rightarrow R, f(x)=x^{2}, \forall x \in R

View answer and explanation

Correct option: D

Neither injection nor surjection

Option D (Neither injection nor surjection) is correct because the function

f(x) = x^2

is neither one-to-one (injective) nor onto (surjective) when mapping from R to R.

467 Easy

467 of 478

Analyzing +4 | -1

Which of the following functions has period

\pi

View answer and explanation

Correct option: A

\left| { - \tan x} \right| + \cos 2\;x

Option A (

\left| { - \tan x} \right| + \cos 2\;x

) is correct because both |tan(x)| and cos(2x) have periods related to or equal to

\pi

, and their combination results in a period of

\pi

468 Easy

468 of 478

Application +4 | -1

Let

\mathrm{R}

be the relation on the set

\mathrm{N}

of natural numbers defined by

R:\{(x, y)\}: x+3 y=12 x \in N, y \in N\}

then domain of

R

View answer and explanation

Correct option: C

\{9,6,3\}

Option C ({9,6,3}) is the correct domain because these values of x satisfy the given relation with y being a natural number.

469 Easy

469 of 478

Application +4 | -1

f(y)=\log y

, then

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

is equal to:

View answer and explanation

Correct option: C

The correct answer is C (0) because the sum of log y and log(1/y) simplifies to log 1, which is 0.

470 Easy

470 of 478

Understanding +4 | -1

x

is real number and

|x| < 3

, then

View answer and explanation

Correct option: B

-3 < x < 3

Option B (-3 < x < 3) is correct because it represents the range of values for x that satisfy the absolute value inequality |x| < 3.

471 Easy

471 of 478

Application +4 | -1

Range of the function

f(x)=\frac{1}{3 \cos x+2}

, is

View answer and explanation

Correct option: C

(-\infty,-1] \cup\left(\frac{1}{5}, \infty\right)

Option C

((-\infty,-1] \cup\left(\frac{1}{5}, \infty\right))

is correct because it represents the range of the function, considering the possible values of the cosine function.

472 Easy

472 of 478

Application +4 | -1

Let

f:(4,6) \rightarrow(6,8)

be a function defined by

f(x)=x+\left[\frac{x}{2}\right]

, where [.] denotes the greatest integer function, then

f^{-1}(x)

is equal to

View answer and explanation

Correct option: C

x-2

Option C (

x-2

) is correct because it is the inverse function of f(x) = x + 2 on the given domain.

473 Easy

473 of 478

Remembering +4 | -1

\sin ^{-1} x=y

, then

View answer and explanation

Correct option: B

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

Option B (

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

) is correct because the range of the principal value of

\sin^{-1} x

[-\frac{\pi}{2}, \frac{\pi}{2}]

474 Easy

474 of 478

Application +4 | -1

If a function

f

, is not identically zero, satisfies

\mathrm{f}(\mathrm{x}+1)=-\mathrm{f}(\mathrm{x}+5), \forall \mathrm{x} \in \mathrm{R}

, then the period of

\mathrm{f}(\mathrm{x})

is :

View answer and explanation

Correct option: C

Option C (8) is correct because it represents the period of the function f(x), derived from the given functional equation f(x+1) = -f(x+5).

475 Easy

475 of 478

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\begin{array}{cccccccccccccccccccc} {0,\;\;\;x = 0}\\ {{x^2}\sin \pi /2x,\;\;\;\left| x \right| < 1} \end{array}}\\ {x\left| x \right|,\;\;\;\left| x \right| \ge 1} \end{array}} \right..

Then,

f\left( x \right)

View answer and explanation

Correct option: B

An odd function

The correct answer is B (An odd function) because f(-x) = -f(x) for all x, satisfying the definition of an odd function.

476 Easy

476 of 478

Application +4 | -1

The domain of definition of the function

f\left( x \right){ = ^{7 - x}}{P_{x - 3}},

View answer and explanation

Correct option: C

\left\{ {3,\;4,\;5} \right\}

Option C (

\left\{ {3,\;4,\;5} \right\}

) is correct because the domain is determined by the conditions 7-x >= 0, x-3 >= 0, and 7-x >= x-3, which lead to x being 3, 4, or 5.

477 Easy

477 of 478

Understanding +4 | -1

Let

A=\{1,2,3,4\}

and

R: A \rightarrow A

be the relation defined by

R=\{(1,1),(2,3),(3,4),(4,2)\}

. The correct statement is :

View answer and explanation

Correct option: C

R

is an onto function.

Option C is correct because the range of R is equal to its codomain A, making it an onto function.

478 Easy

478 of 478

Application +4 | -1

A quadratic function

f(x)

satisfies

f(x) \geq 0

for all real

x

. If

f(2)=0

and

f(4)=12

, then the value of

f(6)

is:

View answer and explanation

Correct option: C

Option C (48) is correct because by using the given conditions, we can find the quadratic function and then evaluate it at x=6.

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