Let $A = \{ x \in R:x$ is not a positive integer}. Define a function $f:A \to R$ as $f(x) = \frac{{2x}}{{x - 1}}$, then $f$ is

Neither injective nor surjective

If $3 < \left| x \right| < $, then $x$ belongs to

$\left( { - 3\;, - 3} \right) \cup \left( {3,6} \right)$

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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 391-420 of 478 questions

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391 Easy

391 of 478

Application +4 | -1

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

, then

(f+g)\left(\frac{\pi}{3}\right)=

View answer and explanation

Correct option: C

\sqrt{3}+\frac{1}{4}

Option C (

\sqrt{3}+\frac{1}{4}

) is correct because it is the result of evaluating

2\sin(x) + \cos^2(x)

x = \frac{\pi}{3}

392 Easy

392 of 478

Analyzing +4 | -1

Let

f: R \rightarrow R, g: R \rightarrow R

be two given functions, such that

f

is injective and

g

is surjective, then which of the following is injêctive?

View answer and explanation

Correct option: D

fof

Option D (

fof

) is correct because the composition of two injective functions is also injective.

393 Easy

393 of 478

Application +4 | -1

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

are defined by

f\left( x \right) = \left| x \right|\;and\;g\left( x \right) = \left[ {x - 3} \right]

for

x \in R

, then

\left\{ {g\left( {f\left( x \right)} \right): - \frac{8}{5} < x < \frac{8}{5}} \right\}

is equal to

View answer and explanation

Correct option: C

\left\{ { - 3,\mathrm{ } - 2} \right\}

Option C (

\left\{ { - 3,\; - 2} \right\}

) is correct because it represents the set of values obtained by evaluating g(f(x)) for the given range of x.

394 Easy

394 of 478

Application +4 | -1

f(x)=2 x

and

g

is identity function, then

View answer and explanation

Correct option: C

(f \circ g)(x)=(g+g)(x)

Option D (

(f \circ g)(x)=(g+g)(x)

) is correct because if

f(x) = 2x

and g is the identity function, then both sides of the equation simplify to

2x

395 Easy

395 of 478

Analyzing +4 | -1

A function

f

from the set of natural numbers to integers defined by

f(n)=\left\{\begin{array}{l} \frac{n-1}{2}, \text { when } n \text { is odd } \\ -\frac{n}{2}, \text { when } n \text { is even } \end{array}\right. \text { is }

View answer and explanation

Correct option: D

one-one and onto both.

Option D (one-one and onto both) is correct because the function maps each natural number to a unique integer, and every integer has a corresponding natural number.

396 Easy

396 of 478

Application +4 | -1

Which of the following function is even function

View answer and explanation

Correct option: B

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

Option B is correct because

f(-x) = f(x)

for the function

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

397 Easy

397 of 478

Understanding +4 | -1

The interval for which

\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}

holds

View answer and explanation

Correct option: C

[0,1]

Option C (

[0, 1]

) is correct because the identity

\sin^{-1}(x) + \cos^{-1}(x) = \frac{{\pi}}{{2}}

holds for all

x

in the interval [-1, 1], but since we have

\sqrt{x}

x

must be non-negative, and

\sqrt{x}

must be in [-1, 1], which means

x

is in [0, 1].

398 Easy

398 of 478

Understanding +4 | -1

x

and

b

are real numbers. If

b > 0

and

|x| > b

, then

View answer and explanation

Correct option: D

x \in(-\infty,-b) \cup(b, \infty)

Option D (

x \in(-\infty,-b) \cup(b, \infty)

) is correct because it represents all values of x whose absolute value is greater than b.

399 Easy

399 of 478

Analyzing +4 | -1

The range of the function

f

defined by

f(x)=\frac{e^x-e^{|x|}}{e^x+e^{|x|}}

View answer and explanation

Correct option: B

(-1,0]

The correct answer is B ((-1,0]) because the range of the function

f(x)=\frac{{e^x-e^{|x|}}}{{e^x+e^{|x|}}}

(-1,0]

, found by considering cases when

x \geq 0

and

x < 0

400 Easy

400 of 478

Application +4 | -1

The domain of the function

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

+\sqrt{4+x}+\sqrt{4-x}

is:

View answer and explanation

Correct option: D

[0,1]

The correct answer is D ([0,1]) because it is the intersection of the domains of all three square root functions.

401 Easy

401 of 478

Understanding +4 | -1

Which of the following are functions?

View answer and explanation

Correct option: B

\{(x, y): y=|x|, x, y \in \mathbf{R}\}

Option B (

\{(x, y): y=|x|, x, y \in \mathbf{R}\}

) is correct because for each x in R, there is a unique y in R such that y = |x|.

402 Easy

402 of 478

Application +4 | -1

f(x+a y, x-a y)=a x y

, then

f(x, y)

is equal to

View answer and explanation

Correct option: C

\frac{x^{2}-y^{2}}{4}

Option C (

\frac{{x^{2}-y^{2}}}{{4}}

) is correct because by substituting

u = x+ay

and

v = x-ay

and solving for x and y, then substituting into the original equation, we get

f(x, y) = \frac{{x^{2}-y^{2}}}{{4}}

403 Easy

403 of 478

Application +4 | -1

Let

f(x)=x^5+2 x^3+3 x+1, x \in R,

and

g(x)

be a function such that

g(f(x))=x

for all

x \in R.

Then

\frac{g(7)}{g^{\prime}(7)}

is equal to:

View answer and explanation

Correct option: D

The correct answer is D (14). The problem involves finding the derivative of an inverse function and evaluating the ratio at a specific point.

404 Easy

404 of 478

Application +4 | -1

Let

f(x)=\sin x-\cos x

and

g(x)=\log \sqrt{5} x

; then the range of

g(\sqrt{2} f(x)+3)

View answer and explanation

Correct option: B

[0,2]

Option D ([0,2]) is correct because it represents the range of the composite function

g(\sqrt{{2}}f(x)+3)

considering the range of

f(x)

and the properties of the logarithm function.

405 Easy

405 of 478

Understanding +4 | -1

If a set

A

has

n

elements then the number of all relations on

A

View answer and explanation

Correct option: A

2^{n^2}

Option A (

2^{n^2}

) is correct because a relation on A is a subset of A x A, and if A has n elements, A x A has

n^2

elements, so there are

2^{n^2}

possible subsets.

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406 Easy

406 of 478

Application +4 | -1

f: \mathbf{R} \rightarrow \mathbf{R}

defined by

f(x)=x^4+2

then the value of

f^{-1}(83)

and

f^{-1}(-2)

respectively are

View answer and explanation

Correct option: B

\{3,-3\}, \phi

Option B (

\left\{3,-3\right\}, \phi

) is correct because

f^{-1}(83)

yields

\pm 3

, and

f^{-1}(-2)

is undefined in real numbers.

407 Easy

407 of 478

Application +4 | -1

f

be the greatest integer function and

g

be the modulus function, then

(g \circ f)\left(-\frac{5}{3}\right)-(f \circ g)\left(-\frac{5}{3}\right)=

View answer and explanation

Correct option: A

Option A (1) is correct because by evaluating the composite functions (gof)(-5/3) and (fog)(-5/3) and subtracting, the result is 1.

408 Easy

408 of 478

Analyzing +4 | -1

Let

A = \{ x \in R:x

is not a positive integer}. Define a function

f:A \to R

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

, then

f

View answer and explanation

Correct option: A

Injective but not surjective

Option A (Injective but not surjective) is correct because the function is one-to-one (injective) but does not cover all possible values in its range (not surjective).

409 Easy

409 of 478

Analyzing +4 | -1

Let

f:N \to N

defined by

f\left( x \right) = {x^2} + x + 1,\;x \in N

then

f

View answer and explanation

Correct option: C

One -one but not onto

Option C is correct. The function is one-to-one because different natural numbers will always produce different outputs. It is not onto because not every natural number is in the range of f.

410 Easy

410 of 478

Analyzing +4 | -1

Let

f(x) = {x^2},x \in R.

For any

A \subseteq R,

define

g(A) = \left\{ {x \in R:f(x) \in A} \right\}.

S = \left[ {0,4} \right],

then which one of the following statements is not true?

View answer and explanation

Correct option: C

g(f(S)) = g(S)

Option C (

g(f(S)) = g(S)

) is correct because it is NOT true. g(f(S)) = [-4, 4] and g(S) = [-2, 2], so they are not equal.

411 Easy

411 of 478

Application +4 | -1

{e^x} = y + \sqrt {1 + {y^2}} ,

then the value of

y

View answer and explanation

Correct option: B

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

Option B is correct because solving the equation for y involves algebraic manipulation and recognizing the hyperbolic sine function.

412 Easy

412 of 478

Application +4 | -1

Domain of the function

\sqrt{\left.\log \left(5 x-x^{2}\right) / 6\right\}}

View answer and explanation

Correct option: B

[2,3]

Option B ([2,3]) is correct because the domain requires the argument of the logarithm to be positive and the logarithm itself to be non-negative.

413 Easy

413 of 478

Understanding +4 | -1

The equivalent function of

\log x^{2}

View answer and explanation

Correct option: B

2 \log |x|

Option B (

2 \log |x|

) is correct because the domain of

\log x^2

includes negative values, which is accounted for by using the absolute value.

414 Easy

414 of 478

Application +4 | -1

y=3[x]+1=2[x-3]+5

, then find the value of

[x+y]

View answer and explanation

Correct option: D

-7

Option D (-7) is correct because solving the equation

3[x]+1=2[x-3]+5

yields

[x] = -2

, and then

y = -5

, so

[x+y] = [-2 - 5] = -7

415 Easy

415 of 478

Application +4 | -1

Let

f(x)=\ln \left(x^2+a x+1\right)

. If

f(x)

is defined

\forall \,x \in R

, then the number of integers in the range of '

a

' is:

View answer and explanation

Correct option: B

Option B is correct because the condition

x^2 + ax + 1 > 0

for all

x

implies

a^2 - 4 < 0

, leading to 3 integers in the range of 'a'.

416 Easy

416 of 478

Analyzing +4 | -1

The number of real solutions of the equation

{3^{ - \left| x \right|}} - {2^{\left| x \right|}} = 0,

View answer and explanation

Correct option: B

Option B (1) is correct because the equation simplifies to a single solution at x=0.

417 Easy

417 of 478

Application +4 | -1

Let

\mathbf{x} \in \mathbf{R}

for which

\left[x+\frac{29}{100}\right]+\left[x+\frac{30}{100}\right]+\ldots+\left[x+\frac{91}{100}\right]=\mathbf{5 4 6}

. Find

\frac{[100 \mathrm{x}]}{170}

where [.] represents G.I.F.

View answer and explanation

Correct option: A

Option A (5) is correct because solving the equation involving the greatest integer function leads to [100x]/170 = 5.

418 Easy

418 of 478

Understanding +4 | -1

3 < \left| x \right| <

, then

x

belongs to

View answer and explanation

Correct option: A

\left( { - 6\;, - 3} \right) \cup \left( {3,6} \right)

Option A (0) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.

419 Easy

419 of 478

Understanding +4 | -1

Suppose

X

contains

m

elements and

Y

contain

n

elements. The number of functions from

X

Y

View answer and explanation

Correct option: B

n^{{m}}

Option B (

n^{{m}}

) is correct because for each of the m elements in X, there are n possible images in Y, leading to

n^m

possible functions.

420 Easy

420 of 478

Understanding +4 | -1

f\left( x \right) = \left| {\sin x} \right|

has an inverse if its domain is

View answer and explanation

Correct option: B

\left[ {0,\;\pi /2} \right]

Option B ([

{0,\\;\pi /2}

]) is correct because the absolute value of sin(x) is one-to-one on this interval, allowing for an inverse function to exist.

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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