The simplified form of $f(x)=|x-2|+|2-x|,-3 \leq x \leq 3$ is

$\left\{\begin{array}{cc}3 x, & 2 \leq x \leq 3 \\ 4, & -2 \leq x<2 \\ -2 x, & -3 \leq x<-2\end{array}\right.$

Let $\left[ x \right]\;$ denote the greatest integer less than or equal to $x$. If $f\left( x \right) = {\sin ^{ - 1}}x,{\rm{g}}\left( x \right) = \left[ {{x^2}} \right]\;and\;h\left( x \right) = 2x,\frac{1}{2} \le x \le \frac{1}{{\sqrt 2 }},$ then

$hofo{\rm{g}} = ho{\rm{g}}\;of$

$fo{\rm{g}}oh\;\left( x \right) = \pi /2$

$fo{\rm{g}}oh\left( x \right) = \pi $

$hofo{\rm{g}} \ne ho{\rm{g}}\;of$

The domain of the function ${\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{{1 + x}}{x}} \right)$ is

$\left( { - 1,\, - \frac{1}{2}} \right] \cup \left( {0,\,\infty } \right)$

$\left[ { - \frac{1}{2},\,0} \right) \cup \left[ {1,\,\infty } \right)$

The domain of the function $f\left( x \right) = {\rm{cose}}{{\rm{c}}^{ - 1}}\left[ {\sin x} \right]$ in $\left[ {0,\;2\;\pi } \right],$ where $\left[ \cdot \right]$ denotes the greatest integer function, is

$\left[ {0,\;\pi /2} \right) \cup \left( {\pi ,3\pi /2} \right]$

$\left( {0,\;\pi } \right] \cup \left\{ {3\;\pi /2} \right\}$

$\left( {\pi /2,\pi } \right) \cup \left( {3\pi /2,2\pi } \right)$

Solution of ${2^x} + {2^{\left| x \right|}} \ge 2\sqrt 2 $ is

$\left( { - \infty ,{{\log }_2}\left( {\sqrt 2 + 1} \right)} \right)$

$\left( {\frac{1}{2},{{\log }_2}\left( {\sqrt 2 - 1} \right)} \right)$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Hard

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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 1-30 of 33 questions

1 Hard

1 of 33

Analyzing +4 | -1

f: \mathbb{R} \rightarrow \mathbb{R}

is a continuous function such that the equation

f(x)=x

has no real solution. Let

g: \mathbb{R} \rightarrow \mathbb{R}

be defined by

g(x)=f(f(x))

. Then the number of real solutions of the equation

g(x)=x

View answer and explanation

Correct option: B

Option B (0) is correct because if f(x)=x has no real solution, then f(x) is always greater than x or always less than x, which implies f(f(x))=x also has no real solution.

2 Hard

2 of 33

Application +4 | -1

Let the function

f\left( x \right) = {x^2} + x + \sin x - \cos x + \log \left( {1 + \left| x \right|} \right)

be defined on the interval

\left[ {0,\;1} \right].

The odd extension of

f\left( x \right)

to the interval

\left[ { - 1,\;1} \right]

View answer and explanation

Correct option: B

- {x^2} + x + \sin x + \cos x - \log \left( {1 + \left| x \right|} \right)

Option B is correct because the odd extension of f(x) involves changing the even terms' signs when x is negative.

3 Hard

3 of 33

Application +4 | -1

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a function such that

f(x+y)=

f(x)+f(y)

for all

x, y

belonging to

\mathbb{R}

. If

m

and

n

are integers, then

f(m / n)

is equal to

View answer and explanation

Correct option: C

\left(\frac{m}{n}\right) f(1)

Option C (

\left(\frac{{m}}{{n}}\right) f(1)

) is correct because it follows from the given functional equation

f(x+y) = f(x) + f(y)

and properties of integers and rational numbers.

4 Hard

4 of 33

Analyzing +4 | -1

The set of values of

x

for which of the function

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

exists is

View answer and explanation

Correct option: C

\phi

Option C (

\phi

) is correct because the intersection of the domains of the individual terms in the function is empty, meaning there's no x value for which the entire function is defined.

5 Hard

5 of 33

Analyzing +4 | -1

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

are given by

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

for each

x \in R,

then

\left\{ {x \in R\;:g\left( {f\left( x \right)} \right) \le f\left( {g\left( x \right)} \right)} \right\} =

View answer and explanation

Correct option: A

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

Option A (

Z \cup ( - \infty ,0)

) is correct because the inequality holds for all integers and all negative real numbers.

6 Hard

6 of 33

Analyzing +4 | -1

{b^2} - 4\;ac = 0,a > 0,

then the domain of the function

f\left( x \right) = \log \left\{ {a{x^3} + \left( {a + b} \right){x^2} + \left( {b + c} \right)x + c)} \right\}

View answer and explanation

Correct option: C

R - \left\{ {\left\{ { - \frac{b}{{2\;a}}} \right\} \cap \left( { - \infty , - 1} \right]} \right\}

Option D (

R - \left\{ {\left\{ { - \frac{{b}}{{2\;a}}} \right\} \cap \left( { - \infty , - 1} \right]} \right\}

) is correct because it represents the domain where the argument of the logarithm is positive, excluding the point where the quadratic term becomes zero and considering the factor (x+1).

7 Hard

7 of 33

Analyzing +4 | -1

The domain of the function

e^{\operatorname{Sin}^{-1} x}+\frac{1}{[x]}+\frac{1}{\sqrt{x+1}}

View answer and explanation

Correct option: C

(-1,0) \cup\{1\}

Option C ((-1,0) ∪ {1}) is correct because it represents the intersection of the domains of the arcsine, reciprocal of the greatest integer function, and reciprocal of the square root functions.

8 Hard

8 of 33

Analyzing +4 | -1

The simplified form of

f(x)=|x-2|+|2-x|,-3 \leq x \leq 3

View answer and explanation

Correct option: A

\left\{\begin{array}{cc}2 x, & 2 \leq x \leq 3 \\ 4, & -2 \leq x<2 \\ -2 x, & -3 \leq x<-2\end{array}\right.

Option D (

R_4=\{(1,3),(2,5),(2,4),(7,9)\}

) is not a relation from

X

Y

because the element (2,4) and (7,9) are not in

X \times Y

9 Hard

9 of 33

Analyzing +4 | -1

Let

\left[ x \right]\;

denote the greatest integer less than or equal to

x

. If

f\left( x \right) = {\sin ^{ - 1}}x,\mathrm{g}\left( x \right) = \left[ {{x^2}} \right]\;and\;h\left( x \right) = 2x,\frac{1}{2} \le x \le \frac{1}{{\sqrt 2 }},

then

View answer and explanation

Correct option: C

hofo\mathrm{g} = ho\mathrm{g}\;of

Option D (

hofo\mathrm{g} = ho\mathrm{g}\;of

) is correct because after evaluating both sides, they are equal for all x in the given interval.

10 Hard

10 of 33

Application +4 | -1

f_1(x)=\frac{x}{4}+6

for all

x \in \mathbb{R}

and

f_n(x)=f_1\left(f_{n-1}(x)\right)

for

n \geq 2

, then

f_n(x)=?

View answer and explanation

Correct option: A

\frac{x-8}{4^n}+8

Option A (

\frac{{x-8}}{{4^n}}+8

) is correct because it represents the nth iteration of the function f1(x).

11 Hard

11 of 33

Analyzing +4 | -1

Let

f: \mathbb{R} \rightarrow \mathbb{R}-\{3\}

be a function such that for some

p>0

f(x+p)=\frac{f(x)-5}{f(x)-3}

for all

x \in \mathbb{R}

. Then, period of

f

View answer and explanation

Correct option: C

4 p

Option C (

4p

) is correct because it is the smallest value for which f(x + 4p) = f(x), indicating the function repeats its values after this interval.

12 Hard

12 of 33

Analyzing +4 | -1

The domain of the function

e^x+\operatorname{Sin}^{-1}\left(\frac{x}{2}-1\right)+\log (\sqrt{x-[x]})

where

[x]

is the integral part of

x

View answer and explanation

Correct option: D

(0,4)-\{1,2,3\}

Option D (

(0,4)-\{1,2,3\}

) is correct because it represents the intersection of the domains of all three terms in the function.

13 Hard

13 of 33

Analyzing +4 | -1

\;{5^x} + {\left( {2\sqrt 3 } \right)^{2x}} \ge {13^x},\;

then the solution set for

x

View answer and explanation

Correct option: C

( - \infty ,2]

Option C,

( - \infty ,2]

, is correct because the inequality holds for x less than or equal to 2, based on the properties of trigonometric functions and exponents.

14 Hard

14 of 33

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.

15 Hard

15 of 33

Analyzing +4 | -1

f\left( x \right) = \sqrt {\cos \left( {\sin x} \right)} + \sqrt {\sin \left( {\cos x} \right)} ,

then range of

f\left( x \right)

View answer and explanation

Correct option: B

\left[ {\sqrt {\cos 1} ,1 + \sqrt {\sin 1} } \right]

Option D (None of these) is incorrect. The correct range is

\left[ {\sqrt {\cos 1} ,1 + \sqrt {\sin 1} } \right]

Explore Other Chapters in MATHEMATICS

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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

16 Hard

16 of 33

Application +4 | -1

The domain of the function

\mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\frac{{1 + x}}{x}} \right)

View answer and explanation

Correct option: D

\left[ { - \frac{1}{2},\,\infty } \right) - \left\{ 0 \right\}

Option D (

\left[ { - \frac{1}{2},\,\infty } \right) - \left\{ 0 \right\}

) is correct because it represents the values of x for which the argument of the inverse cosecant function is greater than or equal to 1 in absolute value.

17 Hard

17 of 33

Application +4 | -1

Let

f:N \to N

be a function such that

f(m + n) = f(m) + f(n)

for every

m,\,\,n \in N.

f(6) = 18,

then

f(2).f(3)

is equal to

View answer and explanation

Correct option: B

Option B (54) is correct because by using the given functional equation and the value of f(6), we can find f(2) and f(3) and then multiply them.

18 Hard

18 of 33

Analyzing +4 | -1

Let

A=\{x \in \mathbb{R}:[x+3]+[x+4] \leq 3\}

B=\left\{x \in \mathbb{R}: 3^{x}\left(\sum\limits_{{r=1}}^{\infty} \frac{3}{10^{r}}\right)^{x-3}<3^{-3 x}\right\}

, where

[t]

denotes greatest integer function. Then,

View answer and explanation

Correct option: D

A=B

Option D (A=B) is correct because both sets A and B are equal to the interval (-

\infty

, -1) after simplification.

19 Hard

19 of 33

Application +4 | -1

Let

g(x)=\frac{e^x-e^{-x}}{2}

and

g(f(x))=x

, then

f\left(\frac{e^{22}-1}{2 e^{11}}\right)-5=

View answer and explanation

Correct option: A

Option A (6) is correct because finding the inverse of g(x) and substituting the given value results in f(x) = 11, thus f(x) - 5 = 6.

20 Hard

20 of 33

Application +4 | -1

Let a and b be two integers such that

10a + b = 5\;and\;P\left( x \right) = x + ax + b.

The integer n such that

P\left( {10} \right).P\left( {11} \right) = P\left( n \right)

View answer and explanation

Correct option: A

Option A (15) is correct because it satisfies the given conditions for integers a and b, and the polynomial P(x).

21 Hard

21 of 33

Understanding +4 | -1

f\left( { - x} \right) = - f\left( x \right),

then

f\left( x \right)

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because the given condition

f(-x) = -f(x)

is the definition of an odd function.

22 Hard

22 of 33

Application +4 | -1

Suppose

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

is defined by

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

then

\left\{ {x \in \left[ { - 2,2} \right]:x \le 0\;\mathrm{and}\;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 it is the only value that satisfies the given conditions and falls within the specified interval.

23 Hard

23 of 33

Application +4 | -1

[x]

be the greatest integer less than or equal to

x,

then

\sum\limits_{n = 8}^{100} {\left[ {\frac{{{{( - 1)}^n}n}}{2}} \right]}

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because the sum telescopes, leaving only the first few terms after cancellations.

24 Hard

24 of 33

Analyzing +4 | -1

Let

x = \left[ {\frac{{a + 2b}}{{a + b}}} \right]

and

y = \frac{a}{b}

, where

a

and

b

are positive integers. If

{y^2} > 2

then

View answer and explanation

Correct option: B

{x^2} < 2

Option B (

{x^2} < 2

) is correct because given the conditions, the value of x is such that its square is less than 2.

25 Hard

25 of 33

Analyzing +4 | -1

Set of values of

x

lying in

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

satisfying the inequality

|\sin x|\, > 2{\sin ^2}x

contains

View answer and explanation

Correct option: A

\left( {\;0,\,\,\frac{\pi }{6}} \right)\,\, \cup \,\,\left( {\;\pi ,\,\frac{{7\pi }}{6}} \right)

Option A (

\left( {\;0,\,\frac{\pi }{{6}}} \right)\\ \cup \\ \left( {\;\pi ,\frac{{7\pi }}{{6}}} \right)

) is correct because it represents the intervals where the absolute value of sin x is between 0 and 1/2, excluding points where sin x is 0.

26 Hard

26 of 33

Analyzing +4 | -1

The domain of the function

f\left( x \right) = \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sin x} \right]

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

where

\left[ \cdot \right]

denotes the greatest integer function, is

View answer and explanation

Correct option: B

\left( {\pi ,2\;\pi } \right) \cup \left\{ {\pi /2} \right\}

Option B (

\left( {\pi ,2\;\pi } \right) \cup \left\{ {\pi /2} \right\}

) is correct because it identifies the intervals where [sin x] is either 1 or -1, which are the only values for which csc^-1[sin x] is defined.

27 Hard

27 of 33

Understanding +4 | -1

The function

f: N \rightarrow N

(

N

is the set of natural numbers) defined by

f(n)=2 n+3

, is

View answer and explanation

Correct option: B

Injective

Option B (Injective) is correct because for every distinct n in N, f(n) produces a distinct value, meaning the function is one-to-one.

28 Hard

28 of 33

Analyzing +4 | -1

Let

f(x)=|x-1|

. Then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because none of the given functional relationships hold true for the function f(x) = |x-1|.

29 Hard

29 of 33

Analyzing +4 | -1

Suppose

a,b

and

c

are real numbers such that

\frac{a}{b} > 1

and

\frac{a}{c} < 0

. Which one of the following is true?

View answer and explanation

Correct option: C

\left( {a - c} \right)\left( {b - c} \right) > 0

The correct answer is C. Given the conditions, we can deduce the signs of a, b, and c, and thus determine that (a-c) and (b-c) have the same sign, making their product positive.

30 Hard

30 of 33

Analyzing +4 | -1

Solution of

{2^x} + {2^{\left| x \right|}} \ge 2\sqrt 2

View answer and explanation

Correct option: D

( - \infty ,{\log _2}\left( {\sqrt 2 - 1} \right)) \cup \left[ {\left. {\frac{1}{2},\infty } \right)} \right.

Option D (

(- \infty, {\log_2}(\sqrt{2} - 1)) \cup [\frac{1}{2}, \infty)

) is correct because it represents the complete solution set obtained by considering both positive and negative values of x.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 33 single-correct-answer MCQs on FUNCTIONS.
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