Range of the function $f\left( x \right) = \frac{{{x^2}}}{{{x^2} + 1}}$ is

$\left( { - 1,{\rm{ }}0} \right)$

$\left( { - 1,{\rm{ }}1} \right)$

If $ f(x)=\frac{x}{x-1} $, then $ \frac{f(a)}{f(a+1)} $ is equal to:

$ \mathrm{f}\left(\frac{1}{\mathrm{a}}\right) $

$ \mathrm{f}\left[\frac{-\mathrm{a}}{\mathrm{a}-1}\right] $

Back to FUNCTIONS

Your Progress

0 / 353 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →

Showing 151-180 of 353 questions

Preparing practice and saved progress…

151 Medium

151 of 353

Application +4 | -1

g(x)= x^2+x-2

and

\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2

, then

f(x)

is equal to:

View answer and explanation

Correct option: A

2 x-3

The correct answer is A (2x-3) because it is the function f(x) that satisfies the given composition equation.

152 Medium

152 of 353

Application +4 | -1

f:R \to S

defined by

f\left( x \right) = \sin \,\,x - \sqrt 3 \,\,\cos \,\,x + 1

, is onto, then the interval of

S

View answer and explanation

Correct option: D

\left[ { - 1,\;3} \right]

Option D is correct because the range of the function is determined by finding the minimum and maximum values of the sine and cosine expression.

153 Medium

153 of 353

Understanding +4 | -1

The function

f:R \to R

is defined by

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

. Observe the following statements

I. f is one-one II. f is onto

III. f is a decreasing function

Out of these, true statement are

View answer and explanation

Correct option: C

Only I, III

Option C (Only I, III) is correct because the function is one-to-one and decreasing, but not onto.

154 Medium

154 of 353

Application +4 | -1

f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|<1

, then

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

is equal to

View answer and explanation

Correct option: A

2 f(x)

Option A (

2 f(x)

) is correct because substituting

\frac{{2x}}{{1+x^2}}

into the function and simplifying using logarithm properties results in

2f(x)

155 Medium

155 of 353

Application +4 | -1

Assertion: If

(x+1, y-2)=(3,1)

, then

x=2

and

y=3

.

Reason: Two ordered pairs are equal, if their corresponding elements are equal.

View answer and explanation

Correct option: A

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion

Option A (Assertion is correct, Reason is correct; Reason is a correct explanation for assertion) is correct because the assertion is true, and the reason correctly explains why ordered pairs are equal.

156 Medium

156 of 353

Analyzing +4 | -1

The domain of

f(x)=\frac{\log _{(x+1)}(x-2)}{e^{2 \log _{e} x}-(2 x+3)}, x \in \mathbb{R}

View answer and explanation

Correct option: D

(2, \infty)-\{3\}

Option D (

(2, \infty)-\{3\}

) is correct because it satisfies all the conditions for the domain of the given function.

157 Medium

157 of 353

Analyzing +4 | -1

\log _{\frac{1}{3}}(|a|+1) > -1

, then the domain of the function

f(x)=\sqrt{2 x^4+a x^3-6 x^2-4 a x-8}

View answer and explanation

Correct option: C

(-\infty,-2] \cup[2, \infty)

Option C ((-∞,-2] ∪[2, ∞)) is correct because it represents the domain where the expression under the square root is non-negative, after determining the range of 'a'.

158 Medium

158 of 353

Application +4 | -1

The number of function

{f}: {A} \rightarrow

where

{A}=\{1,2,3,4,5\}

{B}=\left\{{a}_1, {a}_2, {a}_3, {a}_4, {a}_5\right\}

such that

{f}({i}) \neq {a}_{{i}}, {i}=1,2,3,4,5

View answer and explanation

Correct option: C

1024

Option D (192) is correct because it correctly calculates the number of four-digit odd numbers with distinct digits from the given set, considering the restrictions on the first and last digits.

159 Medium

159 of 353

Analyzing +4 | -1

Range of the function

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

View answer and explanation

Correct option: C

\left[ {0,\mathrm{ }1} \right)

Option C (

\left[ {0,\\;1} \right)

) is correct because the function is always non-negative and less than 1, approaching 1 as x approaches infinity.

160 Medium

160 of 353

Understanding +4 | -1

The domain of the function

f(x)=\frac{\left(x^2+1\right)}{x^2-3 x+3}

View answer and explanation

Correct option: C

R

Option C (

R

) is correct because the denominator

x^2-3x+3

has no real roots, so it is never zero, and the numerator is defined for all real numbers.

161 Medium

161 of 353

Application +4 | -1

Let

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

and

f: A \rightarrow A

satisfy

f(1)=2, f(2)=3, f(3)=4, f(4)=1

.
Suppose

g: A \rightarrow A

satisfies

g(1)=3

and

f o g=g of,

then

g=

View answer and explanation

Correct option: B

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

Option B is correct. By using the given conditions

f(1)=2, f(2)=3, f(3)=4, f(4)=1, g(1)=3

and

f o g=g o f

, we can find the values of

g(2), g(3),

and

g(4)

, which leads to

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

162 Medium

162 of 353

Analyzing +4 | -1

Let

f(x)=(x+1)(x+2)(x+3) \ldots .(x+100)

and.

g(x)=f(x) \cdot f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2

, then

g(x)=0

, has

View answer and explanation

Correct option: A

No solution

Option A (

0 < x < 1

) is correct because it satisfies the domain restrictions imposed by the logarithmic functions and the fourth root.

163 Medium

163 of 353

Application +4 | -1

Let

f

and

g

be two real functions defined by

f(x)=\frac{1}{x+4}, x \neq-4

and

g(x)=(x+4)^3

, then

f / g

is equal to

View answer and explanation

Correct option: B

\frac{1}{(x+4)^4}

Option B (

\frac{1}{(x+4)^4}

) is correct because it represents the division of f(x) by g(x).

164 Medium

164 of 353

Understanding +4 | -1

The domain and range of the function

f

given by

f(x)=2-|x-5|

View answer and explanation

Correct option: B

Domain

=R

, Range

=(-\infty, 2]

Option B (Domain

=R

, Range

=(-\infty, 2]

) is correct because the absolute value function is defined for all real numbers, and the maximum value of the function is 2.

165 Medium

165 of 353

Analyzing +4 | -1

f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}, x>0

, then the least value of

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

View answer and explanation

Correct option: C

The correct answer is C (2) because f(f(x)) = x, leading to f(f(x)) + f(f(4/x)) = x + 4/x, and the minimum value of x + 4/x for x > 0 is 2.

Explore Other Chapters in MATHEMATICS

View All →

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

166 Medium

166 of 353

Understanding +4 | -1

f:R \to R

is defined by

f\left( x \right) = \sin x\;and\,g:\left( {1,\;\infty } \right) \to R

is defined by

g\left( x \right) = \sqrt {{x^2} - 1}

, then

gof\left( x \right)

View answer and explanation

Correct option: D

Not defined

Option D (Not defined) is correct because the range of f(x) is not a subset of the domain of g(x), so the composite function gof(x) is not defined.

167 Medium

167 of 353

Application +4 | -1

g(x)=x^{2}+x-1

and

(g \circ f)(x)=4 x^{2}-10 x+5

, then

f\left(\frac{5}{4}\right)

is equal to:

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (

-\frac{1}{2}

) is correct because it is the solution to the quadratic equation obtained by substituting the given value into the composite function.

168 Medium

168 of 353

Analyzing +4 | -1

Let

f: R \rightarrow R

be a function defined by

f(x)=2 x^3-21 x^2+78 x+24

the number of integers in the solution set of

x

satisfying the inequality

f\left(f\left(f(x)-2 x^3\right)\right) \geq f\left(f\left(2 x^3-f(x)\right)\right)

View answer and explanation

Correct option: C

The correct answer is 5 because the inequality simplifies to comparing x to x, which means we need to find where f(x) is increasing, and then solve the inequality x <= 4, resulting in 5 integer solutions.

169 Medium

169 of 353

Application +4 | -1

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

then

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

is equal to:

View answer and explanation

Correct option: A

Option A (0) is correct because the expression simplifies to 0 using trigonometric identities and the given function definition.

170 Medium

170 of 353

Understanding +4 | -1

The domain of the function

f(x)=\frac{|x+3|}{x+3}

View answer and explanation

Correct option: B

\mathrm{R}-\{-3\}

Option B is correct because the function is defined for all real numbers except x = -3, where the denominator becomes zero.

171 Medium

171 of 353

Analyzing +4 | -1

The domain of the function

f(x)=\sqrt{x^2-3 x+2}+\frac{1}{\sqrt{3+2 x-x^2}}

View answer and explanation

Correct option: B

(-1,1] \cup[2,3)

Option B ((-1,1] ∪ [2,3)) is correct because it represents the intersection of the domains of the square root and the reciprocal square root functions.

172 Medium

172 of 353

Application +4 | -1

Range of the function,

f(x)=\frac{\left(1+x+x^2\right)\left(1+x^4\right)}{x^3}

, for

x>0

is:

View answer and explanation

Correct option: D

[6, \infty)

Option D ([6, ∞)) is correct because the minimum value of the function is 6, found by applying AM-GM inequality.

173 Medium

173 of 353

Application +4 | -1

\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}

are defined by

\mathrm{f}(\mathrm{x})=5 \mathrm{x}-3, \mathrm{~g}(\mathrm{x})=x^2+3

, then

\left(\mathrm{go} f^{-1}\right)(3)=

View answer and explanation

Correct option: B

\frac{111}{25}

Option B (

\frac{111}{25}

) is correct because it is the result of evaluating

g(f^{-1}(3))

after finding the inverse of f(x).

174 Medium

174 of 353

Analyzing +4 | -1

The range of the function

f(x)=2 \sqrt{x-2}+\sqrt{4-x}

View answer and explanation

Correct option: D

[\sqrt{2}, \sqrt{10}]

Option D ([√2, √10]) is correct because it represents the range of the function, considering the domain and critical points.

175 Medium

175 of 353

Analyzing +4 | -1

If [.] denotes the greatest integer function, then the domain of the function

f(x)=\frac{1}{\sqrt{|[|x|-1]|-5}}

View answer and explanation

Correct option: B

(-\infty,-7] \cup[7, \infty)

Option B

((-\infty,-7] \cup[7, \infty))

is correct because it represents the values of x for which the expression inside the square root is positive.

176 Medium

176 of 353

Application +4 | -1

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

, then

\frac{f(a)}{f(a+1)}

is equal to:

View answer and explanation

Correct option: A

\mathrm{f}\left(\mathrm{a}^{2}\right)

Option A (

f(a^2)

) is correct because

\frac{f(a)}{f(a+1)} = \frac{a^2}{a^2-1} = f(a^2)

when

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

177 Medium

177 of 353

Application +4 | -1

The total number of functions

f:\{1,2,3,4\} \rightarrow\{5,6,7,8\}

such that

\sum\limits_ {{k}=1}^4 {f}({k})

is an odd number is

View answer and explanation

Correct option: D

128

Option D (128) is correct because it considers the two possible cases where the sum of the function values is odd: one odd and three even, or three odd and one even.

178 Medium

178 of 353

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.

179 Medium

179 of 353

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

180 Medium

180 of 353

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.

Revision notes and key takeaways

Review FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

Showing 151-180 of 353 questions

Your Progress

Functions practice · Medium

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice