If $g(f(x))=|\sin x|$ and $f(g(x))=(\sin \sqrt{x})^2$, then

$f$ and $g$ cannot be determined

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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Showing 1-30 of 353 questions

1 Medium

1 of 353

Analyzing +4 | -1

g(f(x))=|\sin x|

and

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

, then

View answer and explanation

Correct option: A

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

Option A (7) is correct because by finding the center and radius of the circle using the given tangent and normal equations, the value of

\alpha - \beta + r

is 7.

2 Medium

2 of 353

Application +4 | -1

Let

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

be defined by

f\left( x \right) = x - \left[ x \right]

, then

{f^{ - 1}}\left( x \right)

equals

View answer and explanation

Correct option: D

x + 2

Option D (

x + 2

) is correct because the inverse function is found by solving for x in terms of y.

3 Medium

3 of 353

Application +4 | -1

Domain of the function

f(x)=\left[\log _{10}\left(\frac{5 x-x^{2}}{4}\right)\right]^{1 / 2}

View answer and explanation

Correct option: B

1 \leq x \leq 4

Option B (1 ≤ x ≤ 4) is correct because the domain requires the argument of the logarithm to be positive and the logarithm itself to be non-negative.

4 Medium

4 of 353

Analyzing +4 | -1

Let

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

. The range of

f(x)

View answer and explanation

Correct option: B

[\sqrt{5}, \sqrt{10}]

Option B (

[\sqrt{5}, \sqrt{10}]

) is correct because it accurately determines the range of the function by finding its critical points and endpoints.

5 Medium

5 of 353

Application +4 | -1

f\left( x \right) = \frac{{2x - 1}}{{x + 5}}\left( {x \ne - 5} \right)

, then

{f^{ - 1}}\left( x \right)

is equal to

View answer and explanation

Correct option: B

\frac{{5x + 1}}{{2 - x}},\;x \ne 2

Option B (

\frac{{5x + 1}}{{2 - x}},\;x \ne 2

) is correct because it represents the inverse function obtained by swapping x and y and solving for y.

6 Medium

6 of 353

Analyzing +4 | -1

Let

f:\left( {e,\infty } \right) \to R

be defined by

f\left( x \right) = \log \left[ {\log \left( {\log x} \right)} \right],

then

View answer and explanation

Correct option: C

f is both one-one and onto

Option C (f is both one-one and onto) is correct because the function is strictly increasing and its range covers all real numbers.

7 Medium

7 of 353

Analyzing +4 | -1

Assertion(A): A function

f

satisfies the condition

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

(where

{T}

is a fixed positive number) is periodic with period

2 {T}

.
Reason(R): If

f(x+2 T)=f(x)

, then period of

f(x)

2{T}

View answer and explanation

Correct option: A

Both (A) & (R) are individually true & (R) is correct explanation of (A).

The correct answer is A because both the assertion and the reason are true, and the reason correctly explains the assertion. The function satisfies

f(x+2T) = f(x)

, indicating a period of

2T

8 Medium

8 of 353

Analyzing +4 | -1

The function

f(x)

satisfies

f(10+x)=f(10-x)

and

f(20-x)=-f(20+x)

, then which of the following statements hold?

View answer and explanation

Correct option: B

f(x)

is an odd periodic function

The correct answer is

f(x)

is an odd periodic function because the given conditions lead to

f(x) = f(x+40)

(periodic) and

f(-x) = -f(x)

(odd).

9 Medium

9 of 353

Analyzing +4 | -1

If the function

f(x)=2 \tan x+(2 a+1) \ln |\sec x|+(a-2) x

is increasing in

\left(0, \frac{\pi}{2}\right)

then range of '

a

' equal to

View answer and explanation

Correct option: D

[0, \infty)

Option D ([0, ∞)) is correct because for f(x) to be increasing, its derivative must be greater than or equal to 0, which leads to the condition a ≥ 0.

10 Medium

10 of 353

Analyzing +4 | -1

The domain of the function

f(x)=\log _2\left(-\log _{1 / 2}\left(1+\frac{1}{\sqrt[4]{x}}\right)-1\right)

View answer and explanation

Correct option: A

0 < x < 1

Option A (No solution) is correct because g(x) is always negative, thus it can never be equal to zero.

11 Medium

11 of 353

Application +4 | -1

[x]^2-5[x]+6=0

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

View answer and explanation

Correct option: D

x \in[2,4)

Option D (

x \in [2,4)

) is correct because it encompasses both possible solutions for

x

derived from the given equation involving the greatest integer function.

12 Medium

12 of 353

Analyzing +4 | -1

The domain of the function

f\left( x \right)

given by

f\left( x \right) = \sqrt {\frac{{ - {{\log }_{0.3}}\left( {x - 1} \right)}}{{ - {x^2} + 3x + 18}}} ,\;\;\mathrm{is}

, is

View answer and explanation

Correct option: B

\left( {2,\;6} \right)

Option B

\left( {2,\;6} \right)

is correct because it satisfies the logarithmic and rational function domain restrictions.

13 Medium

13 of 353

Analyzing +4 | -1

For

x \in R, x \neq 0

, let

f_{0}(x)=\frac{1}{1-x}

and

f_{n+1}(x)=f_{0}\left(f_{n}(x)\right)

\mathrm{n}=0,1,2, \ldots

Then the value of

\mathrm{f}_{100}(3)+\mathrm{f}_{1}\left(\frac{2}{3}\right)+\mathrm{f}_{2}\left(\frac{3}{2}\right)

is equal to :

View answer and explanation

Correct option: C

\frac{5}{3}

Option C (5/3) is correct because the function values repeat in a cycle of 3, and substituting the values and simplifying gives 5/3.

14 Medium

14 of 353

Analyzing +4 | -1

A=\{x \in R ;-1 \leq x \leq 1\}=B

then the mapping

f: A \rightarrow B

given by

f(x)=x|x|

View answer and explanation

Correct option: C

Bijective

Option C (Bijective) is correct because the function f(x) = x|x| is both injective (one-to-one) and surjective (onto) over the given domain.

15 Medium

15 of 353

Application +4 | -1

f: R \rightarrow R

satisfies

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

, for all

x

y \in R

and

f(1)=7

, then

\sum_{r=1}^n f(r)

View answer and explanation

Correct option: A

\frac{7 n(n+1)}{2}

Option A (

\frac{{7 n(n+1)}}{{2}}

) is correct because the sum of the function values from 1 to n is a multiple of 7 times the sum of the first n natural numbers.

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16 Medium

16 of 353

Application +4 | -1

Let

\mathrm{R}

be a relation on

\mathrm{N}

defined by

x+2 y=8

. The domain of

\mathrm{R}

View answer and explanation

Correct option: C

\{2,4,6\}

Option C ({2,4,6}) is correct because the domain of the relation R consists of all possible x values that satisfy the equation x + 2y = 8, where x and y are natural numbers.

17 Medium

17 of 353

Application +4 | -1

f(x)=x^2+1

, then

f^{-1}(17)

and

f^{-1}(-3)

will be

View answer and explanation

Correct option: D

None of these

Option D is correct because

f^{-1}(17) = \pm 4

and

f^{-1}(-3)

is not a real number.

18 Medium

18 of 353

Understanding +4 | -1

Consider the following statements.

Statement-I: An onto function

{f}:\{1,2,3\} \rightarrow\{1,2,3\}

is always one-one.

Statement-II: A one-one function

{f}:\{1,2,3\} \rightarrow\{1,2,3\}

must be onto.

View answer and explanation

Correct option: C

Both I and II are true

Option C (Both I and II are true) is correct because for a function from a finite set to itself, being onto implies being one-to-one, and vice versa.

19 Medium

19 of 353

Application +4 | -1

Suppose that

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

and

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

, then

f(x)

View answer and explanation

Correct option: B

2+x^2

Option B (

2+x^2

) is correct because substituting

g(x)

into

f(g(x))

and solving for

f(x)

yields

2+x^2

20 Medium

20 of 353

Application +4 | -1

The domain of the function

{\sin ^ - }\left( {{{\log }_2}\frac{{{x^2}}}{2}} \right)

View answer and explanation

Correct option: B

\left[ { - 2,\mathrm{ }2} \right] - \left( { - 1,\mathrm{ }1} \right)

Option A (

\frac{{{x^2} - {y^2}}}{8}

) is correct because it represents the function

f(x, y)

obtained by solving the system of equations and substituting into the given equation.

21 Medium

21 of 353

Application +4 | -1

f\left( x \right) = {2^x},

then

f\left( 0 \right),f\left( 1 \right),f\left( 2 \right), \ldots

are in

View answer and explanation

Correct option: B

Option B (GP) is correct because the ratio between consecutive terms

f(n+1)

and

f(n)

is constant.

22 Medium

22 of 353

Analyzing +4 | -1

Let

f: R-\left\{\frac{\alpha}{6}\right\} \rightarrow R

be definced by

f(x)=\frac{5 x+3}{6 x-\alpha}

. Then the value of

\alpha

for which

(f o f)(x)=x

, for all

x \in R-\left\{\frac{\alpha}{6}\right\}

, is :

View answer and explanation

Correct option: D

Option D (5) is correct because when α=5, the composite function (f o f)(x) simplifies to x, satisfying the given condition for all x in the domain.

23 Medium

23 of 353

Analyzing +4 | -1

f:R \to R

is defined by

f\left( x \right) = \frac{{{e^{{x^2}}} - {e^{ - {x^2}}}}}{{{e^{{x^2}}} + {e^{ - {x^2}}}}},

View answer and explanation

Correct option: A

One-one but not onto

Option D is correct. The function is neither one-one nor onto because

f(-x) = f(x)

, making it many-to-one, and the range is a subset of R, not all of R, making it 'into'.

24 Medium

24 of 353

Application +4 | -1

A quadratic

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)

View answer and explanation

Correct option: C

Option C (48) is correct because the quadratic can be determined uniquely given the vertex and another point.

25 Medium

25 of 353

Understanding +4 | -1

Assertion: If

f

is even function,

g

is odd function, then

\frac{f}{g}

({g} \neq 0)

is an odd function.

Reason: If

f(-x)=-f(x)

for every

x

of its domain, then

f(x)

is called an odd function and if

f(-x)=f(x)

for every

x

of its domain, then

{f}({x})

is called an even function.

View answer and explanation

Correct option: A

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

Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion about even and odd functions.

26 Medium

26 of 353

Application +4 | -1

f: R \rightarrow R

satisfies

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

\forall x, y \in R

and

f(1)=7

, then

\sum\limits_{{r=1}}^n f(r)

View answer and explanation

Correct option: D

\frac{7 n(n+1)}{2}

Option D (

\frac{{7 n(n+1)}}{{2}}

) is correct because the sum of the function values from 1 to n is derived using the given functional equation and the sum of the first n natural numbers.

27 Medium

27 of 353

Application +4 | -1

For

x \in\left(0, \frac{3}{2}\right), \operatorname{let} f(x)=\sqrt{x}, g(x)=\tan x

and

h(x)=\frac{1-x^{2}}{1+x^{2}}

. If

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

, then

\phi\left(\frac{\pi}{3}\right)

is equal to :

View answer and explanation

Correct option: B

\tan \frac{11 \pi}{12}

The correct answer is B (tan(11π/12)) because it involves evaluating composite functions and using trigonometric identities.

28 Medium

28 of 353

Understanding +4 | -1

Let

f\left( x \right) = x,g\left( x \right) = 1/x\;and\;h\left( x \right) = f\left( x \right)g\left( x \right).

Then,

h\left( x \right) = 1,

View answer and explanation

Correct option: B

x

is a non-zero real number

Option B (

x

is a non-zero real number) is correct because

h(x) = f(x)g(x) = x * (1/x) = 1

is only true when

x

is a non-zero real number, as division by zero is undefined.

29 Medium

29 of 353

Application +4 | -1

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

has extremum at

x = 1\,and\,x = 3

, then

View answer and explanation

Correct option: A

a = - 3/4,b = - 1/8

Option A (

a = - 3/4,b = - 1/8

) is correct because these values satisfy the conditions for extrema at x = 1 and x = 3, derived from the first derivative test.

30 Medium

30 of 353

Application +4 | -1

Inverse of the function

y=2 x-3

View answer and explanation

Correct option: A

\frac{x+3}{2}

Option A is correct because solving for x in terms of y and swapping x and y gives

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

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 353 single-correct-answer MCQs on FUNCTIONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
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