If $ f(x)=\log \frac{1+x}{1-x} $, then $ f(x) $ is

$ f\left(x_{1}\right) f\left(x_{2}\right)=f\left(x_{1}+x_{2}\right) $

$ \frac{f\left(x_{1}\right)}{f\left(x_{2}\right)}=f\left(x_{1}-x_{2}\right) $

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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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 451-480 of 650 questions

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451 Easy

451 of 650

Application +4 | -1

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

are defined by

f\left( x \right) = 2x + 3\;and\;g\left( x \right) = {x^2} + 7,

then the values of x such that

g\left( {f\left( x \right)} \right) = 8

are

View answer and explanation

Correct option: C

- 1, - 2

Option C (-1, -2) is correct because these values satisfy the equation g(f(x)) = 8.

452 Easy

452 of 650

Analyzing +4 | -1

Let

f\left( x \right) = \left| {x - 1} \right|

Then,

View answer and explanation

Correct option: D

None of these

Option D None of these is correct because none of the given statements hold true for the function.

453 Medium

453 of 650

Application +4 | -1

Which of the following is an even function?

View answer and explanation

Correct option: A

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

The correct answer is A (

x(\frac{a^x-1}{a^x+1})

) because substituting -x into the function results in the same function, indicating even symmetry.

454 Easy

454 of 650

Application +4 | -1

Let

f

and

g

be two functions defined by

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

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

. Then

(f \circ g)^{-1}(x)

is equal to

View answer and explanation

Correct option: A

x

Option A (x) is correct because the composite function f(g(x)) simplifies to x, making the inverse function also x.

455 Easy

455 of 650

Application +4 | -1

Let

f:\left( { - \infty ,2} \right] \to \left( { - \infty ,4} \right]

be a function defined by

f\left( x \right) = 4x - {x^2}.

Then,

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

View answer and explanation

Correct option: A

2 - \sqrt {4 - x}

Option B (

\left[ { - 1 - \sqrt 3 \le x \le - 1 + \sqrt 3 } \right]

) is correct because it represents the values of x for which the expression inside the square root is non-negative.

456 Medium

456 of 650

Application +4 | -1

r

is a real number such that

\left| r \right| < 1

and if

a = 5\left( {1 - r} \right),

then

View answer and explanation

Correct option: C

0 < a < 10

Option C

0 < a < 10

is correct because it represents the range of values that

a

can take given the condition on

r

457 Medium

457 of 650

Analyzing +4 | -1

The function

f:R \to R

defined by

f\left( x \right) = {6^x} + {6^{\left| x \right|}},

View answer and explanation

Correct option: C

One-one and into

Option C (One-one and into) is correct because the function is one-to-one (injective) but not onto (surjective) since its range is not the entire real number set.

458 Easy

458 of 650

Application +4 | -1

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

, then

y=

View answer and explanation

Correct option: B

\frac{e^x-e^{-x}}{2}

Option B (

\frac{{e^x-e^{-x}}}{{2}}

) is correct because solving the given equation for

y

involves squaring and rearranging terms to isolate

y

459 Easy

459 of 650

Application +4 | -1

f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}

for

x>2

, then

f(11)=

View answer and explanation

Correct option: C

6 / 7

Option C (

6 / 7

) is correct because simplifying the expression and substituting x=11 leads to the value 6/7.

460 Easy

460 of 650

Understanding +4 | -1

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

, then

f(x)

View answer and explanation

Correct option: D

Odd function

Option D (Odd function) is correct because

f(-x) = -f(x)

, which is the definition of an odd function.

461 Easy

461 of 650

Understanding +4 | -1

The function

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

, is

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because

f(-x) = -f(x)

, satisfying the definition of an odd function.

462 Medium

462 of 650

Analyzing +4 | -1

The second degree polynomial

f\left( x \right)

, satisfying

f\left( 0 \right) = 0,\;f\left( 1 \right) = 1,\;f'\left( x \right) > 0\;for\;all\;x \in \left( {0,\;1} \right)

View answer and explanation

Correct option: C

f\left( x \right) = ax + \left( {1 - a} \right){x^2},\;a \in \left( {0,\;2} \right)

Option C (

f(x) = ax + (1 - a)x^2, a \in (0, 2)

) is correct because it satisfies the given conditions: f(0) = 0, f(1) = 1, and f'(x) > 0 for x in (0, 1).

463 Medium

463 of 650

Analyzing +4 | -1

Let

X\;and\,Y

be subsets of R, the set of all real numbers. The function

f:X \to Y

defined by

f\left( x \right) = {x^2}\;for\,x \in X

is one-one but not onto, if (Here,

{R^ + }

is the set of all positive real numbers)

View answer and explanation

Correct option: C

X = {R^ + },Y = R

Option C (

X = {R^ + },Y = R

) is correct because it defines a function that is one-to-one but not onto.

464 Medium

464 of 650

Application +4 | -1

Let

f(x)

be a function such that

f(x+y)=f(x) \cdot f(y)

for all

x, y \in \mathbb{N}

. If

f(1)=3

and

\sum\limits_{{k=1}}^{n} f(k)=3279

, then the value of

n

View answer and explanation

Correct option: D

Option D (7) is correct because it involves solving a functional equation and a geometric series.

465 Easy

465 of 650

Application +4 | -1

If the functions

f

and

g

are defined by

f\left( x \right) = 3x - 4,\;g\left( x \right) = 3x + 2

, respectively then

{g^{ - 1}}\left( {{f^{ - 1}}\left( 5 \right)} \right) =

View answer and explanation

Correct option: C

1/3

Option C (1/3) is correct because it is the result of evaluating the composite inverse function.

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466 Medium

466 of 650

Understanding +4 | -1

f:R \to R

given by

f\left( x \right) = 5 - 3\sin x,

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is neither one-to-one nor onto.

467 Hard

467 of 650

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.

468 Easy

468 of 650

Application +4 | -1

f(x+y)=f(x y) \, \forall \, x, y \in \mathbf{R}

, and

f(2013)

=2013

, then

f(-2013)

equals

View answer and explanation

Correct option: A

2013

Option A (2013) is correct because f(x+y) = f(xy) implies that f is a constant function, and since f(2013) = 2013, then f(-2013) must also be 2013.

469 Medium

469 of 650

Application +4 | -1

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

is defined for all

x \in R,

then the value of

{x^2} - 2

can be

View answer and explanation

Correct option: D

-1

Option D (-1) is correct because by substituting -x for x and adding the equations, we find that x² = 1, and thus x² - 2 = -1.

470 Easy

470 of 650

Application +4 | -1

If the number of elements in

(A-B)-C,(B-C)

\sim A,(C-A)-B

and

A \cap B \cap C

10,15,20

, and 5 respectively then the number of elements in

(A \Delta B) \Delta C

View answer and explanation

Correct option: B

Option B (50) is correct because (A Δ B) Δ C represents the elements that belong to exactly one or all three of the sets, which sums to 50.

471 Medium

471 of 650

Application +4 | -1

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

, then

f(f(x))

equals:

View answer and explanation

Correct option: A

x

Option A (

x

) is correct because when the function

f(x)

is applied to itself, the result simplifies back to

x

472 Medium

472 of 650

Application +4 | -1

f(x)=\left\{\begin{array}{ll}x^3+1 & x<0 \\ x^2+1, & x \geq 0\end{array}, g(x)= \begin{cases}(x-1)^{1 / 3}, & x<1 \\ (x-1)^{1 / 2}, & x \geq 1\end{cases}\right.

then

(g \circ f)(x)

is equal to

View answer and explanation

Correct option: A

x, \forall x \in R

Option A (

x, \forall x \in R

) is correct because when you compute

(g \circ f)(x)

for both cases of

f(x)

and

g(x)

, the result simplifies to

x

for all real numbers.

473 Easy

473 of 650

Understanding +4 | -1

A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}

are two sets and function

f: A \rightarrow B

is defined by

f(x)=x+2 ; \forall x \in A

, then the function

f

View answer and explanation

Correct option: C

One-one

Option C (One-one) is correct because each element in set A maps to a unique element in set B, satisfying the condition for an injective function.

474 Medium

474 of 650

Application +4 | -1

The period of the function

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

is:

View answer and explanation

Correct option: A

6 \pi

The correct answer is A (6π) because the period of cos(kx) is 2π/k, and in this case, k = 1/3.

475 Easy

475 of 650

Analyzing +4 | -1

Let

f(x)=\sin \sqrt{x}

, then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because

f(x) = \sin \sqrt{x}

does not satisfy the periodicity condition

f(x+T) = f(x)

for any constant

T

476 Easy

476 of 650

Analyzing +4 | -1

Which of the following functions is (are) not an injective map(s)?

View answer and explanation

Correct option: B

\mathrm{g}\left( x \right) = x + \frac{1}{x},x \in \left( {0,\infty } \right)

The correct answer is

\mathrm{g}\left( x \right) = x + \frac{{1}}{{x}},x \in \left( {0,\infty } \right)

because it is not injective; g(2) = g(1/2) = 5/2, demonstrating that different inputs can yield the same output.

477 Medium

477 of 650

Analyzing +4 | -1

The domain of function

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

View answer and explanation

Correct option: C

\left( { - 3, - 2} \right) \cup \left( { - 2, - 1} \right) \cup \left( {1,\infty } \right)

Option C ((

- 3, - 2) \cup ( - 2, - 1) \cup (1,\infty )

) is correct because it satisfies the conditions for the logarithm to be defined.

478 Easy

478 of 650

Understanding +4 | -1

Let

f: R \rightarrow R

be defined as

f(x)=3 x

. Choose the correct answer.

View answer and explanation

Correct option: A

f

is one-one onto

Option A (

f

is one-one onto) is correct because the function

f(x) = 3x

is both injective (one-to-one) and surjective (onto) over the real numbers.

479 Easy

479 of 650

Application +4 | -1

The range of

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

is given by

View answer and explanation

Correct option: D

\left[-\frac{1}{2}, \frac{1}{2}\right]

Option D (

\left[-\frac{1}{2}, \frac{1}{2}\right]

) is correct because by setting

f(x) = y

and analyzing the discriminant of the resulting quadratic equation, we find the range of

y

to be between -1/2 and 1/2.

480 Hard

480 of 650

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).

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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