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

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

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

If $f(x)=3 x-5$, then $f^{-1}(x)$

does not exist because $f$ is not one-one

does not exist because $f$ is not onto

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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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 601-630 of 650 questions

601 Easy

601 of 650

Analyzing +4 | -1

Let

A = \{ 1,\,\,2,\,\,3,\,\,...,\,\,10\}

and

f:A \to A

be defined as

f(k) = \left\{ \begin{array}{l} k + 1,\quad if\,\,k\,\,is\,\,odd\\ k,\quad {\kern 1pt} {\kern 1pt} \quad if\,\,k\,\,is\,\,even \end{array} \right.

Then, the number of possible functions

g:A \to A,

such that

gof = f

View answer and explanation

Correct option: A

{10^5}

Option A is correct. The number of possible functions

g

is found by recognizing that

g(x) = x

for even

x

, and for odd

x

g(x)

can take any value in

A

, leading to

10^5

possibilities.

602 Medium

602 of 650

Application +4 | -1

The domain of the function

f(x)=\sqrt{\log _{10} \frac{\left(3 x-x^2\right)}{2}}

View answer and explanation

Correct option: D

[1,2]

Option D ([1,2]) is correct because it satisfies both conditions: the argument of the logarithm must be positive, and the logarithm itself must be non-negative.

603 Easy

603 of 650

Analyzing +4 | -1

The domain of the function

f(x)=\sin ^{-1}\left[2 x^2-3\right]+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right),

where

[t]

is the greatest integer function, is

View answer and explanation

Correct option: C

\left(1, \frac{5-\sqrt{5}}{2}\right)

Option C

\left(1, \frac{{5 - \sqrt 5 }}{{2}}\right)

is correct because it represents the intersection of the domains of the inverse sine and logarithmic functions, considering the greatest integer function.

604 Medium

604 of 650

Analyzing +4 | -1

f\left( x \right) = \frac{{1 - x}}{{1 + x}},x \ne 0, - 1\;and\;\alpha = f\left( {f\left( x \right)} \right) + f\left( {f\left( {\frac{1}{x}} \right)} \right),

then

View answer and explanation

Correct option: C

\left| \alpha \right| > 2

Option C is correct because after simplifying

f(f(x))

and

f(f(1/x))

, we find that

\alpha = x + 1/x

, and thus

|\alpha| = |x + 1/x| \ge 2

605 Easy

605 of 650

Analyzing +4 | -1

Let

f(x)=9^x /\left(9^x+3\right)

for all

x \in \mathbb{R}

. Then the value of

\sum_{r=1}^{2008} f(r / 2009)

View answer and explanation

Correct option: A

1004

The correct answer is A (1004) because the sum can be simplified by pairing terms

f(x)

and

f(1-x)

, which sum to 1, and then summing these pairs.

606 Medium

606 of 650

Application +4 | -1

Let

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

. The domain of the relation on

A

defined by

R=\{(x, y): y=2 x-1\}

View answer and explanation

Correct option: A

\{1,2,3\}

Option A (

\{1,2,3\}

) is correct because the domain of the relation consists of the x-values for which y is also in set A.

607 Medium

607 of 650

Analyzing +4 | -1

{3^x} + {2^{2x}} \ge {5^x},

then the solution set for

x

View answer and explanation

Correct option: A

\left( { - \infty ,2} \right]

Option A (

(- \infty, 2]

) is correct because the inequality holds true for x less than or equal to 2.

608 Easy

608 of 650

Understanding +4 | -1

Range of the function

f(x)=\sin ^{2}\left(x^{4}\right)+\cos ^{2}\left(x^{4}\right)

View answer and explanation

Correct option: B

\{1\}

Option B (

\left\\{1\\}\right.

) is correct because

\sin^2(x^4) + \cos^2(x^4)

is always equal to 1 due to the trigonometric identity

\sin^2(\theta) + \cos^2(\theta) = 1

609 Medium

609 of 650

Application +4 | -1

Let

f: R \rightarrow R

be continuous and periodic with period

T > 0

. Then

View answer and explanation

Correct option: A

f\left(x_0+T / 2\right)=f\left(x_0\right)

for some

x_0 \in[k, k+T / 2], k \in R

Option A (

f\left(x_0+T / 2\right)=f\left(x_0\right)

for some

x_0 \in[k, k+T / 2], k \in R

) is correct because it applies the Intermediate Value Theorem to show the existence of a point where the function's value at x_0 + T/2 equals its value at x_0.

610 Easy

610 of 650

Application +4 | -1

f(x)=x^2+\lambda x+1

for all real

x

and

f

is an even function, then the value of

\lambda

View answer and explanation

Correct option: D

Option D (0) is correct because for a function to be even,

f(x) = f(-x)

for all x, which implies that the coefficient of the linear term must be zero.

611 Easy

611 of 650

Application +4 | -1

f\left( x \right) = {\left( {25 - {x^4}} \right)^{1/4}}\;for\;0 < x < \sqrt 5 ,

then

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

View answer and explanation

Correct option: D

{2^{ - 1}}

Option D (

{2^{ - 1}}

) is correct because f(f(x)) = x, so f(f(1/2)) = 1/2, which is 2^(-1).

612 Easy

612 of 650

Understanding +4 | -1

The mapping

f:N \to N

given

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

where

N

is the set of natural number, is

View answer and explanation

Correct option: C

One-to-one but not onto

Option C (One-to-one but not onto) is correct because each natural number maps to a unique natural number, but not all natural numbers are in the range.

613 Easy

613 of 650

Analyzing +4 | -1

The function

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

View answer and explanation

Correct option: B

Even

Option B (Even) is correct because the function can be shown to satisfy the condition f(x) = f(-x), indicating even symmetry.

614 Easy

614 of 650

Application +4 | -1

Given that

f\left( x \right)

is a differentiable function of

x

and that

f\left( x \right).\mathrm{ }f\left( y \right) = f\left( x \right) + f\left( y \right) + f\left( {xy} \right) - 2

and that

f\left( 2 \right) = 5

. Then,

f\left( 3 \right)

is equal to

View answer and explanation

Correct option: A

Option A (10) is correct because by substituting y = 1/x into the given functional equation, we can determine that f(x) = x² + 1, and therefore f(3) = 10.

615 Medium

615 of 650

Understanding +4 | -1

Let

f(x)

and

g(x)

are two functions defined in

(-a, a)

such that

f(x)

and

g(x)

are both odd function. Then which one of the following is correct?

View answer and explanation

Correct option: D

fof

(x)

is an odd function

Option D (fof

(x)

is an odd function) is correct because the composition of two odd functions results in an odd function.

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616 Easy

616 of 650

Understanding +4 | -1

Let

f:R \to R

be a function defined by

f\left( x \right) = \cos \left( {5x + 2} \right).

Then,

f

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is neither injective nor surjective, as it's a periodic function with a limited range.

617 Easy

617 of 650

Analyzing +4 | -1

f\;:R \to R

is defined by

f\left( x \right) = x - \left[ x \right] - \frac{1}{2}

for all

x \in R,

, where

\left[ x \right]

denotes the greatest integer function, then

\left\{ {x \in R\;:f\left( x \right) = \frac{1}{2}} \right\}

is equal to

View answer and explanation

Correct option: C

\phi

The correct answer is C (

\phi

) because the equation f(x) = 1/2 leads to a contradiction, implying there are no real solutions for x.

618 Medium

618 of 650

Application +4 | -1

Let

f=\{(0,-1),(-1,3),(2,3),(3,5)\}

be a function from

Z

Z

defined by

f(x)=a x+b

. Then,

View answer and explanation

Correct option: C

a=2, b=-1

Option C (

a=2, b=-1

) is correct because substituting these values into the function

f(x) = ax + b

satisfies the given conditions.

619 Easy

619 of 650

Application +4 | -1

The function

f(x)=(1 / 4) x^3-\sin \pi x+3

[-2,2]

takes the value

View answer and explanation

Correct option: A

Option A (1) is correct because substituting x=-2 into the function gives f(-2) = 1.

620 Easy

620 of 650

Analyzing +4 | -1

The function

f\left( x \right) = {\log _{2x - 5}}\left( {{x^2} - 3x - 10} \right)

is defined for all

x

belonging to

View answer and explanation

Correct option: B

\left( {5,\;\infty } \right)

Option B (

\left( {5,\\;\infty } \right)

) is correct because the argument of the logarithm must be positive, the base must be positive, and the base cannot be equal to 1.

621 Medium

621 of 650

Understanding +4 | -1

Which of the following functions is periodic?

View answer and explanation

Correct option: A

[x]-x

Option A (

[x]-x

) is correct because it is a periodic function with a period of 1.

622 Easy

622 of 650

Application +4 | -1

The domain of the function

\sqrt{\log \left(x^2-6 x+6\right)}

is:

View answer and explanation

Correct option: C

(-\infty, 1] \cup[5, \infty)

The correct answer is C ((−∞,1] ∪ [5,∞)) because it represents the values of x for which the logarithmic expression is greater than or equal to 1.

623 Hard

623 of 650

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.

624 Hard

624 of 650

Application +4 | -1

Let the function

f\left( x \right) = 3{x^2} - 4x + 5\log \left( {1 + \left| x \right|} \right)

be defined on the interval

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

The even extension of f(x) to the interval

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

View answer and explanation

Correct option: A

3{x^2} + 4x + 8\log \left( {1 + \left| x \right|} \right)

The correct answer is A (

3{x^2} + 4x + 8\log ( {1 + \left| x \right|} )

) because the even extension involves replacing x with -x for negative values and simplifying.

625 Easy

625 of 650

Application +4 | -1

A polynomial function

f\left( x \right)

satisfies the condition

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

f\left( {10} \right) = 1001,

then

f\left( {20} \right) =

View answer and explanation

Correct option: C

8001

Option C (8001) is correct because by substituting f(x) = x^n + 1 into the given condition and using f(10) = 1001, we find n = 3, thus f(20) = 20³ + 1 = 8001.

626 Easy

626 of 650

Understanding +4 | -1

Which of the following is one-one function?

View answer and explanation

Correct option: A

e^x

Option A (

e^x

) is correct because it is a strictly increasing function, meaning each x-value maps to a unique y-value.

627 Easy

627 of 650

Analyzing +4 | -1

Let

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

for all

x \in R.

Then, the function

\phi \left( x \right)

satisfying

{\left[ {\phi \left( x \right) - f\left( x \right)} \right]^2} + {\left[ {\phi \left( x \right) - g\left( x \right)} \right]^2} = 0

View answer and explanation

Correct option: A

\phi \left( x \right) = x,x \in \left[ {0,\;\infty } \right)

Option A is correct. The given condition implies

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

. Since

f(x) = x

and

g(x) = |x|

, this holds only for

x \ge 0

, thus

\phi(x) = x

for

x \in [0, \infty)

628 Easy

628 of 650

Application +4 | -1

f(x)=3 x-5

, then

f^{-1}(x)

View answer and explanation

Correct option: B

is given by

\frac{x+5}{3}

Option B (is given by

\frac{x+5}{3}

) is correct because it represents the inverse function of f(x) = 3x - 5.

629 Medium

629 of 650

Understanding +4 | -1

R=\left\{(x, y): x, y \in Z, x^2+y^2 \leq 4\right\}

is a relation in

Z

, then domain of

R

View answer and explanation

Correct option: C

\{-2,-1,0,1,2\}

Option C ({-2, -1, 0, 1, 2}) is correct because the domain of R consists of all integer values of x for which there exists an integer y such that

x^2 + y^2 \leq 4

630 Medium

630 of 650

Application +4 | -1

Given the function

f(x)=\frac{a^{x}+a^{-x}}{2},(a>2)

. Then

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

View answer and explanation

Correct option: A

2 f(x) \cdot f(y)

Option A (

2 f(x) \cdot f(y)

) is correct because the expression

f(x+y) + f(x-y)

simplifies to

2f(x)f(y)

using the given function definition.

Revision notes and key takeaways

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