The inverse of the function $y = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}$ is

$\frac{1}{2}{\log _{10}}\left( {\frac{{2 + x}}{{2 - x}}} \right)$

If ${\log _{\cos x}}\sin x > 2\;{\rm{and}}\;0 < x < 3\pi ,\;$ then $\sin x\;$ lies in the interval

$\left[ {\frac{{\sqrt 5 - 1}}{2},1} \right]$

$\left[ {0,\frac{1}{2}} \right]$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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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 121-150 of 353 questions

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121 Medium

121 of 353

Analyzing +4 | -1

f(x)=\left[x^2\right]-[x]^2

, where [ ] denote the greatest integer function and

x \in[0, n], n \in N

then the number of elements in the range of

f(x)

are

View answer and explanation

Correct option: D

2 n-1

Option D (2n-1) is correct because the range of the function f(x) consists of integers from 0 to 2n-2, resulting in 2n-1 elements.

122 Medium

122 of 353

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because by expanding f(x+y)f(x-y) and using the definition of f(x), we can simplify the expression to match the given option.

123 Medium

123 of 353

Analyzing +4 | -1

The domain of the function

f(x)=\frac{x|x-3|}{\left(x^2-x-6\right)|x|}, x \in \mathbb{R}

View answer and explanation

Correct option: D

\mathbb{R}-\{-2,0,3\}

Option D (

\mathbb{R}-\{-2,0,3\}

) is correct because the function is undefined when the denominator is zero, which occurs at x = -2, 0, and 3.

124 Medium

124 of 353

Application +4 | -1

The inverse of the function

y = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}

View answer and explanation

Correct option: A

\frac{1}{2}{\log _{10}}\left( {\frac{{1 + x}}{{1 - x}}} \right)

Option A is correct because the inverse function is found by swapping x and y, then solving for y, using componendo and dividendo and logarithmic properties.

125 Medium

125 of 353

Analyzing +4 | -1

\sum\limits_{{k=0}}^{21} f\left(\frac{k}{11}+2 x\right)=

constant and

\forall x \in R

f(x)

is periodic, then period of

f(x)

View answer and explanation

Correct option: B

Option B (2) is correct because by replacing x with x+1/22 and subtracting the original equation, we find that f(2x) = f(2x+2), implying a period of 2.

126 Medium

126 of 353

Application +4 | -1

f: R \rightarrow R

is defined by

f(x)=2 x+|x|

, then

f(3 x)-f(-x)-4 x

equals

View answer and explanation

Correct option: D

2 f(x)

Option D (2 f(x)) is correct because after evaluating f(3x) and f(-x) and simplifying, the expression equals 2f(x).

127 Medium

127 of 353

Application +4 | -1

The domain of the function

f(x)=\sqrt{|x-1|-1}

View answer and explanation

Correct option: B

(-\infty, 0] \cup[2, \infty)

Option B

((-\infty, 0] \cup[2, \infty))

is correct because the expression inside the square root must be non-negative, leading to the inequality

|x-1| \geq 1

128 Medium

128 of 353

Application +4 | -1

f\;:N \to N

is defined by

f\left( n \right) =

the sum of positive divisors of

n

, then

f\left( {{2^k} \times 3} \right),

where

k

is a positive integer, is

View answer and explanation

Correct option: C

3\left( {{2^{k + 1}} - 1} \right)

Option C (

3\left( {{2^{k + 1}} - 1} \right)

) is correct because it represents the sum of the positive divisors of

2^k \times 3

129 Medium

129 of 353

Analyzing +4 | -1

Let

A = R - \left\{ 3 \right\},B = R - \left\{ 1 \right\}.

Let

f:A \to B

be defined by

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

.Then,

View answer and explanation

Correct option: A

f is bijective

Option A (f is bijective) is correct because the function f is both injective (one-to-one) and surjective (onto), thus bijective.

130 Medium

130 of 353

Application +4 | -1

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

, then

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

View answer and explanation

Correct option: D

f\left( x \right)

Option D (

f(x)

) is correct because after applying logarithm properties and trigonometric identities, the expression simplifies to

f(x)

131 Medium

131 of 353

Analyzing +4 | -1

If the function

f:R \to A

given by

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

is a surjection, then

A =

View answer and explanation

Correct option: D

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

Option D

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

is correct because the range of the function, excluding 1, makes it a surjection.

132 Medium

132 of 353

Analyzing +4 | -1

If n is an integer, the domain of the function

\sqrt {\sin 2x}

View answer and explanation

Correct option: B

\left[ {n\pi ,\;n\pi + \frac{\pi }{4}} \right]

The correct answer is B ([nπ, nπ + π/4]) because it represents the interval where sin(2x) is non-negative, ensuring the square root is defined for all integer values of n.

133 Medium

133 of 353

Analyzing +4 | -1

Let

f(x)=p x^2+q x+r ; p, q, r \in R

and if

f(x)-x=0

has no real roots then

f(f(x))-x=0

has

View answer and explanation

Correct option: C

No real root

Option C (No real root) is correct because if f(x) - x = 0 has no real roots, then f(x) is always greater or always less than x, implying f(f(x)) - x = 0 also has no real roots.

134 Medium

134 of 353

Analyzing +4 | -1

f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in \mathbb{R},

then

f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\ldots+f\left(\frac{2022}{2023}\right)

is equal to

View answer and explanation

Correct option: C

1011

Option C (1011) is correct because the sum can be simplified by pairing terms f(x) and f(1-x), where f(x) + f(1-x) = 1.

135 Medium

135 of 353

Analyzing +4 | -1

{\log _{\cos x}}\sin x > 2\;\mathrm{and}\;0 < x < 3\pi ,\;

then

\sin x\;

lies in the interval

View answer and explanation

Correct option: B

\left[ {0,\frac{{\sqrt 5 - 1}}{2}} \right]

Option B

\left[ {0,\frac{{\sqrt 5 - 1}}{2}} \right]

is correct because it satisfies the logarithmic inequality and the domain restrictions for sine and cosine.

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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

136 Medium

136 of 353

Application +4 | -1

Let

f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}

and

f(4)=133

f(5)=255

. Then the sum of all the positive integer divisors of

(f(3)-f(2))

View answer and explanation

Correct option: C

Option C (60) is correct because the sum of divisors of the calculated value of (f(3)-f(2)) is 60.

137 Medium

137 of 353

Analyzing +4 | -1

Let

A=[-1,1], B=[-1,1], C=[0, \infty)

. If

R_1=\left\{(x, y) \in A \times B: x^2+y^2=1\right\}

and

R_2=\left\{(x, y) \in A \times C: x^2+y^2=1\right\}

, then

View answer and explanation

Correct option: D

R_2

defines a function from

A

onto

C

Option C (

f(x)=x^m, g(x)=x^n, m \neq n, m, n \in I

) is correct because the composition of these functions is commutative, i.e.,

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

138 Medium

138 of 353

Analyzing +4 | -1

The range of

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

View answer and explanation

Correct option: C

[0,2 \pi)

Option C (

[0,2 \pi)

) is correct because the range of the function is determined by the range of the arcsine function, which is limited to less than 1.

139 Medium

139 of 353

Analyzing +4 | -1

Let

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

be a function such that

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

. Then

View answer and explanation

Correct option: B

f(x)

is one-one in

[1, \infty)

but not in

(-\infty, \infty)

Option B (

f(x)

is one-one in

[1, \infty)

but not in

(\infty, \infty)

) is correct because the function is one-to-one in the specified interval, but not over the entire real number line.

140 Medium

140 of 353

Application +4 | -1

If a function

g(x)

is the inverse of

f(x)

and

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

, then

g^{\prime}(x)=\ldots \ldots

View answer and explanation

Correct option: C

1+[g(x)]^3

Option C (

1+[g(x)]^3

) is correct because it is obtained by differentiating the inverse function relationship and substituting appropriately.

141 Medium

141 of 353

Application +4 | -1

The function

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

View answer and explanation

Correct option: B

Odd function

Option B is correct because

f(-x) = -f(x)

for the given function, indicating it's an odd function.

142 Medium

142 of 353

Application +4 | -1

Let

f: \mathbb{R}-\{0,1\} \rightarrow \mathbb{R}

be a function such that

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

. then

f(2)

is equal to

View answer and explanation

Correct option: B

\frac{9}{4}

Option B (

\frac{9}{4}

) is correct because it is the value of f(2) obtained by solving the system of equations derived from the given functional equation.

143 Medium

143 of 353

Application +4 | -1

f(x)=4 x-x^{2}, x \in R

, then

f(a+1)-f(a-1)

is equal to

View answer and explanation

Correct option: B

4(2-a)

Option B (4(2-a)) is correct because it is the simplified expression after evaluating f(a+1) - f(a-1) using the given function.

144 Medium

144 of 353

Understanding +4 | -1

The Domain of the function

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

View answer and explanation

Correct option: A

[-2,3]

Option A ([-2,3]) is correct because it represents the interval where both square roots are defined (non-negative).

145 Medium

145 of 353

Application +4 | -1

Let

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

be given by

g(x)=3+4 {x}

. If

g^{{n}}({x})=\underbrace{{ gogo } \ldots \ldots \ldots og({x})}_{n \, {\text {times }}}

then

g^{-{n}}({x})=

(where

g^{-n}(x)

denotes inverse of

g^{{n}}({x})

)

View answer and explanation

Correct option: B

(x+1) 4^{-n}-1

Option B (

(x+1) 4^{-n}-1

) is the correct answer because it represents the inverse of the nth composition of the function g(x), derived by finding the inverse function and composing it n times.

146 Medium

146 of 353

Application +4 | -1

f\left( x \right) = {x^n},n \in N\;and\;\mathrm{g}of\left( x \right) = n\;\mathrm{g}\left( x \right),

then

\mathrm{g}\left( x \right)

can be

View answer and explanation

Correct option: D

\log \left| x \right|

Option D (

\log \left| x \right|

) is correct because it satisfies the given functional equation

\mathrm{g}of\left( x \right) = n\;\mathrm{g}\left( x \right)

147 Medium

147 of 353

Analyzing +4 | -1

The range of the function

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

is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the range of the function is determined by finding the minimum and maximum values within its domain.

148 Medium

148 of 353

Analyzing +4 | -1

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

and

f(x) \neq 0

, then

\{f(-2)\}=

____ Here

\{{x}\}

represents fractional part of

{x}

View answer and explanation

Correct option: A

2 / 3

Option A (2/3) is correct because after substituting and simplifying, the fractional part of f(-2) is 2/3.

149 Medium

149 of 353

Application +4 | -1

If the function

f, g, h

are defined from the sets of real numbers

R

R

such that

f(x)=x^2-1

g(x)=\sqrt{x^2+1}, h(x)=\left\{\begin{array}{ll}0, & \text { if } x \leq 0 \\ x, & \text { if } x \geq 0\end{array}\right.

, then the composite function hofog

(x)

View answer and explanation

Correct option: B

\begin{cases}0, & x=0 \\ x^2, & x \neq 0\end{cases}

Option B

\begin{cases}0, & x=0 \\ x^2, & x \neq 0\end{cases}

is correct because by evaluating the composite function hofog(x), we find that it equals 0 when x=0 and x² when x is not 0.

150 Medium

150 of 353

Analyzing +4 | -1

For real

x

, let

f(x)=x^3+5 x+1

, then

View answer and explanation

Correct option: B

f

is one-one and onto

\mathrm{R}

Option B is correct. The function

f(x)

is strictly increasing on R, making it one-to-one, and its range is also R, making it onto.

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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