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 · 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 211-240 of 650 questions

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211 Medium

211 of 650

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.

212 Easy

212 of 650

Application +4 | -1

f\left( x \right) = {x^2} - 1\;and\;g\left( x \right) = {\left( {x + 1} \right)^2}

then

\left( {gof} \right)\left( x \right)

View answer and explanation

Correct option: C

{x^4}

Option C (

{x^4}

) is correct because substituting f(x) into g(x) and simplifying results in x⁴.

213 Easy

213 of 650

Analyzing +4 | -1

The domain of definition of the function

f(x)=

\log _2\left(\log _{1 / 2}\left(x^2+4 x+3\right)\right)+\sin ^{-1}\left(2[x]^2-3\right),[x]

denotes the greatest integer

\leq x

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the domain is determined by the intersection of the logarithmic and inverse sine conditions, resulting in

-1 < x < -2+\sqrt{{2}}

214 Medium

214 of 650

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).

215 Easy

215 of 650

Application +4 | -1

f: R \rightarrow S

, defined by

f(x)=\sin x-\sqrt{3} \cos x+1

, is onto, then the interval of

S

View answer and explanation

Correct option: A

[-1,3]

Option A ([-1,3]) is correct because the range of the function f(x) = sin(x) - √3cos(x) + 1 is [-1, 3], and since f is onto, the interval of S is the range of f(x).

216 Medium

216 of 650

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

217 Easy

217 of 650

Analyzing +4 | -1

f(x)=\sin 2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)

and

g\left(\frac{5}{4}\right)=1

, then

(g o f)(x)

is equal to:

View answer and explanation

Correct option: B

Option B (1) is correct because simplifying the expression for f(x) results in a constant value of 5/4, and since g(5/4) = 1, (g o f)(x) = 1.

218 Medium

218 of 650

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

219 Easy

219 of 650

Application +4 | -1

f(x)=e^{2 x}

and

g(x)=\log \sqrt{x}(x > 0)

, then

f \circ g(x)

is equal to

View answer and explanation

Correct option: D

x

Option D (

x

) is correct because

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

220 Easy

220 of 650

Analyzing +4 | -1

The largest interval for which

{x^{12}} - {x^9} + {x^4} - x + 1 > 0\;

View answer and explanation

Correct option: D

0 < x < \infty

The correct answer is D. The inequality holds for all x > 0. By considering different intervals, we can prove that the expression is always positive.

221 Medium

221 of 650

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.

222 Easy

222 of 650

Application +4 | -1

Domain of the function

\frac{1}{\sqrt{x^2-1}}

is:

View answer and explanation

Correct option: A

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

The correct answer is A ((−∞,−1) ∪ (1,∞)) because the expression inside the square root must be positive.

223 Easy

223 of 650

Application +4 | -1

{\log _2}\left( {{x^2} - 3x + 18} \right) < 4,

then

x

belongs to

View answer and explanation

Correct option: A

\left( {1,2} \right)

Option D (None of these) is correct. The correct solution set is (1, 2).

224 Easy

224 of 650

Application +4 | -1

The inverse of the function

f\left( x \right) = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}} + 1

is given by

View answer and explanation

Correct option: A

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

Option A (

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

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

225 Medium

225 of 650

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)

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226 Medium

226 of 650

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.

227 Medium

227 of 650

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.

228 Easy

228 of 650

Application +4 | -1

f: R \rightarrow R, g: R \rightarrow R

be two given functions, then

f(x)=2 \min (f(x)-g(x), 0)

equals

View answer and explanation

Correct option: D

f(x)-g(x)-|g(x)-f(x)|

Option D (

f(x)-g(x)-|g(x)-f(x)|

) is correct because it correctly represents 2min(f(x)-g(x), 0) using the absolute value function.

229 Easy

229 of 650

Understanding +4 | -1

Let

A=(1,2,3\}

and

B=\{1,2,4\}

, then

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

is a

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the given set F does not represent a function from A to B as the element 1 in A maps to two different elements in B.

230 Medium

230 of 650

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.

231 Medium

231 of 650

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.

232 Easy

232 of 650

Application +4 | -1

f(y)=\frac{y}{\sqrt{\left(1-y^2\right)}}, g(y)=\frac{y}{\sqrt{\left(1+y^2\right)}}

, then

(f o g) y

is equal to

View answer and explanation

Correct option: C

y

Option C (y) is correct because the composite function (f o g)(y) simplifies to y after substituting and simplifying.

233 Easy

233 of 650

Application +4 | -1

Let the function

f,\;g,\;h

are defined from the set of real numbers R to R such that

f\left( x \right) = {x^2} - 1,\;g\left( x \right) = \sqrt {\left( {{x^2} + 1} \right)}

and

h\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {0,\;if\;x < 0}\\ {x,\;if\;x \ge 0} \end{array}} \right.,

then

ho\left( {fog} \right)\left( x \right)

is defined by

View answer and explanation

Correct option: B

{x^2}

Option B (

x^2

) is correct because the composition of the functions simplifies to

x^2

234 Easy

234 of 650

Understanding +4 | -1

Let

I

be the set of integer and

f: I \rightarrow I

be defined as

f(x)=x^2, x \in I

, the function is

View answer and explanation

Correct option: D

None of these

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

235 Medium

235 of 650

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.

236 Easy

236 of 650

Understanding +4 | -1

The range of

f(x)=\cos (x / 3)

View answer and explanation

Correct option: B

[-1,1]

Option B ([-1,1]) is correct because the range of the cosine function is always between -1 and 1, regardless of the argument.

237 Medium

237 of 650

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.

238 Medium

238 of 650

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).

239 Medium

239 of 650

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.

240 Medium

240 of 650

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)

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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