Let $f\left( x \right) = {\sin ^{ - 1}}x$ and $g\left( x \right) = \frac{{{x^2} - x - 2}}{{2{x^2} - x - 6}}$. If $g(2) = \mathop {\lim }\limits_{x \to 2} g(x)$, then the domain of the function $fog$ is

$( - \infty ,\, - 2] \cup \left[ { - \frac{3}{2},\,\infty } \right)$

$( - \infty ,\, - 2] \cup \left[ { - 1,\,\infty } \right)$

$( - \infty ,\, - 1] \cup \left[ { 2,\,\infty } \right)$

Let $\left[ x \right]\;$ denote the greatest integer less than or equal to $x$. If $f\left( x \right) = {\sin ^{ - 1}}x,{\rm{g}}\left( x \right) = \left[ {{x^2}} \right]\;and\;h\left( x \right) = 2x,\frac{1}{2} \le x \le \frac{1}{{\sqrt 2 }},$ then

$hofo{\rm{g}} = ho{\rm{g}}\;of$

$fo{\rm{g}}oh\;\left( x \right) = \pi /2$

$fo{\rm{g}}oh\left( x \right) = \pi $

$hofo{\rm{g}} \ne ho{\rm{g}}\;of$

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

Is given by $ \frac{1}{3 x-5} $

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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121 Easy

121 of 650

Application +4 | -1

f(x)=\frac{x-|x|}{|x|}

, then

f(-1)=

View answer and explanation

Correct option: B

-2

Option B (-2) is correct because substituting x = -1 into the function

f(x)=\frac{{x-|x|}}{{|x|}}

results in -2.

122 Easy

122 of 650

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: A

2f(x)

Option A (

2f(x)

) is correct because substituting

\frac{{2x}}{{1 + {x^2}}}

into the function and simplifying yields twice the original function.

123 Medium

123 of 650

Analyzing +4 | -1

Let

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

and

g\left( x \right) = \frac{{{x^2} - x - 2}}{{2{x^2} - x - 6}}

. If

g(2) = \lim\limits_{x \to 2} g(x)

, then the domain of the function

fog

View answer and explanation

Correct option: C

( - \infty ,\, - 2] \cup \left[ { - \frac{4}{3},\,\infty } \right)

Option C (

(- \infty ,\, - 2] \cup \left[ { - \frac{4}{3},\,\infty } \right)

) is correct because it represents the domain where |g(x)| <= 1, ensuring the composite function fog(x) is defined.

124 Medium

124 of 650

Analyzing +4 | -1

Let

\mathrm{g}\,\,\left( x \right) = \int \nolimits_0^x \,\,f\left( t \right)\,\,dt

where

\frac{1}{2}\,\, \le \,\,f\left( t \right)\,\, \le 1,\,\,t \in \,\,\left[ {0,1} \right]

and

0\,\, \le \,\,f\,\,\left( t \right)\,\, \le \,\,\frac{1}{2}

for

t\,\, \in \,\,\left( {1,2} \right]\,

, then

View answer and explanation

Correct option: B

0\,\, \le \,\,\mathrm{g}\,\,\left( 2 \right) < 2

Option B (

0 \le g(2) < 2

) is correct because it provides the correct bounds for the integral based on the given inequalities for f(t).

125 Easy

125 of 650

Analyzing +4 | -1

The function

f: \mathbf{R} \rightarrow \mathbf{R}

given by

f(x)=3-2 \sin x

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the function

f(x) = 3 - 2\sin x

is neither one-to-one nor onto, as its range is [1, 5] and it's periodic.

126 Easy

126 of 650

Analyzing +4 | -1

Let

f:R - \{ 3\} \to R - \{ 1\}

be defined by

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

. Let

g:R \to R

be given as

g(x) = 2x - 3

. Then, the sum of all the values of

x

for which

{f^{ - 1}}(x) + {g^{ - 1}}(x) = \frac{{13}}{2}

is equal to

View answer and explanation

Correct option: C

Option C (5) is correct because by finding the inverse functions and solving the given equation, the sum of the solutions is 5.

127 Medium

127 of 650

Application +4 | -1

The domain of definition of the function

f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{4}{{3 + 2\cos x}}} \right),\mathrm{is}

View answer and explanation

Correct option: A

\left[ {2n\pi - \frac{\pi }{6},2n\pi + \frac{\pi }{6}} \right],n \in Z

Option A is correct because the domain is found by solving -1 <= 4/(3+2cos(x)) <= 1, which simplifies to cos(x) >= 1/2.

128 Easy

128 of 650

Analyzing +4 | -1

f:(0, \pi) \rightarrow \mathbf{R}

is given by

f(x)=\sum\limits_{{k=1}}^n[1+\sin k x],[x]

denotes the greatest integer function, then the range of

f(x)

View answer and explanation

Correct option: C

\{n, n+1\}

Option C (

\left\{n, n+1\right\}

) is correct because the function f(x) can take two values, n when all

[\sin kx] = 0

, and n+1 when one of the

[\sin kx] = 1

and the rest are 0.

129 Easy

129 of 650

Application +4 | -1

The range of

f(x)=\cos x-\sin x

View answer and explanation

Correct option: D

[-\sqrt{2}, \sqrt{2}]

Option D ([-

\sqrt{2}

\sqrt{2}

]) is correct because the range of f(x) = cos(x) - sin(x) is between -√2 and √2.

130 Easy

130 of 650

Application +4 | -1

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

, then

(f \circ f)(2)

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because when you evaluate the composite function

f(f(x))

x = 2

, the result is 2.

131 Easy

131 of 650

Application +4 | -1

Let the function

f

be defined by

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

, then

f^{-1}(x)

View answer and explanation

Correct option: A

\frac{x-1}{3 x+2}

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

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

132 Easy

132 of 650

Application +4 | -1

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

then

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

equals

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because

f(x^2)

does not equal any of the given expressions.

133 Medium

133 of 650

Application +4 | -1

Let

A

and

B

be two sets such that

n(A \times B)=6

. If three elements of

A \times B

are

(3,2),(7,5)

(8,5)

, then

View answer and explanation

Correct option: A

A=\{3,7,8\}

Option A (

A=\{3,7,8\}

) is correct because these are the first elements of the ordered pairs in

A \times B

134 Medium

134 of 650

Application +4 | -1

The domain of definition of the function

f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{{x - 3}}{2}} \right) - {\log _{10}}\left( {4 - x} \right),\mathrm{is}

View answer and explanation

Correct option: C

1 \le x < 4

Option C (

1 \le x < 4

) is correct because it represents the intersection of the domains of the arcsine function and the logarithmic function, ensuring both are defined.

135 Medium

135 of 650

Application +4 | -1

Let

f(x)=x-3

and

g(x)= \begin{cases}\frac{x^2-9}{x+3} & \text { if } x \neq-3 \\ k & \text { if } x=-3\end{cases}

f(x)=g(x)

for all real

x

, then the value of

k

View answer and explanation

Correct option: C

-6

Option C (-6) is correct because for f(x) to equal g(x) for all x, g(-3) must equal f(-3), which is -6.

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136 Medium

136 of 650

Application +4 | -1

Find the range of

f(x)=1+3 \cos 2 x

View answer and explanation

Correct option: C

[-2,4]

Option C ([-2,4]) is correct because the range of the cosine function is [-1, 1], and this is used to find the range of the given function.

137 Easy

137 of 650

Application +4 | -1

f\left( x \right)

is an odd periodic function with period 2, then

f\left( 4 \right)

equals

View answer and explanation

Correct option: A

The correct answer is A (0) because an odd periodic function with period 2 has the property f(0) = 0, and f(4) = f(0) due to periodicity.

138 Medium

138 of 650

Application +4 | -1

If a function

f\left( x \right)

satisfies the condition

f\left( {x + \frac{1}{x}} \right) = {x^2} + \frac{1}{{{x^2}}},x \ne 0,\;\mathrm{then\;}f\left( x \right)\mathrm{\;equals}

View answer and explanation

Correct option: B

{x^2} - 2

for all

x

satisfying

\left| x \right| \ge 2

Option B (

{x^2} - 2

for all

x

satisfying

\left| x \right| \ge 2

) is correct because the function f(x) is defined as x² - 2 only when |x| >= 2, which is the range of x + 1/x.

139 Easy

139 of 650

Application +4 | -1

The domain of the function

f\left( x \right) = \frac{{{{\sin }^1}\left( {x - 3} \right)}}{{\sqrt {9 - {x^2}} }}

View answer and explanation

Correct option: B

\left[ {2,\mathrm{ }3} \right)

Option B ([2, 3)) is correct because it satisfies both the domain of the arcsine function and the square root function.

140 Hard

140 of 650

Analyzing +4 | -1

Let

\left[ x \right]\;

denote the greatest integer less than or equal to

x

. If

f\left( x \right) = {\sin ^{ - 1}}x,\mathrm{g}\left( x \right) = \left[ {{x^2}} \right]\;and\;h\left( x \right) = 2x,\frac{1}{2} \le x \le \frac{1}{{\sqrt 2 }},

then

View answer and explanation

Correct option: C

hofo\mathrm{g} = ho\mathrm{g}\;of

Option D (

hofo\mathrm{g} = ho\mathrm{g}\;of

) is correct because after evaluating both sides, they are equal for all x in the given interval.

141 Medium

141 of 650

Analyzing +4 | -1

If the domain of the function

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

is the interval

(\alpha ,\,\beta )

, then

\alpha + \beta

is equal to

View answer and explanation

Correct option: A

\frac{3}{2}

Option A (

\frac{3}{2}

) is correct because finding the domain of the given function involves considering the domains of inverse cosine and inverse sine functions, and their intersection gives the interval

(\frac{1}{2}, 1]

, so

\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}

142 Easy

142 of 650

Application +4 | -1

Let

n\left( A \right) = 4\,and\,n\left( B \right) = 6

. The number of one to one functions from A to B is

View answer and explanation

Correct option: D

360

Option D (360) is correct because it is the number of ways to choose 4 elements from 6 and arrange them.

143 Easy

143 of 650

Application +4 | -1

If a function F is such that

F\left( 0 \right) = 2,\;F\left( 1 \right) = 3,\;F\left( {n + 2} \right) = 2F\left( n \right) - F\left( {n + 1} \right)\;for\;n \ne 0

, then

F\left( 5 \right)

is equal to

View answer and explanation

Correct option: D

Option D (13) is correct because by iteratively applying the given recurrence relation, we find that F(5) equals 13.

144 Easy

144 of 650

Application +4 | -1

g\left( x \right) = 1 + \sqrt x \;and\;f\left( {g\left( x \right)} \right) = 3 + 2\sqrt x + x

then,

f\left( x \right)

is equal to

View answer and explanation

Correct option: B

2 + {x^2}

Option B is correct because by substituting g(x) into f(g(x)), we can solve for f(x) to find it equals 2 + x².

145 Easy

145 of 650

Application +4 | -1

a f(x+1)+b f\left(\frac{1}{x+1}\right)=x, x \neq-1, a \neq b

, then

f(2)

is equal to

View answer and explanation

Correct option: A

\frac{2 a+b}{2\left(a^2-b^2\right)}

Option A (

\frac{{2a+b}}{{2(a^2-b^2)}}

) is correct because it is the value of

f(2)

obtained by solving the given functional equation.

146 Easy

146 of 650

Understanding +4 | -1

Assertion: The function

f: R \rightarrow R

given by

f(x)=x^3

is injective.

Reason: The function

f: X \rightarrow Y

is injective, if

f(x)=f(y)

\Rightarrow {x}={y}

for all

{x},{y} \in {X}

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 why the function f(x) = x³ is injective.

147 Easy

147 of 650

Application +4 | -1

3 < 3t - 18 \le 18

, then which one of the following is true?

View answer and explanation

Correct option: C

8 \le t + 1 \le 13

The correct answer is C (

8 \le t + 1 \le 13

) because it is derived by solving the given compound inequality and then adding 1 to each part.

148 Medium

148 of 650

Analyzing +4 | -1

Let

f:N \to R

be a function such that

f(x + y) = 2f(x)f(y)

for natural numbers

x

and

y

. If

f(1) = 2,\,

then the value of

\alpha

for which

\sum\limits_{k = 1}^{10} {f(\alpha + k)} = \frac{{512}}{3}({2^{20}} - 1)

holds, is:

View answer and explanation

Correct option: C

Option C (4) is correct because by substituting the function

f(x)

and simplifying the summation, we can find the value of

\alpha

to be 4.

149 Medium

149 of 650

Analyzing +4 | -1

If the functions are defined as

f(x) = \sqrt x

and

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

, then what is the common domain of the following functions?

f + g,\,\,f - g,\,\,f/g,\,\,g/f,\,\,g - f

, where

(f \pm g)(x) = f(x) \pm g(x),(f/g)(x) = \frac{{f(x)}}{{g(x)}}

View answer and explanation

Correct option: C

0 < x < 1

Option C (0 < x < 1) is correct because it represents the intersection of the domains of f(x) and g(x), excluding points where either function is undefined.

150 Easy

150 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 correct because it represents the inverse function

f^{-1}(x)

obtained by solving for x in terms of y and swapping x and y.

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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