The domain of ${\sin ^{ - 1}}\left( {{{\log }_3}x} \right)$ is

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

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Easy

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151 Easy

151 of 478

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.

152 Easy

152 of 478

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}}

153 Easy

153 of 478

Analyzing +4 | -1

Set of all values of

p

for which the function

f(x)=p x+\sin x

is bijective is

View answer and explanation

Correct option: B

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

Option B

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

is correct because for the function to be bijective, its derivative must be either always non-positive or always non-negative.

154 Easy

154 of 478

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).

155 Easy

155 of 478

Analyzing +4 | -1

\log _{(2 x+3)} x^2 < \log _{(2 x+3)}(2 x+3)

then value of

x

View answer and explanation

Correct option: B

x \in\left(\frac{-3}{2},-1\right) \cup(-1,3)-\{0\}

The correct answer is B (

x \in\left(\frac{-3}{2},-1\right) \cup(-1,3)-\{0\}

) because it represents the solution set for x that satisfies the given logarithmic inequality, considering the base and argument restrictions.

156 Easy

156 of 478

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.

157 Easy

157 of 478

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

158 Easy

158 of 478

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.

159 Easy

159 of 478

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.

160 Easy

160 of 478

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).

161 Easy

161 of 478

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.

162 Easy

162 of 478

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.

163 Easy

163 of 478

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.

164 Easy

164 of 478

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.

165 Easy

165 of 478

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

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166 Easy

166 of 478

Application +4 | -1

If the function

f(x)=x^3+e^{x / 4}

and

g(x)=f-1(x)

, then the value of

g^{\prime}(1)=

View answer and explanation

Correct option: D

Option D (4) is correct because it is the value of the derivative of the inverse function g'(1) calculated using the formula g'(f(x)) = 1/f'(x).

167 Easy

167 of 478

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).

168 Easy

168 of 478

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.

169 Easy

169 of 478

Analyzing +4 | -1

The function

f: R \rightarrow R

defined by

f(x)=(x-1)(x-2)(x-3)

View answer and explanation

Correct option: B

onto but not one-one

Option B (onto but not one-one) is correct because the function has multiple inputs that map to the same output, and its range equals its codomain.

170 Easy

170 of 478

Application +4 | -1

Let

f

be a function satisfying

f(x y)=\frac{f(x)}{y}

for all positive real number

x

and

y

. If

f(30)=20

then the value

f(40)

View answer and explanation

Correct option: A

Option A (15) is correct because using the given functional equation

f(xy) = \frac{{f(x)}}{{y}}

and the value

f(30) = 20

, we can find

f(40)

171 Easy

171 of 478

Analyzing +4 | -1

If the function

f:R - \left\{ {1, - 1} \right\} \to A

defined by

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

, is surjective, then

A

is equal to

View answer and explanation

Correct option: C

R - \left[ { - 1,\,0} \right)

Option C (

R - [-1, 0)

) is correct because the range of the function

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

R - [-1, 0)

, and for a surjective function, the range equals the codomain.

172 Easy

172 of 478

Application +4 | -1

The domain of

{\sin ^{ - 1}}\left( {{{\log }_3}x} \right)

View answer and explanation

Correct option: D

\left[ {\frac{1}{3},\;3} \right]

Option D ([{

\frac{1}{3}

,\;3}]) is correct because it represents the values of x for which the argument of the inverse sine function, log base 3 of x, lies between -1 and 1.

173 Easy

173 of 478

Analyzing +4 | -1

Let

[t]

denote the greatest integer

\le t.

Then the equation

x,[{x^2}] + 2[x + 2] - 7 = 0

has

View answer and explanation

Correct option: A

Infinitely many solutions

Option A (Infinitely many solutions) is correct because the equation simplifies to a condition on [x], leading to intervals of solutions.

174 Easy

174 of 478

Understanding +4 | -1

Let

f(x)=\cos (p x)+\sin x

be periodic, then

p

must be:

View answer and explanation

Correct option: C

Rational

Option C (Rational) is correct because for the sum of cosine and sine functions to be periodic, the ratio of their frequencies must be rational.

175 Easy

175 of 478

Analyzing +4 | -1

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

is a function from

R\;\;\mathrm{to}\;R

, then

f\left( x \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function simplifies to either 0 or 2x, depending on the value of x, and is therefore not injective, surjective, or bijective.

176 Easy

176 of 478

Application +4 | -1

The domain of the function

f\left( x \right) = {\log _2}\left( {{{\log }_3}({{\log }_4}x} \right))

View answer and explanation

Correct option: B

\left( {4,\;\infty } \right)

The correct answer is B ((4,∞)) because the argument of each logarithm must be positive, leading to the condition x > 4.

177 Easy

177 of 478

Application +4 | -1

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

, then

f\left[\frac{2 x}{1+x^{2}}\right]

is equal to

View answer and explanation

Correct option: C

2 f(x)

Option C (

2 f(x)

) is correct because substituting

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

into

f(x)

and simplifying using logarithm properties results in

2f(x)

178 Easy

178 of 478

Understanding +4 | -1

The function

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

View answer and explanation

Correct option: A

Periodic with period

2\;\pi .

Option A (Periodic with period

2\\;\pi .

) is correct because the sine function has a period of

2\pi

, and the given function involves the sine function in the exponent.

179 Easy

179 of 478

Analyzing +4 | -1

The function

f\left( x \right) = \frac{{{{\sin }^4}x + \;\mathrm{co}{\mathrm{s}^4}\;x}}{{{x^3} + {x^4}\tan x}}

View answer and explanation

Correct option: B

Odd

Option B (Odd) is correct because the function satisfies the condition f(-x) = -f(x), indicating odd symmetry.

180 Easy

180 of 478

Analyzing +4 | -1

If a function

f\;:\left[ {2,\;\infty } \right) \to B

defined by

f\left( x \right) = {x^2} - 4\;x + 5

is a bijection, then

B =

View answer and explanation

Correct option: B

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

Option B is correct. Since f is a bijection, B must be the range of f. Completing the square,

f(x) = (x-2)^2 + 1

. For

x \in [2, \infty)

, the range is

[1, \infty)

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