Domain of the function $ f(x)=\sqrt{2-2 x-x^{2}} $ is

$ -\sqrt{3} \leq x \leq \sqrt{3} $

$ -2+\sqrt{3} \leq x \leq-2-\sqrt{3} $

The domain of $f\left( x \right) = \log \left| {{{\log }_e}x} \right|,$ is

$\left( { - \infty ,1} \right)$

The domain of the real function $f(x)=\frac{1}{\sqrt{4-x^2}}$ is

the set of all positive real numbers

If $|x+2| \leq 9$, then

$x \in(-\infty,-7) \cup(11, \infty)$

$x \in(-\infty,-7) \cup[11, \infty)$

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Easy

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Showing 181-210 of 478 questions

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181 Easy

181 of 478

Application +4 | -1

A real valued function

f(x)

satisfies the functional equation

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

where a is a given constant and

f(0)=0, f(2 a-x)

is equal to

View answer and explanation

Correct option: A

-f(x)

Option A (

-f(x)

) is correct because substituting specific values into the functional equation and using the given condition f(0) = 0 leads to f(2a-x) = -f(x).

182 Easy

182 of 478

Analyzing +4 | -1

The number of solutions of

\left[ {\sin x + \cos x} \right] = 3 + \left[ { - \sin x} \right] + \left[ { - \cos x} \right]

in the internal

\left[ {0,\;2\;\pi } \right]

is (where [.] denotes the greatest integer function)

View answer and explanation

Correct option: C

Infinite

Option C (Infinite) is correct because the two expressions are equivalent, leading to infinite solutions within the given interval.

183 Easy

183 of 478

Application +4 | -1

Let

f:R \to R

be a function which satisfies

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)\forall x,\,\,\,y \in R.\,\,

f\left( 1 \right) = 2

and

g\left( n \right) = \sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)},

n \in N,

then the value of

n,

for which

g\left( n \right) = 20

View answer and explanation

Correct option: A

The correct answer is A (5) because it is the value of n that satisfies the given functional equation and summation condition.

184 Easy

184 of 478

Understanding +4 | -1

The domain of definition of the function

f\left( x \right) = \sqrt[3]{{\frac{{2x + 1}}{{{x^2} - 10x - 11}}}},\;\mathrm{is}

View answer and explanation

Correct option: C

R - \left\{ { - 1,\;11} \right\}

Option C (

R - \left\{ { - 1,\\;11} \right\}

) is correct because the function is defined for all real numbers except where the denominator is zero.

185 Easy

185 of 478

Application +4 | -1

Let

A = \left\{ {x \in R:x \le 1} \right\}\;and\,f:A \to A

be defined as

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

Then,

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

View answer and explanation

Correct option: B

1 - \sqrt {1 - x}

Option B (

1 - \sqrt {1 - x}

) is correct because by setting f(x) = y and solving for x, considering the domain restriction x <= 1, we find the inverse function.

186 Easy

186 of 478

Analyzing +4 | -1

The domain of definition of

f\left( x \right) = \sqrt {\frac{{{{\log }_{0.3}}\left| {x - 2} \right|}}{{\left| x \right|}}} ,

View answer and explanation

Correct option: A

\left[ {1,\;2} \right) \cup \left( {2,\;3} \right]

Option A (

[ {1,\\;2} ) \cup \left( {2,\\;3} ]

) is correct because it satisfies the conditions for the logarithm and the square root to be defined.

187 Easy

187 of 478

Application +4 | -1

Let

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

, then

f\left( x \right)

is equal to

View answer and explanation

Correct option: C

{x^2} - 2

Option C (

x^2 - 2

) is correct because substituting

x + \frac{1}{x}

into the function results in

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

188 Easy

188 of 478

Analyzing +4 | -1

Which of the four statements given below is different from others?

View answer and explanation

Correct option: B

f: x \rightarrow f(x)

Option B (

f: x \rightarrow f(x)

) is correct because it represents the mapping of an element to its image, while others describe the function itself.

189 Easy

189 of 478

Application +4 | -1

Let

R

be a relation in

N

defined by

R=\{(x, y): x+2 y=8\}

. The range of

R

View answer and explanation

Correct option: B

\{1,2,3\}

Option B (

\left\{1,2,3\right\}

) is correct because it represents the set of all possible

y

values (range) for which

x + 2y = 8

and

x, y

are natural numbers.

190 Easy

190 of 478

Understanding +4 | -1

If the mappings

f: A \rightarrow B

and

g: B \rightarrow C

are both bijective, then the mapping

g o f: A \rightarrow C

View answer and explanation

Correct option: B

One-one and onto

Option B (One-one and onto) is correct because the composition of two bijective functions is also bijective.

191 Easy

191 of 478

Application +4 | -1

Domain of the function

f(x)=\sqrt{2-2 x-x^{2}}

View answer and explanation

Correct option: B

-1-\sqrt{3} \leq x \leq-1+\sqrt{3}

Option B (-1-√3 ≤ x ≤ -1+√3) is correct because the domain requires the expression inside the square root to be non-negative.

192 Easy

192 of 478

Application +4 | -1

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

, then

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

will be equal to

View answer and explanation

Correct option: D

2f\left( x \right)

Option D (

2f\left( x \right)

) is correct because substituting

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

into

f(x)

and simplifying results in

2f(x)

193 Easy

193 of 478

Understanding +4 | -1

The function

f: R \rightarrow R

defined by

f(x)=e^{x}

View answer and explanation

Correct option: C

One-one and into

Option C (One-one and into) is correct because the exponential function

e^x

is one-to-one over the real numbers and its range is a subset of the codomain R.

194 Easy

194 of 478

Application +4 | -1

f\left( x \right)

satisfies the relation

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

for all real x, then

f\left( x \right)

View answer and explanation

Correct option: B

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

The correct answer is B (

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

) because it is the solution obtained by substituting (1-x) into the original equation and solving the resulting system of equations.

195 Easy

195 of 478

Analyzing +4 | -1

The domain of

f\left( x \right) = \log \left| {{{\log }_e}x} \right|,

View answer and explanation

Correct option: C

\left( {0,1} \right) \cup \left( {1,\infty } \right)

Option C (

\left( {0,1} \right) \cup \left( {1,\infty } \right)

) is correct because the argument of the outer logarithm must be positive, and the argument of the inner logarithm must also be positive and not equal to 1.

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196 Easy

196 of 478

Understanding +4 | -1

Which one of the following functions is not invertible?

View answer and explanation

Correct option: B

f: \mathbb{R} \rightarrow[0, \infty), f(x)=x^2

Option B (

f: \mathbb{{R}} \rightarrow[0, \infty), f(x)=x^2

) is correct because this function is not one-to-one over the entire real numbers since

f(x) = f(-x)

, and therefore not invertible.

197 Easy

197 of 478

Understanding +4 | -1

A = \left\{ {1,\;2,\;3,\;4} \right\},\;B\left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}

are two sets, and function

f:A \to B

is defined by

f\left( x \right) = x + 2\;\forall x \in Ab

, then the function f is

View answer and explanation

Correct option: C

One-one

Option C (One-one) is correct because each element in set A maps to a unique element in set B, and the function is into.

198 Easy

198 of 478

Application +4 | -1

If range of function

f(x)

whose domain is set of all real numbers is

[-2,4]

, then range of function

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

is equal to:

View answer and explanation

Correct option: B

[-1,2]

Option B ([-1,2]) is correct because it is half of the range of f(x) due to the multiplication by 1/2 in g(x).

199 Easy

199 of 478

Application +4 | -1

R = \left\{ {\left( {x,y} \right):x,y \in Z,{x^2} + 3{y^2} \le 8} \right\}

is a relation on the set of integers Z, then the domain of

{R^{ - 1}}

View answer and explanation

Correct option: A

\left\{ { - 1,0,1} \right\}

The correct answer is A (

\left\{ { - 1,0,1} \right\}

) because it represents the domain of the inverse relation, which is the range of the original relation R.

200 Easy

200 of 478

Application +4 | -1

g(f(x))=|\sin x|

and

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

, then

f

and

g

may be given by

View answer and explanation

Correct option: A

f(x)=\sin ^2 x, g(x)=\sqrt{x}

Option A (

f(x)=\sin ^2 x, g(x)=\sqrt{x}

) is correct because it satisfies both given functional equations:

g(f(x))=|\sin x|

and

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

201 Easy

201 of 478

Application +4 | -1

Let

A=\{1,0,1,2\}, B=\{4,2,0,2\}

and

f, g: A \rightarrow B

be functions defined by

f(x)=x^2-x

and

g(x)=2\left|x-\frac{1}{2}\right|-1

. Then,

View answer and explanation

Correct option: A

f=g

Option A (

f=g

) is correct because the functions

f(x)

and

g(x)

yield the same output for all elements in set A.

202 Easy

202 of 478

Analyzing +4 | -1

The domain of the function

f(x)=\log _{10} \frac{x-5}{x^2-10 x+24}-3 \sqrt{x+5}

View answer and explanation

Correct option: D

(4,5) \cup(6, \infty)

Option D ((4,5) ∪ (6, ∞)) is correct because it satisfies both conditions: the argument of the logarithm must be positive, and the expression inside the square root must be non-negative.

203 Easy

203 of 478

Application +4 | -1

The inverse of the function

f:R \to R

given by

f\left( x \right) = {\log _a}\left( {x + \sqrt {{x^2} + 1} } \right)\left( {a > 0,a \ne 1} \right),

View answer and explanation

Correct option: B

\frac{1}{2}\left( {{a^x} - {a^{ - x}}} \right)

Option B (

\frac{1}{2}\left( {{a^x} - {a^{ - x}}} \right)

) is correct because it represents the inverse function of the given logarithmic function.

204 Easy

204 of 478

Analyzing +4 | -1

Let

X=Y=\mathbf{R} \sim\{1\}

. The function

f: X \rightarrow Y

defined by

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

View answer and explanation

Correct option: D

one-one and onto

Option D (one-one and onto) is correct because the function

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

is both injective (one-to-one) and surjective (onto) when

X = Y = \mathbf{{R}} \setminus \{1\}

205 Easy

205 of 478

Understanding +4 | -1

The domain of the real function

f(x)=\frac{1}{\sqrt{4-x^2}}

View answer and explanation

Correct option: C

(-2,2)

Option C ((-2,2)) is correct because the function is defined only when the expression inside the square root is positive, i.e.,

4 - x^2 > 0

, which gives

-2 < x < 2

206 Easy

206 of 478

Application +4 | -1

If D_30 is the set of the divisors of 30,

x,y \in {D_{30}}

we define

x + y = \mathrm{LCM}\left( {x,y} \right),x.y = \gcd \left( {x,y} \right),x' = \frac{{30}}{x}\;\;and\;\;f\left( {x,y,z} \right) = \left( {x + y} \right).\left( {y' + z} \right),

then

f\left( {2,5,15} \right)

is equal to

View answer and explanation

Correct option: C

Option C (10) is correct because it follows from the given definitions of +, ., and ' within the set D_30.

207 Easy

207 of 478

Analyzing +4 | -1

Let

f(x) = |x - 2|\mathrm{ }\,\,\mathrm{and}\,\,\mathrm{ }g(x) = f(f(x)),x \in [0,4].

Then

\int_0^3 {(g(x) - f(x))\,dx}

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because evaluating the integral of the difference between g(x) and f(x) over the given interval yields 1.

208 Easy

208 of 478

Analyzing +4 | -1

Let

f: X \rightarrow[1,27]

be a function by

f(x)=5 \sin x

+12 \cos x+14

. The set

X

so that

f

is one-one and onto is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is not one-to-one due to its periodic nature, even though its range is within [1, 27].

209 Easy

209 of 478

Application +4 | -1

|x+2| \leq 9

, then

View answer and explanation

Correct option: B

x \in[-11,7]

Option B (x ∈ [-11,7]) is correct because the absolute value inequality |x+2| ≤ 9 implies that -9 ≤ x+2 ≤ 9.

210 Easy

210 of 478

Understanding +4 | -1

The natural domain of the real valued function defined by

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

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because

x^2 - 1

must be non-negative for the square root to be real, so

x^2 \ge 1

, which means

x \le -1

x \ge 1

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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