If $\sqrt {9{x^2} + 6x + 1} < \left( {2 - x} \right),$ then

$x \in \left( { - \frac{3}{2},\;\frac{1}{4}} \right)$

$x \in \left[ { - \frac{3}{2},\;\frac{1}{4}} \right]$

Domain of the function $ f(x)=\sin ^{-1}\left(1+3 x+2 x^{2}\right) $ is

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

Back to FUNCTIONS

Your Progress

0 / 650 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 61-90 of 650 questions

61 Easy

61 of 650

Application +4 | -1

Let

f:R \to ,\;g:R \to R

be two functions given by

f\left( x \right) = 2\;x - 3,\;g\left( x \right) = {x^3} + 5.

Then,

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

is equal to

View answer and explanation

Correct option: D

{\left( {\frac{{x - 7}}{2}} \right)^{1/3}}

Option D (

{\left( {\frac{{x - 7}}{2}} \right)^{1/3}}

) is correct because it is the correct inverse function of the composite function

fog(x)

62 Medium

62 of 650

Application +4 | -1

The domain of the function

f\left( x \right) = \exp \left( {\sqrt {5x - 3 - 2{x^2}} } \right)

View answer and explanation

Correct option: B

\left[ {1,\;3/2} \right]

Option B ([1, 3/2]) is correct because the expression inside the square root must be non-negative, leading to the inequality 5x - 3 - 2x² >= 0, which simplifies to 1 <= x <= 3/2.

63 Easy

63 of 650

Analyzing +4 | -1

Consider function

f:A \to B

and

g:B \to C(A,\,B,C \subseteq R)

such that

{(gof)^{ - 1}}

exists, then

View answer and explanation

Correct option: C

f

is one-one and

g

is 0 into

Option C (

f

is one-one and

g

is 0 into) is correct because for the inverse of a composite function to exist, the first function must be injective (one-one) and the second function can be into.

64 Medium

64 of 650

Analyzing +4 | -1

The number of real solutions of the equation

1 - x = \left[ {\mathrm{cos}x} \right]

, is

View answer and explanation

Correct option: B

Option B (2) is correct because the graphs of the two functions intersect at two points.

65 Medium

65 of 650

Application +4 | -1

Let

f: N \rightarrow Y

be a function defined as

f(x)=4 x+3

where

Y=\{y \in N: y=4 x+3

for some

x \in N\}

. If

f

is invertible, then its inverse is

View answer and explanation

Correct option: A

g(y)=\frac{y-3}{4}

Option A (

g(y)=\frac{{y-3}}{{4}}

) is correct because it is found by solving the equation y = 4x + 3 for x in terms of y, which gives the inverse function.

66 Medium

66 of 650

Application +4 | -1

Given

f(g(x))=1+x-2 x^2+5 x^3

and

f(x)=1+3 x \, \forall \, x \in R

, then the value of

3 g(1)

is equal to

View answer and explanation

Correct option: C

The correct answer is C (4) because by substituting f(x) into f(g(x)) and solving for g(x), then evaluating 3g(1) yields 4.

67 Medium

67 of 650

Application +4 | -1

f(x)=3-x,-4 \leq x \leq 4

, then the domain of

\log _e(f(x))

View answer and explanation

Correct option: D

[-4,3)

Option D ([-4,3)) is correct because the argument of the logarithm must be positive, and x is bounded by -4 and 4.

68 Easy

68 of 650

Application +4 | -1

Domain of

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

View answer and explanation

Correct option: A

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

Option A ((-∞,-1) ∪(1, ∞)) is correct because the domain requires the expression inside the square root to be positive and the denominator to be non-zero.

69 Hard

69 of 650

Analyzing +4 | -1

The set of values of

x

for which of the function

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

exists is

View answer and explanation

Correct option: C

\phi

Option C (

\phi

) is correct because the intersection of the domains of the individual terms in the function is empty, meaning there's no x value for which the entire function is defined.

70 Medium

70 of 650

Analyzing +4 | -1

Which of the following is not periodic?

View answer and explanation

Correct option: B

\cos \sqrt x + {\cos ^2}x

Option B (

\cos \sqrt x + {\cos ^2}x

) is correct because

\cos \sqrt x

is not a periodic function, making the entire expression non-periodic.

71 Easy

71 of 650

Analyzing +4 | -1

Let

f:R \to R

be a function defined by

f\left( x \right) = \frac{{{x^2} - 8}}{{{x^2} + 2}}

Then,

f

View answer and explanation

Correct option: D

Neither one-one nor onto

Option D (Neither one-one nor onto) is correct because the function is neither injective (not one-to-one) nor surjective (not onto).

72 Medium

72 of 650

Understanding +4 | -1

The range of the function ⁡

\sin \left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right),\;\left| x \right| \le 1

View answer and explanation

Correct option: D

\left\{ 1 \right\}

Option D ({1}) is correct because the expression simplifies to sin(π/2), which equals 1.

73 Medium

73 of 650

Application +4 | -1

\sqrt {9{x^2} + 6x + 1} < \left( {2 - x} \right),

then

View answer and explanation

Correct option: D

x < \frac{1}{4}

The correct answer is D (

x < \frac{1}{4}

) because solving the inequality after removing the square root leads to this solution.

74 Hard

74 of 650

Analyzing +4 | -1

f\;:R \to R\;and\,g\;:R \to R

are given by

f\left( x \right) = \left| x \right|\;and\,g\left( x \right) = \left[ x \right]

for each

x \in R,

then

\left\{ {x \in R\;:g\left( {f\left( x \right)} \right) \le f\left( {g\left( x \right)} \right)} \right\} =

View answer and explanation

Correct option: A

Z \cup \left( { - \infty ,0} \right)

Option A (

Z \cup ( - \infty ,0)

) is correct because the inequality holds for all integers and all negative real numbers.

75 Medium

75 of 650

Understanding +4 | -1

Let

f(x)=x, g(x)=\frac{1}{x}

and

h(x)=f(x) g(x)

. Then,

h(x)=1

if and only if

View answer and explanation

Correct option: D

x

is a non-zero real number

Option D is correct because the function is only equal to 1 when x is a non-zero real number, as g(x) is undefined at x=0.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

76 Medium

76 of 650

Application +4 | -1

f: R \rightarrow R, f(x)=2 x-1

and

g: R \rightarrow R, g(x)=x^2

then

(g \circ f)(x)

equals?

View answer and explanation

Correct option: B

(2 x-1)^2

The correct answer is B ((2x-1)²) because it is the result of substituting f(x) into g(x).

77 Medium

77 of 650

Analyzing +4 | -1

The domain of definition of

f\left( x \right) = {\log _3}\left| {{{\log }_e}x} \right|,

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the domain is (0, 1) U (1, infinity), which is not covered by the other options.

78 Easy

78 of 650

Application +4 | -1

If the domain of the function

f(x)=\sqrt{3 \cos ^{-1}(4 x)-\pi}

[a, b]

, then the value of

(4 a+64 b)

View answer and explanation

Correct option: A

Option A (9) is correct because using the properties of binomial expansion and the given condition y=12x, we can solve for n.

79 Easy

79 of 650

Application +4 | -1

Domain of the function

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

View answer and explanation

Correct option: C

\left[-\frac{3}{2}, 0\right]

Option C ([-3/2, 0]) is correct because the domain requires the argument of the arcsin function to be between -1 and 1.

80 Medium

80 of 650

Analyzing +4 | -1

The real valued function

f(x) = \frac{{\mathrm{cose}{\mathrm{c}^{ - 1}}x}}{{\sqrt {x - [x]'} }}

where

\left[ x \right]

denotes the greatest integer less than or equal to

x

, is defined for all

x

belonging to

View answer and explanation

Correct option: B

all non-integers expect the interval [-1, 1]

Option B is correct because the function is defined when |x| >= 1 and x is not an integer, excluding the interval [-1, 1].

81 Medium

81 of 650

Application +4 | -1

Let for

a \neq a_1 \neq 0, \quad f(x)=a x^2+b x+c

g(x)=a_1 x^2+b_1 x+c_1

and

p(x)=f(x)-g(x)

. If

p(x)=0

only for

x=-1

and

p(-2)=2

, then the value of

p(2)

View answer and explanation

Correct option: A

Option A (18) is correct because by using the given information about p(x) and its root, we can determine the quadratic expression and then evaluate p(2).

82 Medium

82 of 650

Application +4 | -1

Suppose

f:\left[ { - 2,\;2} \right] \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} { - 1,\;for - 2 \le x \le 0}\\ {x - 1\;for\;0 \le x \le 2} \end{array}} \right.

, then

\left\{ {x \in \left[ { - 2,\;2} \right]:x \le 0\;\mathrm{and}\;f\left( {\left| x \right|} \right) = x} \right\}

is equal to

View answer and explanation

Correct option: C

\left\{ { - \frac{1}{2}} \right\}

Option C (

\left\{ { - \frac{1}{2}} \right\}

) is correct because it represents the solution to the equation

f(|x|) = x

within the given domain and condition.

83 Easy

83 of 650

Understanding +4 | -1

The number of bijective functions from set A to itself when A contains 106 elements is

View answer and explanation

Correct option: C

106\;!

Option C (

106\\;!

) is correct because the number of bijective functions from a set A to itself when A contains n elements is n!.

84 Easy

84 of 650

Understanding +4 | -1

f: R \rightarrow R

, then

f(x)=|x|

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function is neither one-to-one nor onto, as it is many-to-one and its range is a subset of the codomain.

85 Medium

85 of 650

Application +4 | -1

f:R \to R

is given by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} { - 1,\;when\;\;x\;\;is\;\;rational}\\ {1,\;when\;\;x\;\;is\;\;irrational} \end{array}} \right.,

Then

\left( {fof} \right)\left( {1 - \sqrt 3 } \right)

is equal to

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because the function f(x) returns -1 for rational numbers and 1 for irrational numbers. Since 1 - √3 is irrational, f(1 - √3) = 1, which is rational, so f(1) = -1.

86 Easy

86 of 650

Application +4 | -1

-3 x+17 < -13

, then

View answer and explanation

Correct option: A

x \in(10, \infty)

Option A (x ∈ (10, ∞)) is correct because solving the inequality -3x + 17 < -13 results in x > 10.

87 Easy

87 of 650

Understanding +4 | -1

Domain and range of

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

are respectively

View answer and explanation

Correct option: B

R-\{3\},\{1,-1\}

Option B (

R - \\{3\\}, \\{1, -1\\}

) is correct because the domain excludes x=3 where the denominator is zero, and the range consists of 1 and -1 depending on the sign of (x-3).

88 Medium

88 of 650

Application +4 | -1

Let

R

be the set of real numbers and the functions

f: R \rightarrow R

and

g: R \rightarrow R

be defined by

f(x)=x^2+2 x-3

and

g(x)=x+1

. Then, the value of

x

for which

g(f(x))=f(g(x))

View answer and explanation

Correct option: A

-1

Option A (-1) is correct because substituting x = -1 into the equation g(f(x)) = f(g(x)) satisfies the equality.

89 Easy

89 of 650

Understanding +4 | -1

Let

f(x)=\log _e(\sqrt{x-3}+\sqrt{5-x}), x \in R

. Then domain of

f(x)

is -

View answer and explanation

Correct option: A

[3,5]

Option A ([3,5]) is correct because the domain of the function is determined by the intersection of the domains of the square roots, ensuring non-negative values inside the square roots.

90 Hard

90 of 650

Analyzing +4 | -1

{b^2} - 4\;ac = 0,a > 0,

then the domain of the function

f\left( x \right) = \log \left\{ {a{x^3} + \left( {a + b} \right){x^2} + \left( {b + c} \right)x + c)} \right\}

View answer and explanation

Correct option: C

R - \left\{ {\left\{ { - \frac{b}{{2\;a}}} \right\} \cap \left( { - \infty , - 1} \right]} \right\}

Option D (

R - \left\{ {\left\{ { - \frac{{b}}{{2\;a}}} \right\} \cap \left( { - \infty , - 1} \right]} \right\}

) is correct because it represents the domain where the argument of the logarithm is positive, excluding the point where the quadratic term becomes zero and considering the factor (x+1).

Revision notes and key takeaways

Review FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 650 questions

Continue Your Practice

📚 Switch Set (ALL)

Chapter Sets

ALL 650 MCQs EASY 478 MCQs HARD 33 MCQs MEDIUM 353 MCQs

0 /650

0✓ 0✗

0% Accuracy