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

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

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

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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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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.

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Showing 31-60 of 353 questions

31 Medium

31 of 353

Application +4 | -1

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

and the composite function

g\left( {f\left( x \right)} \right) = \left| {\sin x} \right|

, then the function

g\left( x \right)

is equal to

View answer and explanation

Correct option: B

\sqrt x

Option B (

\sqrt x

) is correct because it satisfies the condition g(f(x)) = |sin x|.

32 Medium

32 of 353

Application +4 | -1

f(x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x

, where

[x]

stands for the greatest integer function, then

View answer and explanation

Correct option: C

f(-\pi)=0

Option C (

f(-\pi)=0

) is correct because substituting

-\pi

into the function and using the greatest integer function properties yields 0.

33 Medium

33 of 353

Analyzing +4 | -1

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

, then domian of the function

f\left( x \right) = \log \left\{ {\left( {a{x^2} + bx + c} \right)\left( {x + 1} \right)} \right\}

View answer and explanation

Correct option: C

\left( { - 1,\;\infty } \right) - \left\{ { - \frac{b}{{2a}}} \right\}

Option C (

\left( { - 1,\\;\infty } \right) - \left\\{ { - \frac{{b}}{{{{2a}}}}} \right\\}

) is correct because the domain requires the argument of the logarithm to be positive, and the quadratic has a repeated root.

34 Medium

34 of 353

Analyzing +4 | -1

The domain of the function

f(x) = \frac{1}{{\sqrt {\left| x \right| - x} }}

is:

View answer and explanation

Correct option: B

R^{-}

Option B (

R^{-}

) is correct because the function is defined only when

|x| - x > 0

, which occurs when x is negative.

35 Medium

35 of 353

Analyzing +4 | -1

A function

f\left( x \right)

is given by

f\left( x \right) = \frac{{{5^x}}}{{{5^x} + 5}}

, then the sum of the series

f\left( {\frac{1}{{20}}} \right) + f\left( {\frac{2}{{20}}} \right) + f\left( {\frac{3}{{20}}} \right) + ... + f\left( {\frac{{39}}{{20}}} \right)

is equal to

View answer and explanation

Correct option: C

\frac{{39}}{2}

Option C (39/2) is correct because by pairing terms of the series using the property f(x) + f(2-x) = 1, the sum simplifies to 39/2.

36 Medium

36 of 353

Application +4 | -1

Find the range, domain and inverse of the function

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

View answer and explanation

Correct option: A

R-\{2\}, R-\{1\},(x+3 / x-2)

Option A (R-{2}, R-{1}, (x+3 / x-2)) is correct because it correctly identifies the domain, range, and inverse function of the given function f(x).

37 Medium

37 of 353

Analyzing +4 | -1

If { } denotes the fractional part of

x

, the range of the function

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

View answer and explanation

Correct option: D

\{0\}

Option C (

\pi

) is correct because by differentiating the given relation implicitly and substituting x=0, the second derivative y''(0) is found to be equal to

\pi

38 Medium

38 of 353

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.

39 Medium

39 of 353

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.

40 Medium

40 of 353

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.

41 Medium

41 of 353

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.

42 Medium

42 of 353

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.

43 Medium

43 of 353

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.

44 Medium

44 of 353

Application +4 | -1

Let

\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}

is a function such that

\mathrm{f}^{-1}(2)=3, \mathrm{f}^{-1}(-3)=6

, where

\mathrm{f}^{-1}(\mathrm{x})

is the inverse of function

\mathrm{f}

then

\mathrm{a}+\mathrm{b}

is equal to

View answer and explanation

Correct option: B

\frac{16}{3}

Option B (

\frac{16}{3}

) is correct because by using the given inverse function values, we can find the values of

a

and

b

and then calculate

a+b

45 Medium

45 of 353

Analyzing +4 | -1

f(x)=\sqrt{3|x|-x}, g(x)=1-2 \sin x

, then the domain of definition of

f \circ g(x)

View answer and explanation

Correct option: B

R

Option B (R) is correct because the composition f(g(x)) is defined for all real numbers, as the condition inside the square root is always non-negative.

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46 Medium

46 of 353

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.

47 Medium

47 of 353

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.

48 Medium

48 of 353

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.

49 Medium

49 of 353

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).

50 Medium

50 of 353

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.

51 Medium

51 of 353

Application +4 | -1

The value of

\sin \left[2 \sin ^{-1}(0.8)\right]

is equal to -

View answer and explanation

Correct option: B

0.96

Option B (0.96) is correct because it is the value of

\sin[2 \sin^{-1}(0.8)]

calculated using the double angle formula.

52 Medium

52 of 353

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].

53 Medium

53 of 353

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).

54 Medium

54 of 353

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.

55 Medium

55 of 353

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.

56 Medium

56 of 353

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.

57 Medium

57 of 353

Analyzing +4 | -1

The domain of the function

f\left( x \right) = \frac{{{{\cos }^{ - 1}}x}}{{\left[ x \right]}}

View answer and explanation

Correct option: A

\left[ { - 1,\;0} \right) \cup \left\{ 1 \right\}

Option A (

\left[ { - 1,\;0} \right) \cup \left\{ 1 \right\}

) is correct because it considers the domain of the inverse cosine function and the restriction that the denominator of the fraction cannot be zero.

58 Medium

58 of 353

Analyzing +4 | -1

f\left( {x + 2y,x,x - 2y} \right) = xy

, then

f\left( {x,\;y} \right)

equals

View answer and explanation

Correct option: A

\frac{{{x^2} - {y^2}}}{8}

Option A (

\frac{{{x^2} - {y^2}}}{8}

) is correct because by substituting

x + 2y = u

and

x - 2y = v

, we can express

f(u, v)

in terms of

u

and

v

, leading to the given expression.

59 Medium

59 of 353

Analyzing +4 | -1

Let

f(x)=|x-1|, x \in R

, then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because none of the given equations hold true for the given function

f(x)=|x-1|

60 Medium

60 of 353

Application +4 | -1

Let

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

, then

\underbrace{f o f o f o \ldots \ldots o f}_{n \text { times }}(x)

is:

View answer and explanation

Correct option: B

\frac{x}{\sqrt{1+\left(\sum\limits_{{r=1}}^n 1\right) x^2}}

Option B is correct. By repeatedly applying the function

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

, we observe a pattern that leads to

\underbrace{f o f o f o \ldots \ldots o f}_{n \text { times }}(x) = \frac{x}{\sqrt{1+\left(\sum\limits_{{r=1}}^n 1\right) x^2}}

Revision notes and key takeaways

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