Let $f\left( x \right) = \mathop \smallint \limits_0^x \frac{{\cos t}}{t}dt$ Then, at $x = \left( {2n + 1} \right)\frac{\pi }{2},\;f\left( x \right)$ has

Maxima when $n = - 2,\; - 4,\; - 6$ … and minima when $n = - 1,\; - 3,\; - 5, \ldots $

Minima when $n = 0,\;2,\;4, \ldots $ and maxima when $n = 1,\;3,\;5, \ldots $

The function $f\left( x \right) = {x^{1/x}}$ is

Increasing in $\left( {1,\;\infty } \right)$

Decreasing in $\left( {1,\;\infty } \right)$

Decreasing in $\left( {1,\;e} \right)$ and increasing in $\left( {e,\;\infty } \right)$

Consider the function $f:R \to R$ defined by $f(x) = \left\{ {\begin{array}{*{20}{c}} {\left( {2 - \sin \left( {\frac{1}{x}} \right)} \right)|x|,}&{x \ne 0}\\ 0&{,x \ne 0} \end{array}} \right\}$. Then $f$ is

monotonic on $(0,\,\infty )$ only

monotonic on $( - \infty ,\,0)$ only

The function $f(x)=2 x^2-\log |x|$ is

decreasing in $\left(\frac{1}{2}, \infty\right)$

increasing in $\left(0, \frac{1}{2}\right)$

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

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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Mixed difficulty

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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 91-120 of 200 questions

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91 Easy

91 of 200

Analyzing +4 | -1

The function

f\left( x \right)

given by

f\left( x \right)\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&1&1\\ 1&{x + 1}&1\\ 1&1&{x + 1} \end{array}} \right|

is increasing on

View answer and explanation

Correct option: C

R - \left[ { - 2,\;0} \right]

Option C (

R - \left[ { - 2,\;0} \right]

) is correct because the function is increasing when its derivative is positive, which occurs outside the interval [-2, 0].

92 Medium

92 of 200

Analyzing +4 | -1

f: R \rightarrow R

be definend by

f(x)=2 x+\cos x

, then

f

View answer and explanation

Correct option: D

is an increasing function

Option D (is an increasing function) is correct because the derivative of the function, f'(x) = 2 - sin(x), is always positive, indicating that the function is always increasing.

93 Medium

93 of 200

Application +4 | -1

The function

f\left( x \right) = 2{x^3} - 3{x^2} + 90x + 174

is increasing in the interval

View answer and explanation

Correct option: D

- \infty < x < \infty

Option D (

- \infty < x < \infty

) is correct because the derivative is always positive, indicating an increasing function.

94 Hard

94 of 200

Analyzing +4 | -1

The function

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

is monotonically decreasing on

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the function's derivative is negative in this interval, indicating monotonic decrease.

95 Medium

95 of 200

Analyzing +4 | -1

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

, then

f\left( x \right)

View answer and explanation

Correct option: A

Increasing on

\left[ { - \frac{1}{2},\;1} \right]

Option A (Increasing on

\left[ { - \frac{1}{2},\;1} \right]

) is correct because the derivative of the function is positive in this interval, indicating an increasing function.

96 Medium

96 of 200

Analyzing +4 | -1

f(x)=4 \log _e(x-1)-2 x^2+4 x+5, x>1

which one of the following is NOT correct?

View answer and explanation

Correct option: A

f

is increasing in

(1,2)

and decreasing in

(2

Option A is NOT correct because the function f(x) is increasing in (1,2) and decreasing in (2, infinity) based on the derivative analysis.

97 Easy

97 of 200

Analyzing +4 | -1

The function

f(x)=\tan ^{-1}(\sin x+\cos x)

is an increasing function in

View answer and explanation

Correct option: D

\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)

Option D (

(-\frac{{\pi}}{{2}}, \frac{{\pi}}{{4}})

) is correct because the function

f(x)=\tan^{-1}(\sin x + \cos x)

is increasing when its derivative is positive, which occurs on the interval

(-\frac{{\pi}}{{2}}, \frac{{\pi}}{{4}})

98 Medium

98 of 200

Analyzing +4 | -1

Let

f\left( x \right) = \int \limits_0^x \frac{{\cos t}}{t}dt

Then, at

x = \left( {2n + 1} \right)\frac{\pi }{2},\;f\left( x \right)

has

View answer and explanation

Correct option: B

Maxima when

n = - 1,\; - 3,\; - 5

...and minima when

n = 1,\;3,\;5, \ldots

The correct answer is Maxima when

n = - 1, - 3, - 5

...and minima when

n = 1, 3, 5, ...

because the sign of the second derivative determines whether the function has a maxima or minima at the given points.

99 Easy

99 of 200

Analyzing +4 | -1

Let

f(x) = {e^x} - x

and

g(x) = {x^2} - x,\forall x \in R.

Then, the set of all

x \in R,

where function

h(x) = (fog)(x)

is increasing, is

View answer and explanation

Correct option: A

\left[ {0,\frac{1}{2}} \right] \cup \left[ 1 \right.,\left. \infty \right)

Option A (

{0,\frac{1}{2}} \cup \left[ 1 \right.,\left. \infty \right)

) is correct because it represents the intervals where the derivative of h(x) is greater than or equal to zero, indicating an increasing function.

100 Easy

100 of 200

Analyzing +4 | -1

The function

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

is increasing in the interval

View answer and explanation

Correct option: B

\left( {0,\;e} \right)

Option B is correct because the derivative of the function, f'(x), is positive in the interval (0, e), indicating that the function is increasing in that interval.

101 Hard

101 of 200

Analyzing +4 | -1

The function

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

View answer and explanation

Correct option: C

Increasing in

\left( {1,\;e} \right)

and decreasing in

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

Option C (Increasing in (1, e) and decreasing in (e,

\infty

)) is correct because the derivative of the function is positive in (1, e) and negative in (e,

\infty

102 Easy

102 of 200

Application +4 | -1

The difference between the greatest and the least values of the function

f(x)=\int\limits_{{0}}^{x}\left(a t^{2}+1+\cos t\right) d t, a > 0

for

x \in[2,3]

View answer and explanation

Correct option: A

\frac{19}{3} a+1+(\sin 3-\sin 2)

Option A (

\frac{19}{3} a+1+(\sin 3-\sin 2)

) is correct because it represents the difference between the function's values at the upper and lower limits of the interval [2,3], obtained by integrating the given expression.

103 Easy

103 of 200

Analyzing +4 | -1

The function

f\left( x \right) = \frac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}},

View answer and explanation

Correct option: B

Decreasing on

\left[ {0,\infty } \right)

Option B is correct because the derivative of the function, f'(x), is negative on the interval [0, ∞), indicating that the function is decreasing.

104 Medium

104 of 200

Analyzing +4 | -1

Let

f(x)

and

g(x)

be defined and differentiable for all

x \geq x_0

and

f\left(x_{0})=g\left(x_0\right), f^{\prime}(x) > g^{\prime}(x)\right.

for

x > x_0

, then

View answer and explanation

Correct option: C

f(x) > g(x), x > x_0

The correct answer is C (

f(x) > g(x), x > x_0

) because since

f'(x) > g'(x)

for

x > x_0

and

f(x_0) = g(x_0)

f(x)

increases faster than

g(x)

after

x_0

105 Easy

105 of 200

Analyzing +4 | -1

The function

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

strictly

View answer and explanation

Correct option: D

Decreases in the interval

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

Option D (Decreases in the interval

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

) is correct because the derivative of the function is negative in that interval, indicating a decreasing function.

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

106 Medium

106 of 200

Analyzing +4 | -1

Let

g\left( x \right) = f\left( x \right) - 2{\left\{ {f\left( x \right)} \right\}^2} + 9{\left\{ {f\left( x \right)} \right\}^3}

for all

x \in R.

Then,

View answer and explanation

Correct option: A

g\left( x \right)

and

f\left( x \right)

increases and decrease together

Option A (

g\left( x \right)

and

f\left( x \right)

increases and decrease together) is correct because the derivative of g(x) has the same sign as the derivative of f(x).

107 Easy

107 of 200

Analyzing +4 | -1

0 < x < \frac{\pi}{2}

, then

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because for

0 < x < \frac{{\pi}}{{2}}

, the correct inequality is

\sin x < x < \tan x

108 Medium

108 of 200

Analyzing +4 | -1

Let

f:(0,1) \rightarrow R

be a function defined by

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

, and

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

. Consider two statements

(I)

g

is an increasing function in

(0,1)

(II)

g

is one-one in

(0,1)

Then,

View answer and explanation

Correct option: B

Both (I) and (II) are true

Option B is correct because both statements are true: g(x) is increasing, and it is one-to-one, as shown by its positive derivative.

109 Medium

109 of 200

Analyzing +4 | -1

Let

f

be differentiable on

\mathbb{R}

and

h(x)=f(x)-[f(x)]^2+[f(x)]^3

Then

View answer and explanation

Correct option: A

h

is increasing whenever

f

is increasing

Option A is correct because h'(x) is non-negative whenever f'(x) is non-negative, indicating that h is increasing when f is increasing.

110 Easy

110 of 200

Analyzing +4 | -1

Consider the function

f:R \to R

defined by

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left( {2 - \sin \left( {\frac{1}{x}} \right)} \right)|x|,}&{x \ne 0}\\ 0&{,x \ne 0} \end{array}} \right\}

. Then

f

View answer and explanation

Correct option: B

not monotonic on

( - \infty ,\,0) \cup (0,\,\infty )

Option B is correct. The function f(x) is not monotonic on

( - \infty ,\,0) \cup (0,\,\infty )

because its derivative oscillates.

111 Medium

111 of 200

Analyzing +4 | -1

The function

f

defined by

f(x)=x^3-3 x^2+5 x+7

, is :

View answer and explanation

Correct option: A

increasing in

\mathrm{R}

Option A (increasing in R) is correct because the derivative of the function is always positive, indicating that the function is strictly increasing.

112 Easy

112 of 200

Analyzing +4 | -1

The function

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

increases on the interval

View answer and explanation

Correct option: C

(e, \infty)

Option C (

(e, \infty)

) is correct because the function

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

increases when its derivative is positive, which occurs on the interval

(e, \infty)

113 Easy

113 of 200

Analyzing +4 | -1

The function

y = x - {\cot ^{ - 1}}x - \log \left( {x + \sqrt {{x^2} + 1} } \right)

is increasing on

View answer and explanation

Correct option: D

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

Option D is correct because the derivative of the function is non-negative for all real numbers, indicating that the function is always increasing.

114 Easy

114 of 200

Analyzing +4 | -1

\int_0^x 2 x f^2(t) d t=\left(\int_0^x 2 f(x-t) d t\right)^2

for

f(1)=1

and

f(x)

is continuous function for

x>0

and

\left\{a_n\right\}

is a sequence such that

a_{n+1}=a_n+\sqrt{1+a_n^2}

for

a=0

; if

f(x)

is an increasing function, then

\lim _{n \rightarrow \infty} \frac{a_k}{2^{n-1}}=

(where

\left.k=f\left(n^{\sqrt{2}-1}\right)\right)

View answer and explanation

Correct option: B

4 / \pi

Option B (

4 / \pi

) is correct. After solving the integral equation for

f(x)

and using the recurrence relation for

a_n

, we find the limit to be

4/\pi

115 Medium

115 of 200

Application +4 | -1

The set of values of a for which the function

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

is increasing on

R

, is

View answer and explanation

Correct option: A

\left( {0,\;\infty } \right)

Option A (

\left( {0,\;\infty } \right)

) is correct because it represents the set of values for 'a' that ensures the function's derivative is always positive, making the function increasing on R.

116 Easy

116 of 200

Analyzing +4 | -1

The function

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

is decreasing on the set

View answer and explanation

Correct option: C

R - \left\{ 0 \right\}

Option C (

R - \{ 0 \}

) is correct because the derivative of the function is negative for all real numbers except 0, indicating that the function is decreasing on that set.

117 Easy

117 of 200

Analyzing +4 | -1

Let

f(x)=\left\{\begin{array}{ll}x e^{a x}, & x \leq 0 \\ x+a x^2-x^3, & x > 0\end{array}\right.

where

a

is a positive constant. Then, the interval in which

f^{\prime}(x)

is increasing, is

View answer and explanation

Correct option: C

(-2 / a, a / 3)

Option C ((-2 / a, a / 3)) is correct because it represents the interval where the second derivative of

f(x)

is positive, ensuring that

f'(x)

is increasing.

118 Easy

118 of 200

Analyzing +4 | -1

Let

f(x)

be the function given by

f(x)=3 x^5-5 x^3+21 x+3 \sin x+4 \cos x+5.

Then,

View answer and explanation

Correct option: A

f(x)

is increasing on

R

and

f(x)=0

has exactly one negative root

Option A (

f(x)

is increasing on

R

and

f(x)=0

has exactly one negative root) is correct because the derivative of

f(x)

is always positive, indicating it's increasing, and it crosses the x-axis only once in the negative region.

119 Medium

119 of 200

Analyzing +4 | -1

The function

f(x)=2 x^2-\log |x|

View answer and explanation

Correct option: B

decreasing in

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

Option B is correct because the function's derivative is negative in the interval (0, 1/2), indicating a decreasing function.

120 Hard

120 of 200

Understanding +4 | -1

Let

f\left( x \right) = {\cot ^{ - 1}}\left\{ {g\left( x \right)} \right\},

where

g\left( x \right)

is an increasing function on the interval

\left( {0,\pi } \right).

Then,

f\left( x \right)

View answer and explanation

Correct option: B

Decreasing on

\left( {0,\pi } \right)

Option B (Decreasing on (0,

\pi

)) is correct because the derivative of f(x) is negative on the interval (0,

\pi

), indicating that the function is decreasing.

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