If $f\left( x \right) = \frac{1}{{x + 1}} - \log \left( {1 + x} \right),\;x > 0$, then $f$ is

both increasing and decreasing function

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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Mixed difficulty

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Showing 31-60 of 200 questions

31 Medium

31 of 200

Analyzing +4 | -1

The function

f(x)=3 \cos ^4 x+10 \cos ^3 x+6 \cos ^2 x-3

0 \leq x \leq \pi

View answer and explanation

Correct option: B

increasing in

\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right)

Option B is correct because the first derivative of the function is positive in the interval (

\frac{{\pi}}{{2}}

\frac{{2 \pi}}{{3}}

), indicating that the function is increasing in that interval.

32 Medium

32 of 200

Analyzing +4 | -1

y = {\left\{ {x\left( {x - 3} \right)} \right\}^2}

increases for all values of

x

lying in the interval

View answer and explanation

Correct option: A

0 < x < \frac{3}{2}

Option A (

0 < x < \frac{3}{2}

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

33 Medium

33 of 200

Application +4 | -1

The longest distance of the point

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

from the curve

2{x^2} + {y^2} - 2\;x = 0

, is given by

View answer and explanation

Correct option: D

\sqrt {1 - 2\;a + 2\;{a^2}}

Option D (

\sqrt {1 - 2\;a + 2\;{a^2}}

) is correct because it represents the maximum distance calculated by finding the critical point of the distance function.

34 Easy

34 of 200

Analyzing +4 | -1

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

and

g(x)=\frac{x}{\tan x}

, where

0 < x \leq 1

, then in this interval

View answer and explanation

Correct option: C

f(x)

is an increasing function

Option C is correct.

f(x)

is an increasing function because its derivative is positive in the interval

(0, 1]

35 Medium

35 of 200

Analyzing +4 | -1

Let

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

for all

x \in R

, and let

g(x)=\int\limits_{1}^x\left\{\frac{2(t-1)}{t+1}-\log _e t\right\} f(t)\, dt

for all

x \in(1, \infty)

. Then, which of the following is true?

View answer and explanation

Correct option: B

g

is decreasing on

(1, \infty)

The correct answer is that g is decreasing on (1, ∞) because the derivative of g(x) is negative for all x > 1, indicating a decreasing function.

36 Easy

36 of 200

Application +4 | -1

The interval of increase of the function

F\left( x \right) = x - {e^x} + \tan \left( {\frac{{2\pi }}{7}} \right)

View answer and explanation

Correct option: B

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

Option B (

\frac{\pi }{2}

) is correct because it is the value of c within the given interval that satisfies Lagrange's Mean Value Theorem.

37 Easy

37 of 200

Analyzing +4 | -1

f(x)=\left|x \log _e x\right|, x > 0

, is monotonically decreasing in

View answer and explanation

Correct option: C

[1 / e, 1]

Option C

[1 / e, 1]

is correct because the function's derivative is negative in this interval, indicating that the function is monotonically decreasing.

38 Easy

38 of 200

Application +4 | -1

The value of a in order that

f\left( x \right) = \sin x - \cos x - ax + b

decreases for all real values of

x

is given by

View answer and explanation

Correct option: A

a \ge \sqrt 2

Option A (

a \ge \sqrt 2

) is correct because it ensures that the derivative of the function is always negative, indicating a decreasing function for all real x.

39 Easy

39 of 200

Understanding +4 | -1

f^{\prime}(x)

is zero in the interval

(a, b)

, then in this interval it is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because if the derivative is zero in an interval, the function is constant, not necessarily increasing or decreasing.

40 Easy

40 of 200

Understanding +4 | -1

Let

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

, where

g(x)

is monotonically increasing for

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

. Then,

f(x)

, is

View answer and explanation

Correct option: A

Increasing on

(0, \pi / 2)

Option D (Ordinate of that point) is correct because the product of the subnormal and subtangent at any point on a curve equals the square of the ordinate of that point.

41 Medium

41 of 200

Analyzing +4 | -1

Let

f

and

g

be two differentiable functions on

R

such that

f^{\prime}(x)>0

and

g^{\prime}(x)<0

for all

x \in R

. Then for all

\mathrm{x}

View answer and explanation

Correct option: B

f(g(x))>f(g(x+1))

Option C (Statement-1 is true; Statement-2 is false) is correct because while the given function is increasing, the reasoning provided in statement-2 is incorrect.

42 Medium

42 of 200

Application +4 | -1

f(x)=a \sin x+x+2

is strictly increasing function then number of possible integral values of '

a

' are

View answer and explanation

Correct option: A

Option A (3) is correct because the condition for f(x) to be strictly increasing leads to the inequality -1 <= a <= 1, resulting in three possible integer values for 'a': -1, 0, and 1.

43 Easy

43 of 200

Analyzing +4 | -1

The function

f(x)=x e^{x(1-x)}, x \in \mathbb{R}

, is

View answer and explanation

Correct option: A

Increasing in

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

Option A (Increasing in

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

) is correct because the derivative

f'(x)

is positive in this interval, indicating that the function is increasing.

44 Medium

44 of 200

Analyzing +4 | -1

Let

f(x)=\int e^x(x-1)(x-2) d x

, then

f(x)

decreases in the interval

View answer and explanation

Correct option: C

(1,2)

Option C (

(1,2)

) is correct because

f(x)

decreases when

f'(x) < 0

, which occurs when

e^x(x-1)(x-2) < 0

, leading to the interval

(1,2)

45 Hard

45 of 200

Analyzing +4 | -1

If the function

f\left( x \right) = \left( {2a - 3} \right)\left( {x + 2\sin 3} \right) + \left( {a - 1} \right)\left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \log 2

does not possess critical points, then

View answer and explanation

Correct option: A

a \in \left( { - \infty ,\;4/3} \right) \cup \left( {2,\;\infty } \right)

Option A (

a \in (-\infty, 4/3) \cup (2, \infty)

) is correct because for the function to have no critical points, the derivative must not equal zero, leading to the given range for 'a'.

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46 Easy

46 of 200

Analyzing +4 | -1

Let

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

and

f^{\prime \prime}(x) < 0

for all

x \in(0,2)

. Then,

g(x)

, is

View answer and explanation

Correct option: C

Increasing on

(0,4 / 3)

and decreasing on

(4,3,2)

Option C (Increasing on

(0,4 / 3)

and decreasing on

(4,3,2)

) is correct because

g'(x)

is positive on

(0, 4/3)

and negative on

(4/3, 2)

, indicating increasing and decreasing behavior respectively.

47 Easy

47 of 200

Application +4 | -1

The function

f(x)=e^{a x}+e^{a x}, a > 0

is monotonically increasing for

View answer and explanation

Correct option: D

x > 0

Option D (

x > 0

) is correct because the derivative of the function is positive for

x > 0

, indicating that the function is monotonically increasing in that interval.

48 Easy

48 of 200

Analyzing +4 | -1

Let

f(x)= \begin{cases}-55 x, x < -5 \\ 2 x^3-3 x^2-120 x,-5 \leq x < 4 \\ 2 x^3-3 x^2-36 x+10, x \geq 4\end{cases}

Then the interval in which

f(x)

is monotonically increasing, is

View answer and explanation

Correct option: A

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

Option A ((-5,-4)

\cup

(4,

\infty

)) is correct because the derivative of the function is positive in these intervals, indicating a monotonically increasing function.

49 Medium

49 of 200

Analyzing +4 | -1

\phi \left( x \right)

is continuous at

x = \alpha

such that

\phi \left( \alpha \right) < 0

and

f\left( x \right)

is a function such that

f'\left( x \right) = \left( {ax - {a^2} - {x^2}} \right)\phi \left( x \right)

for all

x

, then

f\left( x \right)

View answer and explanation

Correct option: A

Increasing in the neighbourhood of

x = \alpha

Option A is correct because the derivative of the function, f'(x), is positive in the neighborhood of x = α, indicating that the function is increasing.

50 Medium

50 of 200

Analyzing +4 | -1

f: R \rightarrow R

defined by

f(x)=3 x+2 a \cos x-5

is invertible, then '

a

' belongs to

View answer and explanation

Correct option: A

[-3 / 2,3 / 2]

The correct answer is A (

[-3 / 2,3 / 2]

) because for

f(x)

to be invertible, its derivative must be either always non-negative or always non-positive, leading to the condition

|a| \leq \frac{3}{2}

51 Medium

51 of 200

Analyzing +4 | -1

f(x)

is a function whose derivative is

e^{-\left(x^2+1\right)^2}(2 x)-e^{-x^4}(2 x)

then

f(x)

increases in

View answer and explanation

Correct option: D

(-\infty, 0)

Option D is correct because the derivative is positive for x < 0, indicating that the function is increasing in the interval (-∞, 0).

52 Medium

52 of 200

Analyzing +4 | -1

f^{\prime}\left(x^2-4 x+3\right) > 0

for all

x \in (2,3)

, then

f(\sin x)

is increasing on

View answer and explanation

Correct option: A

\underset{n \in Z}{\cup} \in\left(2 n \pi,(4 n+1) \frac{\pi}{2}\right)

Option A (

\underset{{n \in Z}}{{\cup}} \in\left(2 n \pi,(4 n+1) \frac{{\pi}}{{2}}\right)

) is correct because

f(\sin x)

is increasing when

f'(\sin x) \cos x > 0

, which occurs on the given interval.

53 Easy

53 of 200

Analyzing +4 | -1

f(x)=\int\limits_{x}^{x^2} \frac{1}{(\log t)^2} d t, x \neq 0, x \neq \mathbb{1}

then

f(x)

is monotomically

View answer and explanation

Correct option: A

Increasing on

(2, \infty)

The correct answer is A (Increasing on

(2, \infty)

) because the derivative

f'(x)

is positive for

x > 2

, indicating that

f(x)

is increasing on that interval.

54 Easy

54 of 200

Analyzing +4 | -1

f\left( x \right) = \frac{1}{{4{x^2} + 2x + 1}}

, then its maximum value is

View answer and explanation

Correct option: A

\frac{4}{3}

Option A (

\frac{4}{3}

) is correct because it represents the maximum value of the function, found by taking the derivative and setting it to zero.

55 Easy

55 of 200

Application +4 | -1

The least value of

k

for which the function

x^2+k x+1

is an increasing function in the interval

1 < x < 2

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because it is the least value of k for which the derivative of the function is greater than 0 in the given interval, ensuring the function is increasing.

56 Medium

56 of 200

Analyzing +4 | -1

f\left( x \right) = \frac{1}{{x + 1}} - \log \left( {1 + x} \right),\;x > 0

, then

f

View answer and explanation

Correct option: B

a decreasing function

Option D (None of these) is correct because the maximum slope occurs at x = 5, which is at the boundary of the interval.

57 Medium

57 of 200

Analyzing +4 | -1

The set of all

x

for which

\log (1+x) \leq x

, is

View answer and explanation

Correct option: B

(-1, \infty)

Option B (

(-1, \infty)

) is correct because the inequality

\log (1+x) \leq x

holds for all

x

greater than

-1

, which is the domain of the logarithm.

58 Easy

58 of 200

Analyzing +4 | -1

Let

f

be any function continuous on

[a,\,b]

and twice differentiable on

(a,\,b)

. If for all

x \in \left( {a,\,b} \right),f'\left( x \right) > 0

and

f''\left( x \right) < 0,

then for any

c \in \left( {a,\,b} \right),\frac{{f\left( c \right) - f\left( a \right)}}{{f\left( b \right) - f\left( c \right)}}

is greater than

View answer and explanation

Correct option: C

\frac{{c - a}}{{b - c}}

Option D (

{\pi ^2}

) is correct because it is the result of evaluating the definite integral using properties of definite integrals.

59 Medium

59 of 200

Analyzing +4 | -1

g(x)=2 f\left(2 x^3-3 x^2\right)+f\left(6 x^2-4 x^3-3\right), \forall x \in R

and

f^{\prime \prime}(x)>0, \forall x \in R

, then

g(x)

is increasing in

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the intervals where the derivative of g(x) is positive, indicating that g(x) is increasing.

60 Medium

60 of 200

Analyzing +4 | -1

The function

f\left( x \right) = {\sin ^4}x + {\cos ^4}x

increases, if

View answer and explanation

Correct option: B

\pi /4 < x < 3\pi /8

Option B (

\pi /4 < x < 3\pi /8

) is correct because the derivative of the function is positive in this interval.

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