The function $f\left( x \right) = \frac{x}{{1 + \left| x \right|}}$ is

Neither increasing nor decreasing

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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Easy

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61 Easy

61 of 101

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

62 Easy

62 of 101

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.

63 Easy

63 of 101

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.

64 Easy

64 of 101

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.

65 Easy

65 of 101

Application +4 | -1

Given that

f^{\prime}(x) > g^{\prime}(x)

for all

x \in R

and

f(0)=g(0)

, then

View answer and explanation

Correct option: A

f(x) > g(x)

for all

x \in(0, \infty)

and

f(x) < g(x)

for all

x \in(-\infty, 0)

Option A (f(x) > g(x) for all x in (0, infinity) and f(x) < g(x) for all x in (-infinity, 0)) is correct because h(x) = f(x) - g(x) is increasing and h(0) = 0.

66 Easy

66 of 101

Analyzing +4 | -1

If the function

f(x)=3 \cos |x|-6ax+b

increases for all

x \in R

, then the range of values of

a

is given by

View answer and explanation

Correct option: B

(-\infty,-1 / 2)

Option B ((-∞,-1 / 2)) is correct because it ensures that the derivative of

f(x)

is always positive, making

f(x)

an increasing function for all

x \in R

67 Easy

67 of 101

Analyzing +4 | -1

Let

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

be a decreasing function on the set

R

. Then,

a

and

b

satisfy

View answer and explanation

Correct option: B

a^2+3 b+15 < 0

Option B (

a^2+3 b+15 < 0

) is correct because for the function to be decreasing, its derivative must be negative, leading to the given inequality.

68 Easy

68 of 101

Application +4 | -1

The function

f

defined by

f\left( x \right) = {x^3} - 6{x^2} - 36x + 7

is increasing, if

View answer and explanation

Correct option: D

x < - 2

and also

x > 6

Option D (

x < - 2

and also

x > 6

) is correct because it identifies the intervals where the derivative of the function is positive, indicating that the function is increasing.

69 Easy

69 of 101

Analyzing +4 | -1

For all

x \in(0,1)

View answer and explanation

Correct option: B

\log _e(1+x) < x

Option B (

\log _e(1+x) < x

) is correct because the derivative of

g(x)=\log _e(1+x)-x

is negative for

x \in (0,1)

, implying

g(x) < 0

70 Easy

70 of 101

Analyzing +4 | -1

For the function

f(x)=x \cos \frac{1}{x}, x \geq 1

, which one of the following is incorrect?

View answer and explanation

Correct option: A

For at least one

x

in the interval

[1, \infty), f(x+2)-f(x) < 2

The correct answer is A (For at least one

x

in the interval

[1, \infty), f(x+2)-f(x) < 2

) because

f(x+2) - f(x) > 2

for all

x

in the interval, making the statement that it's less than 2 for at least one x incorrect.

71 Easy

71 of 101

Analyzing +4 | -1

f(x)

and

g(x)

are two positive and increasing functions, then

View answer and explanation

Correct option: D

f(x)>1

, then

(f(x))^{g(x)}

is increasing

Option D is correct because if

f(x) > 1

, then

\log(f(x)) > 0

, and since

f(x)

and

g(x)

are positive and increasing,

(f(x))^{g(x)}

is also increasing.

72 Easy

72 of 101

Analyzing +4 | -1

The set of values of

a

for which the function

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

increases on

R

, is

View answer and explanation

Correct option: B

R-[-3,1]

Option B (

R-[-3,1]

) is correct because it represents the set of values of

a

for which the derivative of

f(x)

is always positive, ensuring that

f(x)

is increasing on

R

73 Easy

73 of 101

Analyzing +4 | -1

Let

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

Then

View answer and explanation

Correct option: A

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

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

Option A (

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

[\frac{{-1}}{{2}}, 1]

) is correct because the derivative is non-negative in the given interval, as shown by the sign analysis of

f'(x)

74 Easy

74 of 101

Analyzing +4 | -1

The function

f(x)=\tan x-4 x

is strictly decreasing on

View answer and explanation

Correct option: A

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

Option A ((-π/3, π/3)) is correct because the derivative of the function is negative within this interval, indicating a strictly decreasing function.

75 Easy

75 of 101

Analyzing +4 | -1

The function

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

View answer and explanation

Correct option: A

Strictly increasing

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

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76 Easy

76 of 101

Analyzing +4 | -1

f(x)=\left\{\begin{array}{ll}3 x^2+12 x-1, & -1 \leq x \leq 2 \\ 37-x, & 2 < x \leq 3\end{array}\right.

, then

View answer and explanation

Correct option: A

f(x)

is increasing on

[-1,2]

Option A (

f(x)

is increasing on

[-1,2]

) is correct because the derivative of

f(x)

is positive on this interval, indicating that

f(x)

is increasing.

77 Easy

77 of 101

Analyzing +4 | -1

Select the correct statement from (a), (b), (c), (d) The function

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

View answer and explanation

Correct option: D

Strictly decreasing in the interval

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

Option D is correct because the derivative of the function is negative in the interval (1, ∞), indicating that the function is strictly decreasing.

78 Easy

78 of 101

Application +4 | -1

If the function

f: R \rightarrow R

given by

f(x)=x^3+a x^2+5 x+\sin 2 x

is invertible, then

View answer and explanation

Correct option: B

a \in(-3,3)

Option B (a in (-3,3)) is correct because for the function to be invertible, its derivative must be strictly positive or strictly negative, leading to the condition -3 < a < 3.

79 Easy

79 of 101

Analyzing +4 | -1

Let

f^{\prime}(x) > 0

and

g^{\prime}(x) < 0

for all

x \in R

. Then,

View answer and explanation

Correct option: A

f\{g(x)\} > f\{g(x+1)\}

Option A (

f\{g(x)\} > f\{g(x+1)\}

) is correct because

f(x)

is an increasing function and

g(x)

is a decreasing function.

80 Easy

80 of 101

Application +4 | -1

f(x)

is a differentiable function satistying

f^{\prime}(x)<2

for all

x \in R, f(1)=2

, then greatest possible integral value of

f(3)

View answer and explanation

Correct option: C

Option C (5) is correct because it represents the greatest possible integral value of f(3) based on the given conditions and the Mean Value Theorem.

81 Easy

81 of 101

Analyzing +4 | -1

The function

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

is increasing on

View answer and explanation

Correct option: A

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

Option A

((0, \infty))

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

x > 0

, indicating an increasing function.

82 Easy

82 of 101

Analyzing +4 | -1

The function,

f(x) = (3x - 7){x^{2/3}},\,\,x \in R

, is increasing for all

x

lying in

View answer and explanation

Correct option: A

\left( { - \infty ,\,0} \right) \cup \left( {\frac{{14}}{{15}},\,\infty } \right)

Option A (

\left( { - \infty ,\,0} \right) \cup \left( {\frac{{14}}{{15}},\,\infty } \right)

) is correct because the function is increasing when its derivative is greater than or equal to zero, which occurs in the specified intervals.

83 Easy

83 of 101

Analyzing +4 | -1

The function

f(x)=\tan x-x

View answer and explanation

Correct option: A

always increases

Option A (always increases) is correct because the derivative of the function, f'(x) = sec²(x) - 1, is always non-negative, indicating that the function is always increasing.

84 Easy

84 of 101

Analyzing +4 | -1

The function

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

increases in the interval

View answer and explanation

Correct option: A

(0, 2)

Option A is correct because the derivative of the function, f'(x), is positive in the interval (0, 2), indicating that the function is increasing in that interval.

85 Easy

85 of 101

Application +4 | -1

f\left( x \right) = {x^3} - 6{x^2} + 9x + 3

be a decreasing function, then

x

lies in

View answer and explanation

Correct option: B

\left( {1,\;3} \right)

Option B (

\left( {1,\;3} \right)

) is correct because the function is decreasing when its derivative is negative, which occurs in this interval.

86 Easy

86 of 101

Analyzing +4 | -1

The value of

x

for which the polynomial

2{x^3} - 9{x^2} + 12x + 4

is a decreasing function of

x

, is

View answer and explanation

Correct option: D

1 < x < 2

Option D (

1 < x < 2

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

87 Easy

87 of 101

Analyzing +4 | -1

The function

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

View answer and explanation

Correct option: B

Decreases in

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

Option B (Decreases in

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

) is correct because the derivative of the function is negative for all x in the real numbers, indicating a decreasing function.

88 Easy

88 of 101

Analyzing +4 | -1

The function

f

defined by

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

in increasing for

View answer and explanation

Correct option: D

x > \frac{1}{2}

Option D (

x > \frac{1}{2}

) is correct because the function

f(x) = 4x^4 - 2x + 1

is increasing when its derivative

f'(x) = 16x^3 - 2

is positive, which occurs for

x > \frac{1}{2}

89 Easy

89 of 101

Analyzing +4 | -1

The function

{x^x}

is increasing, when

View answer and explanation

Correct option: A

x > \frac{1}{e}

Option A (

x > \frac{1}{e}

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

x

is greater than

\frac{1}{e}

90 Easy

90 of 101

Analyzing +4 | -1

f: R \rightarrow R

is the function defined by

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

, then

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the function is neither strictly increasing nor onto.

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