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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Easy

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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 1-30 of 101 questions
1 Easy
1 of 101
Application +4 | -1
The value of aa in order that f(x)=f(x)= sin⁡x−cos⁡x−ax+b\sin x-\cos x-a x+b decreases for all real values is given by
View answer and explanation
Correct option: A
a≥2a \geq \sqrt{2}
Option A (a≥2a \geq \sqrt{2}) is correct because it ensures that the derivative of the function is always negative, indicating that the function is decreasing for all real values.
2 Easy
2 of 101
Analyzing +4 | -1
The function f(x)=xex(1−x),x∈Rf(x)=x e^{x(1-x)}, x \in R, is
View answer and explanation
Correct option: A
increasing in (−12,1)\left(-\frac{1}{2}, 1\right)
Option A (increasing in (-1/2, 1)) is correct because the derivative of the function is positive in this interval, indicating that the function is increasing.
3 Easy
3 of 101
Analyzing +4 | -1
If ff and gg are two increasing functions such that fogfog is defined, then
View answer and explanation
Correct option: A
gofgof is an increasing function
Option A (gofgof is an increasing function) is correct because the composition of two increasing functions is also an increasing function.
4 Easy
4 of 101
Analyzing +4 | -1
Let g: (0,∞)→Rg:\,(0,\infty ) \to R be a differentiable function such that
∫(x(cos⁡x−sin⁡x)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=xg(x)ex+1+c\int {\left( {\frac{{x(\cos x - \sin x)}}{{{e^x} + 1}} + \frac{{g(x)\left( {{e^x} + 1 - x{e^x}} \right)}}{{{{\left( {{e^x} + 1} \right)}^2}}}} \right)} dx = \frac{{xg(x)}}{{{e^x} + 1}} + c, for all x>0,x > 0, where cc is an arbitrary constant. Then:
View answer and explanation
Correct option: D
g−g′g - g' is increasing in (0,π2)\left( {0,\frac{\pi }{2}} \right)
Option D (g−g′g - g' is increasing in (0,π2)\left( {0,\frac{\pi }{2}} \right)) is correct because by differentiating the given integral equation and analyzing the sign of J′(x)=g′(x)−g′′(x)J'(x) = g'(x) - g''(x), it can be shown that g−g′g - g' is increasing in the specified interval.
5 Easy
5 of 101
Analyzing +4 | -1
The function f(x)=sin⁡4x+cos⁡4xf(x)=\sin ^4 x+\cos ^4 x increasing, if
View answer and explanation
Correct option: B
π/4<x<3π/8\pi /4 < x < 3\pi /8
Option B (π/4<x<3π/8\pi /4 < x < 3\pi /8) is correct because the derivative of the function is positive in this interval, indicating an increasing function.
6 Easy
6 of 101
Analyzing +4 | -1
If a<0,f(x)=eax+e−axa < 0, f(x)=e^{a x}+e^{-a x} and S={x:f(x)S=\{x: f(x) is monotonically increasing}\}, then SS equals
View answer and explanation
Correct option: A
{x:x>0}\{x: x > 0\}
The correct answer is A ({x: x > 0}) because the condition for f(x)f(x) to be monotonically increasing, given a<0a < 0, leads to x>0x > 0.
7 Easy
7 of 101
Understanding +4 | -1
For the every value of xx the function f(x)=15xf(x)=\frac{1}{5^x} is
View answer and explanation
Correct option: A
Decreasing
Option A (Decreasing) is correct because the derivative of the function is negative for all values of x, indicating a decreasing function.
8 Easy
8 of 101
Analyzing +4 | -1
The function y=x3−3  x2+6  x−17y = {x^3} - 3\;{x^2} + 6\;x - 17
View answer and explanation
Correct option: A
Increases everywhere
Option A (Increases everywhere) is correct because the derivative of the function is always positive, indicating an increasing function.
9 Easy
9 of 101
Application +4 | -1
Let y=x2e−x,y = {x^2}{e^{ - x}}, then the interval in which y increases with respect to xx is
View answer and explanation
Correct option: D
(0,  2)\left( {0,\;2} \right)
Option D ((0, 2)) is correct because it represents the interval where the derivative of the function is positive, indicating that the function is increasing.
10 Easy
10 of 101
Application +4 | -1
The value of bb for which the function f(x)=sin⁡x−bx+cf\left( x \right) = \sin x - bx + c is decreasing in the interval (−∞,  ∞)\left( { - \infty ,\;\infty } \right) is given by
View answer and explanation
Correct option: C
b>1b > 1
Option C (b>1b > 1) is correct because for the function to be decreasing, its derivative must be negative for all x, implying b>cos⁡xb > \cos x for all x, hence b>1b > 1.
11 Easy
11 of 101
Analyzing +4 | -1
Let f(x)=sin⁡4x+cos⁡4xf(x)=\sin ^4 x+\cos ^4 x. Then ff is increasing function in the interval
View answer and explanation
Correct option: A
(π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right)
Option A ((π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right)) is correct because the derivative of the function is positive in this interval, indicating an increasing function.
12 Easy
12 of 101
Analyzing +4 | -1
If f(x)=sin⁡x,x∈[−π/2,π/2],f(x)=\sin x, x \in[-\pi / 2, \pi / 2], then which one of the following is not correct?
View answer and explanation
Correct option: C
fof(x)fof(x) is decreasing on [−π/2,0][-\pi / 2,0] and is increasing on [0,π/2][0, \pi / 2]
The correct answer is C because fof(x)fof(x) is increasing on the entire interval [−π/2,π/2][-\pi / 2, \pi / 2], not decreasing on [−π/2,0][-\pi / 2,0] and increasing on [0,π/2][0, \pi / 2].
13 Easy
13 of 101
Analyzing +4 | -1
The function f(x)=sin⁡xxf(x)=\frac{\sin x}{x} is decreasing in the interval
View answer and explanation
Correct option: B
(0,π/2)(0, \pi / 2)
Option B ((0,π/2)(0, \pi / 2)) is correct because the derivative of the function is negative in this interval, indicating a decreasing function.
14 Easy
14 of 101
Analyzing +4 | -1
Let the function g:(−∞,∞)→(−π/2,π/2)g:(-\infty, \infty) \rightarrow(-\pi / 2, \pi / 2) be given by g(u)=2tan⁡−1(eu)−π2g(u)=2 \tan ^{-1}\left(e^{u }\right)-\frac{\pi}{2}. Then, gg is
View answer and explanation
Correct option: C
Odd and is strictly increasing in (−∞,∞)(-\infty, \infty)
The correct answer is C (Odd and is strictly increasing in (−∞,∞)(-\infty, \infty)) because the derivative g′(u)g'(u) is always positive, indicating that g(u)g(u) is strictly increasing, and g(−u)=−g(u)g(-u) = -g(u), showing that g(u)g(u) is odd.
15 Easy
15 of 101
Application +4 | -1
The value of xx for which the polynomial 2x3−9x2+12x+42{x^3} - 9{x^2} + 12x + 4 is a decreasing function of xx, is
View answer and explanation
Correct option: D
1<x<21 < x < 2
Option D (1<x<21 < x < 2) is correct because the function is decreasing when its derivative is negative, which occurs in the interval 1<x<21 < x < 2.

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16 Easy
16 of 101
Understanding +4 | -1
Which of the following function is decreasing on (0,π2)\left(0, \frac{\pi}{2}\right) ?
View answer and explanation
Correct option: C
cos⁡x\cos x
Option C (cos x) is correct because its derivative, -sin x, is negative in the interval (0, π/2), indicating a decreasing function.
17 Easy
17 of 101
Analyzing +4 | -1
If f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡xf\left( x \right) = \frac{{a\sin x + b\cos x}}{{c\sin x + d\cos x}} is decreasing for all xx, then
View answer and explanation
Correct option: B
ad−bc<0ad - bc < 0
Option B (ad−bc<0ad - bc < 0) is correct because for the function to be decreasing, its derivative must be negative, leading to this condition on the coefficients.
18 Easy
18 of 101
Analyzing +4 | -1
If the function f(x)=cos⁡∣x∣−2  ax+bf\left( x \right) = \cos \left| x \right| - 2\;ax + b increases along the entire number scale, the range of a is given by
View answer and explanation
Correct option: C
a<−12a < - \frac{1}{2}
Option C (a<−12a < - \frac{1}{2}) is correct because it ensures that the derivative of the function f(x)f(x) is always positive, making the function increasing along the entire number scale.
19 Easy
19 of 101
Analyzing +4 | -1
If f(x)=xsin⁡xf(x)=\frac{x}{\sin x} and g(x)=xtan⁡xg(x)=\frac{x}{\tan x}, where 0<x≤10 < x \leq 1, then in this interval
View answer and explanation
Correct option: C
f(x)f(x) is an increasing function
Option C is correct. f(x)f(x) is an increasing function because its derivative is positive in the interval (0,1](0, 1].
20 Easy
20 of 101
Application +4 | -1
The interval of increase of the function F(x)=x−ex+tan⁡(2π7)F\left( x \right) = x - {e^x} + \tan \left( {\frac{{2\pi }}{7}} \right) is
View answer and explanation
Correct option: B
(−∞,  0)\left( { - \infty ,\;0} \right)
Option B (π2\frac{\pi }{2}) is correct because it is the value of c within the given interval that satisfies Lagrange's Mean Value Theorem.
21 Easy
21 of 101
Analyzing +4 | -1
f(x)=∣xlog⁡ex∣,x>0f(x)=\left|x \log _e x\right|, x > 0, is monotonically decreasing in
View answer and explanation
Correct option: C
[1/e,1][1 / e, 1]
Option C [1/e,1][1 / e, 1] is correct because the function's derivative is negative in this interval, indicating that the function is monotonically decreasing.
22 Easy
22 of 101
Application +4 | -1
The value of a in order that f(x)=sin⁡x−cos⁡x−ax+bf\left( x \right) = \sin x - \cos x - ax + b decreases for all real values of xx is given by
View answer and explanation
Correct option: A
a≥2a \ge \sqrt 2
Option A (a≥2a \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.
23 Easy
23 of 101
Understanding +4 | -1
If f′(x)f^{\prime}(x) is zero in the interval (a,b)(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.
24 Easy
24 of 101
Understanding +4 | -1
Let f(x)=tan⁡−1(g(x))f(x)=\tan ^{-1}(g(x)), where g(x)g(x) is monotonically increasing for 0<x<π20 < x < \frac{\pi}{2}. Then, f(x)f(x), is
View answer and explanation
Correct option: A
Increasing on (0,π/2)(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.
25 Easy
25 of 101
Analyzing +4 | -1
The function f(x)=xex(1−x),x∈Rf(x)=x e^{x(1-x)}, x \in \mathbb{R}, is
View answer and explanation
Correct option: A
Increasing in (−12,1)\left(-\frac{1}{2}, 1\right)
Option A (Increasing in (−12,1)\left(-\frac{1}{2}, 1\right)) is correct because the derivative f′(x)f'(x) is positive in this interval, indicating that the function is increasing.
26 Easy
26 of 101
Analyzing +4 | -1
Let g(x)=2f(x2)+f(2−x)g(x)=2 f\left(\frac{x}{2}\right)+f(2-x) and f′′(x)<0f^{\prime \prime}(x) < 0 for all x∈(0,2)x \in(0,2). Then, g(x)g(x), is
View answer and explanation
Correct option: C
Increasing on (0,4/3)(0,4 / 3) and decreasing on (4,3,2)(4,3,2)
Option C (Increasing on (0,4/3)(0,4 / 3) and decreasing on (4,3,2)(4,3,2)) is correct because g′(x)g'(x) is positive on (0,4/3)(0, 4/3) and negative on (4/3,2)(4/3, 2), indicating increasing and decreasing behavior respectively.
27 Easy
27 of 101
Application +4 | -1
The function f(x)=eax+eax,a>0f(x)=e^{a x}+e^{a x}, a > 0 is monotonically increasing for
View answer and explanation
Correct option: D
x>0x > 0
Option D (x>0x > 0) is correct because the derivative of the function is positive for x>0x > 0, indicating that the function is monotonically increasing in that interval.
28 Easy
28 of 101
Analyzing +4 | -1
Let f(x)={−55x,x<−52x3−3x2−120x,−5≤x<42x3−3x2−36x+10,x≥4f(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)f(x) is monotonically increasing, is
View answer and explanation
Correct option: A
(−5,−4)∪(4,∞)(-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.
29 Easy
29 of 101
Analyzing +4 | -1
If f(x)=∫xx21(log⁡t)2dt,x≠0,x≠1f(x)=\int\limits_{x}^{x^2} \frac{1}{(\log t)^2} d t, x \neq 0, x \neq \mathbb{1} then f(x)f(x) is monotomically
View answer and explanation
Correct option: A
Increasing on (2,∞)(2, \infty)
The correct answer is A (Increasing on (2,∞)(2, \infty)) because the derivative f′(x)f'(x) is positive for x>2x > 2, indicating that f(x)f(x) is increasing on that interval.
30 Easy
30 of 101
Analyzing +4 | -1
If f(x)=14x2+2x+1f\left( x \right) = \frac{1}{{4{x^2} + 2x + 1}}, then its maximum value is
View answer and explanation
Correct option: A
43\frac{4}{3}
Option A (43\frac{4}{3}) is correct because it represents the maximum value of the function, found by taking the derivative and setting it to zero.
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