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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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61 Easy
61 of 200
Analyzing +4 | -1
Let λ∗{\lambda ^*} be the largest value of λ\lambda for which the function fλ(x)=4λx3−36λx2+36x+48{f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48 is increasing for all x∈R.x \in R. Then fλ∗(1)+fλ∗(−1){f_\lambda }^*(1) + {f_\lambda }^*( - 1) is equal to:
View answer and explanation
Correct option: D
72
Option D (72) is correct because after finding the largest value of λ\lambda for which the function is increasing, substituting λ∗\lambda^* into the function and evaluating at x=1 and x=-1 gives a sum of 72.
62 Medium
62 of 200
Analyzing +4 | -1
The length of the longest interval, in which the function 3sin⁡x−4sin⁡3x3\sin x - 4{\sin ^3}x is increasing, is
View answer and explanation
Correct option: A
π3\frac{\pi }{3}
Option A (π3\frac{\pi}{3}) is correct because the function can be simplified to sin⁡(3x)\sin(3x), which increases over an interval of length π3\frac{\pi}{3}.
63 Medium
63 of 200
Analyzing +4 | -1
If f(x)=3x4+4x3−12x2+12f\left( x \right) = 3{x^4} + 4{x^3} - 12{x^2} + 12, then f(x)f\left( x \right) is
View answer and explanation
Correct option: B
Increasing in (−2,0)\left( { - 2,\mathrm{ }0} \right) and in (1,  ∞)\left( {1,\;\infty } \right)
Option B is correct because the function is increasing where its derivative is positive, which occurs in the intervals (-2, 0) and (1, ∞).
64 Medium
64 of 200
Analyzing +4 | -1
Let f(x)=xcos⁡−1(sin⁡(−∣x∣),x∈(−π2,π2)f(x)=x \cos ^{-1}\left(\sin (-|x|), x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right. Then,
View answer and explanation
Correct option: D
f′(x)f^{\prime}(x) is decreasing in (−π2,0)\left(-\frac{\pi}{2}, 0\right) and increasing in (0,π2)\left(0, \frac{\pi}{2}\right)
Option D (f′(x)f^{\prime}(x) is decreasing in (−π2,0)\left(-\frac{\pi}{2}, 0\right) and increasing in (0,π2)\left(0, \frac{\pi}{2}\right)) is correct because the second derivative is negative for negative x and positive for positive x.
65 Easy
65 of 200
Analyzing +4 | -1
The function f(x)=x3−3xf(x)=x^3-3 x, is
View answer and explanation
Correct option: A
Increasing on (−∞,−1]∪[1,∞)(-\infty,-1] \cup[1, \infty) and decreasing on (−1,1)(-1,1)
Option A (Increasing on (−∞,−1]∪[1,∞)(-\infty,-1] \cup[1, \infty) and decreasing on (−1,1)(-1,1)) is correct because the derivative is positive on (−∞,−1]∪[1,∞)(-\infty,-1] \cup[1, \infty) and negative on (−1,1)(-1,1).
66 Easy
66 of 200
Application +4 | -1
If f(x)=x3−ax2+48x+7f(x)=x^{3}-a x^{2}+48 x+7 is increasing then a lies between
View answer and explanation
Correct option: A
±12\pm 12
Option A (±12\pm 12) is correct because for the function to be increasing, its derivative must be greater than 0, leading to the condition that a lies between -12 and 12.
67 Easy
67 of 200
Analyzing +4 | -1
Let f:(1,3)→Rf:(1,3) \rightarrow R be a function given by f(x)=x[x]x2+1f(x)=\frac{x[x]}{x^2+1}, where [⋅][\cdot] denotes the greatest integer function, then range of f(x)f(x) is
View answer and explanation
Correct option: B
(25,12)∪(35,45]\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]
Option B ((25,12)∪(35,45]\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]) is correct because it represents the range of the function considering the greatest integer function and the given interval.
68 Easy
68 of 200
Analyzing +4 | -1
Let f(x)f(x) be a function defined by
f(x)=(ab−a2−2)x−∫0x(cos⁡4t+sin⁡2t−2)dtf(x)=\left(a b-a^2-2\right) x-\int\limits_{0}^x\left(\cos ^4 t+\sin ^2 t-2\right) d t

If f(x)f(x) is a decreasing function for all x∈Rx \in R and a∈Ra \in R, where aa is independent of xx, then
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the condition for f(x)f(x) to be decreasing leads to b∈(−3,3)b \in(-\sqrt{3}, \sqrt{3}), which is not covered by the other options.
69 Medium
69 of 200
Analyzing +4 | -1
Let f(x)=cos⁡x−1+(x2/2)f(x)=\cos x-1+\left(x^2 / 2\right). Then
View answer and explanation
Correct option: D
ff decreases in (−∞,0](-\infty, 0] and increases in [0,∞)[0, \infty)
Option D is correct because the first derivative is negative for x < 0 and positive for x > 0, indicating decreasing and increasing behavior respectively.
70 Medium
70 of 200
Analyzing +4 | -1
How many real solutions does the equation x7+14x5+16x3+30x−560=0{x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0 have?
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Correct option: C
1
Option C (1) is correct because the function's derivative is always positive, indicating it's strictly increasing and thus has only one real root.
71 Medium
71 of 200
Analyzing +4 | -1
(1+x)n≤1+xn,{\left( {1 + x} \right)^n} \le 1 + {x^n}, where
View answer and explanation
Correct option: B
0≤n≤1 and x>00 \le n \le 1\,\mathrm{and}\,x > 0
Option B (0≤n≤1 and x>00 \le n \le 1\,\mathrm{and}\,x > 0) is correct because it satisfies the inequality based on the properties of the function's derivative.
72 Medium
72 of 200
Analyzing +4 | -1
Consider the following two statements.

I: If f(x)f(x) and g(x)g(x) are continuous and monotonic functions on the real line R\mathbb{R}, then f(x)+g(x)f(x)+g(x) is also a monotonic function.

II: If f(x)f(x) is a continuous decreasing function for all x>0x>0 and f(1)f(1) is positive, then f(x)=0f(x)=0 exactly at one value of xx.
Then
View answer and explanation
Correct option: B
both I and II are false
Option B is correct because both statements I and II are false, as demonstrated by counterexamples.
73 Medium
73 of 200
Analyzing +4 | -1
The function f(x)=2log⁡(x−2)−x2+4x+1f(x)=2 \log (x-2)-x^2+4 x+1
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Correct option: A
increases in (2,3)(2,3)
Option A is correct because the function's derivative is positive in the interval (2, 3), indicating an increasing function.
74 Easy
74 of 200
Analyzing +4 | -1
The function f(x)=xxf(x)=x^x decreases on the interval
View answer and explanation
Correct option: C
(0,1/e)(0,1 / e)
Option C ((0,1/e)(0,1 / e)) is correct because the function f(x)=xxf(x)=x^x decreases when its derivative is negative, which occurs on the interval (0,1/e)(0,1 / e).
75 Easy
75 of 200
Analyzing +4 | -1
Let f :(a,b) →Rf\,:(a,b)\, \to R be twice differentiable function such that f(x) = ∫axg(t) dtf(x)\, = \,\int_a^x {g(t)\,dt} for a differentiable function g(x)g(x). If f(x) = 0f(x)\, = \,0 has exactly five distinct roots in (a,b)(a,b), then g(x) g′(x) = 0g(x)\,g'(x)\, = \,0 has at least
View answer and explanation
Correct option: C
seven roots in (a,b)(a,b)
Option C (seven roots in (a,b)(a,b)) is correct because g(x)g′(x)=0g(x)g'(x) = 0 has at least 7 roots, derived from the properties of f(x)f(x) and its derivatives.
76 Easy
76 of 200
Analyzing +4 | -1
If g(x)g(x) is a continuous function at x=ax=a such that g(a)>0g(a) > 0 and f′(x)=g(x)(x2−ax+a2)f^{\prime}(x)=g(x)\left(x^2-a x+a^2\right) for all x∈Rx \in R, then f(x)f(x), is
View answer and explanation
Correct option: A
Increasing in the neighbourhood of x=ax=a
Option A (Increasing in the neighbourhood of x=ax=a) is correct because f′(x)>0f'(x) > 0 in the neighborhood of x=ax=a, indicating that f(x)f(x) is increasing.
77 Easy
77 of 200
Analyzing +4 | -1
Let f(x)f(x) and g(x)g(x) be increasing and decreasing functions respectively from [0,∞)[0, \infty) to [0,∞)[0, \infty), Let h(x)=fog(x)h(x)=fog(x). If h(0)=0h(0)=0, then h(x)h(x), is
View answer and explanation
Correct option: A
Always 0
Always 0. Since h(x)=f(g(x))h(x) = f(g(x)), ff is increasing, and gg is decreasing, h(x)h(x) is decreasing. Given h(0)=0h(0) = 0 and h(x)h(x) maps to [0,∞)[0, \infty), h(x)h(x) must always be 0.
78 Easy
78 of 200
Analyzing +4 | -1
f(x)=(e2x−1e2x+1)f(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right) is
View answer and explanation
Correct option: A
an increasing function
Option A (an increasing function) is correct because the derivative of the function is positive for all x in R, indicating an increasing function.
79 Easy
79 of 200
Analyzing +4 | -1
Consider the polynomial f(x)=1+2x+3x2+4x3.f(x)=1+2 x+3 x^2+4 x^3. Let ss be the sum of all distinct real roots of f(x)f(x). Then ss lies in the interval
View answer and explanation
Correct option: C
(−3/4,−1/2)(-3 / 4,-1 / 2)
The correct answer is C ((-3 / 4,-1 / 2)) because f(x)f(x) has a real root in this interval since f(−3/4)<0f(-3/4) < 0 and f(−1/2)>0f(-1/2) > 0, and f(x)f(x) is strictly increasing.
80 Medium
80 of 200
Application +4 | -1
Set of all possible values of kk for which f(x)=sin⁡x−cos⁡x−kx+bf(x)=\sin x-\cos x-k x+b decreases for all real values of xx, is:
View answer and explanation
Correct option: C
[2,∞)[\sqrt{2}, \infty)
Option C ([2,∞)[\sqrt{2}, \infty)) is correct because for f(x) to be decreasing, its derivative must be less than or equal to zero for all x, which leads to k≥2k \geq \sqrt{2}.
81 Medium
81 of 200
Analyzing +4 | -1
In [0,1],f[0,1], f is twice differentiable and f′′(x)<0f^{\prime \prime}(x)<0. Let

ϕ(x)=f(x)+f(1−x)\phi(x)=f(x)+f(1-x)
for x∈[0,1]x \in[0,1]. Then
View answer and explanation
Correct option: D
ϕ\phi decreases on [1/2,1][1 / 2,1]
Option D is correct because the derivative of φ(x) is negative in the interval [1/2, 1], indicating a decreasing function.
82 Easy
82 of 200
Analyzing +4 | -1
The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t)dt−(x2−x+1)ex,x∈Rf(x)=\int\limits_0^x e^{x-t} f^{\prime}(t) \mathrm{dt}-\left(x^2-x+1\right) e^x, x \in \mathbb{R}, is
View answer and explanation
Correct option: A
−2e-\frac{2}{\sqrt{e}}
Option A (-2e\frac{{2}}{{\sqrt{{e}}}}) is correct because it represents the minimum value of the function f(x) found by analyzing its derivatives.
83 Medium
83 of 200
Analyzing +4 | -1
Let f:[0,2]→Rf:[0,2] \rightarrow R be a twice differentiable function such that f′′(x)>0f^{\prime \prime}(x)>0, for all x∈(0,2)x \in(0,2).If ϕ(x)=f(x)+f(2−x)\phi(x)=f(x)+f(2-x), then ϕ\phi is
View answer and explanation
Correct option: C
Decreasing on (0,1)(0,1) and increasing on (1,2)(1,2)
Option D (If assertion is incorrect, but reason is correct) is correct because the function f(x) is differentiable at x=1, -1, and 0, making the assertion false, while the reason about the differentiability of |x| and |x2-1| is true.
84 Easy
84 of 200
Application +4 | -1
Let f(x)=x3+6x2+ax+2f(x)=x^{3}+6 x^{2}+a x+2, if (−3,−1)(-3,-1) is the largest possible interval for which f(x)f(x) is decreasing function, then a=a=
View answer and explanation
Correct option: B
9
Option B (9) is correct because using the derivative and the given interval, we can determine the value of 'a'.
85 Medium
85 of 200
Analyzing +4 | -1
If f(x)=∫x2x2+1e−t2dt,f\left( x \right) = \int \nolimits_{{x^2}}^{{x^2} + 1} {e^{ - {t^2}}}dt, then f(x)f\left( x \right) increases on
View answer and explanation
Correct option: C
(−∞,0)\left( { - \infty ,0} \right)
Option C is correct because the function f(x)f(x) increases where its derivative f′(x)f'(x) is positive, which occurs for x∈(−∞,0)x \in (-\infty, 0).
86 Easy
86 of 200
Analyzing +4 | -1
The maximum slope of the curve y=12x4−5x3+18x2−19xy = \frac{1}{2}{x^4} - 5{x^3} + 18{x^2} - 19x occurs at the point
View answer and explanation
Correct option: A
(2, 2)
Option A (2, 2) is correct. The maximum slope occurs where the second derivative of the curve is zero and the third derivative is negative, which happens at x=2. Substituting x=2 into the original equation gives y=2.
87 Medium
87 of 200
Analyzing +4 | -1
The function f(x)=tan⁡x−xf\left( x \right) = \tan x - x
View answer and explanation
Correct option: A
Always increases
Option A is correct because the derivative of the function is always non-negative, indicating that the function is always increasing.
88 Easy
88 of 200
Analyzing +4 | -1
If f(x)=x3+bx2+cx+df(x)=x^3+b x^2+c x+d and 0<b2<c0 < b^2 < c, then in (−∞,∞)(-\infty, \infty)
View answer and explanation
Correct option: A
f(x)f(x) is strictly increasing function
The correct answer is A (f(x)f(x) is strictly increasing function) because the derivative f′(x)f'(x) is always positive, indicating that f(x)f(x) is strictly increasing.
89 Easy
89 of 200
Analyzing +4 | -1
Let the function f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)=\tan ^{-1}(\sin x+\cos x) be defined on [0,2π][0,2 \pi]. Then, f(x)f(x) is
View answer and explanation
Correct option: A
Increasing on [0,π/4)∪(5π/4,2π][0, \pi / 4) \cup(5 \pi / 4,2 \pi]
Option A is correct. The function is increasing where its derivative is positive, which occurs on the intervals [0,π/4)∪(5π/4,2π][0, \pi / 4) \cup(5 \pi / 4,2 \pi].
90 Easy
90 of 200
Application +4 | -1
If f(x)=k  x3−9  x2+9  x+3f\left( x \right) = k\;{x^3} - 9\;{x^2} + 9\;x + 3 is increasing on RR, then
View answer and explanation
Correct option: B
k>3k > 3
Option B (k>3k > 3) is correct because for the function to be increasing on R, its derivative must be greater than 0, leading to the condition k > 3.
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