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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Medium

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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 31-60 of 92 questions
31 Medium
31 of 92
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?
View answer and explanation
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.
32 Medium
32 of 92
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.
33 Medium
33 of 92
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.
34 Medium
34 of 92
Analyzing +4 | -1
The function f(x)=2log⁡(x−2)−x2+4x+1f(x)=2 \log (x-2)-x^2+4 x+1
View answer and explanation
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.
35 Medium
35 of 92
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}.
36 Medium
36 of 92
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.
37 Medium
37 of 92
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.
38 Medium
38 of 92
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).
39 Medium
39 of 92
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.
40 Medium
40 of 92
Analyzing +4 | -1
If f:R→Rf: R \rightarrow R be definend by f(x)=2x+cos⁡xf(x)=2 x+\cos x, then ff
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.
41 Medium
41 of 92
Application +4 | -1
The function f(x)=2x3−3x2+90x+174f\left( x \right) = 2{x^3} - 3{x^2} + 90x + 174 is increasing in the interval
View answer and explanation
Correct option: D
−∞<x<∞- \infty < x < \infty
Option D (−∞<x<∞- \infty < x < \infty) is correct because the derivative is always positive, indicating an increasing function.
42 Medium
42 of 92
Analyzing +4 | -1
If f(x)=xex(1−x)f\left( x \right) = x{e^{x\left( {1 - x} \right)}}, then f(x)f\left( x \right) is
View answer and explanation
Correct option: A
Increasing on [−12,  1]\left[ { - \frac{1}{2},\;1} \right]
Option A (Increasing on [−12,  1]\left[ { - \frac{1}{2},\;1} \right]) is correct because the derivative of the function is positive in this interval, indicating an increasing function.
43 Medium
43 of 92
Analyzing +4 | -1
f(x)=4log⁡e(x−1)−2x2+4x+5,x>1f(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
ff is increasing in (1,2)(1,2) and decreasing in (2(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.
44 Medium
44 of 92
Analyzing +4 | -1
Let f(x)=∫0xcos⁡ttdtf\left( x \right) = \int \limits_0^x \frac{{\cos t}}{t}dt Then, at x=(2n+1)π2,  f(x)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,  −5n = - 1,\; - 3,\; - 5 ...and minima when n=1,  3,  5,…n = 1,\;3,\;5, \ldots
The correct answer is Maxima when n=−1,−3,−5n = - 1, - 3, - 5 ...and minima when n=1,3,5,...n = 1, 3, 5, ... because the sign of the second derivative determines whether the function has a maxima or minima at the given points.
45 Medium
45 of 92
Analyzing +4 | -1
Let f(x)f(x) and g(x)g(x) be defined and differentiable for all x≥x0x \geq x_0 and f(x0)=g(x0),f′(x)>g′(x)f\left(x_{0})=g\left(x_0\right), f^{\prime}(x) > g^{\prime}(x)\right. for x>x0x > x_0, then
View answer and explanation
Correct option: C
f(x)>g(x),x>x0f(x) > g(x), x > x_0
The correct answer is C (f(x)>g(x),x>x0f(x) > g(x), x > x_0) because since f′(x)>g′(x)f'(x) > g'(x) for x>x0x > x_0 and f(x0)=g(x0)f(x_0) = g(x_0), f(x)f(x) increases faster than g(x)g(x) after x0x_0.
46 Medium
46 of 92
Analyzing +4 | -1
Let g(x)=f(x)−2{f(x)}2+9{f(x)}3g\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∈R.x \in R. Then,
View answer and explanation
Correct option: A
g(x)g\left( x \right) and f(x)f\left( x \right) increases and decrease together
Option A (g(x)g\left( x \right) and f(x)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).
47 Medium
47 of 92
Analyzing +4 | -1
Let f:(0,1)→Rf:(0,1) \rightarrow R be a function defined by f(x)=11−e−xf(x)=\frac{1}{1-e^{-x}}, and g(x)=(f(−x)−f(x))g(x)=(f(-x)-f(x)). Consider two statements

(I) gg is an increasing function in (0,1)(0,1)

(II) gg is one-one in (0,1)(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.
48 Medium
48 of 92
Analyzing +4 | -1
Let ff be differentiable on R\mathbb{R} and

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

Then
View answer and explanation
Correct option: A
hh is increasing whenever ff 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.
49 Medium
49 of 92
Analyzing +4 | -1
The function ff defined by f(x)=x3−3x2+5x+7f(x)=x^3-3 x^2+5 x+7, is :
View answer and explanation
Correct option: A
increasing in R\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.
50 Medium
50 of 92
Application +4 | -1
The set of values of a for which the function
f(x)=2ex−ae−x+(2a+1)x−3f\left( x \right) = 2{e^x} - a{e^{ - x}} + \left( {2a + 1} \right)x - 3
is increasing on RR, is
View answer and explanation
Correct option: A
(0,  ∞)\left( {0,\;\infty } \right)
Option A ((0,  ∞)\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.
51 Medium
51 of 92
Analyzing +4 | -1
The function f(x)=2x2−log⁡∣x∣f(x)=2 x^2-\log |x| is
View answer and explanation
Correct option: B
decreasing in (0,12)\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.
52 Medium
52 of 92
Analyzing +4 | -1
If f(x)=∫x2x2+1e−t2dtf(x)=\int\limits_{{x^2}}^{x^2+1} e^{-t^2} d t, then f(x)f(x) increases in
View answer and explanation
Correct option: D
(−∞,0)(-\infty, 0)
Option D ((−∞,0))((-\infty, 0)) is correct because the derivative f′(x)f'(x) is positive only when x<0x < 0, indicating that f(x)f(x) is increasing in this interval.
53 Medium
53 of 92
Analyzing +4 | -1
If a,ba, b, cc be real, then f(x)=∣x+a2abacabx+b2bcacbcx+c2∣f(x)=\left|\begin{array}{ccc}x+a^2 & a b & a c \\ a b & x+b^2 & b c \\ a c & b c & x+c^2\end{array}\right|, is decreasing on
View answer and explanation
Correct option: A
(−23(a2+b2+c2),0)\left(-\frac{2}{3}\left(a^2+b^2+c^2\right), 0\right)
Option A ((−23(a2+b2+c2),0)\left(-\frac{{2}}{{3}}\left(a^2+b^2+c^2\right), 0\right)) is correct because the function f(x)f(x) is decreasing when its derivative is negative, which occurs on the given interval.
54 Medium
54 of 92
Analyzing +4 | -1
The function f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡xf(x)=\frac{4 x^3-3 x^2}{6}-2 \sin x+(2 x-1) \cos x is
View answer and explanation
Correct option: A
Increases in (1/2,∞)(1 / 2, \infty)
Option A is correct. The function increases in (1/2,∞)(1 / 2, \infty) because its derivative is positive in that interval.
55 Medium
55 of 92
Analyzing +4 | -1
If f(x)=(ab+b2+1)x+∫0x(cos⁡4θ+sin⁡4θ)dθf(x)=\left(a b+b^2+1\right) x+\int\limits_{0}^x\left(\cos ^4 \theta+\sin ^4 \theta\right) d \theta is an increasing function of xx for all x∈Rx \in R and b∈R,bb \in R, b being independent of xx, then
View answer and explanation
Correct option: B
a∈(−6,6)a \in(-\sqrt{6}, \sqrt{6})
Option B (a∈(−6,6)a \in(-\sqrt{6}, \sqrt{6})) is correct because it ensures that the derivative of f(x)f(x) is always positive, making f(x)f(x) an increasing function.
56 Medium
56 of 92
Analyzing +4 | -1
The function f(x)=x1/xf\left( x \right) = {x^{1/x}} is increasing in the interval
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the function f(x)=x1/xf(x) = {x^{1/x}} is increasing in the interval (0, e), which is not listed in the options.
57 Medium
57 of 92
Analyzing +4 | -1
If f(x)=x+sin⁡x,g(x)=e−x,u=c+1−cf(x)=x+\sin x, g(x)=e^{-x}, u=\sqrt{c+1}-\sqrt{c}, v=c−c−1,(c>1)v=\sqrt{c}-\sqrt{c-1},(c > 1), then
View answer and explanation
Correct option: C
gof(u)>g∘f(v)g o f(u) > g \circ f(v)
Option C (g o f(u) > g o f(v)) is correct because f is increasing, g is decreasing, and u < v, implying f(u) < f(v) and thus g(f(u)) > g(f(v)).
58 Medium
58 of 92
Analyzing +4 | -1
Suppose f(x)f(x) is a real-valued differentiable function such that f(x)f′(x)<0f(x) f^{\prime}(x)<0 for all real xx. Then
View answer and explanation
Correct option: D
∣f(x)∣|f(x)| is a decreasing function
Option D is correct because the condition f(x)f'(x) < 0 implies that |f(x)| is decreasing, as the square of the function is decreasing.
59 Medium
59 of 92
Application +4 | -1
The number of integral values of a for which f(x)=x3+(a+2)x2+3ax+5f(x)=x^{3}+(a+2) x^{2}+3 a x+5 is monotonic in ∀x∈R\forall x \in R.
View answer and explanation
Correct option: B
4
Option B (4) is correct because the function is monotonic when its derivative has a discriminant less than or equal to zero, leading to a range of [1, 4] for 'a', which includes 4 integers.
60 Medium
60 of 92
Analyzing +4 | -1
If f(x)f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1, 0, 1,x = - 1,\,0,\,1, then the set S={x∈R:f(x)=f(0)}S = \left\{ {x \in R:f(x) = f(0)} \right\} contains exactly
View answer and explanation
Correct option: D
two irrational and one rational number
The correct answer is D (two irrational and one rational number) because the polynomial equation f(x)=f(0)f(x) = f(0) has solutions x=0,±2x = 0, \pm \sqrt{2}, resulting in one rational and two irrational roots.
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