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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Hard

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Practice format and chapter quiz

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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1 Hard
1 of 19
Analyzing +4 | -1
If the total maximum value of the function f(x)=(3e2sin⁡x)sin⁡2x,x∈(0,π2),f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right), is ke\frac{k}{e}, then (ke)8+k8e5+k8\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8} is equal to
View answer and explanation
Correct option: A
e3+e6+e11e^{3}+e^{6}+e^{11}
Option A (e3+e6+e11e^{3}+e^{6}+e^{11}) is correct because the maximum value of the function is found to be e3/8e^{3/8}, leading to the given expression.
2 Hard
2 of 19
Analyzing +4 | -1
If f′(x)=(x−a)2n(x−b)2  p+1f'\left( x \right) = {\left( {x - a} \right)^{2n}}{\left( {x - b} \right)^{2\;p + 1}} where nn and pp are positive integers, then
View answer and explanation
Correct option: C
x=ax = a is not a point of maximum or minimum
Option C (x=ax = a is not a point of maximum or minimum) is correct because the sign of the derivative does not change as xx passes through aa.
3 Hard
3 of 19
Analyzing +4 | -1
A function ff is defined by f(x)=∣x∣m∣x−1∣n,∀x∈Rf(x)=|x|^m|x-1|^n, \forall x \in R. The maximum value of the function is (m,n∈N)(m, n \in N)
View answer and explanation
Correct option: C
mm⋅nn(m+n)m+n\frac{m^m \cdot n^n}{(m+n)^{m+n}}
Option C (mm⋅nn(m+n)m+n\frac{m^m \cdot n^n}{(m+n)^{m+n}}) is correct. By finding the critical points of the function and evaluating the function at those points, we find the maximum value to be mm⋅nn(m+n)m+n\frac{m^m \cdot n^n}{(m+n)^{m+n}}.
4 Hard
4 of 19
Analyzing +4 | -1
If a>b>0a > b > 0,then maximum value of ab(a2−b2)sin⁡xcos⁡xa2sin⁡2x+b2cos⁡2x,  x∈(0,  π/2)\frac{{ab\left( {{a^2} - {b^2}} \right)\sin x\cos x}}{{{a^2}{{\sin }^2}x + {b^2}{{\cos }^2}x}},\;x \in \left( {0,\;\pi /2} \right) is
View answer and explanation
Correct option: B
a2−b22\frac{{{a^2} - {b^2}}}{2}
Option B (a2−b22\frac{{{a^2} - {b^2}}}{2}) is correct because it represents the maximum value of the given expression, achieved when the denominator is minimized.
5 Hard
5 of 19
Analyzing +4 | -1
A population p(t)p\left( t \right) of 1000 bacteria introduced into nutrient medium grows according to the relation p(t)=1000+1000t100+t2p\left( t \right) = 1000 + \frac{{1000t}}{{100 + {t^2}}}. The maximum size of this bacterial population is
View answer and explanation
Correct option: C
1050
Option C (1050) is correct because it represents the maximum size of the bacterial population, found by analyzing the derivative of the population function.
6 Hard
6 of 19
Analyzing +4 | -1
The minimum & maximum value of f(x)=sin⁡(cos⁡x)+cos⁡(sin⁡x)f(x)=\sin (\cos x)+\cos (\sin x) for −π2≤x≤π2-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} are respectively
View answer and explanation
Correct option: A
cos⁡1\cos 1 and 1+sin⁡11+\sin 1
Option A, cos⁡1\cos 1 and 1+sin⁡11+\sin 1, is correct because the maximum of the function occurs at x=0 and the minimum occurs at x=π/2\pi/2 due to the decreasing nature of the function in the given interval.
7 Hard
7 of 19
Analyzing +4 | -1
The functionf(x)=ax+bx,  b,  x>0f\left( x \right) = ax + \frac{b}{x},\;b,\;x > 0 takes the least value at xx equal to
View answer and explanation
Correct option: D
ba\sqrt {\frac{b}{a}}
Option D is correct because the function has a minimum at x = b/a\sqrt{{b/a}}, which is found by setting the derivative equal to zero and confirming a positive second derivative.
8 Hard
8 of 19
Analyzing +4 | -1
The maximum value of f(x)=x4+x+x2f\left( x \right) = \frac{x}{{4 + x + {x^2}}} on [-1,1] is
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because it represents the maximum value of the function on the given interval, found by evaluating the function at critical points and endpoints.
9 Hard
9 of 19
Analyzing +4 | -1
The maximum value of (1x)2x2{\left( {\frac{1}{x}} \right)^{2{x^2}}}, is
View answer and explanation
Correct option: B
ee\sqrt[e]{e}
Option B (ee\sqrt[e]{e}) is correct because it represents the maximum value of the function, found by setting the derivative to zero and confirming it's a maximum.
10 Hard
10 of 19
Analyzing +4 | -1
Let f(x)=1+2x2+22x4  +...+210x20f\left( x \right) = 1 + 2{x^2} + {2^2}{x^4}\; + ... + {2^{10}}{x^{20}}. Then, f(x)f\left( x \right) has
View answer and explanation
Correct option: A
More than one minimum
Option A (More than one minimum) is correct because the derivative of the function has multiple roots, indicating multiple minima.
11 Hard
11 of 19
Analyzing +4 | -1
The greatest value of f(x)=(x+1)1/3−(x−1)1/3f\left( x \right) = {\left( {x + 1} \right)^{1/3}} - {\left( {x - 1} \right)^{1/3}} on [0,1]is
View answer and explanation
Correct option: C
2
Option C (2) is correct. By finding the critical point and evaluating the function at the endpoints and critical point, we find the greatest value is 2.
12 Hard
12 of 19
Understanding +4 | -1
If f(x)=∫x2x2dtlog⁡t,x>0f(x)=\int\limits_{{x^2}}^{x^2} \frac{d t}{\log t}, x>0, then
View answer and explanation
Correct option: D
f(x)f(x) is neither maximum nor minimum at x=1x=1
Option D (f(x)f(x) is neither maximum nor minimum at x=1x=1) is correct because the first derivative does not change sign at x=1.
13 Hard
13 of 19
Analyzing +4 | -1
The set of all values of the parameter aa for which the points of minimum of the function y=1+a2x−x3y = 1 + {a^2}x - {x^3} satisfy the inequality x2+x+2x2+5x+6≤0\frac{{{x^2} + x + 2}}{{{x^2} + 5x + 6}} \le 0, is
View answer and explanation
Correct option: C
(23,  33)\left( {2\sqrt 3 ,\;3\sqrt 3 } \right)
The correct answer is C because the minimum point of the function must satisfy the given inequality, leading to the interval (2√3, 3√3).
14 Hard
14 of 19
Application +4 | -1
The minimum value of px+qypx + qy when xy=r2xy = {r^2}, is
View answer and explanation
Correct option: A
2rpq2r\sqrt {pq}
Option A (2rpq2r\sqrt {pq}) is correct because it represents the minimum value obtained by minimizing the expression px+qypx + qy subject to the constraint xy=r2xy = r^2 using calculus.
15 Hard
15 of 19
Application +4 | -1
A cubic f(x)f\left( x \right) vanishes at x=−2x = - 2 and has relative minimum/maximum at x=−1 and x=13x = - 1\,and\,x = \frac{1}{3} such that ∫−11f(x)dx=143\int \limits_{ - 1}^1 f\left( x \right)dx = \frac{{14}}{3} Then, f(x)f\left( x \right) is
View answer and explanation
Correct option: C
x3+x2−x+2{x^3} + {x^2} - x + 2
Option D (x3+x2−x+2{x^3} + {x^2} - x + 2) is correct because it satisfies the given conditions: vanishes at x = -2, has relative extrema at x = -1 and x = 1/3, and the integral from -1 to 1 is 14/3.
16 Hard
16 of 19
Analyzing +4 | -1
A wire of length 20m20 \mathrm{m} to be cut into two pieces. A piece of length l1l_{1} is bent to make a square of area A1A_{1} and the other piece of length l2l_{2} is made into a circle of area A2A_{2}. If 2A1+3A22 A_{1}+3 A_{2} is minimum then (πl1):l2\left(\pi l_{1}\right): l_{2} is equal to
View answer and explanation
Correct option: C
6:16: 1
Option C is correct because the ratio of π l_1 to l_2 is calculated to be 6:1 when the expression 2A_1 + 3A_2 is minimized.
17 Hard
17 of 19
Analyzing +4 | -1
The maximum value of f(x)=x4+x+x2f\left( x \right) = \frac{x}{{4 + x + {x^2}}} on [-1, 1] is
View answer and explanation
Correct option: D
16\frac{1}{6}
Option D (16\frac{1}{6}) is correct because it is the maximum value of the function f(x)f(x) on the given interval [-1, 1].
18 Hard
18 of 19
Analyzing +4 | -1
If ax+bx≥cax + \frac{b}{x} \ge c for all positive values of xx and a,b,c are positive constants, then
View answer and explanation
Correct option: A
ab≥c24ab \ge \frac{{{c^2}}}{4}
Option A (ab≥c24ab \ge \frac{{{c^2}}}{4}) is correct because the minimum value of the function ax+bxax + \frac{b}{x} must be greater than or equal to c, leading to the inequality ab≥c24ab \ge \frac{{{c^2}}}{4}.
19 Hard
19 of 19
Analyzing +4 | -1
The function f(x)=2x3−15x2+36x+4f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 4 is maximum at
View answer and explanation
Correct option: A
x=2x = 2
Option A (x=2x = 2) is correct because the function's first derivative is zero and the second derivative is negative at this point, indicating a maximum.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 19 single-correct-answer MCQs on MAXIMA & MINIMA.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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