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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Mixed difficulty

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Showing 1-30 of 174 questions
1 Medium
1 of 174
Analyzing +4 | -1
Let f(x)=(1+b2)x2+2bx+1f(x)=\left(1+b^2\right) x^2+2 b x+1. Let m(b)m(b) be the minimum value of f(x)f(x) (that is least value). As bb varies, the range of m(b)m(b) is
View answer and explanation
Correct option: D
(0,1](0,1]
Option D ((0,1]) is correct because the range of the minimum value function m(b) is calculated using calculus and analysis of the function's behavior.
2 Easy
2 of 174
Analyzing +4 | -1
Let f:R→Rf: R \rightarrow R be a function defined by f(x)=(x−3)n1(x−5)n2,n1,n2∈Nf(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N. Then, which of the following is NOT true?
View answer and explanation
Correct option: C
For n1=3,n2=5n_1=3, n_2=5, there exists α∈(3,5)\alpha \in(3,5) where ff attains local maxima.
The correct answer is C. For n1=3,n2=5n_1=3, n_2=5, there exists α∈(3,5)\alpha \in(3,5) where ff attains local maxima is NOT true because with both powers odd, there will be a local minimum.
3 Medium
3 of 174
Analyzing +4 | -1
The equation ∣2ax−3∣+∣ax+1∣+∣5−ax∣=12|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2} possesses
View answer and explanation
Correct option: D
no real solutions for all a∈Ra \in R
Option D (no real solutions for all a∈Ra \in R) is correct. The inequality derived from the triangle inequality shows that the equation has no real solutions for any value of aa.
4 Medium
4 of 174
Analyzing +4 | -1
Let the function f(x)=2x3+(2p−7)x2+3(2p−9)xf(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x -6 have a maxima for some value of x<0x<0 and a minima for some value of x>0x>0. Then, the set of all values of pp is
View answer and explanation
Correct option: D
(−∞,92)\left(-\infty, \frac{9}{2}\right)
Option D ((−∞,92)\left(-\infty, \frac{9}{2}\right)) is correct because the condition for maxima at x < 0 and minima at x > 0 implies that f′(0)<0f'(0) < 0, leading to p<92p < \frac{9}{2}.
5 Medium
5 of 174
Analyzing +4 | -1
The least value of the f(x)f\left( x \right) given by f(x)=tan⁡−1x−12log⁡exf\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x in the interval [1/3,  3]\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right], is
View answer and explanation
Correct option: B
π3−14log⁡e3\frac{\pi }{3} - \frac{1}{4}{\log _e}3
Option B (π3−14log⁡e3\frac{\pi }{3} - \frac{1}{4}{\log _e}3) is correct because it represents the least value of the function within the specified interval, determined by evaluating the function at critical points and endpoints.
6 Hard
6 of 174
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.
7 Easy
7 of 174
Application +4 | -1
The sum of two numbers is 20. If the product of the square of one number and cube of the other is maximum, then the numbers are
View answer and explanation
Correct option: A
12, 8
Option A (12, 8) is correct because these numbers maximize the product of the square of one number and the cube of the other, given their sum is 20.
8 Medium
8 of 174
Analyzing +4 | -1
Let f(x)=∣x2−5x+6∣,x∈[0,2.4]f(x)=\left|x^2-5 x+6\right|, x \in[0,2.4]. Then, the sum of the maximum and minimum values of ff on [0[0, 2.4] is
View answer and explanation
Correct option: A
6
Option A (6) is correct because the sum of the maximum and minimum values of f on [0, 2.4] is 6.
9 Easy
9 of 174
Analyzing +4 | -1
The coordinates of the point on the curve x3=y(x−a)2,a>0x^3=y(x-a)^2, a>0, where the ordinate is minimum
View answer and explanation
Correct option: C
(3a,27a4)\left(3 a, \frac{27 a}{4}\right)
Option C (3a,27a4)\left(3 a, \frac{{27 a}}{{4}}\right) is correct because it represents the point where the first derivative is zero and the second derivative is positive, indicating a minimum ordinate.
10 Medium
10 of 174
Analyzing +4 | -1
Absolute value of difference of maximum and minimum values of x3−18x2+96x{x^3} - 18{x^2} + 96x in interval (0,9)(0, 9) is _____
View answer and explanation
Correct option: A
32
Option A is correct because the absolute difference between the maximum and minimum values of the function in the given interval is 32.
11 Medium
11 of 174
Analyzing +4 | -1
The maximum value of P=2a2+1−2b2+1+3c2+1P=\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1} is ______ where a,b,c>0{a}, {b}, {c}>0 and abc+a+c=ba b c+a+c=b
View answer and explanation
Correct option: D
103\frac{10}{3}
Option D is correct because the maximum value is found by using trigonometric substitutions and maximizing the resulting expression using calculus or completing the square.
12 Medium
12 of 174
Analyzing +4 | -1
The greatest value of the function
f(x)=xax2+bf(x)=\frac{x}{a x^2+b}
is
View answer and explanation
Correct option: B
12ab\frac{1}{2 \sqrt{a b}}
Option B (12ab\frac{1}{2 \sqrt{a b}}) is correct because it represents the maximum value of the function, found by using calculus to determine the critical points.
13 Medium
13 of 174
Application +4 | -1
Let x,y,zx, y, z be positive numbers such that
x+y+z=26 and y=3xx+y+z=26 \text { and } y=3 x
If x2+y2+z2x^2+y^2+z^2 is to be least, then the value of xx is
View answer and explanation
Correct option: B
4
The correct answer is B (4) because the minimum value of x2+y2+z2x^2 + y^2 + z^2 occurs when x=4x = 4, which is found by minimizing the function using calculus.
14 Medium
14 of 174
Analyzing +4 | -1
If a differentiable function f(x)f\left( x \right) has a relative minimum at x=0x = 0, then the function ϕ(x)=f(x)+ax+b\phi \left( x \right) = f\left( x \right) + ax + b has a relative minimum at x=0x = 0 for
View answer and explanation
Correct option: B
All b if a=0a = 0
Option B (All b if a = 0) is correct because if a=0, the derivative of φ(x) at x=0 is determined by f'(x), which has a relative minimum at x=0.
15 Medium
15 of 174
Analyzing +4 | -1
Let f:R→Rf: \mathrm{R} \rightarrow \mathrm{R} be a continuous function defined by
f(x)=1ex+2e−xf(x)=\frac{1}{e ^ x+2 e ^ {-x}}
Statement -1: f(c)=13f(c)=\frac{1}{3}, for some c∈Rc \in \mathrm{R}.
Statement -2:0 < f(x) ≤122\leq \frac{1}{2 \sqrt{2}}, for all x∈Rx \in \mathrm{R}
View answer and explanation
Correct option: D
Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1 .
Statement - 1 is true, Statement - 2 is true, and Statement -2 is a correct explanation for Statement -1. The maximum value of f(x) is 1/(2*sqrt(2)), and since 1/3 is less than this maximum, f(c) = 1/3 for some c.
16 Easy
16 of 174
Analyzing +4 | -1
On the interval [5π4,4π3]\left[\frac{5 \pi}{4}, \frac{4 \pi}{3}\right], the least value of the function f(x)=∫5π/4x(3sin⁡t+4cos⁡t)dtf(x)=\int_{5 \pi / 4}^x(3 \sin t+4 \cos t) d t is
View answer and explanation
Correct option: A
32+12−23\frac{3}{2}+\frac{1}{\sqrt{2}}-2 \sqrt{3}
Option A is correct because the function f(x)f(x) is decreasing on the given interval, so the least value occurs at the right endpoint, x=4π3x = \frac{4\pi}{3}.
17 Easy
17 of 174
Analyzing +4 | -1
If ∣a2∣>∣a1+a3∣\left|a_2\right| > \left|a_1+a_3\right|, then atleast one root of the equation a1x4+a2x3+a3=0a_1 x^4+a_2 x^3+a_3=0 lies in the interval
View answer and explanation
Correct option: B
(−1,1)(-1,1)
Option B ((-1,1)) is correct because the condition |a_2| > |a_1+a_3| implies that there is a sign change in the function within the interval (-1, 1), indicating a root.
18 Medium
18 of 174
Application +4 | -1
f:[0,4]→Rf:[0,4] \rightarrow \mathrm{R} is a differentiable function. If there exist a,b∈(0,4)a, b \in(0,4) such that (f(4))2−(f(0))2=kf′(a)f(b)(f(4))^2-(f(0))^2=k f^{\prime}(a) f(b), then k=……k=\ldots \ldots.
View answer and explanation
Correct option: A
8
Option A (8) is correct because by applying the Mean Value Theorem and manipulating the resulting equations, the value of k is found to be 8.
19 Medium
19 of 174
Application +4 | -1
The sum of an infinitely decreasing geometric progression is equal to the least value of the function
f(x)=3x2−x+2512f(x)=3 x^2-x+\frac{25}{12}
and the first term of the progression is equal to the square of its common ratio. Then, the common ratio is
View answer and explanation
Correct option: B
3−1\sqrt{3}-1
Option B (3−1\sqrt{{3}}-1) is correct because it is the common ratio of the geometric progression, derived by minimizing the function and solving the resulting quadratic equation.
20 Medium
20 of 174
Application +4 | -1
Let 2p+3q+4r=152 p+3 q+4 r=15 and if the expression p3q5r7p^3 q^5 r^7 takes the maximum value, then p:q:rp: q: r is equal to (where p,q,r∈R+{p}, {q}, {r} \in {R}^{+})-
View answer and explanation
Correct option: A
18:20:2118: 20: 21
Option A (18:20:21) is correct because to maximize the expression p3q5r7 subject to the constraint 2p+3q+4r=15, we need to set 2p/3 = 3q/5 = 4r/7, leading to the ratio 18:20:21.
21 Medium
21 of 174
Application +4 | -1
α∈(0,π2)\alpha \in\left(0, \frac{\pi}{2}\right); then minimum value of x2+x+tan⁡2αx2+x\sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}} is
View answer and explanation
Correct option: C
2tan⁡α2 \tan \alpha
Option C (2tan⁡α2 \tan \alpha) is correct because it represents the minimum value obtained by applying the AM-GM inequality to the given expression.
22 Medium
22 of 174
Analyzing +4 | -1
The function f(x)=x+tan⁡xf(x)=x+\tan x has
View answer and explanation
Correct option: D
neither maximum nor minimum
Option D (neither maximum nor minimum) is correct because the derivative of the function is always positive, indicating that the function is strictly increasing.
23 Easy
23 of 174
Analyzing +4 | -1
Let f(x)={∣x3+x2+3x+sin⁡x∣(3+sin⁡1x),  x≠00,                                                                                                        x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left| {{x^3} + {x^2} + 3x + \sin x} \right|\left( {3 + \sin \frac{1}{x}} \right),\;x \ne 0}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 0} \end{array}} \right. then number of points (where f(x)f\left( x \right) attains its minimum value) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the function attains its minimum value of 0 only at x=0, due to the absolute value and the term (3 + sin(1/x)).
24 Medium
24 of 174
Analyzing +4 | -1
The maximum value of x1/x{x^{1/x}} is
View answer and explanation
Correct option: C
e1/e{e^{1/e}}
Option C (e1/e{e^{1/e}}) is correct because it is the maximum value of the function x1/x{x^{1/x}}.
25 Hard
25 of 174
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.
26 Medium
26 of 174
Analyzing +4 | -1
The number of values of xx at which the function f(x)=(x−1)x2/3f(x)=(x-1) x^{2 / 3} has extremum values is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the function has two critical points, x=0 and x=2/5, where it attains extremum values.
27 Medium
27 of 174
Analyzing +4 | -1
Let f(x)=3sin⁡x−4sin⁡3xf(x)=3 \sin x-4 \sin ^3 x. The length of the longest interval in which f′(x)≥0f^{\prime}(x) \geq 0 is
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
Option A (π3\frac{π}{3}) is correct because it represents the length of the interval where the derivative of f(x) is non-negative.
28 Hard
28 of 174
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}}.
29 Medium
29 of 174
Analyzing +4 | -1
The maximum value of x4e−x2x^4 e^{-x^2} is:
View answer and explanation
Correct option: D
4e−24 e^{-2}
Option D (4e−24 e^{-2}) is correct because it represents the maximum value of the function obtained by finding the critical points and using the second derivative test.
30 Easy
30 of 174
Analyzing +4 | -1
If the absolute maximum value of the function f(x)=f(x)= (x2−2x+7)e(4x3−12x2−180x+31)\left(x^2-2 x+7\right) e^{\left(4 x^3-12 x^2-180 x+31\right)} in the interval [−3,0][-3,0] is f(α)f(\alpha), then
View answer and explanation
Correct option: B
α=−3\alpha=-3
Option B (α=−3\alpha=-3) is correct because the absolute maximum value of the function occurs at x = -3 within the given interval [-3, 0].
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Key Takeaways

  • ✓ This set contains 174 single-correct-answer MCQs on MAXIMA & MINIMA.
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