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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Mixed difficulty

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121 Easy
121 of 174
Analyzing +4 | -1
Let a function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be defined as : f(x)={∫0x(5−∣t−3∣)dt,x>4x2+bx,x≤4f(x)=\left\{\begin{array}{cc}\int_0^x(5-|t-3|) d t, & x>4 \\ x^2+b x, & x \leq 4\end{array}\right.
where b∈Rb \in \mathbb{R}. If ff is continuous at x=4x=4 then which of the following statements is NOT true?
View answer and explanation
Correct option: C
ff is increasing in (−∞,18)∪(8,∞)\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)
Option C (288125\frac{288}{125}) is correct because the sum of the given infinite series is calculated using a method of differences, resulting in the value 288125\frac{288}{125}.
122 Medium
122 of 174
Application +4 | -1
If the cubic polynomial y=ax3+bx2+cx+d(a,b,c,d∈R)y=a x^{3}+b x^{2}+c x+d(a, b, c, d \in R) has only one critical point in its entire domain and ac=2a c=2, then the value of ∣b∣|b| is :
View answer and explanation
Correct option: D
6\sqrt{6}
Option D (6\sqrt{6}) is correct because a single critical point implies the derivative has only one root, meaning the discriminant of the derivative is zero, leading to ∣b∣=6|b| = \sqrt{6}.
123 Easy
123 of 174
Application +4 | -1
A covered box of volume 72 cm372 \mathrm{~cm}^3 and the base sides in a ratio of 1:21: 2 is to be made. The length all sides so that the total surface area is the least possible is
View answer and explanation
Correct option: D
6,3,46,3,4
Option D (6,3,4) is correct because these dimensions minimize the surface area of the box given the volume constraint and the ratio of base sides.
124 Easy
124 of 174
Analyzing +4 | -1
The largest value of 2x3−3x2−12x+5  2{x^3} - 3{x^2} - 12x + 5\; for −2≤x≤4- 2 \le x \le 4 occurs at xx is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because the largest value of the function within the given interval occurs at x = 4, as determined by evaluating the function at critical points and endpoints.
125 Medium
125 of 174
Analyzing +4 | -1
Local maximum value of the function log⁡ xx\frac{{\log \,x}}{x} is
View answer and explanation
Correct option: C
1e\frac{1}{e}
Option C is correct because the local maximum value is found by taking the derivative, setting it to zero, and then evaluating the function at that point.
126 Medium
126 of 174
Analyzing +4 | -1
If y=sin⁡(x+a)sin⁡(x+b),a≠b,y = \frac{{\sin \left( {x + a} \right)}}{{\sin \left( {x + b} \right)}},a \ne b, then yy is
View answer and explanation
Correct option: C
Neither minima nor maxima at x=0x = 0
Option C (Neither minima nor maxima at x=0x = 0) is correct because the first derivative is non-zero, indicating a point of inflection rather than a local extremum.
127 Medium
127 of 174
Analyzing +4 | -1
The function f(x)=x2e−2x,  x>0f\left( x \right) = {x^2}{e^{ - 2x}},\;x > 0 Then, the maximum value of f(x)f\left( x \right) is
View answer and explanation
Correct option: C
1e2\frac{1}{{{e^2}}}
Option C (1e2\frac{1}{{{e^2}}}) is correct because it represents the maximum value of the function f(x)=x2e−2xf(x) = x^2e^{-2x} for x>0x > 0, found by analyzing its first and second derivatives.
128 Medium
128 of 174
Application +4 | -1
The corner points of the feasible region determined by the following system of linear inequalities:
2x+y≤10,x+3y≤15,x,y≥02 x+y \leq 10, x+3 y \leq 15, x, y \geq 0 are (0,0),(5,0),(3,4)(0,0),(5,0),(3,4) and (0,5)(0,5). Let Z=px+qyZ=p x+q y, where p,q>0p, q > 0. Condition on pp and qq so that the maximum of ZZ occurs at both (3,4)(3,4) and (0,5)(0,5) is
View answer and explanation
Correct option: D
q=3pq=3 \mathrm{p}
Option D (q=3p) is correct because the maximum value of Z occurs at both (3,4) and (0,5), implying that Z has the same value at these points.
129 Medium
129 of 174
Analyzing +4 | -1
If f(x)={2x2+2x2,0<∣x∣≤23,x>2f(x)=\left\{\begin{array}{ll}2 x^2+\frac{2}{x^2}, & 0<|x| \leq 2 \\ 3, & x>2\end{array}\right., then
View answer and explanation
Correct option: B
x=1,−1x=1,-1 are the points of local minima
Option B (x=1,−1x=1,-1 are the points of local minima) is correct because the second derivative is positive at these points.
130 Easy
130 of 174
Analyzing +4 | -1
Let k\mathrm{k} and K\mathrm{K} be the minimum and the maximum values of the function f(x)=(1+x)0.61+x0.6\mathrm{f}(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}} in [0,1][0,1] respectively, then the ordered pair (k,K)(\mathrm{k}, \mathrm{K}) is equal to :
View answer and explanation
Correct option: A
(2−0.4,1)\left(2^{-0.4}, 1\right)
Option A ((2−0.4,1)\left(2^{-0.4}, 1\right)) is correct. The function is decreasing in the interval [0,1], so the maximum value is at x=0 and the minimum value is at x=1.
131 Medium
131 of 174
Analyzing +4 | -1
The number of values of xx where f(x)=cos⁡x+cos⁡2x  f\left( x \right) = \cos x + \cos \sqrt 2 x\; attains its maximum is
View answer and explanation
Correct option: A
1
Option A is correct because the function attains its maximum only once at x = 0.
132 Easy
132 of 174
Application +4 | -1
If sin⁡θ+cos⁡θ=1\sin \theta+\cos \theta=1, then the minimum value of (1+cosec⁡θ)(1+sec⁡θ)(1+\operatorname{cosec} \theta)(1+\sec \theta) is
View answer and explanation
Correct option: D
9
Option D (9) is correct because it is the minimum value obtained by using AM ≥ GM inequality and the given condition sin⁡θ+cos⁡θ=1\sin \theta + \cos \theta = 1.
133 Hard
133 of 174
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).
134 Medium
134 of 174
Analyzing +4 | -1
If xy=a2  and  S=b2x+c2yxy = {a^2}\;\mathrm{and}\;S = {b^2}x + {c^2}y where a,b and ca,b\,and\,c are constants, then the minimum value of SS is
View answer and explanation
Correct option: C
2abc2abc
Option C (2abc) is correct because it represents the minimum value of S, obtained by using calculus to find the critical point and verifying it's a minimum.
135 Easy
135 of 174
Analyzing +4 | -1
If y=y(x)y = y\left( x \right) is the solution of the differential equation xdydx+2y=xex,  y(1)=0x\frac{{dy}}{{dx}} + 2y = x{e^x},\,\,y\left( 1 \right) = 0 then the local maximum value of the function z(x)=x2y(x)−ex, x∈Rz\left( x \right) = {x^2}y\left( x \right) - {e^x},\,x \in R is
View answer and explanation
Correct option: D
4e−e\frac{4}{e} - e
Option D (4e−e\frac{4}{e} - e) is correct because it represents the local maximum value of the function z(x)z(x) obtained by solving the differential equation and finding the critical point.
136 Hard
136 of 174
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.
137 Easy
137 of 174
Application +4 | -1
If f(x)=Kx3−9x2+9x+3f(x)=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 f(x)f(x) to be increasing on RR, its derivative f′(x)f'(x) must be greater than 0, which leads to the condition K>3K>3.
138 Easy
138 of 174
Analyzing +4 | -1
The maximum value of f(x)=3cos⁡2x+4sin⁡2x+cos⁡x2+sin⁡x2f\left( x \right) = 3{\cos ^2}x + 4{\sin ^2}x + \cos \frac{x}{2} + \sin \frac{x}{2}, is
View answer and explanation
Correct option: C
4+24 + \sqrt 2
Option C (4+24 + \sqrt 2) is correct because it represents the maximum value of the function found by analyzing its derivatives.
139 Medium
139 of 174
Application +4 | -1
If x+y=8x + y = 8, then maximum value of x2y{x^2}y is
View answer and explanation
Correct option: D
204827\frac{{2048}}{{27}}
Option D (204827\frac{{2048}}{{27}}) is correct because it represents the maximum value of the expression x2y{x^2}y under the constraint x+y=8x + y = 8, found using calculus to optimize the function.
140 Medium
140 of 174
Analyzing +4 | -1
The function f(x)=x+1xf\left( x \right) = x + \frac{1}{x} has
View answer and explanation
Correct option: B
A local minima at x=1x = 1 and a local maxima at x=−1x = - 1
Option B is correct because the function has a local minima at x = 1 and a local maxima at x = -1, determined by the first and second derivative tests.
141 Medium
141 of 174
Analyzing +4 | -1
If a,b∈Ra, b \in R distinct numbers satisfying ∣a−1∣+∣b−1∣=∣a∣+∣b∣=∣a+1∣+∣b+1∣|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|, then the minimum value of ∣a−b∣|a-b| is
View answer and explanation
Correct option: D
2
Option D (2) is correct because the minimum distance between a and b, given the conditions, occurs when -1 and 1 are included in the interval [a, b].
142 Medium
142 of 174
Application +4 | -1
Let MM and mm be respectively the absolute maximum and the absolute minimum values of the function, f(x)=2x3−9x2+12x+5f(x)=2 x^3-9 x^2+12 x+5 in the interval [0,3][0,3]. Then M−mM-m is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the difference between the absolute maximum and minimum values of the function in the given interval is calculated correctly.
143 Hard
143 of 174
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.
144 Medium
144 of 174
Analyzing +4 | -1
Let f(x)f(x) be a polynomial of degree 5 such that x=±1x= \pm 1 are its critical points. If =4=4, then which one of the following is not true ?
View answer and explanation
Correct option: D
x=1x=1 is a point of minima and x=−1x=-1 is a point of maxima of ff
Option D (x=1x=1 is a point of minima and x=−1x=-1 is a point of maxima of ff) is correct. The function has a local maximum at x=1 and a local minimum at x=-1, which is opposite to the statement.
145 Easy
145 of 174
Analyzing +4 | -1
If a2x4+b2y4=c6{a^2}{x^4} + {b^2}{y^4} = {c^6}, then maximum value of xyxy is&nbsp;
View answer and explanation
Correct option: C
c32ab\frac{{{c^3}}}{{\sqrt {2ab} }}&nbsp;
Option C (c32ab\frac{{{c^3}}}{{\sqrt {2ab} }}) is correct because it represents the maximum value of xyxy derived using calculus and optimization techniques.
146 Hard
146 of 174
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.
147 Medium
147 of 174
Application +4 | -1
If mm and MM respectively denote the minimum and maximum of f(x)=(x−1)2+3  for  x∈[−3,1],f\left( x \right) = {\left( {x - 1} \right)^2} + 3\;for\;x \in \left[ { - 3,1} \right], then the ordered pair (m,Mm,M) is equal to
View answer and explanation
Correct option: B
(3,19)\left( {3,19} \right)
Option B ((3,19)) is correct because it represents the minimum and maximum values of the function within the given interval, found by evaluating the function at the critical points and endpoints.
148 Medium
148 of 174
Analyzing +4 | -1
For x∈(0,π)x \in(0, \pi), the least value of f(x)=(2+cos⁡xsin⁡x)2f(x)=\left(\frac{2+\cos x}{\sin x}\right)^2 is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the least value of the function is 3.
149 Easy
149 of 174
Application +4 | -1
The maximum value of xyxy when x+2y=8x + 2y = 8 is
View answer and explanation
Correct option: D
8
Option D (8) is correct because it is the maximum value of the product xyxy when x+2y=8x + 2y = 8, found using calculus.
150 Easy
150 of 174
Analyzing +4 | -1
If mm and nn are positive integers and f(x)=∫1x(t−a)2n(t−b)2m+1dt,a≠bf(x)=\int_1^x(t-a)^{2 n}(t-b)^{2 m+1} d t, a \neq b, then
View answer and explanation
Correct option: A
x=bx=b is a point of local minimum
Option A (x=bx=b is a point of local minimum) is correct because the first derivative changes sign from negative to positive at x=b.
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