If $f(x)=\left\{\begin{array}{ll}2 x^2+\frac{2}{x^2}, & 0 2\end{array}\right.$, then

$x=1,-1$ are the points of global minima

$x=0$ is the points of local minima

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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima 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 61-90 of 97 questions

61 Medium

61 of 97

Analyzing +4 | -1

A closed circular cylinder of volume

k

cubic units is to be formed with minimum amount of material. The ratio of its height to the radius of the base is

View answer and explanation

Correct option: C

Option C (2) is correct because the ratio of height to radius that minimizes the material used for a closed cylinder of fixed volume is 2.

62 Medium

62 of 97

Analyzing +4 | -1

Given a function

y=x^x, x>0

and

0 < x < 1

. The values of

x

for which the function attain values exceeding the values of its inverse are

View answer and explanation

Correct option: A

0 < x < 1

Option A (

0 < x < 1

) is correct because the graph of

y=x^x

lies above the line

y=x

for

0 < x < 1

63 Medium

63 of 97

Application +4 | -1

The maximum distance from the origin of a point on the curve

x = a\sin t - \sin \left( {\frac{{at}}{b}} \right),\;y = a\cos t - b\cos \left( {\frac{{at}}{b}} \right)

,both

a,\;b > 0

View answer and explanation

Correct option: B

a + b

Option B (

a + b

) is correct because it represents the maximum distance from the origin of a point on the given curve.

64 Medium

64 of 97

Application +4 | -1

xy = {c^2}

, then minimum value of

ax + by

View answer and explanation

Correct option: B

2c\sqrt {ab}

Option B (

2c\sqrt {ab}

) is correct because it is the minimum value of

ax + by

subject to the constraint

xy = c^2

, found using calculus or AM-GM inequality.

65 Medium

65 of 97

Analyzing +4 | -1

The minimum value of

\frac{\log x}{x}

in the interval

[2, \infty)

is:

View answer and explanation

Correct option: D

Does not exist

The minimum value of log(x)/x in the interval [2, infinity) does not exist because the function has a maximum at x=e, and it decreases as x approaches infinity.

66 Medium

66 of 97

Analyzing +4 | -1

The maximum value of

\frac{{\log x}}{x}

View answer and explanation

Correct option: A

\frac{1}{e}

Option A (

\frac{1}{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.

67 Medium

67 of 97

Analyzing +4 | -1

If the function

f\left( x \right) = \frac{{ax + b}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}

has an extremum at

P\left( {2,\; - 1} \right)

then

View answer and explanation

Correct option: C

a = 1,\;b = 0

Option C (a = 1, b = 0) is correct because substituting these values into the function and its derivative satisfies the given conditions of having an extremum at P(2, -1).

68 Medium

68 of 97

Analyzing +4 | -1

The absolute maximum of

{x^{40}} - {x^{20}}

on the interval [0, 1] is

View answer and explanation

Correct option: B

Option B (0) is correct because it represents the absolute maximum value of the function on the given interval, considering both critical points and endpoints.

69 Medium

69 of 97

Analyzing +4 | -1

f'\left( x \right) = {\left( {x - a} \right)^{2n}}{\left( {x - b} \right)^{2m + 1}}

where

m,\;n \in N

, then

View answer and explanation

Correct option: A

x = b

is a point of minimum

Option A is correct because the sign of the derivative changes from negative to positive at x = b, indicating a point of minimum.

70 Medium

70 of 97

Analyzing +4 | -1

The maximum value of function

f\left( x \right) = \sin x\left( {1 + \cos x} \right),\;x \in R

View answer and explanation

Correct option: A

\frac{{{3^{3/2}}\;\;}}{4}

Option A (

\frac{{{{3^{{3/2}}}}\\;}}{4}

) is correct because it represents the maximum value of the function, found by differentiation and checking the second derivative.

71 Medium

71 of 97

Application +4 | -1

If the cubic polynomial

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

a c=2

, then the value of

|b|

is :

View answer and explanation

Correct option: D

\sqrt{6}

Option D (

\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| = \sqrt{6}

72 Medium

72 of 97

Analyzing +4 | -1

Local maximum value of the function

\frac{{\log \,x}}{x}

View answer and explanation

Correct option: C

\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.

73 Medium

73 of 97

Analyzing +4 | -1

y = \frac{{\sin \left( {x + a} \right)}}{{\sin \left( {x + b} \right)}},a \ne b,

then

y

View answer and explanation

Correct option: C

Neither minima nor maxima at

x = 0

Option C (Neither minima nor maxima at

x = 0

) is correct because the first derivative is non-zero, indicating a point of inflection rather than a local extremum.

74 Medium

74 of 97

Analyzing +4 | -1

The function

f\left( x \right) = {x^2}{e^{ - 2x}},\;x > 0

Then, the maximum value of

f\left( x \right)

View answer and explanation

Correct option: C

\frac{1}{{{e^2}}}

Option C (

\frac{1}{{{e^2}}}

) is correct because it represents the maximum value of the function

f(x) = x^2e^{-2x}

for

x > 0

, found by analyzing its first and second derivatives.

75 Medium

75 of 97

Application +4 | -1

The corner points of the feasible region determined by the following system of linear inequalities:

2 x+y \leq 10, x+3 y \leq 15, x, y \geq 0

are

(0,0),(5,0),(3,4)

and

(0,5)

. Let

Z=p x+q y

, where

p, q > 0

. Condition on

p

and

q

so that the maximum of

Z

occurs at both

(3,4)

and

(0,5)

View answer and explanation

Correct option: D

q=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.

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76 Medium

76 of 97

Analyzing +4 | -1

f(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,-1

are the points of local minima

Option B (

x=1,-1

are the points of local minima) is correct because the second derivative is positive at these points.

77 Medium

77 of 97

Analyzing +4 | -1

The number of values of

x

where

f\left( x \right) = \cos x + \cos \sqrt 2 x\;

attains its maximum is

View answer and explanation

Correct option: A

Option A is correct because the function attains its maximum only once at x = 0.

78 Medium

78 of 97

Analyzing +4 | -1

xy = {a^2}\;\mathrm{and}\;S = {b^2}x + {c^2}y

where

a,b\,and\,c

are constants, then the minimum value of

S

View answer and explanation

Correct option: C

2abc

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.

79 Medium

79 of 97

Application +4 | -1

x + y = 8

, then maximum value of

{x^2}y

View answer and explanation

Correct option: D

\frac{{2048}}{{27}}

Option D (

\frac{{2048}}{{27}}

) is correct because it represents the maximum value of the expression

{x^2}y

under the constraint

x + y = 8

, found using calculus to optimize the function.

80 Medium

80 of 97

Analyzing +4 | -1

The function

f\left( x \right) = x + \frac{1}{x}

has

View answer and explanation

Correct option: B

A local minima at

x = 1

and a local maxima at

x = - 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.

81 Medium

81 of 97

Analyzing +4 | -1

a, b \in R

distinct numbers satisfying

|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|

, then the minimum value of

|a-b|

View answer and explanation

Correct option: D

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].

82 Medium

82 of 97

Application +4 | -1

Let

M

and

m

be respectively the absolute maximum and the absolute minimum values of the function,

f(x)=2 x^3-9 x^2+12 x+5

in the interval

[0,3]

. Then

M-m

is equal to

View answer and explanation

Correct option: A

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.

83 Medium

83 of 97

Analyzing +4 | -1

Let

f(x)

be a polynomial of degree 5 such that

x= \pm 1

are its critical points. If

=4

, then which one of the following is not true ?

View answer and explanation

Correct option: D

x=1

is a point of minima and

x=-1

is a point of maxima of

f

Option D (

x=1

is a point of minima and

x=-1

is a point of maxima of

f

) is correct. The function has a local maximum at x=1 and a local minimum at x=-1, which is opposite to the statement.

84 Medium

84 of 97

Application +4 | -1

m

and

M

respectively denote the minimum and maximum of

f\left( x \right) = {\left( {x - 1} \right)^2} + 3\;for\;x \in \left[ { - 3,1} \right],

then the ordered pair (

m,M

) is equal to

View answer and explanation

Correct option: B

\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.

85 Medium

85 of 97

Analyzing +4 | -1

For

x \in(0, \pi)

, the least value of

f(x)=\left(\frac{2+\cos x}{\sin x}\right)^2

View answer and explanation

Correct option: B

Option B (3) is correct because the least value of the function is 3.

86 Medium

86 of 97

Analyzing +4 | -1

Let

f\left( x \right) = 1 + 2{x^2} + {2^2}{x^4} + . \ldots \ldots + {2^{10}}{x^{20}}

.Then,

f\left( x \right)

has

View answer and explanation

Correct option: B

Exactly one minimum

Option B (Exactly one minimum) is correct because the first derivative has only one real root (x=0) and the second derivative is positive at that point, indicating a minimum.

87 Medium

87 of 97

Analyzing +4 | -1

The smallest value of the polynomial

x^3-18 x^2+96 x

[0,9]

View answer and explanation

Correct option: B

Option B (0) is correct because it is the smallest value obtained by evaluating the polynomial at the critical points and endpoints of the interval [0, 9].

88 Medium

88 of 97

Application +4 | -1

x \in(0,1)

the greatest root of the equation

\sin 2 \pi x=\sqrt{2} \cos \pi x

is -

View answer and explanation

Correct option: A

\frac{3}{4}

Option A (3/4) is correct because solving the trigonometric equation within the given interval yields x = 1/4, 1/2, and 3/4, with 3/4 being the greatest.

89 Medium

89 of 97

Analyzing +4 | -1

Consider the function

f(x)=x^3-3 x+3

on the interval

[-3,3 / 2]

. Let

M=\operatorname{Max} f(x)

and

m=\operatorname{Min} f(x)

[-3,3 / 2]

. Then

View answer and explanation

Correct option: B

M=5, m=-15

Option B (

M=5, m=-15

) is correct because the maximum and minimum values of the function on the interval are 5 and -15, respectively.

90 Medium

90 of 97

Application +4 | -1

If the point

(1,3)

serves as the point of inflection of the curve

y=a x^3+b x^2,

then the value of

a

and

b

are

View answer and explanation

Correct option: D

a=-\frac{3}{2}

and

b=\frac{9}{2}

Option D (

a=-\frac{3}{2}

and

b=\frac{9}{2}

) is correct because the point (1,3) lies on the curve and the second derivative at x=1 is zero, indicating a point of inflection.

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