Let $\mathrm{k}$ and $\mathrm{K}$ be the minimum and the maximum values of the function $\mathrm{f}(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}}$ in $[0,1]$ respectively, then the ordered pair $(\mathrm{k}, \mathrm{K})$ is equal to :

$\left(2^{-0.4}, 2^{0.6}\right)$

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$, then maximum value of $xy$ is

$\frac{{{c^2}}}{{\sqrt {ab} }}$

Themaximum value of $[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1$ is

$\left(\frac{1}{3}\right)^{\frac{1}{3}}$

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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Easy

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Showing 31-58 of 58 questions

31 Easy

31 of 58

Analyzing +4 | -1

Consider a cuboid of sides

2x,\,4x\,\,{\text{and}}\,{\text{5x}}

and a closed hemisphere of radius

r.

If the sum of their surface areas is a constant

k,

then the ratio

x:r,

for which the sum of their volumes is maximum, is

View answer and explanation

Correct option: B

19:45

Option B (

19:45

) is correct because it represents the ratio

x:r

that maximizes the sum of the volumes given the constraint on the surface areas.

32 Easy

32 of 58

Application +4 | -1

x

is real, then the minimum value of

x^2-8 x+17

View answer and explanation

Correct option: C

Option C (1) is correct because the minimum value of the quadratic function

x^2 - 8x + 17

occurs at its vertex, which can be found by completing the square or using calculus.

33 Easy

33 of 58

Analyzing +4 | -1

Let

f: N \rightarrow N

be such that

f(n+1)>f(f(n))

for all

n \in N

, then

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct. The function f(n+1) > f(f(n)) implies f(n) = n, but none of the given options satisfy this condition.

34 Easy

34 of 58

Application +4 | -1

Let

f(x)

be a differentiable function and

\mathrm{G}

is the graph of

f(x)

. If

\mathrm{P}(a, f(a))

\mathrm{G}

is the point nearest to

(0,0)

, then

f(a) f^{\prime}(a)=

View answer and explanation

Correct option: B

-a

Option B (

-a

) is correct because the point nearest to (0,0) implies the distance function is minimized, leading to the given relation.

35 Easy

35 of 58

Analyzing +4 | -1

Let

a_1, a_2, a_3, \ldots

be an A. P. with

a_6=2

. Then the common difference of this A.P., which maximises the product

a_1 a_4

a_5

, is :

View answer and explanation

Correct option: B

\frac{8}{5}

Option B (

\frac{8}{5}

) is correct. By expressing the product in terms of the common difference and maximizing it using calculus, the common difference is found to be

\frac{8}{5}

36 Easy

36 of 58

Analyzing +4 | -1

Let

f(x)=x^2-2 x

and

g(x)=f(f(x)-1)+f(5-f(x))

, then

View answer and explanation

Correct option: D

g(x) \geq 0, \forall x \in R

Option D (

g(x) \geq 0, \forall x \in R

) is correct. By finding the derivative of

g(x)

and analyzing its critical points, we can determine that

g(x)

is always non-negative.

37 Easy

37 of 58

Analyzing +4 | -1

The number of points of local maxima of the function

f(x)=(x-2)^{2 / 3}(2 x+1)

View answer and explanation

Correct option: A

Option A (1) is correct because the function has one local maximum at x=1. The first derivative test confirms this.

38 Easy

38 of 58

Analyzing +4 | -1

If,

f\left( x \right) = 2{x^3} - 21{x^2} + 36x - 30

, then which one of the following is correct

View answer and explanation

Correct option: C

f\left( x \right)\mathrm{has\;maximum}\;at\;x = 1

Option C (

f(x) \text{ has maximum at } x = 1

) is correct because the first derivative is zero and the second derivative is negative at

x=1

39 Easy

39 of 58

Analyzing +4 | -1

Let a function

f: \mathbb{R} \rightarrow \mathbb{R}

be defined as :

f(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 \in \mathbb{R}

. If

f

is continuous at

x=4

then which of the following statements is NOT true?

View answer and explanation

Correct option: C

f

is increasing in

\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)

Option C (

\frac{288}{125}

) is correct because the sum of the given infinite series is calculated using a method of differences, resulting in the value

\frac{288}{125}

40 Easy

40 of 58

Application +4 | -1

A covered box of volume

72 \mathrm{~cm}^3

and the base sides in a ratio of

1: 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

6,3,4

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.

41 Easy

41 of 58

Analyzing +4 | -1

The largest value of

2{x^3} - 3{x^2} - 12x + 5\;

for

- 2 \le x \le 4

occurs at

x

is equal to

View answer and explanation

Correct option: D

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.

42 Easy

42 of 58

Analyzing +4 | -1

Let

\mathrm{k}

and

\mathrm{K}

be the minimum and the maximum values of the function

\mathrm{f}(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}}

[0,1]

respectively, then the ordered pair

(\mathrm{k}, \mathrm{K})

is equal to :

View answer and explanation

Correct option: A

\left(2^{-0.4}, 1\right)

Option A (

\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.

43 Easy

43 of 58

Application +4 | -1

\sin \theta+\cos \theta=1

, then the minimum value of

(1+\operatorname{cosec} \theta)(1+\sec \theta)

View answer and explanation

Correct option: D

Option D (9) is correct because it is the minimum value obtained by using AM ≥ GM inequality and the given condition

\sin \theta + \cos \theta = 1

44 Easy

44 of 58

Analyzing +4 | -1

y = y\left( x \right)

is the solution of the differential equation

x\frac{{dy}}{{dx}} + 2y = x{e^x},\,\,y\left( 1 \right) = 0

then the local maximum value of the function

z\left( x \right) = {x^2}y\left( x \right) - {e^x},\,x \in R

View answer and explanation

Correct option: D

\frac{4}{e} - e

Option D (

\frac{4}{e} - e

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

z(x)

obtained by solving the differential equation and finding the critical point.

45 Easy

45 of 58

Application +4 | -1

f(x)=K x^3-9 x^2+9 x+3

is increasing on

R

, then -

View answer and explanation

Correct option: B

K>3

Option B (

K>3

) is correct because for

f(x)

to be increasing on

R

, its derivative

f'(x)

must be greater than 0, which leads to the condition

K>3

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46 Easy

46 of 58

Analyzing +4 | -1

The maximum value of

f\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 + \sqrt 2

Option C (

4 + \sqrt 2

) is correct because it represents the maximum value of the function found by analyzing its derivatives.

47 Easy

47 of 58

Analyzing +4 | -1

{a^2}{x^4} + {b^2}{y^4} = {c^6}

, then maximum value of

xy

View answer and explanation

Correct option: C

\frac{{{c^3}}}{{\sqrt {2ab} }}

Option C (

\frac{{{c^3}}}{{\sqrt {2ab} }}

) is correct because it represents the maximum value of

xy

derived using calculus and optimization techniques.

48 Easy

48 of 58

Application +4 | -1

The maximum value of

xy

when

x + 2y = 8

View answer and explanation

Correct option: D

Option D (8) is correct because it is the maximum value of the product

xy

when

x + 2y = 8

, found using calculus.

49 Easy

49 of 58

Analyzing +4 | -1

m

and

n

are positive integers and

f(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=b

is a point of local minimum

Option A (

x=b

is a point of local minimum) is correct because the first derivative changes sign from negative to positive at x=b.

50 Easy

50 of 58

Analyzing +4 | -1

P(x)

is polynomial satisfying

P\left(x^2\right)=x^2 P(x)

and

P(0)=-2, P^{\prime}(3 / 2)=0

and

P(1)=0

.
The maximum value of

P(x)

View answer and explanation

Correct option: B

-\frac{1}{4}

Option B (

-\frac{1}{4}

) is correct. By using the given conditions to determine the polynomial

P(x)

, we find its maximum value to be

-\frac{1}{4}

51 Easy

51 of 58

Application +4 | -1

If sum of two numbers is 6, the minimum value of the sum of their reciprocals is

View answer and explanation

Correct option: C

\frac{2}{3}

Option C (

\frac{2}{3}

) is correct because it represents the minimum value of the sum of the reciprocals of two numbers whose sum is 6, found using calculus to minimize the function.

52 Easy

52 of 58

Analyzing +4 | -1

Let

f(x)=\left\{\begin{array}{l}1+\sin x, \quad x<0 \\ x^2-x+1, x \geq 0\end{array}\right.

Then,

View answer and explanation

Correct option: A

f

has a local maximum at

x=0

Option A (

f

has a local maximum at

x=0

) is correct because the function is continuous at x=0, and the derivative changes from positive to negative at x=0.

53 Easy

53 of 58

Analyzing +4 | -1

a{x^2} + \frac{b}{x} \ge c

for all positive

x

, where ,

a,b, > 0

then

View answer and explanation

Correct option: A

27a{b^2} \ge 4{c^3}

Option A (

27a{b^2} \ge 4{c^3}

) is correct because it represents the condition for the inequality to hold for all positive x, derived using calculus and optimization.

54 Easy

54 of 58

Analyzing +4 | -1

Let

x=2

be a local minima of the function

f(x)=2 x^{4}

-18 x^{2}+8 x+12, x \in(-4,4)

. If

{M}

is local maximum value of the function

f

(-4,4)

, then

M=

View answer and explanation

Correct option: A

12 \sqrt{6}-\frac{33}{2}

Option A (

12 \sqrt{6}-\frac{33}{2}

) is correct because we find the critical points by setting the derivative to zero, and then evaluate the function at the local maxima within the given interval.

55 Easy

55 of 58

Analyzing +4 | -1

Let

f(x)=x^{n+1}+a \cdot x^n

, where

a

is a positive real number. Then,

x=0

is a point of

View answer and explanation

Correct option: C

local minimum if

n

is an even integer

Option D (local minimum if

n

is an even integer) is correct because when n is even, the first derivative changes from negative to positive at x=0.

56 Easy

56 of 58

Analyzing +4 | -1

a>b>0

and

f(\theta)=\frac{\left(a^2-b^2\right) \cos \theta}{a-b \sin \theta}

, then the maximum value of

f(\theta)

, is

View answer and explanation

Correct option: C

\sqrt{a^2-b^2}

Option C (

\sqrt{{a^2-b^2}}

) is correct because the maximum value of

f(\theta)

is obtained when

h(\theta)

is minimized, which occurs when

\sin \theta = \frac{{b}}{{a}}

57 Easy

57 of 58

Analyzing +4 | -1

Themaximum value of

[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1

View answer and explanation

Correct option: C

Option C (1) is correct because it represents the maximum value of the function within the given interval, found by analyzing the critical points and endpoints.

58 Easy

58 of 58

Analyzing +4 | -1

The positive difference between the local maximum value and the local minimum value of the function

f(x)=x^3-3 x-1, \forall x \in[-2,3]

View answer and explanation

Correct option: B

Option B (4) is correct because the difference between the local maximum and minimum values of the function within the given interval is 4.

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