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 · Mixed difficulty

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151 Medium

151 of 174

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.

152 Medium

152 of 174

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].

153 Medium

153 of 174

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.

154 Medium

154 of 174

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.

155 Easy

155 of 174

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}

156 Easy

156 of 174

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.

157 Medium

157 of 174

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.

158 Hard

158 of 174

Analyzing +4 | -1

The maximum value of

f\left( x \right) = \frac{x}{{4 + x + {x^2}}}

on [-1, 1] is

View answer and explanation

Correct option: D

\frac{1}{6}

Option D (

\frac{1}{6}

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

f(x)

on the given interval [-1, 1].

159 Easy

159 of 174

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.

160 Easy

160 of 174

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.

161 Easy

161 of 174

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.

162 Hard

162 of 174

Analyzing +4 | -1

ax + \frac{b}{x} \ge c

for all positive values of

x

and a,b,c are positive constants, then

View answer and explanation

Correct option: A

ab \ge \frac{{{c^2}}}{4}

Option A (

ab \ge \frac{{{c^2}}}{4}

) is correct because the minimum value of the function

ax + \frac{b}{x}

must be greater than or equal to c, leading to the inequality

ab \ge \frac{{{c^2}}}{4}

163 Hard

163 of 174

Analyzing +4 | -1

The function

f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 4

is maximum at

View answer and explanation

Correct option: A

x = 2

Option A (

x = 2

) is correct because the function's first derivative is zero and the second derivative is negative at this point, indicating a maximum.

164 Medium

164 of 174

Analyzing +4 | -1

Let

f\left( x \right) = {e^x}\sin x

, slope of the curve

y = f\left( x \right)

is maximum at

x = a

, if ‘a’ equals

View answer and explanation

Correct option: C

\frac{\pi }{2}

Option C (

\frac{\pi}{2}

) is correct because the slope's derivative is zero and its second derivative is negative at this point, indicating a maximum slope.

165 Medium

165 of 174

Application +4 | -1

The maximum value of xy subject to

x + y = 8

, is

View answer and explanation

Correct option: B

Option B (16) is correct because it represents the maximum value of xy when x + y = 8, found by optimization.

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166 Medium

166 of 174

Analyzing +4 | -1

For

x \in\left(0, \frac{5 \pi}{2}\right)

, define

f(x)=\int_0^x \sqrt{t} \sin t d t

. Then

f

has

View answer and explanation

Correct option: C

local maximum at

\pi

and local minimum at

2 \pi

The function f(x) has a local maximum at

\pi

and a local minimum at

2\pi

because the sign of f'(x) changes from positive to negative at

\pi

and from negative to positive at

2\pi

167 Easy

167 of 174

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.

168 Easy

168 of 174

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}}

169 Medium

169 of 174

Application +4 | -1

The minimum value of

2x + 3y

, when

xy = 6

, is

View answer and explanation

Correct option: A

Option A (12) is correct because the minimum value is found by differentiating the function and substituting the value of x.

170 Medium

170 of 174

Application +4 | -1

The maximum area of a triangle whose one vertex is at

(0,0)

and the other two vertices lie on the curve

y=-2 x^2+54

at points

(x, y)

and

(-x, y)

where

y > 0

, is :

View answer and explanation

Correct option: D

108

The correct answer is D (108) because the maximum area is found by expressing the area as a function of x, taking the derivative, and finding the critical points.

171 Easy

171 of 174

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.

172 Easy

172 of 174

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.

173 Medium

173 of 174

Application +4 | -1

A function

y=f(x)

has a second order derivative

f^{\prime \prime}(x)=6(x-1)

. If its graph passes through the point

(2,1)

and at that point the tangent to the graph is

y=3 x-5

, then the function is :

View answer and explanation

Correct option: B

(x-1)^{3}

Option B (

(x-1)^{3}

) is correct because integrating the second derivative and using the given conditions leads to this function.

174 Medium

174 of 174

Analyzing +4 | -1

The minimum value of

\left( {1 + \frac{1}{{{{\sin }^n}\alpha }}} \right)\left( {1 + \frac{1}{{{{\cos }^n}\alpha }}} \right)

, is

View answer and explanation

Correct option: C

{\left( {1 + {2^{n/2}}} \right)^2}

Option C is correct because the minimum value is found by setting the derivative equal to zero, solving for alpha, and substituting back into the original function.

Revision notes and key takeaways

Review MAXIMA & MINIMA in your textbook or class notes, then use this set to check your understanding.

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