
\[ {\text{Consider the following statements in respect of the function }}f(x) = {x^3} - 1, \\
x \in [ - 1,1] \\
{\text{I}}{\text{. }}f(x){\text{ is increasing in }}[ - 1,1] \\
{\text{II}}f'(x){\text{ has no root in (}} - 1,1]. \\
{\text{Which of the statements given above is/are correct?}} \\
{\text{A}}{\text{. only I}} \\
{\text{B}}{\text{. only II}} \\
{\text{C}}{\text{. Both I and II}} \\
{\text{D}}{\text{. Neither I nor II}} \\
\]
Answer
633.9k+ views
$ {\text{Solution: - }} \\
{\text{To check a function either it is increasing or decreasing we have to double differentiate the function}} \\
{\text{and check the function in their domain either it is increasing or decreasing}} \\
{\text{in this question our function is }}f(x) = {x^3} - 1 \\
{\text{ so let's find the first derivative }}f'(x) = 3{x^2} \\
{\text{Now the second derivative is }}f''(x) = 6x{\text{ , we check the function for }}x \in [ - 1,1] \\
{\text{here }}f''(x) \in [ - 6,6]{\text{ }}\therefore {\text{ the function }}f(x){\text{ is increasing }}{\text{.}} \\
{\text{II}}{\text{. To find the root of }}f'(x){\text{ we have to equate }}f'(x) = 0. \\
\Rightarrow 3{x^2} = 0{\text{ }} \Rightarrow x = 0{\text{ }} \\
{\text{there is one root of }}f'(x){\text{ in ( - 1,1]}}{\text{.}} \\
\therefore {\text{Statement I is correct and II is incorrect }} \\
{\text{Answer is A}}{\text{.}} \\
{\text{Note: - To check a function either it is increasing or decreasing we have to differentiate the function}} \\
{\text{ when first derivative is always positive in the given domain then it is strictly increasing}}{\text{.}} \\
{\text{ }} \\
$
{\text{To check a function either it is increasing or decreasing we have to double differentiate the function}} \\
{\text{and check the function in their domain either it is increasing or decreasing}} \\
{\text{in this question our function is }}f(x) = {x^3} - 1 \\
{\text{ so let's find the first derivative }}f'(x) = 3{x^2} \\
{\text{Now the second derivative is }}f''(x) = 6x{\text{ , we check the function for }}x \in [ - 1,1] \\
{\text{here }}f''(x) \in [ - 6,6]{\text{ }}\therefore {\text{ the function }}f(x){\text{ is increasing }}{\text{.}} \\
{\text{II}}{\text{. To find the root of }}f'(x){\text{ we have to equate }}f'(x) = 0. \\
\Rightarrow 3{x^2} = 0{\text{ }} \Rightarrow x = 0{\text{ }} \\
{\text{there is one root of }}f'(x){\text{ in ( - 1,1]}}{\text{.}} \\
\therefore {\text{Statement I is correct and II is incorrect }} \\
{\text{Answer is A}}{\text{.}} \\
{\text{Note: - To check a function either it is increasing or decreasing we have to differentiate the function}} \\
{\text{ when first derivative is always positive in the given domain then it is strictly increasing}}{\text{.}} \\
{\text{ }} \\
$
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
A large number of liquid drops each of radius r coalesce class 11 physics CBSE

The period of a conical pendulum in terms of its length class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

In a fight of 600km an aircraft was slowed down du-class-11-maths-CBSE

State and prove Bernoullis theorem class 11 physics CBSE

A solution of a substance X is used for white washing class 11 chemistry CBSE

