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

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

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Easy

1 of 1

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.

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. Then, the set of all values of

p

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f\left( x \right)

given by

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

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