JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Application +4 | -1

Let

f

be differentiable for all

x

. If

f\left( 1 \right) = - 2

and

f'\left( x \right) \ge 2

for all

x \in \left[ {1,\;6} \right],

, then

View answer and explanation

Correct option: D

f\left( 6 \right) \ge 8

Option D (

f(6) \ge 8

) is correct because by Lagrange's mean value theorem, the minimum value of

f(6)

is 8 given the conditions.

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f(x)

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f(x)

f(x)

f(x)=|\sin | x ||

\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]

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[3,5]

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f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)