JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

1 Easy

1 of 1

Application +4 | -1

Let

f

be continuous on [1, 5] and differentiable in

\left( {1,5} \right)

. If

f\left( 1 \right) = - 3

and

f'\left( x \right) \ge 9

for all

x \in \left( {1,\;5} \right)

, then

View answer and explanation

Correct option: A

f\left( 5 \right) \ge 33

Option A (

f\left( 5 \right) \ge 33

) is correct because it represents the minimum possible value of f(5) based on the Mean Value Theorem and the given conditions.

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g(x)

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f(x)

f(x)

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

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

f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26

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\int\limits_3^5 {f\left( x \right)} \,dx

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