JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Application +4 | -1

Suppose

f

is differentiable on

\mathbb{R}, f(0)=0

and

1 \leq f^{\prime}(x) \leq 2

for all

x \in \mathbb{R}

. Then

View answer and explanation

Correct option: A

x \leq f(x) \leq 2 x

for all

x \geq 0

Option A (

x \leq f(x) \leq 2x

for all

x \geq 0

) is correct because it uses the Lagrange's Mean Value Theorem and the given bounds on

f'(x)

to establish the inequality.

More Questions from MEAN VALUES THEOREM

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f(x)

g(x)

[0,1]

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0 < \mathrm{c} < 1

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

f'\left( x \right)

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- 1 \le x \le 1

f(x)

[3,5]

\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)