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VITEEE 2019 Paper

Section: Mathematics

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Question : 117 of 125

Marks: +1, -0

The value of x in the interval [4,9] at which the function

f(x) = \sqrt{x}

satisfies the mean value theorem is

$\frac{13}{4}$
$\frac{17}{4}$
$\frac{21}{4}$
$\frac{25}{4}$

Solution:

(i)

f(x) = \sqrt{x}

is continuous in [4,9]

(ii)

f'(x)=\frac{1}{2\sqrt{x}}

Thus f(x) is differentiable in (4,9)

(iii)

f(4) \neq f(9)

. All the three conditions of LMV theorem satisfied then there exist at least one

c \in (4,9)

such that.

f'(c)=\frac{f(b)-f(a)}{b-a}

\Rightarrow\frac{1}{2\sqrt{c}}=\frac{1}{5}

\Rightarrow c=\frac{25}{4}

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