5. Continuity and Differentiation
medium

Let $f(x) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} ;} $ $1 < x < 26$ be real valued function. Then $f\,'(x)$ for $1 < x < 26$ is

A

$0$

B

${1 \over {\sqrt {x - 1} }}$

C

$2\sqrt {x - 1} - 5$

D

None of these

Solution

(a) From Rolle’s theorem in  $(1, 26),$  $f(1) = f(26) = 5$. 

In given interval, function satisfy all the conditions of Rolle's theorem,

therefore in $ [1, 26], $ at least, there is a point for which $f'(x) = 0$.

Standard 12
Mathematics

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