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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