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5. Continuity and Differentiation
easy
In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$ is
A
$8$
B
$5.25$
C
$4$
D
$6.25$
Solution
(d) $f(x) = \sqrt x $
$\therefore \,\,f(a) = \sqrt 4 = 2,$ $f(b) = \sqrt 9 = 3$ ; $f'(x) = \frac{1}{{2\sqrt x }}$
Also, $f'(c) = \frac{{f(b) – f(a)}}{{b – a}} = \frac{{3 – 2}}{{9 – 4}} = \frac{1}{5}$
$\therefore $ $\frac{1}{{2\sqrt c }} = \frac{1}{5}$ ==> $c = \frac{{25}}{4} = 6.25$.
Standard 12
Mathematics
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