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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
$1 - {{\sqrt {15} } \over 6}$
$1 + \sqrt {15} $
$1 - {{\sqrt {21} } \over 6}$
$1 + \sqrt {21} $
Solution
(c) From mean value theorem $f'(c) = \frac{{f(b) – f(a)}}{{b – a}}$
$a = 0,\,f(a) = 0$ ==> $b = \frac{1}{2},\,f(b) = \frac{3}{8}$
$f'(x) = (x – 1)(x – 2) + x(x – 2) + x(x – 1)$
==> $f'(c) = (c – 1)(c – 2) + c(c – 2) + c(c – 1)$
==> $f'(c) = {c^2} – 3c + 2 + {c^2} – 2c + {c^2} – c$
==> $f'(c) = 3{c^2} – 6c + 2$
According to mean value theorem, $f'(c) = \frac{{f(b) – f(a)}}{{b – a}}$
==> $3{c^2} – 6c + 2 = \frac{{(3/8) – 0}}{{(1/2) – 0}}\, = \frac{3}{4}$
==> $3{c^2} – 6c + \frac{5}{4} = 0$
$c = \frac{{6 \pm \sqrt {36 – 15} }}{{2 \times 3}} = \frac{{6 \pm \sqrt {21} }}{6}$
$ = 1 \pm \frac{{\sqrt {21} }}{6}$.
