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Let $f(x)$ satisfy all the conditions of mean value theorem in $[0, 2]. $ If $ f (0) = 0 $ and $|f'(x)|\, \le {1 \over 2}$ for all $x$ in $[0, 2]$ then
$f(x) \le 2$
$|f(x)| \le 1$
$f(x) = 2x$
$f(x) = 3$ for at least one $ x $ in $[0, 2]$
Solution
(b) $\frac{{f(2) – f(0)}}{{2 – 0}} = f'(x) \Rightarrow \frac{{f(2) – 0}}{2} = f'(x)$
$ \Rightarrow \frac{{df(x)}}{{dx}} = \frac{{f(2)}}{2} \Rightarrow f(x) = \frac{{f(2)}}{2}x + c$
$\therefore f(0) = 0 \Rightarrow c = 0$;
$\therefore f(x) = \frac{{f(2)}}{2}x$…..$(i)$
Given $|f'(x)| \le \frac{1}{2} \Rightarrow \left| {\frac{{f(2)}}{2}} \right| \le \frac{1}{2}$…..$(ii)$
$(i)$ ==> $|f(x)| = \left| {\frac{{f(2)}}{2}x} \right| = \left| {\frac{{f(2)}}{2}} \right||x| \le \frac{1}{2}|x|$ [from $(ii)$]
In $ [0, 2],$ for maximum $x(x = 2)$
$|f(x)| \le \frac{1}{2}.\,\,\,2 \Rightarrow |f(x)| \le 1$ .
