Let f(x) satisfy all the conditions of mean v | vedclass

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Question

(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}

Vedclass · Accepted Answer

$|f(x)| \le 1$

Vedclass · Answer

(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$

$	herefore f(0) = 0 \Rightarrow c = 0$; 

$	herefore f(x) = \frac{{f(2)}}{2}x$.....$(i)$

Given $|f'(x)| \le \frac{1}{2} \Rightarrow \left| {\frac{{f(2)}}{2}} ight| \le \frac{1}{2}$&hellip;..$(ii)$

$(i)$ ==> $|f(x)| = \left| {\frac{{f(2)}}{2}x} ight| = \left| {\frac{{f(2)}}{2}} ight||x| \le \frac{1}{2}|x|$   [from $(ii)$]

In $ [0, 2],$  for maximum $x(x = 2)$

$|f(x)| \le \frac{1}{2}.\,\,\,2 \Rightarrow |f(x)| \le 1$ .

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

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