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Let $f(x)$ satisfy all the conditions of the mean value theorem in $[0,2]$. If $f(0)=0$ and $|f(x)| \leq \frac{1}{2}$ for all $x$, in $[0,2]$ then
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Verified Answer
The correct answer is:
$|f(x)| \leq 1$
$\begin{aligned} & \frac{f(2)-f(0)}{2-0}=f^{\prime}(x) \Rightarrow \frac{f(2)-0}{2}=f^{\prime}(x) \\ & \Rightarrow \frac{d f(x)}{d x}=\frac{f(2)}{2} \Rightarrow f(x)=\frac{f(2)}{2} x+c \\ & \therefore f(0)=0 \Rightarrow c=0 . \therefore f(x)=\frac{f(2)}{2} x...(i)\end{aligned}$
Given
$|f(x)| \leq \frac{1}{2} \Rightarrow\left|\frac{f(2)}{2}\right| \leq \frac{1}{2}$
(i) $\Rightarrow$
$|f(x)|=\left|\frac{f(2)}{2} x\right|=\left|\frac{f(2)}{2}\right||x| \leq \frac{1}{2}|x|$
[from (ii)]
In $[0,2]$, for maximum $x(x=2)$
$\left.|f(x)| \leq \frac{1}{2} \cdot 2 \Rightarrow f(x) \right\rvert\, \leq 1$
Given
$|f(x)| \leq \frac{1}{2} \Rightarrow\left|\frac{f(2)}{2}\right| \leq \frac{1}{2}$
(i) $\Rightarrow$
$|f(x)|=\left|\frac{f(2)}{2} x\right|=\left|\frac{f(2)}{2}\right||x| \leq \frac{1}{2}|x|$
[from (ii)]
In $[0,2]$, for maximum $x(x=2)$
$\left.|f(x)| \leq \frac{1}{2} \cdot 2 \Rightarrow f(x) \right\rvert\, \leq 1$
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