A) \[f(x)\le 2\]
B) \[|f(x)|\le 1\]
C) \[f(x)=2x\]
D) \[f(x)=3\]for at least one x in [0, 2]
Correct Answer: B
Solution :
\[\frac{f(2)-f(0)}{2-0}=f'(x)\Rightarrow \frac{f(2)-0}{2}=f'(x)\] \[\Rightarrow \frac{df(x)}{dx}=\frac{f(2)}{2}\Rightarrow f(x)=\frac{f(2)}{2}x+c\] \[\therefore f(0)=0\Rightarrow c=0\]; \[\therefore f(x)=\frac{f(2)}{2}x\] .....(i) Given \[|f'(x)|\le \frac{1}{2}\Rightarrow \left| \frac{f(2)}{2} \right|\le \frac{1}{2}\] ?..(ii) (i) Þ \[|f(x)|=\left| \frac{f(2)}{2}x \right|=\left| \frac{f(2)}{2} \right||x|\le \frac{1}{2}|x|\][from (ii)] In [0, 2], for maximum \[x(x=2)\] \[|f(x)|\le \frac{1}{2}.\,\,\,2\Rightarrow |f(x)|\le 1\] .You need to login to perform this action.
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