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5. Continuity and Differentiation
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Let $f(x)$ satisfy the requirement of lagranges mean value theorem in $[0,2]$ . If $f(x)=0$ ; $\left| {f'\left( x \right)} \right| \leqslant \frac{1}{2}$ for all $x \in \left[ {0,2} \right]$, then-
A
$f\left( x \right) \geqslant 2$
B
$\left| {f\left( x \right)} \right| \leqslant 1$
C
$f\left( x \right) = 2x$
D
$f(x) = 3$ for at least one $x$ in $[0,2]$
Solution
${f^\prime }(c) = \frac{{f(x) – f(0)}}{{x – 0}} = \frac{{f(x)}}{x}$
$\left|\frac{f(x)}{x}\right|=\left|f^{\prime}(c)\right| \leq \frac{1}{2}$
$\Rightarrow|f(\mathrm{x})| \leq \frac{|\mathrm{x}|}{2} \leq 1 \quad \because \mathrm{x} \in[0,2] \therefore|\mathrm{x}|<2$
Standard 12
Mathematics
