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-
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 $[0,2]$
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