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If $f:R\rightarrow R$ is twice differentablr function such that f''(x)>0, for all xε R, and $f(\frac{1}{2})=\frac{1}{2}$ , f(1)=1, then

$f''(1)\leq0$

$f'(1)>1$

$0\lt f'(1) \le \frac{1}{2}$

$\frac{1}{2}\lt f'(1) \le 1$

Option B

f'(x) is increasing

For some x in ( $\frac{1}{2},1$ )

f'(x)=1

f'(1)>1

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