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Let $f$ be differentiable for all $x$. If $f(1)=-2$ and $f'(x)\geq 2$ for $x\in [1,6]$ then

$\begin{array}{1 1}(a)\;f(6)\geq 8&(b)\;f(6)< 8\\(c)\;f(6)< 5&(d)\;f(6)=5\end{array}$

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As $f(1)=-2$ and $f'(x)\geq 2\forall x\in [1,6]$
Applying Langrange's mean value theorem
$\large\frac{f(6)-f(1)}{5}$$=f'(c)\geq 2$
$\Rightarrow f(6)\geq 10+f(1)$
$\Rightarrow f(6)\geq 10-2$
$\Rightarrow f(6)\geq 8$
Hence (a) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 

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