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Let $f^{\prime}(x)$, be differentiable $\forall x$. If $f(1)=-2$ and $f^{\prime}(x) \geq 2 \forall x \in[1,6]$, then
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Verified Answer
The correct answer is:
$f(6) \geq 8$
$f^{\prime}(x)$ is differentiable $\forall x \in[1,6]$
By Lagrange's mean value theorem,
$\begin{array}{l}
\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \\
\mathrm{f}^{\prime}(\mathrm{x}) \geq 2 \forall \mathrm{x} \in[1,6]
\end{array}$
(given)
$\Rightarrow \frac{\mathrm{f}(6)+2}{5} \geq 2 \quad[\because \mathrm{f}(1)=-2]$
$\Rightarrow \mathrm{f}(6) \geq 10-2 \Rightarrow \mathrm{f}(6) \geq 8$
By Lagrange's mean value theorem,
$\begin{array}{l}
\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \\
\mathrm{f}^{\prime}(\mathrm{x}) \geq 2 \forall \mathrm{x} \in[1,6]
\end{array}$
(given)
$\Rightarrow \frac{\mathrm{f}(6)+2}{5} \geq 2 \quad[\because \mathrm{f}(1)=-2]$
$\Rightarrow \mathrm{f}(6) \geq 10-2 \Rightarrow \mathrm{f}(6) \geq 8$
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