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Let $f^{\prime}(0)=-3$ and $f^{\prime}(x) \leq 5$ for all real values of $x$. The $\mathrm{f}(2)$ can have possible maximum value as
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The correct answer is:
7
Applying Lagrange's mean value theorem on interval $[0,2]$, we get there exist atleast one ' $c$ ' $\in(0,2)$ such that
$\begin{array}{ll}
& \frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)-\mathrm{f}(0)=2 \mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)=\mathrm{f}(0)+2 \mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)=-3+2 \mathrm{f}^{\prime}(\mathrm{c})
\end{array}$
Given that $\mathrm{f}^{\prime}(x) \leq 5$ for all $x$
$\begin{array}{ll}
\therefore & \mathrm{f}(2) \leq-3+10 \\
\therefore & \mathrm{f}(2) \leq 7
\end{array}$
$\therefore \quad$ Largest possible value of $\mathrm{f}(2)$ is 7 .
$\begin{array}{ll}
& \frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)-\mathrm{f}(0)=2 \mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)=\mathrm{f}(0)+2 \mathrm{f}^{\prime}(\mathrm{c}) \\
\therefore & \mathrm{f}(2)=-3+2 \mathrm{f}^{\prime}(\mathrm{c})
\end{array}$
Given that $\mathrm{f}^{\prime}(x) \leq 5$ for all $x$
$\begin{array}{ll}
\therefore & \mathrm{f}(2) \leq-3+10 \\
\therefore & \mathrm{f}(2) \leq 7
\end{array}$
$\therefore \quad$ Largest possible value of $\mathrm{f}(2)$ is 7 .
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