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If $f(x)$ is a differentiable function, $f^{\prime}(x) \geq 5 \forall x \in[2,6]$, $f(2)=4$ and $f(3)=15$, then a possible value of $f(6)$
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Verified Answer
The correct answer is:
$=24$
Since $f(3) \neq f(2)$
$\therefore$ Using Mean value theorem
$$
\begin{aligned}
& f^{\prime}(x)=\frac{f(3)-f(2)}{3-2}=\frac{15-4}{3-2} \\
& \Rightarrow f^{\prime}(x)=11 \\
& \Rightarrow f^{\prime}(x) \geq 5 \\
& \Rightarrow \frac{f(6)-f(2)}{6-2} \geq 5 \\
& \Rightarrow f(6)-4 \geq 20 \\
& \Rightarrow f(6) \geq 24 \\
& \Rightarrow \text { Possible values of } f(6)=24 .
\end{aligned}
$$
$\therefore$ Using Mean value theorem
$$
\begin{aligned}
& f^{\prime}(x)=\frac{f(3)-f(2)}{3-2}=\frac{15-4}{3-2} \\
& \Rightarrow f^{\prime}(x)=11 \\
& \Rightarrow f^{\prime}(x) \geq 5 \\
& \Rightarrow \frac{f(6)-f(2)}{6-2} \geq 5 \\
& \Rightarrow f(6)-4 \geq 20 \\
& \Rightarrow f(6) \geq 24 \\
& \Rightarrow \text { Possible values of } f(6)=24 .
\end{aligned}
$$
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