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Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in (1,3) an $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,
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Verified Answer
The correct answer is:
$f(3)-f(1)=5$ is false
Given that $f:[1,3] \rightarrow R$ be a continuous and differentiable in (1,3)
and $f^{\prime}(x)=|f(x)|^{2}+4$
By applying $L M V T,$ there exist at least one point $c \in(1,3)$ such that
$$
\begin{array}{c}
\frac{f(3)-f(1)}{3-1}=f^{\prime}(c)\left[\because f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\right] \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=f^{\prime}(c) \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=\left[f|c|^{2}+4\right] \\
\Rightarrow f^{\prime}(c)=|f(c)|^{2}+4 \\
\Rightarrow \quad f(3)-f(1)=2 \cdot|f(c)|^{2}+8
\end{array}
$$
$\Rightarrow \quad f(3)-f(1) \geq 8$
and $f^{\prime}(x)=|f(x)|^{2}+4$
By applying $L M V T,$ there exist at least one point $c \in(1,3)$ such that
$$
\begin{array}{c}
\frac{f(3)-f(1)}{3-1}=f^{\prime}(c)\left[\because f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\right] \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=f^{\prime}(c) \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=\left[f|c|^{2}+4\right] \\
\Rightarrow f^{\prime}(c)=|f(c)|^{2}+4 \\
\Rightarrow \quad f(3)-f(1)=2 \cdot|f(c)|^{2}+8
\end{array}
$$
$\Rightarrow \quad f(3)-f(1) \geq 8$
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