(B) Given that $f:[1,3] \rightarrow R$ is continuous on $[1,3]$ and differentiable in $(1,3)$ with $f^{\prime}(x)=|f(x)|^{2}+4$. By applying the Lagra

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(B) Given that $f:[1,3] \rightarrow R$ is continuous on $[1,3]$ and differentiable in $(1,3)$ with $f^{\prime}(x)=|f(x)|^{2}+4$. By applying the Lagrange Mean Value Theorem $(LMVT)$, there exists at least one point $c \in (1,3)$ such that: $\frac{f(3)-f(1)}{3-1} = f^{\prime}(c)$ $\frac{f(3)-f(1)}{2} = |f(c)|^{2} + 4$ Since $|f(c)|^{2} \geq 0$, we have $|f(c)|^{2} + 4 \geq 4$. Therefore, $\frac{f(3)-f(1)}{2} \geq 4$, which implies $f(3)-f(1) \geq 8$. Thus, the statement $f(3)-f(1)=5$ is false.

Class 12 Mathematics Continuity and Differentiation Rolle’s theorem, Lagrange's mean value theorem

Medium

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

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A
$f(3)-f(1)=5$ is true
B
$f(3)-f(1)=5$ is false
C
$f(3)-f(1)=7$ is true
D
$f(3)-f(1) < 0$ only at one point of $(1,3)$

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Rolle’s theorem, Lagrange's mean value theorem

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