Let y = f(x) and y = g(x) be two differentiab | vedclass

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(A) Define a function $h(x) = f(x) - k g(x)$.Since $f(x)$ and $g(x)$ are differentiable in $[0, 2]$,$h(x)$ is also differentiable in $[0, 2]$.Given th

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$2$

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(A) Define a function $h(x) = f(x) - k g(x)$.Since $f(x)$ and $g(x)$ are differentiable in $[0, 2]$,$h(x)$ is also differentiable in $[0, 2]$.Given that there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,we have $h'(c) = f'(c) - k g'(c) = 0$.By Rolle's Theorem,if $h'(c) = 0$ for some $c \in (0, 2)$,then $h(0) = h(2)$.Substituting the given values:$h(0) = f(0) - k g(0) = 3 - k(1) = 3 - k$$h(2) = f(2) - k g(2) = 5 - k(2) = 5 - 2k$Equating $h(0) = h(2)$:$3 - k = 5 - 2k$$2k - k = 5 - 3$$k = 2$

Let $y = f(x)$ and $y = g(x)$ be two differentiable functions in $[0, 2]$ such that $f(0) = 3$,$f(2) = 5$,$g(0) = 1$,and $g(2) = 2$. If there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,then $k$ must be:

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