Let f and g be differentiable on the interval | vedclass

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(A, C) Consider the function $h(x) = e^{g(x)} f(x)$.Since $f$ and $g$ are differentiable on $I$,$h(x)$ is also differentiable on $I$.Given $f(a) = 0$

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If $g(a)=0=g(b)$,the equation $g^{\prime}(x)+k g(x)=0$ is solvable in $(a, b), k \in R$

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(A, C) Consider the function $h(x) = e^{g(x)} f(x)$.Since $f$ and $g$ are differentiable on $I$,$h(x)$ is also differentiable on $I$.Given $f(a) = 0$ and $f(b) = 0$,we have $h(a) = e^{g(a)} f(a) = 0$ and $h(b) = e^{g(b)} f(b) = 0$.By Rolle's Theorem,there exists at least one $c \in (a, b)$ such that $h^{\prime}(c) = 0$.$h^{\prime}(x) = e^{g(x)} g^{\prime}(x) f(x) + e^{g(x)} f^{\prime}(x) = e^{g(x)} [f^{\prime}(x) + f(x) g^{\prime}(x)]$.Since $e^{g(x)} 
eq 0$,$h^{\prime}(c) = 0$ implies $f^{\prime}(c) + f(c) g^{\prime}(c) = 0$.Thus,option $(a)$ is correct.Similarly,for option $(c)$,consider $m(x) = e^{kx} g(x)$.Since $g(a) = 0$ and $g(b) = 0$,we have $m(a) = 0$ and $m(b) = 0$.By Rolle's Theorem,there exists $c \in (a, b)$ such that $m^{\prime}(c) = 0$.$m^{\prime}(x) = e^{kx} g^{\prime}(x) + k e^{kx} g(x) = e^{kx} [g^{\prime}(x) + k g(x)]$.Since $e^{kx} 
eq 0$,$m^{\prime}(c) = 0$ implies $g^{\prime}(c) + k g(c) = 0$.Thus,option $(c)$ is also correct.

Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,

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