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Let $f:[a, b] \rightarrow R$ be differentiable on $[a, b]$ and $k \in R$. Let $f(a)=0=f(b)$
Also let $J(x)=f'(x)+k f(x) .$ Then
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Also let $J(x)=f'(x)+k f(x) .$ Then
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Verified Answer
The correct answer is:
$J(x)=0$ has at least one root in $(a, b)$
We have, $f:[a, b] \longrightarrow R$ be differentiable
on $[a, b]$ and $k \in R,$ also $f(a)=0=f(b)$ and $\quad J(x)=f(x)+k f(x)$
Let $g(x)=k x f(x)$ which is continuous in $[a, b]$ and differentiable in $(a, b)$ such that
$$
g(a)=0=g(b)
$$
Then, for every $c \in(a, b), g^{\prime}(c)=0$
(by Rolle's theorem) $\begin{array}{lr}\text { Now, } & g^{\prime}(x)=k f(x)+k x f^{\prime}(x) \\ \Rightarrow \quad & g(c)=k f(c)+k c f^{\prime}(c)\end{array}$
$\Rightarrow \quad k f(c)+k c f(c)=0$
$\Rightarrow \quad f(x)=0,$ for every $x=c \in(a, b)$
$\therefore J(x)=0$ has atleast one root in $(a, b)$
on $[a, b]$ and $k \in R,$ also $f(a)=0=f(b)$ and $\quad J(x)=f(x)+k f(x)$
Let $g(x)=k x f(x)$ which is continuous in $[a, b]$ and differentiable in $(a, b)$ such that
$$
g(a)=0=g(b)
$$
Then, for every $c \in(a, b), g^{\prime}(c)=0$
(by Rolle's theorem) $\begin{array}{lr}\text { Now, } & g^{\prime}(x)=k f(x)+k x f^{\prime}(x) \\ \Rightarrow \quad & g(c)=k f(c)+k c f^{\prime}(c)\end{array}$
$\Rightarrow \quad k f(c)+k c f(c)=0$
$\Rightarrow \quad f(x)=0,$ for every $x=c \in(a, b)$
$\therefore J(x)=0$ has atleast one root in $(a, b)$
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