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Let $f:[a, b] \rightarrow R$ be such that $f$ is differentiable in $(a, b)$ is continuous at $x=a$ and $x=b$ and moreover $f(a)=0=f(b)$. Then
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The correct answer is:
there exists at least one point $c$ in $(a,b)$ such that $f^{\prime}=f$
Let $g(x)=e^{-x} f(x)$
such that $g(a)=0, g(b)=0$
and $g(x)$ is continuous and differentiable.
Then, for atleast one value of $c \in(a, b)$ such that $g(c)=0$
Now, $g(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)$
$\Rightarrow \quad g(c)=e^{-c} f^{\prime}(c)+\left(-e^{-\ell}\right) f(c)=0$
$\Rightarrow e^{-c} f^{\prime}(c)=e^{-c} f(c) \Rightarrow f^{\prime}(c)=f(c)$
such that $g(a)=0, g(b)=0$
and $g(x)$ is continuous and differentiable.
Then, for atleast one value of $c \in(a, b)$ such that $g(c)=0$
Now, $g(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)$
$\Rightarrow \quad g(c)=e^{-c} f^{\prime}(c)+\left(-e^{-\ell}\right) f(c)=0$
$\Rightarrow e^{-c} f^{\prime}(c)=e^{-c} f(c) \Rightarrow f^{\prime}(c)=f(c)$
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