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If $f(x)=x^\alpha \log x$ and $f(0)=0$, then the value of $\alpha$ for which Rolle's theorem can be applied in $[0,1]$ is
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The correct answer is:
$1 / 2$
$f(x)=x^a \log x$ and $f(0)=0$
$\because$ Rolle's theorem is applicable in $[0,1]$.
$\Rightarrow$ It is continuous in closed interval $[0,1]$ and It is differentiable in $(a, b)$ and $f(0)=f(1)$.
$\because f(x)$ is continuous at $x=0$
$\because \quad \lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \quad \lim _{x \rightarrow 0^{+}} x^a \log x=0$
$$
\begin{aligned}
\Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\log x}{x^{-a}} & =0 \\
& {\left[\because \text { It is } \frac{\infty}{\infty} \text { indeterminate form }\right] }
\end{aligned}
$$
Using L' hospital rule
$$
\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-a \cdot x^{-a-1}}=0 \Rightarrow \lim _{x \rightarrow 0^{+}}-\frac{x^a}{a}=0
$$
$\because$ It is possible when $a>0 \Rightarrow a=\frac{1}{2}$
$\because$ Rolle's theorem is applicable in $[0,1]$.
$\Rightarrow$ It is continuous in closed interval $[0,1]$ and It is differentiable in $(a, b)$ and $f(0)=f(1)$.
$\because f(x)$ is continuous at $x=0$
$\because \quad \lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \quad \lim _{x \rightarrow 0^{+}} x^a \log x=0$
$$
\begin{aligned}
\Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\log x}{x^{-a}} & =0 \\
& {\left[\because \text { It is } \frac{\infty}{\infty} \text { indeterminate form }\right] }
\end{aligned}
$$
Using L' hospital rule
$$
\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-a \cdot x^{-a-1}}=0 \Rightarrow \lim _{x \rightarrow 0^{+}}-\frac{x^a}{a}=0
$$
$\because$ It is possible when $a>0 \Rightarrow a=\frac{1}{2}$
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