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Q.
Let $f(x)=\begin{cases}x^\alpha \ln x, x>0 \\ 0, \quad x=0\end{cases}$. Rolle's theorem is applicable to f for $x \in[0,1]$, if $\alpha=$
Application of Derivatives
Solution:
$f ( x )=\begin{cases}x ^a \ln x , x >0 \\0, \quad x =0\end{cases}$
for rolles theorem to be applicable to $f$, for $x \in[0,1]$ we should have
(i) $f(1)=f(0)$; (ii) $f$ is continuous in $[0,1]$ and (iii) $f$ is differentiable in $(0,1)$
for (i) $f (1)=|\alpha \ln |=0 \quad \Rightarrow f (1)= f (0)$
for (ii) $\underset{x \rightarrow 0^{+}}{\text{Lim}} f(x)=\underset{h \rightarrow 0}{\text{Lim}}
(0+h)^\alpha \cdot \ln h$
$=0$ only if $\alpha<0$ only $\alpha=\frac{1}{2}$ can be the choice