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Q.
If $ f(x) = x^\alpha \log x $ and $ f(0) = 0 $ , then the value
of a for which Rolle’s theorem can be applied in $ [0,1] $ is
AMUAMU 2016Continuity and Differentiability
Solution:
We have, $f(x)=x^{\alpha} \log x$ and $f(x)=0$
For Rolle's theorem, in $[0,1]$
$\Rightarrow f(0)=f(1)=0$
$\because$ The function has to be continuous in $[0,1]$.
$\Rightarrow f(0) =\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=0$
$=\displaystyle\lim _{x \rightarrow 0} x^{\alpha} \log x=0 $
$=\displaystyle\lim _{x \rightarrow 0} \frac{\log x}{x^{-\alpha}}=0$
Applying L'Hospital rule, we get
$\displaystyle\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{-\alpha x^{-\alpha-1}}=0$
$ \Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{-x^{\alpha}}{\alpha}=0$
Here, $x$ is very near to zero.
$\therefore$ From given option,
For $\alpha=\frac{1}{2}$, i.e. $(x)^{1 / 2}$ is near to zero.