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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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For Rolle's theorem in $[\mathrm{a}, \mathrm{b}], \mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b}),$
$\operatorname{In}[0,1] \Rightarrow \mathrm{f}(0)=\mathrm{f}(1)=0$
$\because$ the function has to be continuous in [0,1] $\Rightarrow f(0)=\lim _{x \rightarrow 0^{+}} f(x)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{\alpha} \log x=0 \Rightarrow \lim _{x \rightarrow 0} \frac{\log x}{x^{-\alpha}}=0$
Applying L.H. Rule $\lim _{x \rightarrow 0} \frac{1 / x}{-\alpha x^{-\alpha-1}}=0$
$\Rightarrow \lim _{x \rightarrow 0} \frac{-x^{\alpha}}{\alpha}=0 \Rightarrow \alpha>0$
$\operatorname{In}[0,1] \Rightarrow \mathrm{f}(0)=\mathrm{f}(1)=0$
$\because$ the function has to be continuous in [0,1] $\Rightarrow f(0)=\lim _{x \rightarrow 0^{+}} f(x)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{\alpha} \log x=0 \Rightarrow \lim _{x \rightarrow 0} \frac{\log x}{x^{-\alpha}}=0$
Applying L.H. Rule $\lim _{x \rightarrow 0} \frac{1 / x}{-\alpha x^{-\alpha-1}}=0$
$\Rightarrow \lim _{x \rightarrow 0} \frac{-x^{\alpha}}{\alpha}=0 \Rightarrow \alpha>0$
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