If f(x) = x^{alpha} log x, x 0, f(0) = 0 an | vedclass

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(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must be continuous on $[0, 1]$ and differentiable on $(0, 1)$,with $f(0) = f(

Vedclass · Accepted Answer

$1/2$

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(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must be continuous on $[0, 1]$ and differentiable on $(0, 1)$,with $f(0) = f(1)$.Given $f(0) = 0$ and $f(1) = 1^{\alpha} \log(1) = 0$,the condition $f(0) = f(1)$ is satisfied for any $\alpha$.For continuity at $x = 0$,we must have $\lim_{x 	o 0^+} f(x) = f(0) = 0$.$\lim_{x 	o 0^+} x^{\alpha} \log x = \lim_{x 	o 0^+} \frac{\log x}{x^{-\alpha}}$.Using $L$'$H$ôpital's rule,this limit is $\lim_{x 	o 0^+} \frac{1/x}{-\alpha x^{-\alpha-1}} = \lim_{x 	o 0^+} \frac{x^{\alpha}}{-\alpha} = 0$ if $\alpha > 0$.Checking the options,for $\alpha = 1/2$,the limit is $0$,which satisfies the continuity condition.Thus,$\alpha = 1/2$ is the correct value.

If $f(x) = x^{\alpha} \log x, x > 0, f(0) = 0$ and $f(x)$ satisfies Rolle's theorem on $[0, 1]$,then what is the value of $\alpha$?

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