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If \mathrm{ 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

Option: 1

-2


Option: 2

-1


Option: 3

0


Option: 4

\frac{1}{3}


Answers (1)

best_answer

 We must have
(i) f(1)=f(0)=0 (given)
(ii) f(x) is continuous in [0,1]
\mathrm{\text { i.e., } f(0)=\lim _{z \rightarrow 0^{+}} f(z)=0 }
(iii) f(x) is differentiable in (0,1) i.e., \mathrm{f^{\prime}(x)=0 }
\mathrm{\begin{aligned} & f^{\prime}(x)=x^{\alpha-1}(1+\alpha \log x)=0 \Rightarrow 1+\alpha \log x=0 \\ \therefore & \alpha=-\frac{1}{\log x} \\ \therefore \quad & \text { But } \log x \text { is - ve in }[0,1] \\ \therefore \quad & \alpha \text { is positive. } \end{aligned} }

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