Question

If $f\left(x\right)=x^{\alpha }\log x$ and $f\left(0\right)=0,$ then the value of $\alpha$ for which $Roll{e}^{\prime }stheorem$ can be applied in $[0,1],$ is

• A. $-1$
• B. $-2$
• C. $0$
• D. none of these

Answer 1

The correct option is C $0$

For function to satisfy the condition of Rolle's theorem,

it should be contiunous in $[0,1]$

$\begin{array}{c}\Rightarrow {lim}_{x\to 0^{+}}f\left(x\right)=f\left(0\right)\\ \Rightarrow {lim}_{x\to 0^{+}}\frac{\log x}{x^{-\alpha }}=0\\ \Rightarrow {lim}_{x\to 0^{+}}\frac{\frac{1}{x}}{-\alpha x^{-\alpha -1}}=0\\ \Rightarrow \alpha >0\\ Also\forall \alpha >0\\ f\left(x\right)isdifferentiablein\\ \left(0,1\right)andf\left(1\right)=0=f\left(0\right)\end{array}$