Question

If $f\left(x\right)=x^{\alpha }\ell nx$, and $f\left(0\right)=0$ if Rolle's theorem can be applied to $f$ in $[0,1]$, then value of $\alpha$ can be

• A. $-2$
• B. $-1$
• C. $0$
• D. $1/2$

Answer 1

The correct option is D $1/2$

$\underset{x\to 0^{+}}{lim}{x}^{\alpha }\ln \left(x\right)=0$

If $\alpha <0$ then

$\underset{x\to 0^{+}}{lim}{x}^{\alpha }=+\infty ,\underset{x\to 0^{+}}{lim}\ln x=-\infty$

If $\alpha =0$ then

$\underset{x\to 0^{+}}{lim}{x}^{\alpha }=1,\underset{x\to 0^{+}}{lim}1\times x=\infty$

If $\alpha >0$ then

$\underset{x\to 0^{+}}{lim}{x}^{\alpha }\cong 0\underset{x\to 0^{+}}{lim}\ln x=\infty$

then $xy$ limit is defined $\underset{x\to 0^{+}}{lim}{x}^{\alpha }\ln x$ can be equal zero

So $\alpha =\frac{1}{2}$

$D$ is correct.