Question

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

• A. $-2$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Here $f\left(0\right)=0$

$f\left(\Delta \right)=1^r\log 1=0$

$\therefore$ We have only to check continuity of $f\left(x\right)$ in $[0,1]$ for the applicability of Rolle's theorem

$\therefore \underset{x\to 0}{lim}{x}^{\alpha }\log x=0$ [for continuity at $x=0$]

$\underset{x\to 0}{lim}{x}^{\alpha }\log x=\underset{x\to 0}{lim}=\frac{\log x}{x^{-\alpha }}$ (Using $L$-Hospital rule)

$\underset{x\to 0}{lim}\frac{1/x}{-\alpha x^{-\alpha -1}}$

$\underset{x\to 0}{lim}\frac{1}{-\alpha }{x}^{\alpha }$

for above limit to be equal to $0$ we must have

$\alpha >0$

$\therefore \alpha =1/2$ is correct option