Question

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

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

Answer 1

Answer: (B)

We have, $f(x)=x^{\alpha }\log x$ and $f(x)=0$
For Rolle's theorem, in $[0,1]$
$\Rightarrow f(0)=f(1)=0$
$\because$ The function has to be continuous in $[0,1]$.
$\Rightarrow f(0)=\underset{x\to 0^{+}}{\lim }f(x)=0$
$=\underset{x\to 0}{\lim }{x}^{\alpha }\log x=0$
$=\underset{x\to 0}{\lim }\frac{\log x}{x^{-\alpha }}=0$
Applying L'Hospital rule, we get
$\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{-\alpha x^{-\alpha -1}}=0$
$\Rightarrow \underset{x\to 0}{\lim }\frac{-x^{\alpha }}{\alpha }=0$
Here, $x$ is very near to zero.
$\therefore$ From given option,
For $\alpha =\frac{1}{2}$, i.e. $(x{)}^{1/2}$ is near to zero.