Question

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

• A. -2
• B. -1
• C. 0
• D. 1/2

Answer 1

Answer: (D)

For Rolle's theorem in [a, b]
$f(a)=f(b),ln[0,1]\Rightarrow f(0)=f(1)=0$
$\because$ The function has to be continuous in [0, 1]
$\Rightarrow f\left(0\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=0\Rightarrow \underset{x\to 0}{\lim }{x}^{\alpha }\log x=0$
$\Rightarrow \underset{x\to 0}{\lim }\frac{\log x}{x^{-\alpha }}=0$
Applying L' Hospital's Rule
$\underset{x\to 0}{\lim }\frac{1/x}{-a{x}^{-\alpha -1}}=0\Rightarrow \underset{x\to 0}{\lim }\frac{-x^{\alpha }}{\alpha }=0\Rightarrow \alpha >0$