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]
$\Rightarrow f\left(0\right) = \displaystyle \lim_{x \to 0^{+} }f\left(x\right) =0 \Rightarrow \displaystyle \lim_{x\to 0} x^{\alpha }\log x = 0$
$ \Rightarrow \displaystyle \lim_{x\to 0} \frac{\log x}{x^{-\alpha}} = 0$
Applying L' Hospital's Rule
$\displaystyle \lim_{x \to 0} \frac{1/x}{-ax^{-\alpha -1}} = 0 \Rightarrow \displaystyle \lim_{x\to 0} \frac{-x^{\alpha}}{\alpha} = 0 \Rightarrow \alpha > 0$