Question

If $f\left(x\right)=x^{a}logx$ and $f\left(0\right)=0$, then the value of a for which Rolles theorem can be applied in [0, 1] is

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

Answer 1

Answer: (C)

$0$

For Rolles theorem to be applicable function must be continuous in the interval $[0,1]$

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

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

Applying L'Hospital's Rule,

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

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

For the limit to exist $\alpha >0$

For $\alpha >0$. $\underset{x\to 0^{+}}{lim}f\left(x\right)=0$

So, the answer is option (C).