Question

Let $f(x)=\{\begin{array}{l}{x}^{\alpha }\ln x,x>0\\ 0,x=0\end{array}$. Rolle's theorem is applicable to f for $x\in [0,1]$, if $\alpha =$

• A. -2
• B. -1
• C. 0
• D. 1/2

Answer 1

Answer: (D)

$f(x)=\{\begin{array}{l}{x}^a\ln x,x>0\\ 0,x=0\end{array}$
for rolles theorem to be applicable to $f$, for $x\in [0,1]$ we should have
(i) $f(1)=f(0)$; (ii) $f$ is continuous in $[0,1]$ and (iii) $f$ is differentiable in $(0,1)$
for (i) $f(1)=|\alpha \ln |=0\Rightarrow f(1)=f(0)$
for (ii) $\underset{x\to 0^{+}}{\mathrm{Lim}}f(x)=\underset{h\to 0}{\mathrm{Lim}}<br/>(0+h{)}^{\alpha }\cdot \ln h$
$=0$ only if $\alpha <0$ only $\alpha =\frac{1}{2}$ can be the choice