Question

If $f\left(x\right)=x^{\alpha }\sin x$ when $x\ne 0,$ and $f\left(0\right)=0.$ If Rolle's theorem can be applied to $f$ in $[0,\pi ]$ then value(s) of $\alpha$ can be

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

Answer 1

Answer: (C), (D)

$1$

For Rolle's theorem to be applicable : $f$ should be continuous in $[0,\pi ]$ and differentible in $(0,\pi ).$
For continuity at $x=0:$
$R.H.L.=\underset{x\to 0^{+}}{lim}{x}^{\alpha }\sin x=f\left(0\right)$
$\Rightarrow \underset{x\to 0^{+}}{lim}{x}^{\alpha }\cdot \frac{\sin x}{x}\cdot x=0$
$\Rightarrow \underset{x\to 0^{+}}{lim}{x}^{\alpha +1}=0$
$\Rightarrow \alpha +1>0(\because \underset{x\to 0^{+}}{lim}{x}^{\alpha +1}\to \infty ,\text{if}\alpha +1<0\text{and}R.H.L.=1,\text{if}\alpha +1=0)$
$\therefore \alpha >-1$
Also $f\left(x\right)=x^{\alpha }\sin x$ is continuous on $[0,\pi ],$ differentiable in $(0,\pi )$, for $\alpha >-1.$
$\therefore$ Rolle's theorem is applicale on $f$ if $\alpha >-1$