Question

If $f\left(x\right)=x^n\ln x$ and $f\left(0\right)=0$ then value of $\alpha$ for which Rolle's Theorem can be applied in $[0,1]$

• A. $-2$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $0$

$f\left(x\right)=x^n\ln x$

$\Rightarrow f^{\prime }\left(x\right)=n{x}^{n-1}\ln x+\frac{x^n}{x}$

$\Rightarrow f^{\prime }\left(x\right)=n{x}^{n-1}\ln x+x^{n-1}$

$\Rightarrow f^{\prime }\left(x\right)=x^{n-1}(n\ln x+1)$

According to Rolle's Theorem

$\Rightarrow f^{\prime }\left(c\right)=0$

$\Rightarrow c^{n-1}(n\ln c+1)=0$

Thus $c=0$ or $n\ln c+1=0\Rightarrow c^n=e$

But $c$ should lie in $[0,1]$

Therefore, $c=0$