Question

Let $f\left(x\right)=\left\{x^a\ln x,whenx>0and0,whenx=0\right\}$, Rolle.s theorem is applicable to $f$for $x\in \left[0,1\right]$, if $a=$?

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

Answer 1

The correct option is D

$\frac{1}{2}$

Explanation for the correct option:

Given that: $f\left(x\right)=\left\{x^a\ln x,whenx>0and0,whenx=0\right\}$

Rolle's theorem is applicable to $f$ for $x\in \left[0,1\right]$.

This means that:

$f\left(1\right)=f\left(0\right)..........\left(1\right)$

$f$ is continuous in $x\in \left[0,1\right]$ and $f$ is differentiable for $x\in (0,1)$ $...........\left(2\right)$

From $\left(1\right)$and $\left(2\right)$ we get that:

$f\left(1\right)=0$

$0=f\left(0\right)=li{m}_{x\to a^{+}}{x}^a\ln x$ which is true only for positive values of $a$.

Thus, the correct Option is (D).