Question

Let $f\left(x\right)=x^{\alpha }\log x$ for $x>0$ and $f\left(x\right)$ follows Rolle's theorem for $\left[0,1\right]$ then $\alpha$ is :

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

Answer 1

The correct option is C $0$

If $f\left(x\right)=a^x\log x$ and $f\left(0\right)=0$

then value of a for which rolle,s theorem is applicable

Rolle,s Theorem is Applicable in $x$ ∈[0,1]

(1) If function $f\left(x\right)$ is Continuous in $[0,1]$.

(2) If function$f\left(x\right)$ is Differentiable in $(0,1)$

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

Then There exists at least one value of x=c

for which$f\text{'}\left(c\right)=0$

Where $x$=c∈(0,1)

Now ,

$f\left(x\right)=x^a\log x\left(Given\right)$

$a=0,b=1$

By roll’s theorem

$f^{\prime }\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$=\frac{1^a\log 1-1^0\log 0}{1-0}$

$=\frac{0-0}{1}=0$