Question

If Rolle's theorem is applicable to the function, $f\left(x\right)=\frac{\ln x}{x}$ over the interval $[a,b]$, where $a,b\in I^{+}$ then the value of $a+b$ is

Answer 1

We have,

$f\left(x\right)=\frac{\ln x}{x}$

$f\left(a\right)=f\left(b\right)$

but $a\ne b$

Therefore,

$\frac{\ln a}{a}=\frac{\ln b}{b}$

$b\ln a=a\ln b$

$a^b=b^a$

Put $a=2$ and $b=4$

$2^4=4^2$

So, the value of $a$ and $b$ is $2,4$, respectively. (By Hit and trial and these are the only possible integral values. Also, a clear bijection exists here, which means $a=4,b=2$)

Therefore,

The value of $a+b=2+4=6$

Hence, the value of $a+b=6$.