Question

Discuss the applicability of Rolle's theorem to $f\left(x\right)=\log \left[\frac{x^2+ab}{(a+b)x}\right],$ in the interval$[a,b].$

• A. Yes Rolle's theorem is applicable and the stationary point is $x=\sqrt{ab}$
• B. No Rolle's theorem is not applicable due to the discontinuity in the given interval
• C. Yes Rolle's theorem is applicable and the stationary point is $x=ab$
• D. none of these

Answer 1

The correct option is A Yes Rolle's theorem is applicable and the stationary point is $x=\sqrt{ab}$

We have $f\left(a\right)=\log \left[\frac{a^2+ab}{(a+b)a}\right]=\log 1=0$ and $f\left(b\right)=\log \left[\frac{b^2+ab}{(a+b)b}\right]=\log 1=0$
$\Rightarrow f\left(a\right)=f\left(b\right)=0.$ Also, it
can be easily seen that $f\left(x\right)$ is continuous on $[a,b]$ and differentiable on
$[a,b]$.
Thus all the three conditions of Rolle's
theorem are satisfied. Hence $f^{{}^{\prime }}\left(x\right)=0$for at last one value of x in $[a,b]$
Now $f^{{}^{\prime }}\left(x\right)=0=\frac{2x}{x^2+ab}-\frac{1}{x}=0\Rightarrow 2x^2-(x^2+ab)=0=x^2=ab$ or $x=\sqrt{ab}$ which is also known as stationary point.