Question

The constant $c$ of Rolle's theorem for the function $f\left(x\right)=\log \frac{x^2+ab}{(a-b)x}$ in $[a,b]$ where $0\notin [a,b]$ is

• A. $\sqrt{ab}$
• B. $\frac{a+b}{2}$
• C. $\frac{a-b}{2}$
• D. $\frac{b-a}{2}$

Answer 1

The correct option is A $\sqrt{ab}$

We have,

The function $f\left(x\right)=\log \frac{x^2+ab}{(a+b)x}in[a,b]$

Now,

$x\in (a,b)$

$\therefore \frac{x^2+ab}{(a+b)x}\in R^{+ve}$

We know that,

$\left(1\right).$ log with the positive quantities as it domain is continuous.

$\therefore f\left(x\right)$ is continuous on $[a,b].$

$\left(2\right).f\left(x\right)=\log (x^2+ab)-\log (a+b)x$

$=\log (x^2+ab)-\log (a+b)x-\log x$

On differentiation and we get,

$f^{\prime }\left(x\right)=\frac{2x}{x^2+ab}-0-\frac{1}{x}\because x\in (a,b)$

So, $f^{\prime }\left(x\right)$ is exists on $(a,b)$.

$\left(3\right).f\left(a\right)=\log \left(\frac{a^2+ab}{(a+b)a}\right)=\log \left(1\right)=0$

$f\left(b\right)=\log \left(\frac{b^2+ab}{(a+b)b}\right)=\log \left(1\right)=0$

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

Then,

There exists at least one real.

$c\in (a,b)$ such that $f^{\prime }\left(c\right)=0$.

Now,

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

$\Rightarrow \frac{2c}{c^2+ab}-\frac{1}{c}=0$

$\Rightarrow \frac{2c}{c^2+ab}=\frac{1}{c}$

$\Rightarrow 2c^2=c^2+ab$

$\Rightarrow c^2=ab$

$\Rightarrow c=\sqrt{ab}\in R^{+ve}$

Hence, this is the answer.