Question

The value of c in Rolle's theorem for $f\left(x\right)=log\left(\frac{x^2+ab}{x(a+b)}\right)$ in $(a,b)$ where $a>0$ is

• A. A.M. of $a,b$
• B. G.M. of $a,b$
• C. H.M. of $a,b$
• D. $\frac{1}{a}+\frac{1}{b}$

Answer 1

Answer: (B)

G.M. of $a,b$

Rolle's theorem states that if $f\left(x\right)$ be continuous on $[a,b]$, differentiable on $(a,b)$ and $f\left(a\right)=f\left(b\right)$ then there exists some $c$ between $a$ and $b$ such that $f{}^{\prime }\left(c\right)=0$

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

$⟹f^{{}^{\prime }}\left(x\right)=\frac{x(a+b)}{x^2+ab}\left(\frac{x(a+b)2x-(x^2+ab)(a+b)}{\left(x\right(a+b){)}^2}\right)$

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

$⟹f{}^{\prime }\left(x\right)=\frac{x^2-ab}{x(x^2+ab)}$

Therefore, $f{}^{\prime }\left(c\right)=\frac{c^2-ab}{c(c^2+ab)}=0$

$⟹c^2=ab$

$⟹c=\sqrt{ab}$

we know that G.M. of $a,b$ is $\sqrt{ab}$

Therefore, $c$ is the G.M. of $a,b$