Practicing Success
Practicing Success
CUET Preparation Today
From mean value theorem: $f(b)-f(a)=(b-a) f'\left(x_1\right) ; a<x_1<b$ if $f(x)=\frac{1}{x'}$ then $x_1=$ |
$\sqrt{a b}$ $\frac{a+b}{2}$ $\frac{2 a b}{a+b}$ $\frac{b-a}{b+a}$ |
$\sqrt{a b}$ |
Since $f(x)=\frac{1}{x}$ $f'\left(x_1\right)=\frac{f(b)-f(a)}{b-a}$ $\Rightarrow -\frac{1}{x_1^2}=\frac{1 / b-1 / a}{b-a}$ $\Rightarrow x_1=\sqrt{a b}$ |
