Question

If $a,b,c,d$ be the sides of a quadrilateral and $g\left(x\right)=f\left[f\left\{f\left(x\right)\right\}\right],$ where $f\left(x\right)=\frac{1}{1-x}$, then $\frac{d^2}{a^2+b^2+c^2}$ is equal to

• A. greater than $g\left(3\right)$
• B. lesser than $g\left(3\right)$
• C. greater then$g\left(2\right)$
• D. lesser than $g\left(4\right)$

Answer 1

Answer: (B)

lesser than $g\left(3\right)$

$g\left(x\right)=f\left[f\left\{f\left(x\right)\right\}\right]$
$\Rightarrow g\left(x\right)=f\left[f\left\{\frac{1}{1-x}\right\}\right]$
$\Rightarrow g\left(x\right)=f\left(\frac{1}{1-\frac{1}{1-x}}\right)$
$=f\left(\frac{1-x}{-x}\right)$
$=\frac{1}{1-\frac{1-x}{-x}}$
$=\frac{-x}{-x-1+x}=x$
$\because {(a-b)}^2+{(b-c)}^2+{(c-a)}^2\ge 0$
$\Rightarrow 2(a^2+b^2+c^2)\ge 2(ab+bc+ca)$
$\Rightarrow 3(a^2+b^2+c^2)\ge {(a+b+c)}^2>d^2$
$\therefore \frac{d^2}{a^2+b^2+c^2}<3=g\left(3\right)$
$\therefore \frac{d^2}{a^2+b^2+c^2}<g\left(3\right)$