Question

$\text{If}f\left(x\right)=\sqrt{1+x^2}\text{, then}$ ____

• A. $f\left(xy\right)=f\left(x\right)+f\left(y\right)$
• B. $f\left(xy\right)\le f\left(x\right)\times f\left(y\right)$
• C. $f\left(xy\right)\ge f\left(x\right)\times f\left(y\right)$
• D. $f\left(xy\right)=f\left(x\right)\times f\left(y\right)$

Answer 1

The correct option is B $f\left(xy\right)\le f\left(x\right)\times f\left(y\right)$

$\text{Given,}f\left(x\right)=\sqrt{1+x^2}$
$\therefore f\left(y\right)=\sqrt{1+y^2}\text{and}f\left(xy\right)=\sqrt{1+(xy{)}^2}$
$f\left(x\right)\times f\left(y\right)=\sqrt{1+x^2}\times \sqrt{1+y^2}$
$=\sqrt{1+x^2+y^2+x^2{y}^2}$

$\text{As}\sqrt{1+(xy{)}^2}\le \sqrt{1+x^2+y^2+x^2{y}^2}$
$\therefore f\left(xy\right)\le f\left(x\right)\times f\left(y\right)$