Question

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

• A. $f(xy)=f(x).f(y)$
• B. $f\left(xy\right)\ge f\left(x\right).f\left(y\right)$
• C. $f\left(xy\right)\le f\left(x\right).f\left(y\right)$
• D. None of these

Answer 1

Answer: (C)

$f\left(xy\right)=\sqrt{1+x^2{y}^2}$
$f\left(x\right)f\left(y\right)=\sqrt{1+x^2}\sqrt{1+y^2}=\sqrt{1+x^2{y}^2+x^2+y^2}$
$\ge \sqrt{1+x^2{y}^2}=f\left(xy\right)$
$\therefore f\left(xy\right)\le f\left(x\right)f\left(y\right)$