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)

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