Question:
Let $f(x)=\sqrt{x^{2}+1}$. Then, which of the following is correct?
(a) $f(x y)=f(x) f(y)$
(b) $f(x y) \geq f(x) f(y)$
(c) $f(x y) \leq f(x) f(y)$
(d) none of these
Solution:
Given: $f(x)=\sqrt{x^{2}+1}$ ....(1)
Replacing x by y in (1), we get
$f(y)=\sqrt{y^{2}+1}$
$\therefore f(x) f(y)=\sqrt{x^{2}+1} \sqrt{y^{2}+1}$
$=\sqrt{\left(x^{2}+1\right)\left(y^{2}+1\right)}$
$=\sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$
Also, replacing x by xy in (1), we get
$f(x y)=\sqrt{x^{2} y^{2}+1}$
Now,
$x^{2} y^{2}+1 \leq x^{2} y^{2}+x^{2}+y^{2}+1$
$\Rightarrow \sqrt{x^{2} y^{2}+1} \leq \sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$
$\Rightarrow f(x y) \leq f(x) f(y)$
Hence, the correct answer is option (c).