# Let f(x)=

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 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).