Let f(x)=


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    


Given: $f(x)=\sqrt{x^{2}+1}$  ....(1)

Replacing x by in (1), we get


$\therefore f(x) f(y)=\sqrt{x^{2}+1} \sqrt{y^{2}+1}$



$=\sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$

Also, replacing x by xy in (1), we get

$f(x y)=\sqrt{x^{2} y^{2}+1}$


$x^{2} y^{2}+1 \leq x^{2} y^{2}+x^{2}+y^{2}+1$

$\Rightarrow \sqrt{x^{2} y^{2}+1} \leq \sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$


$\Rightarrow f(x y) \leq f(x) f(y)$

Hence, the correct answer is option (c).

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now