Question

Let $f\left(x\right)=\sqrt{x^2+1}$. Then, which of the following is correct?

(a) $f\left(xy\right)=f\left(x\right)f\left(y\right)$ (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

Given: $f\left(x\right)=\sqrt{x^2+1}$ .....(1)

Replacing x by y in (1), we get

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

$$\begin{gathered} \therefore f\left(x\right)f\left(y\right)=\sqrt{x^2+1}\sqrt{y^2+1} \\ =\sqrt{\left(x^2+1\right)\left(y^2+1\right)} \\ =\sqrt{x^2{y}^2+x^2+y^2+1} \end{gathered}$$

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

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

Now,

$$\begin{gathered} x^2{y}^2+1\le x^2{y}^2+x^2+y^2+1 \\ \Rightarrow \sqrt{x^2{y}^2+1}\le \sqrt{x^2{y}^2+x^2+y^2+1} \\ \Rightarrow f\left(xy\right)\le f\left(x\right)f\left(y\right) \end{gathered}$$

Hence, the correct answer is option (c).