Question

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

 

Options

• (a)  \[f\left( xy \right) = f\left( x \right)f\left( y \right)\]

   

• (b)  \[f\left( xy \right) \geq f\left( x \right)f\left( y \right)\]

   

•   (c)  \[f\left( xy \right) \leq f\left( x \right)f\left( y \right)\]

   

• (d) none of these                        

   

Answer 1

Given: 

\[f\left( x \right) = \sqrt{x^2 + 1}\]        .....(1)

Replacing x by y in (1), we get

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

\[\therefore f\left( x \right)f\left( y \right) = \sqrt{x^2 + 1}\sqrt{y^2 + 1}\]
\[ = \sqrt{\left( x^2 + 1 \right)\left( y^2 + 1 \right)}\]
\[ = \sqrt{x^2 y^2 + x^2 + y^2 + 1}\]

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

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

Now,

\[x^2 y^2 + 1 \leq x^2 y^2 + x^2 + y^2 + 1\]
\[ \Rightarrow \sqrt{x^2 y^2 + 1} \leq \sqrt{x^2 y^2 + x^2 + y^2 + 1}\]
\[ \Rightarrow f\left( xy \right) \leq f\left( x \right)f\left( y \right)\]