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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
Solution
Given:
Replacing x by y in (1), we get
\[\therefore f\left( x \right)f\left( y \right) = \sqrt{x^2 + 1}\sqrt{y^2 + 1}\]
\[ = \sqrt{\left( x^2 + 1 \right)\left( y^2 + 1 \right)}\]
\[ = \sqrt{x^2 y^2 + x^2 + y^2 + 1}\]
Also, replacing x by xy in (1), we get
Now,
\[x^2 y^2 + 1 \leq x^2 y^2 + x^2 + y^2 + 1\]
\[ \Rightarrow \sqrt{x^2 y^2 + 1} \leq \sqrt{x^2 y^2 + x^2 + y^2 + 1}\]
\[ \Rightarrow f\left( xy \right) \leq f\left( x \right)f\left( y \right)\]
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