# Let $F\Left( X \Right) = \Sqrt{X^2 + 1}\ ] . Then, Which of the Following is Correct? - Mathematics MCQ 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:

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

Replacing x by 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,

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Q 43 | Page 45