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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                        

     
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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?
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APPEARS IN

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