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  • I need help with - Sets, Relations and Functions - JEE Main-3

The function f\left ( x \right )= \log \left ( x+\sqrt{x^{2}+1} \right ) is

  • Option 1)

    an odd function

  • Option 2)

    a periodic function

  • Option 3)

    neither an even nor an odd function

  • Option 4)

    an even function.

 

Answers (2)

best_answer

As we learnt in

Odd Function -

f(-x)= -f(x)

- wherein

Symmetric about origin

 

 f(x)=log(x+\sqrt{1+x^{2}})

f(x)=log(1-x+\sqrt{1+x^{2}})

f(x)+f(-x)=log(\sqrt{1+x^{2}}+x)+log(\sqrt{1+x^{2}}-x)

=log(1+x^{2}-x^{2})=log1=0

\therefore    It is odd function.

Correct option is 1.

 


Option 1)

an odd function

This is the correct option.

Option 2)

a periodic function

This is an incorrect option.

Option 3)

neither an even nor an odd function

This is an incorrect option.

Option 4)

an even function.

This is an incorrect option.

Posted by

Aadil

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