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)

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.

N neha

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions