Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! - Sets, Relations and Functions - JEE Main

The domain of the function

f(x)=\frac{1}{\sqrt{\left | x \right |-x}}is

  • Option 1)

    (-\infty ,0)\;

  • Option 2)

    \; (-\infty,\infty)-\left \{ 0 \right \}\;

  • Option 3)

    \; \; (-\infty,\infty)\;

  • Option 4)

    \; (0,\infty)

 

Answers (2)

best_answer

As we learnt in

DOMAIN -

The domain of R is the set of all first elements of the ordered pairs in a relation R.

- wherein

eg. R={(a,b),(c,d)}, then domain is {a,c}

 

 f(x)=\frac{1}{\sqrt{|x|-x}},                 |x| - x > 0

    x < |x|  it is only for -ve number. So that n\epsilon(-\infty, 0)

Correct option is 1.

 


Option 1)

(-\infty ,0)\;

This is the correct option.

Option 2)

\; (-\infty,\infty)-\left \{ 0 \right \}\;

This is an incorrect option.

Option 3)

\; \; (-\infty,\infty)\;

This is an incorrect option.

Option 4)

\; (0,\infty)

This is an incorrect option.

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question