Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A discontinuous function  satisfying. <img alt="\mathrm{x^2+y^2=4}" src="htt

A discontinuous function \mathrm{y=f(x)} satisfying. \mathrm{x^2+y^2=4} is given by \mathrm{f(x)=} ? for \mathrm{0 \leq x \leq 2}

Option: 1

\mathrm{-\sqrt{\left(4-x^2\right)}}


Option: 2

\mathrm{4 x}


Option: 3

\mathrm{\sqrt{x^2-4}}


Option: 4

\mathrm{ \sqrt{x^2+4}}


Answers (1)

best_answer

By choosing any arc of the circle \mathrm{x^2+y^2=4} we can define a discontinuous function, one of which is
Hence,\mathrm{\begin{aligned} & f(x)=\sqrt{\left(4-x^2\right) ;-2} \leq x \leq 0 \\ & f(x)=\left\{\begin{array}{l} \sqrt{\left(4-x^2\right)} ;-2 \leq x \leq 0 \\ -\sqrt{\left(4-x^2\right)} ; 0 \leq x \leq 2 \end{array}\right. \end{aligned}}

Posted by

Ritika Jonwal

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