Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The straight line    divides the circular region  <img

The straight line  \mathrm{2 x-3 y=1}  divides the circular region  \mathrm{x^2+y^2 \leq 6}  into two parts.
If \mathrm{S=\left\{\left(2, \frac{3}{4}\right),\left(\frac{5}{2}, \frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8}, \frac{1}{4}\right)\right\}}
then the number of point(s) in S lying inside the smaller part is

Option: 1

2


Option: 2

1


Option: 3

0


Option: 4

3


Answers (1)

best_answer

The smaller region of circle is the region given by \mathrm{x^2+y^2 \leq 6}  and \mathrm{2 x-3 y \geq 1} 
Since only two points  \mathrm{\left(2, \frac{3}{4}\right)}  and \mathrm{\left(\frac{1}{4}, \frac{-1}{4}\right)}  satisfy above inequations
\mathrm{\therefore \quad 2} points in S lie inside the smaller region.

Posted by

vinayak

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 2027 exam dates out; session 1 from January 22
anam-khan

NTA Exam Calendar 2027 Out: JEE Main from January 22, CUET PG in March

Latest Question