# Modulus Plotting – CAT Scan

## 13 thoughts on “Modulus Plotting – CAT Scan”

1. jaspreet |

sir cat 2003 Q,it should be none of these,isnt it? maxima pt will be tending to infinity..minimum however would be 2.5…plz correct me if i’m wrong..

• Yes Jaspreet, it should have been minimum. I will correct it!

regards
J

2. Geet |

Sir I’m extremely, extremely grateful for your posts, this blog has helped me immensely and I’m going to follow it even after my CAT this year because I’m just a smarter person thanks to it! I’m looking forward to the circles plotting post, sir! Would be lovely if you post it anytime soon. I have to write the CAT on the 7th, it would be simply brilliant to take the skills you share with me to the exam.

Sincerely,
A CAT-holics fan.

3. seema |

I did not understand the last question, how did you choose the x and y values to be +/-2? because when we solve the equation:
x+y+y-x = 4
we get y=1,
or
x+y+x-y = 4,
we get x = 1
or
-x-y+y-x = 4,
we get x= -1
or
-x-y-y+x = 4,
we get y = -1
so we get x=y= +/- 1.. plz help

4. x+y+y-x = 4
=> 2y = 4
=> y = 2
Where’s the problem?

regards
J

5. seema |

oh yes.. sorry by mistake..

6. seema |

• binditajoshi |

That would be 0 i think for the regions do not overlap..

• I’m afraid not, as it is a single equation and not two separate equations. I won’t give you the precise answer, but I would urge you to plug in values of x and y and see for yourself (hint: it will be a square)

regards
J

7. d |

Hi J,
As there is difference of two moduli, wont we get an infinite number of integral co-ordinates?

• Not sure which question you’re asking about. Also, can you give me some examples of such infinite points, with larger values of x or y?

regards
J

8. d |

For question : find the area bounded by |x+y| – |x-y| = 4.
(x,y) = (4,2), (5,2), (6,2), ………. ,(32,2), (34,2),……
Also, (-4,-2), (-5,-2), (-6,-2),………,(-32,-2), (-34,-2),…

9. Ohh for that one. I thought you meant the question in the original post. Yeah, my bad, it will be a square for a plus sign in between the two modulus functions, not the minus 🙂 For the minus sign, you’re absolutely right and it won’t be an enclosed figure.

regards
J