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Test of Mathematics Solution Subjective 90 - Graphing Inequality

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 90 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem:

:
Draw the region of points $ {\displaystyle{(x,y)}}$ in the plane, which satisfy $ {\displaystyle{|y| {\le} |x| {\le} 1}}$.


Solution:

$ {\displaystyle{|y| {\le} |x| {\le} 1}}$ region will bounded by lines $ {\displaystyle{x = y}}$, $ {\displaystyle{x = -y}}$, $ {\displaystyle{x = -1}}$ & $ {\displaystyle{x = 1}}$. Why is that?

First note that ( |x| \le 1 ) implies:

Inequality region

Similarly, if we demand ( |y| \le 1 ) (the double shaded zone).

Now if we want ( |y| \le |x| ) . This can be achieved by

  • \( y \le x \) when x and y are both positive (in the first quadrant); that is the region below the line x = y
  • \( y \le -x\) when x is negative and y positive (in the second quadrant); hence the region below the line y = -x
  • \(-y \le -x \) when (x, y) is in the third quadrant.
  • \( -y \le x \) when (x, y) is in fourth quadrant.

Therefore the final region is the following shaded region:

Inequality region

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 90 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem:

:
Draw the region of points $ {\displaystyle{(x,y)}}$ in the plane, which satisfy $ {\displaystyle{|y| {\le} |x| {\le} 1}}$.


Solution:

$ {\displaystyle{|y| {\le} |x| {\le} 1}}$ region will bounded by lines $ {\displaystyle{x = y}}$, $ {\displaystyle{x = -y}}$, $ {\displaystyle{x = -1}}$ & $ {\displaystyle{x = 1}}$. Why is that?

First note that ( |x| \le 1 ) implies:

Inequality region

Similarly, if we demand ( |y| \le 1 ) (the double shaded zone).

Now if we want ( |y| \le |x| ) . This can be achieved by

  • \( y \le x \) when x and y are both positive (in the first quadrant); that is the region below the line x = y
  • \( y \le -x\) when x is negative and y positive (in the second quadrant); hence the region below the line y = -x
  • \(-y \le -x \) when (x, y) is in the third quadrant.
  • \( -y \le x \) when (x, y) is in fourth quadrant.

Therefore the final region is the following shaded region:

Inequality region

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