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# Test of Mathematics Solution Subjective 90 - Graphing Inequality

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:

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:

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:

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: