Test of Mathematics Solution Subjective 126 - Graphs of Absolute Value Functions
This is a Test of Mathematics Solution Subjective 126 (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.
Problem
Sketch, on plain paper, the regions represented, on the plane by the following:
(i) $ |y| = \sin x $
(ii) $ |x| - |y| \ge 1 $
Solution
First we need to understand what |y| signifies. It is the absolute value of y, that is it is +y when y is positive and -y when y is negative.
Lets test with $ x = \frac{\pi}{6} $. Clearly then sin x = 1/2. This implies $ |y| = 1/2 $ or $ y = 1/2, -1/2 $.
Again let us test with $ x = \pi + \frac{\pi}{6} $. Then $ \sin (\pi + \frac{\pi}{6}) = - \frac{1}{2} $ implying $ |y| = - \frac{1}{2} $. But this is impossible as absolute value cannot be negative.
Using these observations we get a clear idea about what is happening.
The values of x where sin (x) is positive (that is when $ \displaystyle {\frac {4k \pi}{2} \le x \le \frac {(4k+2) \pi }{2} }$, where k is any integer), we draw the graph of y = sin x and reflect it about x axis (as y = sin x and y = - sin x both satisfies the equation).
The values of x where sin (x) is negative (that is when $ \displaystyle {\frac {(4k+2) \pi}{2} \le x \le \frac {(4k+4) \pi }{2} }$, where k is any integer), the given relation is not defined as absolute value of y cannot be negative.
For Part (ii)
Notice that |x| gives distance of a point from y axis and |y| gives distance of a point from x axis.
Let us split the problem into cases:
x, y both non negative (first quadrant). Then |x| = x, |y| = y, implying $ |x| - |y| \ge 1 $ is same as $ x - y \ge 1 $
$ x \le 0, y \ge 0 $ implies |x| = -x. Thus in second quadrant the inequality becomes $ -x -y \ge 1$ or $ x+y \le -1 $
$ x \le 0, y \le 0 $ implies |x| = -x, |y| = -y. Thus in third quadrant the inequality becomes $ -x +y \ge 1$ or $ x-y \le -1 $
$ x \ge 0, y \le 0 $ implies |x| = x, |y| = -y. Thus in fourth quadrant the inequality becomes $ x + y \ge 1$
Hence the final picture is:
(only the shaded zones, not the lines).
Chatuspathi:
What is this topic: Graphing of functions, inequalities
What are some of the associated concept: Absolute value functions, inequalities
Book Suggestions: Play with Graphs (Arihant Publication)
Related
This is a Test of Mathematics Solution Subjective 126 (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.
Problem
Sketch, on plain paper, the regions represented, on the plane by the following:
(i) $ |y| = \sin x $
(ii) $ |x| - |y| \ge 1 $
Solution
First we need to understand what |y| signifies. It is the absolute value of y, that is it is +y when y is positive and -y when y is negative.
Lets test with $ x = \frac{\pi}{6} $. Clearly then sin x = 1/2. This implies $ |y| = 1/2 $ or $ y = 1/2, -1/2 $.
Again let us test with $ x = \pi + \frac{\pi}{6} $. Then $ \sin (\pi + \frac{\pi}{6}) = - \frac{1}{2} $ implying $ |y| = - \frac{1}{2} $. But this is impossible as absolute value cannot be negative.
Using these observations we get a clear idea about what is happening.
The values of x where sin (x) is positive (that is when $ \displaystyle {\frac {4k \pi}{2} \le x \le \frac {(4k+2) \pi }{2} }$, where k is any integer), we draw the graph of y = sin x and reflect it about x axis (as y = sin x and y = - sin x both satisfies the equation).
The values of x where sin (x) is negative (that is when $ \displaystyle {\frac {(4k+2) \pi}{2} \le x \le \frac {(4k+4) \pi }{2} }$, where k is any integer), the given relation is not defined as absolute value of y cannot be negative.
For Part (ii)
Notice that |x| gives distance of a point from y axis and |y| gives distance of a point from x axis.
Let us split the problem into cases:
x, y both non negative (first quadrant). Then |x| = x, |y| = y, implying $ |x| - |y| \ge 1 $ is same as $ x - y \ge 1 $
$ x \le 0, y \ge 0 $ implies |x| = -x. Thus in second quadrant the inequality becomes $ -x -y \ge 1$ or $ x+y \le -1 $
$ x \le 0, y \le 0 $ implies |x| = -x, |y| = -y. Thus in third quadrant the inequality becomes $ -x +y \ge 1$ or $ x-y \le -1 $
$ x \ge 0, y \le 0 $ implies |x| = x, |y| = -y. Thus in fourth quadrant the inequality becomes $ x + y \ge 1$
Hence the final picture is:
(only the shaded zones, not the lines).
Chatuspathi:
What is this topic: Graphing of functions, inequalities
What are some of the associated concept: Absolute value functions, inequalities
