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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.

** **Sketch, on plain paper, the regions represented, on the plane by the following:

(i) $ |y| = \sin x $

(ii) $ |x| - |y| \ge 1 $

** **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).

**What is this topic:**Graphing of functions, inequalities**What are some of the associated concept:**Absolute value functions, inequalities**Where can learn these topics:****Book Suggestions:**Play with Graphs (Arihant Publication)

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.

** **Sketch, on plain paper, the regions represented, on the plane by the following:

(i) $ |y| = \sin x $

(ii) $ |x| - |y| \ge 1 $

** **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).

**What is this topic:**Graphing of functions, inequalities**What are some of the associated concept:**Absolute value functions, inequalities**Where can learn these topics:****Book Suggestions:**Play with Graphs (Arihant Publication)

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