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

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

(ii) \(|x| – |y| \ge 1 \)

**Discussion: **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.

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