Question
Let $S=\left\{(x, y): \frac{y(3 x-1)}{x(3 x-2)}<0\right\}$$S^{\prime}=\{(x, y) \in A \times B:-1 \leq A \leq 1,-1 \leq B \leq 1\}$then the area of the region enclosed by all points in $S \cap S^{\prime}$ is(a) 1(b) 2(c) 3(d) 4
Step 1
The set $S$ is defined by the inequality $\frac{y(3x-1)}{x(3x-2)}<0$. This inequality is satisfied when the numerator and denominator have opposite signs. This happens in the intervals $x<\frac{1}{3}$ and $\frac{1}{3}<x<\frac{2}{3}$. Show more…
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