Question
For any two sets of A and B, prove that:$\text{A}'\cup\text{B}=\text{U}\Rightarrow\text{A}\subset\text{B.}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $A$ | Column $B$ | ||
| $a.$ | The polar form of $\text{i}+\sqrt{3}$ is | $i.$ | Perpendicular bisector of segment joining $(– 2, 0)$ and $(2, 0).$ |
| $b.$ | The amplitude of $-1+\sqrt{-3}$ is | $ii.$ | On or outside the circle having centre at $(0, – 4)$ and radius $3.$ |
| $c.$ | If $|z + 2| = |z - 2|,$ then locus of $z$ is | $iii.$ | $\frac{2\pi}{3}$ |
| $d.$ | If $|z + 2i| = |z - 2i|,$ then locus of $z$ is | $iv.$ | Perpendicular bisector of segment joining $(0, – 2)$ and $(0, 2).$ |
| $e.$ | Region represented by $|\text{z}+4\text{i}|\geq3$ is | $v.$ | $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$ |
| $f.$ | Region represented by $|\text{z}+4|\leq3$ is | $vi.$ | On or inside the circle having centre $(– 4, 0)$ and radius $3$ units. |
| $g.$ | Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in | $vii.$ | First quadrant. |
| $h.$ | Reciprocal of $1 - i$ lies in | $viii.$ | Third quadrant. |