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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 2)

Let A, B, C be three sets of complex numbers as defined below :

$$A = \left\{ {z:\,{\mathop{\rm Im}\nolimits} \,\,z\,\, \ge \,1} \right\}$$

$$B = \left\{ {z:\,\,\left| {z - 2 - i} \right| = 3} \right\}$$

$$C = \left\{ {z:\,{\mathop{\rm Re}\nolimits} (1 - i)z) = \sqrt 2 \,} \right\}$$

Let A, B, C be three sets of complex numbers as defined below :

$$A = \left\{ {z:\,{\mathop{\rm Im}\nolimits} \,\,z\,\, \ge \,1} \right\}$$

$$B = \left\{ {z:\,\,\left| {z - 2 - i} \right| = 3} \right\}$$

$$C = \left\{ {z:\,{\mathop{\rm Re}\nolimits} (1 - i)z) = \sqrt 2 \,} \right\}$$

Let A, B, C be three sets of complex numbers as defined below :

$$A = \left\{ {z:\,{\mathop{\rm Im}\nolimits} \,\,z\,\, \ge \,1} \right\}$$

$$B = \left\{ {z:\,\,\left| {z - 2 - i} \right| = 3} \right\}$$

$$C = \left\{ {z:\,{\mathop{\rm Re}\nolimits} (1 - i)z) = \sqrt 2 \,} \right\}$$

The number of elements in the set $$A \cap B \cap C$$ is
0
1
2
$$\infty $$

Explanation

In the Cartesian coordinates sets A, B and C defined the regions given by

$$A:y\ge1,B:(x-2)+(y-1)^2=9$$

B and C being a circle and a straight line intersect in two points, out of which only one satisfies $$y > -1$$.

Thus, the no. of elements in the set A $$\cap$$ B $$\cap$$ C is 1.

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