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Class 12 Relations And Functions · Mcq Single Correct1 Markstypes_of_relations

Class 12 Maths · Relations And Functions · Mcq Single Correct #97

On \(\mathbb C\setminus\{0\}\), define \(z_1Rz_2\) to mean that \(z_1+z_2\ne0\) and \(\frac{z_1-z_2}{z_1+z_2}\in\mathbb R\). Then \(R\) is

✅ Step-by-Step Solution

Final Answer: Option (B): Reflexive and symmetric but not transitive
  1. With the quotient required to be defined, every nonzero z is related to itself, giving zero. Reversal negates a real quotient, so symmetry holds. But 1R2 and 2R(-1), while 1R(-1) fails because its denominator is zero; transitivity fails.

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