Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Let
.
List I | List II | ||
(P) | For each there exists a such that | (I) | True |
(Q) | There exists a such that has no solution in the set of complex numbers. | (II) | False |
(R) | equals | (III) | 1 |
(S) | equals | (IV) | 2 |
Text Solution
Verified by ExpertsThe correct answer is:
C
(P)
Hence for each
there exists
such that
(True)
(Q)
for
&
for
(False)
(R)
are roots of the equation
other then unity, hence

Substituting
, we get
(S)
sum of real parts of roots of
except 1 



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