JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Analyzing +4 | -1

\omega(\neq 1)

is cube root of unity satisfying

\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2 \omega^2

and

\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2 \omega

, then the value of

\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because

\omega

and

\omega^2

are roots of the equation

\frac{{1}}{{a+x}} + \frac{{1}}{{b+x}} + \frac{{1}}{{c+x}} = \frac{{2}}{{x}}

, so substituting

x=1

gives

\frac{{1}}{{a+1}} + \frac{{1}}{{b+1}} + \frac{{1}}{{c+1}} = 2

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