JEE MATHEMATICS MCQ: THEORY OF EQUATION

1 Easy

1 of 1

Application +4 | -1

If the sum of the squares of the reciprocals of the roots

\alpha

and

\beta

of the equation

3{x^2} + \lambda x - 1 = 0

is 15, then

6{\left( {{\alpha ^3} + {\beta ^3}} \right)^2}

is equal to:

View answer and explanation

Correct option: B

Option B (24) is correct because after finding the value of

\lambda

from the given condition, we can calculate the value of

6{(\alpha^3 + \beta^3)}^2

which equals 24.

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