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← JEE Main 2022 · 24 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 7 of 90

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 7

If the sum of the squares of the reciprocals of the roots α\alpha and β\beta of the equation 3x2+λx−1=03x^2+\lambda x-1=0 is 15, then 6(α3+β3)26(\alpha^3+\beta^3)^2 is equal to :

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Answer and worked solution

Official answer: B

Sum and product. α+β=−λ3\alpha+\beta=-\frac\lambda3, αβ=−13\alpha\beta=-\frac13.

Condition. 1α2+1β2=(α+β)2−2αβ(αβ)2=9(λ29+23)=λ2+6=15\dfrac1{\alpha^2}+\dfrac1{\beta^2}=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=9\left(\dfrac{\lambda^2}9+\dfrac23\right)=\lambda^2+6=15, so λ2=9\lambda^2=9.

Cubes. α3+β3=(α+β)3−3αβ(α+β)=−λ327−λ3\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=-\dfrac{\lambda^3}{27}-\dfrac\lambda3, which is ∓2\mp2 for λ=±3\lambda=\pm3.

Value. 6(α3+β3)2=6⋅4=246(\alpha^3+\beta^3)^2=6\cdot4=24.

(B)

Topics:Roots and coefficients
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