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MathematicsSingle correct+3 / -1Question 35 of 84

IIT-JEE 2010 · Paper 1, Mathematics question 35

Let p and q be real numbers such that p≠0p\ne0, p3≠qp^3\ne q and p3≠−qp^3\ne-q. If α\alpha and β\beta are nonzero complex numbers satisfying α+β=−p\alpha+\beta=-p and α3+β3=q\alpha^3+\beta^3=q, then a quadratic equation having αβ\dfrac\alpha\beta and βα\dfrac\beta\alpha as its roots is

Choose one option
Answer and worked solution

Official answer: B

Product αβ. α3+β3=(α+β)3−3αβ(α+β)\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta):
q=−p3+3pαβ  ⟹  αβ=p3+q3p.q=-p^3+3p\alpha\beta\implies\alpha\beta=\frac{p^3+q}{3p}.

Sum of the new roots.
αβ+βα=(α+β)2−2αβαβ=p2αβ−2=3p3p3+q−2=p3−2qp3+q.\frac\alpha\beta+\frac\beta\alpha=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{p^2}{\alpha\beta}-2=\frac{3p^3}{p^3+q}-2=\frac{p^3-2q}{p^3+q}.
Their product is 1.

Equation. (p3+q)x2−(p3−2q)x+(p3+q)=0(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0.

(B)

Topics:Roots and coefficients
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