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JEE Advanced 2019 Paper 1 · Mathematics · Quadratic Equations · hard

Let α,β,γ be the roots of x^3-3x+1=0. Which of the following statements is/are correct?

One or more options correct

Let α,β,γ\alpha,\beta,\gamma be the roots of x3−3x+1=0x^3-3x+1=0. Which of the following statements is/are correct?
  1. A. α4+β4+γ4=18\alpha^4+\beta^4+\gamma^4=18
  2. B. α3+β3+γ3=3\alpha^3+\beta^3+\gamma^3=3
  3. C. ∣(α−β)(β−γ)(γ−α)∣=81\left|(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\right|=81
  4. D. 1α+1β+1γ=−3\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=-3
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Answer: A

Here e1=α+β+γ=0e_1=\alpha+\beta+\gamma=0, e2=−3e_2=-3, e3=−1e_3=-1. Then ∑α2=e12−2e2=6\sum\alpha^2=e_1^2-2e_2=6. Each root satisfies α3=3α−1\alpha^3=3\alpha-1, so ∑α3=3e1−3=−3\sum\alpha^3=3e_1-3=-3; option (B) claims +3+3 and is false. Multiplying by α\alpha, α4=3α2−α\alpha^4=3\alpha^2-\alpha, so ∑α4=3⋅6−0=18\sum\alpha^4=3\cdot6-0=18, hence (A) is true. The discriminant of x3+px+qx^3+px+q is −4p3−27q2=108−27=81-4p^3-27q^2=108-27=81, and it equals [(α−β)(β−γ)(γ−α)]2\left[(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\right]^2, so the modulus of the product is 99, not 8181; (C) is false. Finally ∑1α=e2e3=−3−1=3\sum\frac1\alpha=\frac{e_2}{e_3}=\frac{-3}{-1}=3, not −3-3, so (D) is false.
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