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JEE Advanced 2024 Paper 2 · Mathematics · Quadratic Equations · medium

Let α,β,γ be the roots of x^3-3x+1=0. Which of the following 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 is/are correct?
  1. A. α2+β2+γ2=6\alpha^2+\beta^2+\gamma^2=6
  2. B. α3+β3+γ3=3\alpha^3+\beta^3+\gamma^3=3
  3. C. 1α+1β+1γ=3\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=3
  4. D. α4+β4+γ4=12\alpha^4+\beta^4+\gamma^4=12
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Answer: A, C

From Vieta: ∑α=0\sum\alpha=0, ∑αβ=−3\sum\alpha\beta=-3, αβγ=−1\alpha\beta\gamma=-1. Then ∑α2=0−2(−3)=6\sum\alpha^2=0-2(-3)=6 (A true). Each root satisfies α3=3α−1\alpha^3=3\alpha-1, so ∑α3=3⋅0−3=−3\sum\alpha^3=3\cdot0-3=-3, and the option claiming 33 is false (B false). ∑1α=∑αβαβγ=−3−1=3\sum\frac1\alpha=\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}=\dfrac{-3}{-1}=3 (C true). Multiplying α3=3α−1\alpha^3=3\alpha-1 by α\alpha gives α4=3α2−α\alpha^4=3\alpha^2-\alpha, so ∑α4=3⋅6−0=18≠12\sum\alpha^4=3\cdot6-0=18\ne12 (D false). A numerical check of the three real roots confirms these power sums.
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