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JEE Advanced 2020 Paper 1 · Mathematics · Quadratic Equations · hard

Let a,b,c be the roots of x^3-3x+1=0.

One or more options correct

Let a,b,ca,b,c be the roots of x3−3x+1=0x^{3}-3x+1=0. Which of the following is/are correct?
  1. A. a2+b2+c2=6a^{2}+b^{2}+c^{2}=6
  2. B. a3+b3+c3=3a^{3}+b^{3}+c^{3}=3
  3. C. a2b2+b2c2+c2a2=9a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=9
  4. D. Exactly one root lies in the interval (1,2)(1,2)
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Answer: A, C, D

By Vieta, a+b+c=0a+b+c=0, ab+bc+ca=−3ab+bc+ca=-3, abc=−1abc=-1. (A) ∑a2=(∑a)2−2∑ab=6\sum a^2=(\sum a)^2-2\sum ab=6, true. (B) Each root satisfies a3=3a−1a^3=3a-1, so ∑a3=3∑a−3=−3\sum a^3=3\sum a-3=-3; the claimed value 33 is false. (C) ∑a2b2=(∑ab)2−2abc∑a=9−0=9\sum a^2b^2=(\sum ab)^2-2abc\sum a=9-0=9, true. (D) Let g(x)=x3−3x+1g(x)=x^3-3x+1: g(−2)=−1g(-2)=-1, g(−1)=3g(-1)=3, g(0)=1g(0)=1, g(1)=−1g(1)=-1, g(2)=3g(2)=3, so there is one root in each of (−2,−1)(-2,-1), (0,1)(0,1) and (1,2)(1,2); exactly one root lies in (1,2)(1,2), so (D) is true. Numerically the roots are −1.879, 0.347, 1.532-1.879,\ 0.347,\ 1.532, consistent with 2cos⁡160∘,2cos⁡80∘,2cos⁡40∘2\cos 160^\circ,2\cos80^\circ,2\cos40^\circ.
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