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JEE Advanced 2024 Paper 1 · Mathematics · Quadratic Equations · hard

Then (1)/(2-α^2+(1)/(2-β^2+(1)/(2-γ^2 equals

Single correct option

Let α,β,γ\alpha,\beta,\gamma be the roots of x3−3x+1=0x^3-3x+1=0. Then 12−α2+12−β2+12−γ2\dfrac{1}{2-\alpha^2}+\dfrac{1}{2-\beta^2}+\dfrac{1}{2-\gamma^2} equals
  1. A. 33
  2. B. −3-3
  3. C. −32-\dfrac{3}{2}
  4. D. 32\dfrac{3}{2}
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Answer: (B) −3-3

Write p(x)=x3−3x+1p(x)=x^3-3x+1 and c=2c=\sqrt2. Then 12−α2=122(1c−α+1c+α)\dfrac{1}{2-\alpha^2}=\dfrac{1}{2\sqrt2}\left(\dfrac{1}{c-\alpha}+\dfrac{1}{c+\alpha}\right). For a polynomial with roots αi\alpha_i, ∑1t−αi=p′(t)p(t)\sum \dfrac{1}{t-\alpha_i}=\dfrac{p'(t)}{p(t)}. At t=2t=\sqrt2: p′=3⋅2−3=3p'=3\cdot2-3=3 and p=22−32+1=1−2p=2\sqrt2-3\sqrt2+1=1-\sqrt2, giving 31−2\dfrac{3}{1-\sqrt2}. For ∑1c+αi=−p′(−c)p(−c)\sum\dfrac{1}{c+\alpha_i}=-\dfrac{p'(-c)}{p(-c)} with p′(−2)=3p'(-\sqrt2)=3 and p(−2)=1+2p(-\sqrt2)=1+\sqrt2, giving −31+2-\dfrac{3}{1+\sqrt2}. The bracket is 3(11−2−11+2)=3⋅22−1=−623\left(\dfrac{1}{1-\sqrt2}-\dfrac{1}{1+\sqrt2}\right)=3\cdot\dfrac{2\sqrt2}{-1}=-6\sqrt2. Multiplying by 122\dfrac{1}{2\sqrt2} gives −3-3. A direct numerical check with the roots 1.532, 0.347, −1.8791.532,\,0.347,\,-1.879 also gives −3-3. Forgetting the sign in the second sum leads to the wrong value 33.
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Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake