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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 1

Let α,β\alpha,\beta be the roots of the equation x2−3x+r=0x^2-3x+r=0, and α2,2β\dfrac\alpha2,2\beta be the roots of the equation x2+3x+r=0x^2+3x+r=0. If the roots of the equation x2+6x=mx^2+6x=m are 2α+β+2r2\alpha+\beta+2r and α−2β−r2\alpha-2\beta-\dfrac r2, then mm is equal to :

Choose one option
Answer and worked solution

Official answer: D

Roots. α+β=3\alpha+\beta=3 and α2+2β=−3\frac\alpha2+2\beta=-3 (both products equal αβ=r\alpha\beta=r). Solving, β=−3\beta=-3, α=6\alpha=6, r=αβ=−18r=\alpha\beta=-18.

New roots. 2α+β+2r=12−3−36=−272\alpha+\beta+2r=12-3-36=-27 and α−2β−r2=6+6+9=21\alpha-2\beta-\frac r2=6+6+9=21; their sum −6-6 matches x2+6x−m=0x^2+6x-m=0.

Product. −m=(−27)(21)=−567-m=(-27)(21)=-567, so m=567m=567.

(D)

Topics:Roots and coefficients
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