
α+α2=3a&α×α2=2a2↓(α+α2)3=27a3⇒2a2+4a4+3(3a)(2a2)=27a3⇒2+4a2+18a=27a⇒4a2−9a+2=0⇒4a2−8a−a+2=0⇒(4a−1)(a−2)=0⇒a=2 so 6x2−36x+48=0 ⇒x2−6x+8=0 ...(1) If we take a=41 then α=21 which is not possible
JEE Main 2024 — Mathematics Calculus
If the function f(x)=2x3−9x2+12a2x+1,a>0 has a local maximum at x=α and a local minimum at x=α2, then α and α2 are the roots of the equation :
Held on 8 Apr 2024 · Verified 6 Jul 2026.
x2−6x+8=0
x2+6x+8=0
8x2+6x−1=0
8x2−6x+1=0
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