a+b=t→−1+lim(α+β)=t→−1+lim−(t+2)71−1(t+2)61−1 let t+2=ya+b=y→1+limy1/7−1y1/6−1=6772(a+b)2=723649=98
JEE Main 2025 — Mathematics Algebra
For t>−1, let αt and βt be the roots of the equation ((t+2)71−1)x2+((t+2)61−1) x+((t+2)211−1)=0
If t→−1+limαt=a and t→−1+limβt=b, then 72(a+b)2 is equal to ________.
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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