x→βlim(x−β)2e2(x2+bx+c)−1−2(x2+bx+c)x→βlim(x−β)21(1+1!2(x2+bx+c)+2!22(x2+bx+c)2+...)−1−2(x2+bx+c)x→βlim(x−β)22(x2+bx+c)2x→βlim(x−β)22(x−α)2(x−β)22(β−α)2=2(b2−4c)
If α ,β are the distinct roots of x^2+bx+c=0, then undersetx→ β lim e^2(x^2+bx+c)-1-2(x^2+bx+c) (x-β )^2 is equal to
JEE Main 2021 — Mathematics Calculus
If α,β are the distinct roots of x2+bx+c=0, then x→βlim(x−β)2e2(x2+bx+c)−1−2(x2+bx+c) is equal to
Held on 27 Aug 2021 · Verified 6 Jul 2026.
Options
- A
2(b2+4c)
- B
b2−4c
- C
2(b2−4c)
- D
b2+4c
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