x→0limx(1−2!x2)−(x−2x2…)(1+(a−1)x+2!(a−1)2x2)+2(1−2!b2x2)+(c−2)(1−x+2!x2)=2
x→0lim2x2−2!x3+…(1+2+c−2)+x(a−1−c+2)+x2(2(a−1)2−b2+(2c−2))=2
For which
∵c+1=0⇒c=−1
∵a−c=−1⇒a=−2
∵2(a−1)2−b2+(2c−2)=1
29−b2−23=1⇒b2=2
a2+b2+c2=4+2+1=7
JEE Main 2026 — Mathematics Calculus
If x→0limxcosx−loge(1+x)e(a−1)x+2cos bx+(c−2)e−x=2, then a2+b2+c2 is equal to :
Held on 22 Jan 2026 · Verified 6 Jul 2026.
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