The values of a and b such that limx→0 x(1+acosx)−bsinxx3=1, are
52,32
52,−32
−52,−32
none of these
We have,
limx→0 x(1+a cosx)−b sinxx3=1⇒limx→0 x1+a1−x22!+x44!−x66!+…−bx−x33!+x55!…x3=1⇒limx→0 (1+a−b)+x2b3!−a2!+x4a4!−b5!+…x2=1 … (i)
If 1+a−b≠0, then LHS →∞ as x→0 while RHS =1
∴ 1+a−b=0
From (i), we have
limx→0 x2b3!−a2!+x4a4!−b5!+…x2=1∴ b3!−a2!=1⇒b−3a=6
Solving 1 + a - b = 0 and b - 3 a = 6, we get a=−52,b=−32