f(x) is continuous on R
⇒f(1−)=f(1)=f(1+)
∣a+1+b∣=x→1+limsin(πx)
∣a+1+b∣=0⇒a+b=−1
⇒ Also
f(−1−)=f(−1)=f(−1+)
x→−1lim2sin(2−πx)=∣a−1+b∣
∣a−1+b∣=2
Either a−1+b=2 or a−1+b=−2
a+b=3...(2) or a+b=−1...(3)
from (1) and (2) ⇒a+b=3=−1 (reject)
from (1) and (3) ⇒a+b=−1