In the question, it is given that the equation |x+a|+|x−1|=2 has an infinite number of solutions for any value of x This is possible when x in |x+a| and x in |x−1| cancels out. Case 1: x+a<0,x−1≥0 −a−x+x−1=2 a=−3 Case 2: x+a≥0 and x−1<0 x+a−x+1=2 a=1 Largest value of a is 1. The answer is option C.