f(x)=x−sinx f′(x)=1−cosx≥0 This is increasing function on R−{2nπ:n∈z} ⇒f′(x)>f(x) for all x>0 Thus x>sinx for all x>0 and x<sinx for x≠−2nπ Thus, x−sinx≠0 for all x∈R−2nπ Hence, x−sinx=0 for x=0 only ∴n2=1 We should know that graph of f−1(x) is mirror image of f(x) with respect to y=x as a mirror.
Number of solutions n1=0 (only (0,0) is the solution of this equation) of equation x=|sin−1x| Hence n2−n1=1.