f(x)=(1+px)−(1−px)x,    −1≤x<02x+1x−2,    0≤x≤1 is continuous in the interval [−1,1]then  is equal to

f0+0=limh→020+h+10+h−2=−12f0−0=limh→01−ph−1+ph0−h                      =limh→0  −2ph−h1−ph+1+ph=pf(x) is continuous at x=0 in 1,1 iff0+0=f0−0=f0⇒p=−1/2.Hence  (2) is the correct answer.