The correct option is A f and f' are continous for x + 1 > 0
Let us divide interval into two sub - intervals
I1,−1≤x<0 so that x is -ve and I2,x≥0 so that x is +ve.
For I1,f(x)=∫x−1(−t)dt=−12(x2−1) ⋯(i)
For I2,f(x)=∫0−1(−t)dt+∫x0t dt
=−12[t2]0−1+12[t2]x0=12(1+x2) ⋯(ii)
Hence the function can be defined as the following
f(x)={−12(x2−1), If−1≤x<012(x2+1), if x≥0
For f, L = R = V =12 at x =0, so f' is also continuous at x = 0. Thus both f and f' are continuous at x = 0 and hence both are continuous for x > -1 i.e., x + 1 > 0