The function f(x)=x2−1x2−3x+2+cos⁡(|x|) is not differentiable at:

We  have x2−3x+2=x−1+x−2                                        =1−x2−x,    if x≤1x−12−x,    if 12This function is differentiable  at  all points  except possible  at x=1 and x=2.              Lf(1)  =limx→1−f(x)−f(1)x−1                             =limx→1−x2−1x−2+cosx−cos1x−1                            =  0−1sin1=−sin1.and  Rf(1)  = limx→1+f(x)−f(1)x−1                          =limx→1+−x2−1x−2+cosx−cos1x−1                         = 0−sin1=−sin1∴       f is differentiable at x=1.        Lf(2)=limx→2−f(x)−f(2)x−2                      = limx→2−−x2−1x−1+cosx−cos2x−2                     = −4−12−1−sin2=−3−sin2    Rf(2)= limx→2+f(x)−f(2)x−2                   = limx→2+x2−1x−1+cosx−cos2x−2                  = 22−12−1−sin2=3−sin2As Lf(2)≠Rf(2), f is not differentiable at x=2.Hence (4) is the correct answer.