f(x)=sin|x|−|x|+2(x−π)cos|x| There are two cases, Case (1), x > 0 f(x)=sinx−x+2(x−π)cosx f′(x)=cosx−1+2(1−0)cosx−2sin(x−π) f′(x)=3cosx−2(x−π)sinx−1 Then, function f(x) is differentiable for all x>0 Case (2) x < 0 f(x)=−sinx+x+2(x−π)cosx f′(x)=−cosx+1−2(x−π)sinx+2cosx f′(x)=cosx+1−2(x−π)sinx Then, function f(x) is differentiable for all x<0 Now check for x = 0 f′(0+)R.H.D.=3−1=2 f′(0−)L.H.D.=1+1=2 L.H.D. = R.H.D. Then, function f(x) is differentiable for x=0. So it is differentiable everywhere Hence, k=ϕ