Given that f(x)=xx| and g(x)=sinx So that gof(x)=g(f(x))=g(x|x|)=sinx|x| ={
sin(−x2),
if x<0
sin(x2)
if x≥0
={
−sinx2,
if x<0
sin(x2)
if x≥0
∴(gof)′(x)={
−2xcosx2,
if x<0
2xcosx2,
if x≥0
Here we observe L(gof)′(0)=0=R(gof)′(0) ⇒ go f is differentiable at x=0 and (gof)′ is continuous at x=0
Now (gof)"(x)={
−2cosx2+4x2sinx2,
x<0
2cosx2−4x2sinx2,
x≥0
Here L(gof)"(0)=−2 and R(gof)"(0)=2 ∵L(gof)"(0)≠R(gof)" ⇒ go f (x) is not twice differentiable at x = 0. ∴ Statement - 1 is true but statement -2 is false.