Given that f(x)=x|x| and g(x)=sin‌x So that go f(x)=g(f(x)) =g(x|x|)=sin‌x|x| ={
sin‌(−x2)
‌
‌ if ‌x<0
sin‌(x2)
‌
‌ if ‌x≥0
. ={
−sin‌x2
‌
‌ if ‌x<0
sin‌x2
‌
‌ 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(g∘f)′′(0)=−2 and R(g∘f)′′(0)=2 As L(g∘f)′′(0)≠R(g∘f)′′(0) ⇒gof(x) is not twice differentiable at x=0. ∴ Statement - 1 is true but statement −2 is false.
