AIEEE 2009 Solved Paper

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.