Answer:
Option A,B,C
Explanation:
(a) f(x)=sin[π6sinπ2(sinx)],xϵR
= sin(π6sinθ),θϵ[−π2,π2],
where θ=π2sinx
= sinα,αϵ[−π6,π6],
where α=π6sinθ
∴ f(x)ϵ[−12,12]
Hence, range of f(x)ϵ[−12,12]
So, option (a) is correct.
(b) f{g(x)}=f(t),tϵ[−π2,π2]
⇒ f(t)ϵ[−12,12]
∴ Option (b) is correct.
(c) limx→0f(x)g(x)
=limx→0sin[π6sin(π2sinx)]π2(sinx)
=limx→0sin[π6sin(π2sinx)]π6sin(π2sinx)
π6sin(π2sinx)(π2sinx)
= 1×π6×1=π6
∴ Option (c) is correct.
(d) g{ f(x)}=1
⇒ π2sin{f(x)}=1
⇒ sin{f(x)}=2π .........(i)
But f(x)ϵ[−12,12]⊂[−π6,π6]
∴ sin{f(x)}ϵ[−12,12] ..............(ii)
⇒ sin{f(x)}≠2π
[from Eqs(i) and (ii) ]
i.e, no solution
∴ Option (d) is not correct.