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1)

Let   f(x)=sin[π6sin(π2sinx)]  for all x ε  R and   g(x)=(π/2)sinx4 for all x ε  R . Let (fog)(x)  denotes f(g(x)) and (gof)(x) denotes g{f(x)} . Then, which of the following is /are true?


A) Range of f is [12,12]

B) Range of fog is [12,12]

C) limx0f(x)g(x)=π6

D) There is an an x e R such that (gof)(x)=1

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)   limx0f(x)g(x)

                                        =limx0sin[π6sin(π2sinx)]π2(sinx)

                          =limx0sin[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.