Let `f(x)=sin[pi/6sin(pi/2sinx)]` for all `x in RR`A. Range of f is `[-(1)/(2), (1)/(2)]`B. Range of fog is `[-(1)/(2), (1)/(2)]`C. `lim_(x to 0)(f(x))/(g(x))=(pi)/(6)`D. There is an ` x in R` such that `(gof)(x) = 1`

Let `f(x)=sin[pi/6sin(pi/2sinx)]` for all `x in RR`A. Range of f is `[

1.	Let `f(x)=sin[pi/6sin(pi/2sinx)]` for all `x in RR`A. Range of f is `[-(1)/(2), (1)/(2)]`B. Range of fog is `[-(1)/(2), (1)/(2)]`C. `lim_(x to 0)(f(x))/(g(x))=(pi)/(6)`D. There is an ` x in R` such that `(gof)(x) = 1`
Answer» Correct Answer - A::B::C (a) `f(x)=sin[(pi)/(6)sin((pi)/(2)sin x )], x in R` `=sin((pi)/(6)sin theta), theta in [-(pi)/(2), (pi)/(2)]" where " theta =(pi)/(2) sinx` `=sin alpha, alpha in [-(pi)/(6),(pi)/(6)]," where " alpha =(pi)/(6) sin theta ` ` therefore f(x) in [-(1)/(2),(1)/(2)]` Hence, range of `f(x) in [-(1)/(2),(1)/(2)]` So, option (a) is correct. (b) ` f{g(x)}=f(t), t in [-(pi)/(2),(pi)/(2)] rArr f(t) in [-(1)/(2),(1)/(2)]` `therefore` Option (b) is correct. (c) `lim_(x to 0) (f(x))/(g(x))=lim_(x to 0)(sin[(pi)/(6)sin((pi)/(2)sin x )])/((pi)/(2)(sinx))` `=lim_(x to 0) (sin[(pi)/(6)sin((pi)/(2)sin x )])/((pi)/(6)sin((pi)/(2)sin x ))*((pi)/(6)sin((pi)/(2)sin x ))/(((pi)/(2) sinx))` `=1 xx (pi)/(6) xx 1 = (pi)/(6)` ` therefore` Option (c) is correct. (d) ` g{f(x)}=1` `rArr (pi)/(2) sin {f(x)}=1` `rArr sin{f(x)} =(2)/(pi) " ...(i)" ` But ` f(x) in [-(1)/(2),(1)/(2)] subset [-(pi)/(6),(pi)/(6)] ` ` therefore sin{f(x)} in [-(1)/(2),(1)/(2)] " ...(ii)" ` `rArr sin{f(x)} ne (2)/(pi)," " `[from Eqs. (i) and (ii) ] i.e. No solution. ` therefore` Option (d) is not correct.