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.
