Question

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 1

Correct Answer - A::B::C
We have, `g(x) = (pi)/(2) sin x`
`therefore g(g(x)) = g((pi)/(2) sin x) = (pi)/(2) sin ((pi)/(2) sinx)`
Thus , `f(x) = sin ((1)/(3)g(g(x)))`
We observe that
`-(pi)/(2) le (pi)/(2) sin x le(pi)/(2) ` for all `x in R`
`rArr` Range `(g) = [-(pi)/(2),(pi)/(2)]`
`rArr g(x) in [-(pi)/(2),(pi)/(2)]`
`rArr g(g(x)) in [-(-pi)/(2),(pi)/(2)]`
`rArr (1)/(3)g(g(x)) in [-(pi)/(6),(pi)/(6)]`
`rArr sin ((1)/(3)g(g(x))) in [-(1)/(2),(1)/(2)]`
`rArr` Range of `f=[-(1)/(2),(1)/(2)]`
So, option (a) is ture.
Clearly,`-(pi)/(2) le g(x)le(pi)/(2)` for all `x in R and -(1)/(2) le f(x)le (1)/(2)`for all `x in [-(pi)/(2),(pi)/(2)]`
Therefore, range of fog is `[-(1)/(2),(1)/(2)]`
So, option (b) is ture
Now,
`underset(x to 0)("lim")(f(x))/(g(x)) = underset(x to0)("lim") (sin ((1)/(3)g(g(x))))/((1)/(3)g(g(x))) xx ((1)/(3)g(g(x)))/(g(x))`
`= underset(x to 0)("lim")(sin ((1)/(3)g(g(x))))/((1)/(3)g(g(x)))xx=(1)/(3) underset(x to 0)("lim")(g(g(x)))/(g(x))`
`= (1)/(3) xx(pi)/(2) = (pi)/(6)`
Hence, options (a), (b) and (c) are correct.