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Let f(x)=sin[pi/6sin(pi/2sinx)] for all ...

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

`underset(x to 0)(lim)(f(x))/(g(x))=(pi)/(6)`

D

There is an `x in R` such that `(gof)(x)=1`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`f(x)=sin((pi)/(6) sin((pi)/(2)sinx))`
We know that `-1 le sinx le 1`
`implies -(pi)/(2) le (pi)/(2)sinx le (pi)/(2)`
`implies -1 le sin((pi)/(2)sinx) le 1`
`implies -(pi)/(6) le (pi)/(6) sin((pi)/(2) sinx) le (pi)/(6)`
`implies -(1)/(2) le sin((pi)/(6) sin((pi)/(2)sinx)) le (1)/(2)`
Now, `fog(x)=sin((pi)/(6) sin((pi)/(2)sin((pi)/(2)sinx)))-1 le sin((pi)/(2)sinx) le 1`
`implies (-pi)/(2) le (pi)/(2)(sin((pi)/(2)sinx)) le (pi)/(2)`
`implies -1 le sin ((pi)/(2)(sin((pi)/(2)sinx))) le 1`
`implies (-pi)/(6)le (pi)/(6) sin((pi)/(2)(sin((pi)/(2)sinx))) le (pi)/(6)`
`implies (-1)/(2) le f(x) le (1)/(2)`
Thus, range of fog is also `[-(1)/(2),(1)/(2)].`
Now, `underset(xto0)(lim)(sin((pi)/(6)((pi)/(2)sinx)))/((pi)/(2) sinx)`
`=underset(xto0)(lim)(2)/(pi)(sin((pi)/(6)((pi)/(2)sinx)))/((pi)/(6)sin((pi)/(2) sinx))xx((pi)/(6)((pi)/(2)sinx))/(( sinx)/(x)xx x)`
`underset(x to 0)(lim) (2)/(pi) xx (pi)/(6)xx(sin((pi)/(2) sinx))/((pi)/(2)sinx)xx((pi)/(2)sinx)/(x)`
`=(1)/(3)xx (pi)/(2)=(pi)/(6)`
`gof(x) in [-(pi)/(2) sin((1)/(2)),(pi)/(2)sin((1)/(2))]`
`implies gof(x) ne 1`
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