The domain of the function f(x) is denoted by `D_(f)`

Answer

(A) `rarr` (s); (B) `rarr` (r); (C ) `rarr` (p); (D) `rarr` (q)

Explanation

Let `y=sqrt(3-x)+sin^(-1)((3-2x)/(5))`
For y to be defined `3-xge0" on "-1le(3-2x)/(5)le1xle3" ... (i)"`
`-5le3-2xle5`
and `-1lexle4" ... (ii)"`
From Eqs. (i) and (ii), we get
`x in [-1, 3]`
(B) Let `y = log_(10)` `{1-log_(10)(x^(2)-5x+16)}` for y to be defined `x^(2)-5x+16gt0and1-log_(10)(x^(2)-5x+16)gt0`
`(x-(5)/(2))^(2)+(39)/(4)gt0 and log_(10)(x^(2)-5x+16)lt1`
which is true, `AAx inR" ... (i)"`
`impliesx^(2)-5x+16lt10`
`implies x^(2)-5x+6lt0implies(x-3)(x-2)lt0`
`implies 2ltxlt3" ... (ii)"`
From Eqs. (i) and (ii), `x in(2, 3)`
(C ) Let `y=cos^(-1).(2)/(2+sinx)`, for y to be defined
`-1le(2)/(2+sinx)le1 {:[(because-1ltsinxle1),(1lt2+sinxle3)]:}`
Multiplying by (2+sinx)
`-(2+sinx)le2le2+sinx`
`implies-2-sinxle2|2le2+sinx`
`{:(implies.-sin x le 4),(implies.sinxge-4):}:|:{:(sinxge0),(2npilexle(2n+1)pi.ninz" ... (i)"):}`
We know that sin `x in [-1, 1]`
`therefore x in R" ... (ii)"`
From Eqs. (i) and (ii), `x in [2kpi, (2k+1)pi]`
Domain `underset(k in I)(uu)[2kpi, (2k+1pi)]`
(D) `y=sqrt(sinx)+sqrt(16-x^(2))` for y to be defined
`{:(sinxge0),(x in[2kpi,(2k+1)pi]",k" in"I ... (i)"):}:|:{:(16-x^(2)ge0),(-4lexle4" ... (ii)"):}`
From Eqs. (i) and (ii), we get
`x in [-4, -pi]uu[0, pi]`