1.Find the range of 3sinx+4cosx-5.

2.If [2sinx]+[cosx]=-3,then the range of the fnctn f(x)=sinx+√3cosx in [0,2Π]is;(whr [.] denotes GIF).

(1)

3sin(x) - 4cos(x) = 5 sin(x + arcsin(-4/5))

Thus, the range of the function asin x - bcos x is
± √[(a)^2 + (b)^2]

so here its : ±5 -5

whihc is [-10,0]

(2 ) given [2sinx]+[cosx]=-3

so both sin x and cos x lies on (-0.5 , -1]

it can be written as : 2 sin (x+pi/3)

in (o,2pi ) sin (x+pi/3) range is [0.866,-1]

so its range is [1.732, -2]

--

regards

Ramesh