Question

what is the range of function
y=√(-2cos^2x+3cosx-1)
PLZ help me fast !!!!

Answer 1

y = √( -2cos²x +3cosx -1)

let cosx = P

then, y = √( -2P² +3P -1)

( 2P -1)(P -1) ≤ 0

1/2 ≤ P ≤ 1

domain of function ,

1/2 ≤ cosx ≤ 1

2nπ -π/3 ≤ x ≤ 2nπ + π/3

now,
range ,
y = √(-2P² +3P -1)
y² = -2P² +3P -1
2P² -3P +1 + y² = 0
P = { 3 ±√(1 -8y²) }/2

value is real when ,

1 - 8y² ≥0

-1/2√2 ≤ y ≤ 1/2√2

but y ≥0 becoz y = square root function (which is always positive )
so,
0 ≤ y ≤ 1/2√2

range € [ 0 , 1/2√2 ]

Answer 2

Hey hi friend !!!

Well i have explained each and every step in detail !! please do tell me if you don't get any step :-)


so the given function is
y= √(-2cos²x+3cosx-1)
 i.e = √[-2(cos²x-3/2+1/2)]
i.e = √[-2(cosx-3/4)²-9/16+1/2]
i.e. = √[-2(cos-3/4)²-1/16]
i.e. = √[1/8-3(cosx=3/4)²]-----------(1)

Now here  in this equation is this quantity :-
(cosx=3/4)²----------------(2)   is to it's minimum value then the whole equation
i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa


And we know that cosx-3/4 will be minimum if cosx=3/4
therefore put this in (1) we get
(cosx=3/4)²=0    [ cosx=3/4]
hence the minimum value of the quantity (cosx=3/4)² is 0

put this in equation (1)
we get ,
i.e. = √[1/8-3(cosx=3/4)²]
   =√[1/8-3(0)]        [ because minimum value of of the quantity (cosx=3/4)² is 0 ]
     =√1/8
      =1/(2√2)

this is the maximum value now to find the minimum value

since this is function of root so the value of y will always be ≥0

hence the minimum value of the function y is 0


There fore the range of function y is [0,1/(2√2)]



Hope this helped u !!! :-)