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kotykmax [81]
3 years ago
14

What is the range of function y=√(-2cos^2x+3cosx-1)

Mathematics
1 answer:
Ugo [173]3 years ago
8 0
Hello Friend,here is the solution for your question


<span>so the given function is </span>
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 :-
<span>(cosx=3/4)²----------------(2)   is to it's minimum value then the whole equation </span>
<span>i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa </span>


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

<span>put this in equation (1) </span>
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)

<span>this is the maximum value now to find the minimum value </span>

<span>since this is function of root so the value of y will always be ≥0 </span>

<span>hence the minimum value of the function y is 0 </span>


<span>Therefore, the range of function </span>y is [0,1/(2√2)]


__Well,I have explained explained each and every step,do tell me if you don't understand any step._

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