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._
Answer: c=19 and n=4
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Truest sense that the world of war in my opinion was gon in my own name in
Answer:
x= 15/16
Step-by-step explanation:
