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f(X)=cos2x+sin4x    is a function fraom R-R v   then find f(R)=?
5 months ago

Dear student
You have asked a meaningless Question.
Please repost the proper question or you can also attach an image.
I will be happy to help you.
5 months ago

I think the question is about finding the range of function f(x) = cos2x + sin4x defined from R to R.
f(x) = cos2x + (sin2x)2 = cos2x + (1 – cos2x)2 = cos2x + 1 – 2cos2x + cos4x
= cos4x – cos2x + 1 = (cos2x – 1/2)2 + 3/4
Now, we have to find the range of f(x) = (cos2x – 1/2)2 + 3/4.
Clearly, f(x) is always positive. We know that square of a term is always greater or equal to zero.
So, minimum value of f(x) occurs when the square term (cos2x – 1/2)2 is zero
i.e. minimum value of f(x) will be 3/4.
For maximum value of f(x) the square term (cos2x – 1/2)2 should be maximum i.e.
(cos2x – 1/2) should be maximum. This occurs when cos2x is maximum.
We know that maximum value of cos2x is 1.
So, maximum value of f(x) is (1 – 1/2)2 + 3/4 = 1/4 + 3/4 = 1.
Thus, range of f(x) is [3/4 , 1] or f(R) = [3/4 , 1]
5 months ago
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