The known function f (x) = 2asinx * cosx + 2cos ^ 2x + 1, f (π / 6) = 4 Find the real number a Finding the symmetric center coordinates of the function f (x) image Finding the range of function f (x) on [- π / 4, π / 4]

F (π / 6) = asin π / 3 + 2 (COS π / 6) ^ 2 + 1 = (√ 3 / 2) a + 2 × (3 / 4) + 1 = 4A = √ 3. In this case, f (x) = √ 3sin2x + cos2x + 2 = 2Sin (2x + π / 6) + 22x + π / 6 = k π; X = k π / 2 - π / 12, the symmetry center coordinate is (K π - π / 12,2) when x ∈ [- π / 4, π / 4], 2x + π / 6 ∈ [- π / 3,2 π / 3] 2x + π /

It is known that Y-3 is in positive proportion to 2x-1, and y = 12 when x = 2. It is urgent to find the analytic expression of the function of Y and X in 1 hour!

Let Y-3 = K (2x-1)
Substitute x = 2, y = 12 into the equation
So 12-3 = 4k-k
So k = 3
Substitute k = 3 into the equation
That is, y = 6x

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