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How to find the minimum positive period of function y = 4sin (2x + π / 3) + 1? Please write it down
There are no steps
A knowledge point y = asin (BX + ∮) + C
The minimum positive period is t = 2 π / b
The solution is pi
The period of the function y = cos (x / 2) is
T=2π/(1/2)=4π
The value range of function y = LG [x ^ 2 + (K + 2) x + 5 / 4] is all real numbers, and the value range of K
x²+(k+2)x+5/4>0
Because the domain of definition is all real numbers and the opening is upward, the discriminant is used
△=b^2-4ac
8. If the function y = (K & # 178; - 1) x & # 178; + (k-1) x + 2 is a nonnegative real number, find the value range of real number K
The range of y = (K & # 178; - 1) x & # 178; + (k-1) x + 2 is a nonnegative real number parabola with the opening upward, that is, (K & # 178; - 1) ≥ 0. Here we deduce K ≥ 1 or K ≤ - 1
The range of y = (K & # 178; - 1) x & # 178; + (k-1) x + 2 is a nonnegative real number, that is, (k-1) &# 178; - 4 (K & # 178; - 1) = 0, k = 1, k = - 5 / 3
The intersection of the two answers is k = 1, k = - 5 / 3
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