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Several mathematical function problems 1. Function y = the square of X / the square of X + 1, X belongs to real number, find the function range 2. The maximum value of function y = 1 / 1-x (1-x) 3. If the square of function y = x + AX-1 has the minimum value - 2 in the interval [0,3], then the real number a is equal to? 4.2lg (X-Y) = lgx + lgY, then Y / x equals?
1.y=x2/(x2+1)=(x2+1-1)/(x2+1)=1-1/(x2+1)
∵ x ∈ R, then x2 + 1 ≥ 1
Therefore - 1 ≤ - 1 / (x2 + 1) ＜ 0
0 ≤ 1-1 / (x2 + 1) < 1, that is, the range of the function is [0,1]
2.y=1/[1-x(1-x)]=1/(x2-x+1)
∵x2-x+1=(x-1/2)2+3/4≥3/4
∴0＜1/(x2-x+1)≤4/3
The maximum value of this function is 4 / 3
The third problem is more complicated and can be discussed in different situations: one is that the vertex of the quadratic function is in the interval [0,3] and the ordinate of the vertex is - 2; the second is that when x = 3 is y = - 2 and the function is monotonically decreasing in the interval [0,3], only the first case is consistent with the problem, and a = - 2
4.∵2lg(x-y)=lgx+lgy
Ψ LG (X-Y) 2 = LG (XY), so (X-Y) 2 = XY
After finishing, we get x2-3xy + y2 = 0
XY ≠ 0, divide both sides of the equation by XY at the same time
x/y - 3 + y/x=0
Let Y / x = t, (T > 0)
Then 1 / T - 3 + T = 0, that is T2 - 3T + 1 = 0,
The solution is t = (3 ± root 5) / 2
The value of Y / X is (3 ± 5) / 2
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