It is known that the function y = f (x) = (AX ^ 2 + 1) / (BX + C) (a, B, C ∈ R, and a > 0, b > 0) is odd. When x > 0, f (x) has a minimum value of 2, It is known that the function y = f (x) = (AX ^ 2 + 1) / (BX + C) (a, B, C ∈ R, and a > 0, b > 0) is odd, When x > 0, f (x) has a minimum value of 2, where B ∈ N and f (1)

It is known that the function y = f (x) = (AX ^ 2 + 1) / (BX + C) (a, B, C ∈ R, and a > 0, b > 0) is odd. When x > 0, f (x) has a minimum value of 2, It is known that the function y = f (x) = (AX ^ 2 + 1) / (BX + C) (a, B, C ∈ R, and a > 0, b > 0) is odd, When x > 0, f (x) has a minimum value of 2, where B ∈ N and f (1)

It's an odd function
Then f (- x) = (AX & # 178; + 1) / (- BX + C) = - f (x) = - (AX & # 178; + 1) / (BX + C)
The solution is C = 0
So f (x) = ax / B + 1 / (BX)
When x > 0, a > 0, b > 0
f(x)≥2√(ax/b*1/bx)=2√(a/b²)
That is, f (x) min = 2 √ (A / B & # 178;) = 2
So a = B & # (1)
From the known f (1) = A / B + 1 / b = (a + 1) / b

Let a belong to R, quadratic function f (x) = ax ^ 2-2x-2a, if the solution set of F (x) > 0 is a, and B = (1,3), a ∩ B = empty set
Find the value range of real number a

There are two situations,
1. The solution set greater than 0 is on the left side of 1 and on the right side of 3, and a > 0, f (1)

In a circle with radius r, if the length of string AB is equal to root 2R, then the length of inferior arc AB is?

Connecting OA and ob
∵OA²+OB²=2r² AB²=2r²
∴AB²=OA²+OB²
∴∠O=90°
The length of inferior arc AB = 90 × π × R / 180 = π R / 2