It is known that f (x) is an odd function defined on the real number set R, and when x ∈ (0,1), f (x) = 2 ^ X / (4 ^ x + 1) 1. Find the analytical formula of function f (x) on [- 1,1] 2. Judge the monotonicity of F (x) on (0,1) and prove it 3. When t takes what value, the equation f (x) = t has a real number solution on [- 1,1]

It is known that f (x) is an odd function defined on the real number set R, and when x ∈ (0,1), f (x) = 2 ^ X / (4 ^ x + 1) 1. Find the analytical formula of function f (x) on [- 1,1] 2. Judge the monotonicity of F (x) on (0,1) and prove it 3. When t takes what value, the equation f (x) = t has a real number solution on [- 1,1]

1.-1

It is known that f (x) is an odd function defined on R with period 3. If F1) > 1 and f (2) = (2a + 3) / (a + 1), the value range of real number a This is what the Internet does ∵ f (x) is an odd function on R, f (1) > 1 ∴f(-1)=-f(1)＜-1 And f (x) is a function with period 3 ∴f(2)=f(-1+3)=f(-1)＜-1 ∴(2a-3)/(a+1)＜-1⑤ ∴-1＜a＜2/3 I want to ask ⑤, how can he determine 1 + a > 0 and directly solve this inequality

Not directly determined
If a + 1

Let f (x) be an odd function defined on R with a period of 3, if f (1) > 1, f (2) = 2A − 3 A + 1, then the value range of real number a is () A. (−∞，2 3) B. (−∞，−1)∪(2 3，+∞) C. (−1，2 3) D. (−∞，−1)∪(−1，2 3)

F (x) is an odd function with a period of 3 defined on R,
∴f（-2）=f（-2+3）=f（1）＞1
And f (- 2) = - f (2) = 3 − 2A
a+1＞1
The solution is - 1 < a < 2
three
Therefore, C

Let f (x) be an odd function defined on R with period 3. If f (1) < = 1, f (2) = (2a-3) / (a + 1), the value range of real number a is

The period is 3, so f (2) = f (- 1). From the odd function, f (- 1) = - f (1)

Let f (x) be an odd function defined on R with period 3. If f (1) is less than or equal to 1, f (2) = (2a-3) / (a + 1), then the range of real number a

F (1) = f (- 2) = - f (2) less than or equal to 1
Then f (2) ≥ - 1
（2a-3）/（a+1）≥-1
Solution
A ≤ - 1 or a ≥ 2 / 3

Let function f (x) be an odd function with period 3 defined on R, if f (1) > = 1, f (2) = (3a-4) / (a + 1), the value range of real number a

f(2)=f(-1+3)=f(-1)=-f(1);
-f(2)=f(1)=-(3a-4)/(a+1)>=1;
Solving this inequality yields:
-1 respondent: qyh6010 - probation level II 2009-10-6 12:50
His solution of Fractional Inequality is a typical mistake. First move 1 to the left, divide it all, and then solve it