It is known that the solution set of the inequality ax2-3x + 2 & gt; 0 is {x | X & lt; 1 or X & gt; B} (1) to find the value of a and B; (2) to solve the inequality x2-b (a + C) x + 4C & gt; 0 about X

It is known that the solution set of the inequality ax2-3x + 2 & gt; 0 is {x | X & lt; 1 or X & gt; B} (1) to find the value of a and B; (2) to solve the inequality x2-b (a + C) x + 4C & gt; 0 about X

(1) We know that a & gt; 0 and 1, B are the roots of the equation ax2-3x + 2 = 0, a = 1 and 1 × B = 2A, B = 2 (5 points) (2) inequality can be reduced to x2-2 (c + 1) x + 4C & gt; 0, that is, (x-2c) (X-2) & gt; 0 (7) when 2C & gt; 2 is C & gt; 1, the solution set of inequality is {x | X & lt; 2
(17) We know that the solution set of the inequality where the square of ax - 3x + 2 is greater than 0 is {X / X is less than 1 or X is greater than B} 1. Find the value of a and B and solve the square of the inequality ax - (a +...)
(17) It is known that the solution set of the inequality where the square of ax - 3x + 2 is greater than 0 is {X / X is less than 1 or X is greater than B} 1. It is clear to find the value of a and B and solve the inequality where the square of ax - (a + b) x + B is less than o
Substituting x = 1 into the equation AX ^ 2-3x + 2 = 0
A = 1
So the inequality is x ^ 2-3x + 2 > 0
The solution is x2
So B = 2
2. Inequality ax ^ 2 - (a + b) x + B
When x = 1, ax ^ 2-3x + 2 = 0, that is, A-3 + 2 = 0, a = 1
So x ^ 2-3x + 2 > 0
We get X2, B = 2
The square of ax - (a + b) x + B = x ^ 2-3x + 2
The solution set of known inequality ax ^ 2-3x + 6 is {x | x is less than 1 or X is greater than 6}
(1) Find a, B (2) solution inequality ax ^ 2 - (AC + b) x + BC less than 0
You gave me the direction
That can only be said to be a polynomial
In the quadratic function y = ax & # 178; + BX + C, a × C < 0, then the number of zeros of the function is?
Ac0
That is, y = ax & # 178; + BX + C = 0 has two different roots
So the number of zeros is two
Two.
because
a×c＜0
So both of them can be positive and negative. A is the direction of the opening, C is the position of the vertex on the x-axis (i.e. the position on the y-axis). If the discriminant △ = B & # 178; - 4ac > 0 is used, because ac0. So the number of zeros is two.
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Two.
because
a×c＜0
So both of them can be positive and negative. A is the direction of the opening, C is the position of the vertex on the x-axis (i.e. the position on the y-axis). If the discriminant △ = B & # 178; - 4ac > 0 is used, because ac0. So the number of zeros is two.
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The range of the function y = - 2tan (π / 6 + x) on the interval [- π / 3, π / 6] is?
x∈[-π/3,π/6]
∴π/6+x∈[-π/6,π/3]
tan(π/6+x)∈[-√3/3,√3]
∴-2tan(π/6+x)∈[-2√3,2√3/3]
The range of [√ 2,3] is
Let f (x) = AX2 + BX + C, the two zeros of F (x) = f (x) - X be m, n (m < n). (1) if M = - 1, n = 2, find the solution set of inequality f (x) > 0; (2) if a > 0 and 0 < x < m < n < 1a, compare the size of F (x) and m
(1) From the meaning of the question, f (x) = f (x) - x = a (x-m) (x-n) when m = - 1, n = 2, the inequality f (x) > 0 is a (x + 1) (X-2) > 0. When a > 0, the solution set of inequality f (x) > 0 is {x | x ＜ - 1, or x > 2}; when a < 0, the solution set of inequality f (x) > 0 is {x | - 1 ＜ x < 2}
The range of function y = 1 / TaNx (- 4 / π ≤ x ≤ π / 4, and X ≠ 0)
When 0 ＜ x ≤ π / 4, 0 ＜ TaNx ≤ 1 is obtained
That is 1 / TaNx ≥ 1
That is, y ≥ 1
When π / Tan ≤ 0 - x ≤ 1
That is 1 / TaNx ≤ - 1
That is y ≤ - 1
So the range of the function is [1, positive infinity) ∪ (negative infinity, - 1]
Find the range of denominator first
t=tanx -π/4≤x
Let f (x) = AX2 + BX + C (a is not equal to 0) for any real number, f (2 + T) = f (2-T) holds,
Then the function value f (- 1). The smallest one of F (1) f (2) f (5) cannot be that one
f(2+t)=f(2-t)
The axis of symmetry of F (x) is x = 2
f(x)=a(x-2)^2+d
F (2) is the maximum value = D
f(5)=9a+d
f(1)=a+d
f(-1)=9a+d
A>0
F (2) = D min
A
The smallest one can't be f (1)
For any real number T, f (2 + T) = f (2-T) holds,
The axis of symmetry of the function f (x) = AX2 + BX + C (a ≠ 0) is x = 2,
When a ＜ 0, the smallest of F (- 1), f (1), f (2), f (5) is f (2)
When a > 0, the smallest of the function values f (- 1), f (1), f (2), f (5) is f (- 1) and f (5)
The smallest one can't be f (1)
For any real number T, f (2 + T) = f (2-T) holds,
The axis of symmetry of the function f (x) = AX2 + BX + C (a ≠ 0) is x = 2,
When a ＜ 0, the smallest of F (- 1), f (1), f (2), f (5) is f (2)
When a > 0, the smallest of the function values f (- 1), f (1), f (2), f (5) are f (- 1) and f (5)
(2+t)=f(2-t)
The axis of symmetry of F (x) is x = 2
f(x)=a(x-2)^2+d
F (2) is the maximum value = D
f(-1)=9a+d
It is impossible to be the smallest f (1)
The range of the function y = (1 / 2) ^ | x + 1 | is
Because | x + 1 | > = 0
And because y = (1 / 2) ^ x, this function is a decreasing function
So the maximum value of y = (1 / 2) ^ | x + 1 | is y = (1 / 2) ^ 0 = 1
So the range is (0,1]
Given that the quadratic function f (x) = AX2 BX C (a is not equal to zero, B, C belongs to R) satisfies: for any real number
1) Let f (1) - 1 ≥ 0, and f (1) ≤ (1 + 1, 2) 2 = 1, ｜ f (1) = 1. (2) Let f (x) = AX2 + BX + C, f (- 1) = 0, then A-B + C = 0, f (1) = 1, ｜ a + B + C = 1, then B = 12, a + C = 12