Let the quadratic function f (x) satisfy that f (x + 2) = f (2-x) and the sum of the squares of the two real roots of F (x) = 0 be 10. The image passes through (0,3), and the analytic expression of F (x) is obtained

Let the quadratic function f (x) satisfy that f (x + 2) = f (2-x) and the sum of the squares of the two real roots of F (x) = 0 be 10. The image passes through (0,3), and the analytic expression of F (x) is obtained

Let f (x) = ax ^ 2 + BX + C
f(2a+x)=f(2a-x)
f(x+2)=f(2-x)
So a = 1
If the image passes (0,3), then C = 3
And (B / a) ^ 2-2c / a = 10
B = 4 or - 4
So f (x) = x ^ 2 + 4x + 3 or F (x) = x ^ 2-4x + 3
Let the even function f (x) defined on R satisfy f (x) = x cube-8 (x is greater than or equal to 0), and find the solution set of the inequality f (X-2) greater than 0
When x > = 0, f (x) = x ^ 3-8
Let t > 0, then - t = 2,
f(x-2)=(x-2)^3-8>0
(x-2)^3>8
X>4
(2) If X0
(X-2) ^ 34 or X
Let the even function f (x) defined on R satisfy f (x) = x cube-8 (x is greater than or equal to 0), and find the solution set of the inequality f (X-2) greater than 0
First, we consider the solution set of F (x) greater than 0
If x > = 0, x ^ 3-8 > 0, x > 2,
Because f (x) is an even function defined on R, x0 is x > 2,
Because f (x) is an even function defined on R, x2
x>2+2=4
From even function f (x) about y axis symmetry, X
It is known that the coordinates of a and B are [1; 0]. [2.0]. If the quadratic function y = x ^ 2 + (A-3) x + 3, the image and line AB have a common point a value range
Because there is a root between a and B, and f (0) = 3, then the axis of symmetry of the function must be on the right side of the y-axis, and we get: - (A-3) / 2 > 0, if there is a root, we get: △≥ 0, we get a ≤ 3-2 radical 3, because there is only one root between a and B, we must have: 1, f (1) < 0, f (2) ≥ 0; 2, f (1) ≥ 0, f (2) < 0, 3, if △ = 0, a = 3-2 radical 3, we get (radical 3
-1 ≤ a < - 1 / 2 or a = 3-2 radical 3
1L is correct
I checked this question
Given that f (x) is an even function defined on R and monotonically increasing on [0, + ∞), find the value that makes the inequality f (2) less than or equal to f [(A-1) 1 / 2] hold
a≠1
Because f (x) is an even function 1 / A-1 > 2 or 1 / A-1
If the image of the quadratic function y = - x2 + MX-1 has two different intersections with the line AB whose two ends are a (0,3), B (3,0), then the value range of M is______ .
It is known that the equation of line segment AB is y = - x + 3 (0 ≤ x ≤ 3). Because there are two different intersections between quadratic function image and line segment AB, the equations y = − x2 + MX − 1y = − x + 3, 0 ≤ x ≤ 3 have two different real solutions. The elimination result is: X2 - (M + 1) x + 4 = 0 (0 ≤ x ≤ 3), Let f (x) = X2 - (M + 1) x +
Given that f (x) is an even function on R and a decreasing function on (- ∞, 0), then the solution set of the inequality f (x) ≤ f (3) is______ .
From the meaning of the question, we can get: F (x) is an even function on R and a decreasing function on (- ∞, 0), because the inequality f (x) ≤ f (3), so | x | ≤ 3, so - 3 < x < 3. So the answer is: [- 3, 3]
If we know that the image of a (1,0), B (2,0), quadratic function y = x2 + (A-3) x + 3 has an intersection with line AB, then the value range of a is______
Please be clear,
solution
By combining number with shape, we can get
1+a-3+3≥0
4+2a-6+3≤0
The solution is: - 1 ≤ a ≤ - 1 / 2
lim1/(ln(tanx+1)-1/sinx)
The following is my personal summary. Multiplication or division can be replaced. For example, ln (1 + x) * SiNx \ / x ^ 2 ~ x ^ 2 \ / x ^ 2. Addition and subtraction of infinitesimals of different orders can be replaced. For example, addition and subtraction of infinitesimals of the same order can not be replaced. Taylor expansion or lobita rule are often used
The coordinates of points a and B are (1,0), (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x2 + (A-3) x + 3 and line AB, then the value range of a is______ .
If the vertex of quadratic function is below the x-axis, if YX = 1 ＜ 0 and YX = 2 ≥ 0, that is 1 + (a − 3) + 3 ＜ 04 + 2 (a − 3) + 3 ≥ 0, the solution of the inequality is - 1 ≤ a ＜ - 12; if YX = 2 ＜ 0 and YX = 1 ≥ 0, that is 1 + (a − 3) + 3 ≥ 04 + 2 (a − 3) + 3 ＜ 0, the solution is - 1 ≤ a ＜ - 12
Why ln | (1 + SiNx) / cosx | = ln | secx + TaNx|
Because (1 + SiNx) / cosx = secx + TaNx
It is proved that 1 + SiNx = cosx / cosx + SiNx * cosx / cosx
So (1 + SiNx) / cosx = 1 / cosx + SiNx / cosx = secx + TaNx
Supplement secx = 1 / cosx (1 + SiNx) / cosx = 1 / cosx + SiNx / cosx = secx + TaNx. In fact, it's better to write the denominator separately