∵x 1, x2 is two real roots of the unary quadratic equation x 2-2 (m- 1) x+m+ 1 = 0.
∴[2(m- 1)]^2-4(m+ 1)>； =0
Then m & gt=3 or m.
And Xi+x2 = 2 (m-1) x1x2 = m+1.
Y = f (m) = x12+x 2 = [(x1+x2) 2-2x1x 2] = [2 (m-1)] 2-2 (m =3 or m
2. It is known that the domain of f(x) = x 2-2mx-6 is [- 1, 1]. For any m∈R, find the analytical formula of the minimum value g(m) of the function f (x).
The second question is as shown in the figure:
3. let a = {x |-2 ≤ x ≤ 4}, b = {x | x < a}, A∩B= empty set, and find the range of a.
Solution: Because A∩B= empty set, the maximum value of b = {x | x < a} should be less than or equal to -2, and A happens to be the maximum value of set B, so a < =-2.
4. Given the set a= {x | ax 2-2x+ 1 = 0, x ∈ r}, if there is at most one element in A, then the real number a= value range.
Solution: There is at most one element in A.
That is, the equation AX 2-2x+ 1 = 0 has at most one solution.
Then b 2-4ac
∴a>； = 1
5. let the complete set U=R, a = {x | 3m- 1 < x < 2m}, and b = {x |- 1 < x < 3}. If CuB really contains a, the range of the real number m.
Solution: does CuB really contain a? What do you mean by that?
6. If f(xy)=f(x)*f(y) is true for all real numbers x and y, and f(0)≠0, then f(2009)=?
Solution: ∫f(xy)= f(x)* f(y) holds for all real numbers x and y.
∴f( 1* 1)=f( 1)*f( 1)？ That is, f( 1)=2*f( 1)
So f( 1)=0.
f(2009)= f(2009 * 1)= f(2009)* f( 1)= 0
There seems to be something wrong with this question.
7. Even if the function f (x) defined on R = 1-x is under the root sign, then when x≤0, f(x)=?
Solution: Is there a problem with the problem?
8. If f (x) is a function that is added to [0, +∞), what is the relationship between f(-3/4) and f (-1+a-a 2)? .
There should be something missing from this question.
Solution: Is f(x) a odd function or an even function?
- 1+a-a^2=-(a^2-a+ 1/4)-3/4=-(a- 1/2)^2-3/4<； =-3/4
(1) is an odd function.
∫f(x) increases on [0, +∞]
Then f(x) increases at (-∞, 0).
∵- 1+a-a^2<； =-3/4
∴f(-3/4)>； =f(∵- 1+a-a^2)
(2) is an even function.
∫f(x) increases on [0, +∞]
Then f(x) decreases at (-∞, 0).
∫- 1+A-A 2
∴f(-3/4)<； =f(∵- 1+a-a^2)
9. Write the minimum value of the quadratic function f (x) = x 2+1in the interval [a, a+ 1].
Solution? :( 1)a+ 1 & lt； =0 is a.
Then f(x) decreases in the interval [a, a+ 1].
∴ minimum value f (a+ 1) = a 2+2a+2.
(2)a & lt； 0，a+ 1 & gt； 0? , that is? - 1 & lt； A<0 is
The monotonicity of f(x) in the interval [a, a+ 1] is uncertain, but the minimum value is f(0)= 1.
(3)a & gt； 0
So f(x) increases in the interval [a, a+ 1]?
∴ minimum value f (a) = a 2+1
To sum up: what is the minimum value of f (x) = x 2+1in the interval [a, a+ 1]? a^2+2a+2(a<； =- 1)
1(- 1 & lt； a & lt0)
a^2+ 1(a>； 0)
10. Judge the monotonicity of x in the interval [0, +∞] under the function f(x)= radical sign, and prove it. ?
Solution: Under f(x)= root sign, x monotonically increases in the interval [0, +∞].
Proof: make x1>; x2 & gt=0
Then f (x1)-f (x2) = √ x1-√ x2 =) = (√ x1-√ x2) * (√ x1+√ x2)/(√ x/kloc-.
∵x 1 & gt； x2
∴f(x 1)-f(x2)=(x 1-x2)/(√x 1+√x2)>； 0
That is, f (x 1) >: f(x2)
Therefore, under f(x)= root sign, x monotonically increases in the interval [0, +∞].
There are some questions. Please correct them. I'm doing it.