Urgent!! High School Math Questions, Urgent !! Senior 1 math problems

That is, the minimum value of f(x) on x [-1,+) is greater than or equal to a;

For the maximum-value problem of quadratic functions, the relationship between the axis of symmetry and the given interval is seen; When the axis of symmetry is x=a1)a<-1, it increases over the interval x [-1,+, and when x=-1, f(x) has a minimum value f(-1)=2a+3;

2a+3 a gets: a -3, so: -3 a<-1;

2) A -1, the axis of symmetry is in the given interval, so the minimum value of f(x) is f(a)=-a 2+2;

a 2+2 a i.e.: a 2 + a-2 0, (a+2)(a-1) 0, get: -2 a 1;again a -1;

So:-1 a 1;

In summary, the range of a is: -3 a 1;

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1) When a<-1, it is incremented on the interval x [-1,+, and when x=-1, f(x) has a minimum value f(-1)=2a+3;

2a+3 a gets: a -3, so: -3 a<-1;

2) A -1, the axis of symmetry is in the given interval, so the minimum value of f(x) is f(a)=-a 2+2;

a 2+2 a i.e.: a 2 + a-2 0, (a+2)(a-1) 0, get: -2 a 1;again a -1;

So:-1 a 1;

In summary, the range of a is: -3 a 1;

When x=-1, 1+2a+2 is greater than or equal to a, find a greater than or equal to -3

A is less than or equal to 1

Use the formula of the axis of symmetry to find the x of the lowest point, and then substitute it to get 1 2-2a+2 a, and you can find it.

Let gx=fx-a make gx=0 find δ and divide the interval according to the value of a at δ=0.

Set the length of the front grille to be x meters.

Then the length of the wall is s x

So the total cost: 400*x+450*2*s x+200*s, so there is an inequality 400*x+450*2*s x+200*s<=32000 solution: s<=(320x-4x 2) (9+2x).

Let u=9+2x, so u 2=4x 2+36x+81 get s<=-u+178-1521 u=178-(u+1521 u) we can know that when u=39, then s<=100 (the maximum value of 100 digits) so the maximum area is 100 square meters, and u=39, so x=15 meters are obtained.

d.Solution: Because u=r,a=,b=,then:cub=,cua=

Then: (ancub)=,(bncua)=

So: (ancub)u(bncua)={x|x>0 or x<=-1}

The answer is D, just look at the math textbook for the first year of high school.

1.From a+2b+c=0,5a+4b-4c=0, 3a+4b=0, a=2c

Original = (1 + 27 8-5 2) (3-9 4 + 5 4) = 15 162Let (a+b-c) c=(a-b+c) b=(-a+b+c) a=k, yes.

a+b)/c=k+1

a+c)/b=k+1

b+c)/a=k+1

Original = (k+1) 3

a+b-c=ck

a-b+c=bk

a+b+c=ak

2c=(b+a)k

2b=(a+c)k

2a=(b+c)k

2(a+b+c)=2(a+b+c)k

k = 1 or a + b + c = 0

a+b+c=0, k=-2

Original = -1 or 8

3.(v-2f)/(2f-u)=(v-f)/fvf-2f^2=2fv-2f^2-uv+ufuv=uf+fv

Divide both sides by fuv at the same time to get 1 f = 1 v + 1 u.

The original formula is proven.

Solution: Distributed by 3 7 2 faction: cos >0

From the cosine and the reciprocal of the cosine are the two real roots of the equation about x, xx, kx, and 3

cos * (1 cos) = k square 3

cosα+(1/cosα)=k

Solution: k = -2 cos = 1

cos(3 faction +) sin (pie) cos(pie +) sin (pie) sin -cos

The root number is [1-(cos)square]-cos

First, p is true, q is false, then 4-m>1, m<1-x 2 or m>(1, 2) x+4 get m<1

The second q is a true proposition, p is a false proposition 4-m<1, 1-x 2 m (1 2) x+4 gives 3 a combination, m < 1 or 3

Explain that the PQ is true and false.

p is really 4-m greater than 1, i.e. m is less than 3

Q is true if m is greater than 1 and less than or equal to 4

Therefore, p is true and q false is m less than or equal to 1

p, false, q, true, m, greater than or equal to 3, less than or equal to 4

So it is found that m is less than or equal to 1 or m is greater than or equal to 3 and less than or equal to 4