Two high school math problems. Ask for help with two high school math problems

1) First, where [1, positive infinity] is a subtraction function and log(1 2) is a subtraction function, then x 2-ax+a should be an increasing function, and its axis of symmetry x = a 2, then a 2< = 1, and a < = 2. And x 2-ax+a>0, there is a 2-4a<0,02), since the function has no monotonicity at a certain point, it is the same as the first question. Get 0

Haha, everyone else is in the eighth grade. I'll help you, just give the idea, the idea is your own. That logarithmic function is decreasing, so the quadratic function inside is incremental.

Considering that the quadratic function opens upward, only the right side is taken when increasing, and the symmetry axis and minimum value of the quadratic function are comprehensively considered, and the rest can be figured out by yourself. By the way, I'm a junior, hehe.

First of all, we define that the domain is greater than zero, and we get x 2-ax+a in the interval "0", and then from the monotonicity of the composite function, we can know that if the function decreases in the interval (same increase and decrease), only x 2-ax+a=0 needs to increase in [1, positive infinity], and then it should be noted that the monotonically increasing interval of the function f=x 2-ax+a is not [1, positive infinity, but [1, positive infinity is included in the decreasing interval of the function.

So there is a quadratic function monotonicity, and it can be obtained that the axis of symmetry a 2 is less than 1, and both f(x)=x 2-ax+a is in the interval [1, positive infinity", 0 and a 2 are less than 1, i.e., f(1)>0 and a<2, and the result is a<2

The second question is that since the function is not monotonicity at a certain point, monotonicity is for a region, so.

Same as the first question.

There is no absolute certainty that the answer will be correct, just a casual answer.

1.The number of viruses is 2 (x-1), where x is the number of days (1), 10, 8=2 (x-1), x=27 (because it is a whole number), so it should be taken 27 days at the latest.

2) x=27, then the number of bacteria is 67108864

After killing 98%, there are still 1342177 bacteria (rounded) 10 8 = 1342177 2 y

y=9, so the latest is 9 days later, which is the 36th day.

When a=0, it is an odd function, and when a is not equal to 0, it is a non-odd and non-even function.

2^(1-1)=1,2^(2-1)=2,2^(3-1)=4,2^(4-1)=8,2^(5-1)=16,2^(6-1)=32,2^(7-1)=64

So the relationship between the total number of virus cells n in the body and the number of days x is recorded as follows:

n=2^(x-1)

When the number of viral cells in the mouse exceeds 10 8, the mouse will die: n=2 (x-1)>=108

x>=days can be found.

So the drug should be injected on day 27 at the latest.

2.When x=27, we get n=67108864

After killing 98%, the remaining 67108864*2%=1342177 (rounded) 10 8=1342177 2 x

x=9, so the latest is 9 days later, which is the 36th day.

3.Since f(-x)=a+2 x

So when a=0, it is an odd function.

When a is not equal to 0, it is a non-odd and non-even function.

If the number of bacterial reproduction satisfies: y=2 (x-1)) power(1): 10 8)=(2 (x-1)), x= should be taken 27 days at the latest.

2): When x=27, find y=67108864

After 98% is killed, the remaining bacteria are 1342177 and then y=1342177 2x

In the same way, find x2Is a minus sign after a?

1. It is known that the quadratic function f(x)=ax +bx+c has a unique zero point -1

1) Find the expression f(x).

Crossing the point (0,1) then c=1

And there is a unique zero point -1 f(-1)=a-b+1=0

b/2a=-1

a=1,b=2

f(x)=x²+2x+1

2) When x belongs to [-2,2], find the minimum value g(k) of the function f(x)=f(x)—kx

f（x)=f(x)—kx

x²+2x+1-kx

x²+(2-k)x+1

1. When -2=<-(2-k) 2<=2, that is, when -2=2, that is, when k>6.

The minimum value g(k) = f(2).

4+2(2-k)+1

2k+93, when -(2-k) 2<-2, that is, when k<-2.

The minimum value g(k) = f(2).

4-2(2-k)+1

2k+12, it is known that the set m stone satisfies the following properties of the function f(x): in the definition domain d there is x0, and yes f(x0+1) = f(x0) + f(1) holds.

1) Does the function f(x)=1 x belong to the set m? Explain why.

f(x0+1)=f(x0)+f(1)

1/(x0+1)=1/x0+1

x0=(x0+1)²

x0²+x0+1=0

0, apparently non-existent.

2) If the function f(x)=kx+b belongs to the set m, let try to find the constraints that the real numbers k and b satisfy.

f(x0+1)=f(x0)+f(1)

k(x0+1)+b=kx0+b+k+b

b=0 so the constraint is b=0 and k belongs to r

3) Let the function f(x)=lga x +1 belong to the set m, and find the range of the value of the real number a.

f(x0+1)=f(x0)+f(1)

LGA (x0+1) +1=LGA x0 +1+LGA+1 A is obviously greater than 0

lga-lg(x0+1)²=lga-lgx0²+lga+1

lg(x0+1)²=lgx0²-lga-1

x0+1)²=x0²/(10a)

10a-1)/(10a) x0²+2x0+1=0

The equation has a solution, >=0

4-4(10a-1)/(10a) >=0

1-(10a-1)/(10a) >=0

a>0, so the value range of a is a>0

LG25 and 2LG2

lg25=lg5²=2lg5

2lg2+lg25=2(lg5*2)=2

1,1) knows the quadratic function f(x)=a(x+1) 2 from and has a unique zero point-1, and brings (0,1) into a(0+1) 2=1, which gives a=1, i.e., f(x)=(x+1) 2;

2>f(x)=x^2+(2-k)x+1；The axis of symmetry is x0=(k-2) 2

i) When k>6, x0>2, g(k) = f(2) = 9-2k

ii) When -2<=k<=6, -2<=x0<=2, g(k)=f(k2-1)=-k2 4+k

iii) When k<-2, x0<-2, g(k)=f(-2)=2k+1

21.If f(x) belongs to m, then there is a non-0 x, such that f(x+1) = f(x) + f(1) (because there is no 0 in the defined domain).

That is, 1 (x+1) = 1 x + 1= (x+1) x =》x=(x+1) 2, x 2+x+1=0, and because the quadratic equation = 1-4=3 <0 has no real root, f(x) does not belong to m,2If the function f(x)=kx+b belongs to the set m, then there is x, such that f(x+1)=f(x) +f(1), i.e., k(x+1)+b=kx+b+k*1+b, and the solution is b=0, i.e., as long as b=0, that is, f(x) belongs to m;

f(1) (defines the domain x>0), i.e., lga (x+1) +1=lga x +1+lga+1, lga-lg(x+1) =2lga-lgx +1

LGA=-1-LG [(x+1) x ] =LG [x 10(x+1) Since 0<=x 10(x+1) <=1 10, so 0 your title should be LG25+2LG2=LG5 2+2LG2=2LG5+2LG2=2LG2=2LG2=2LG10=2, Note: The values of LG2 and LG5 and pi (about the same as a constant, but also an irrational number, so usually write LG2 and LG5 directly.

(1) Because the image is over the dot (0, 1), so c = 1

And because there is a unique zero point -1, -b 2a=-1

And (-1,0) brings in f(x)=ax 2+2ax+1A=1, so f(x)=x2+2x+1

2）f(x)=x^2+(2-k)x+1.

f'(x)=2x+2-k.

Order f'(x)=0，k=2x+2

Classification discussion (i) k>6 at 2x+2-k<0,f'(x)<0 decreasing, g(k)=f(2)=9-2k

ii)-2<=k<=6, f(x) decreases first and then increases, g(k)=f(k2-1)=-k2 4+k

iii) 2x+2-k>0,f at k<-2'(x)>0 increments, g(k)=f(-2)=2k+

f(x+1)=f(x) +1

x+1)^2=x^2+1

2x=0x =0

f(x)=x^2 ∈ m

11/(x+1) = 1/x + 1

x+1)/x

x=(x+1)^2

x^2+x+1 = 0

no real root

y=1 x does not belong to m

b/(x+a)

f(x+1) = f(x) +1

b/(x+a+1) = b/(x+a) +1= (b+x+a)/(x+a)

b(x+a)= (b+x+a)(x+a+1)= (x+a)^2+(b+1)(x+a)+b(x+a)^2+(x+a)+b = 0

1- 4b ≥ 0

b 1 4 Copied from somewhere else I hope it helps.

It's not an absolute value, it's the mold length.

1 Let a=(x,2x) be set in this way by parallel.

then x*x+(2x)*(2x)=3*3

x=plus or minus under the root sign.

2 is set to n=(y, -2y) by the vertical can be set as such.

then 5y*y=1

y=plus or minus under the root sign.

There are two answers.

1 Because the vector a is vector b, let a = (x, 2x), then [x 2 + (2x) 2] (1 2) = 3, then x = (9 5) (1 2), a is calculated.

2 is to find a, and the unit vector length is 1. If we find a, let a=(x,y), then, 4x+2y=0, and there are x 2+y 2=1, we get two sets of solutions, x=(1 5) (1 2), y=-2*(1 5) (1 2).

and x=-(1 5) (1 2), y=2*(1 5) (1 2).

(1) Let the coordinates of vector a be (x,y).

x^2+y^2=9

Vector A Vector B

2x-y=0

y=2x5x^2=9

x = plus or minus 3 root number 3 5

The coordinates of vector a are (3 roots, 3 5, 6 roots, 3 5) (-3 roots, 3 5, -6 roots, 3 5).

2) Isn't the length of the unit vector just 1?

1.It is known that the domain of the function f(x)= 1-x 2y=f(x+1) is [-2,3], i.e., -2 x 3 in y=f(x+1).

1≤x+1≤4

So y=f(x) is defined in the domain [-1,4], so:y=f(2x+1).

1≤2x+1≤4

Solution: 1 x 3 2

So y=f(2x+1) is defined in the domain [-1,3 2].

To solve similar problems, grasp a principle:

That is, for the same function f(x), its value range and definition domain are fixed!

That is, no matter what is in (), in short, the value range of () is certain, that is, the definition domain!

Knowing that the domain of y=f(x+1) is [-2,3], when you find the domain of f(x), (x 1) is a whole, which is equivalent to the (x) in f(x) that you require

So the range of () is the range of (x 1)!

The x in y=f(x+1) belongs to [-2,3], and obviously the (x) in f(x) is the range of x+1, which is [-1,4].

Knowing the domain of f(x) [-1,4], when finding the domain of f(2x+1), (2x+1) is a whole, which is equivalent to (x) in f(x).

The value range of (x) in f(x) is [-1,4], so the value range of (2x+1) in f(2x+1) is [-1,4], and the value range of x is the value range of x in f(2x+1), that is, the definition domain of f(2x+1).

2. 1-x^2 >0 10

1-x2^2 >0

x2 > x1

1-x1^2 -1-x2^2 >0

Monotonically decreasing.