2 math problems for senior 1 and several math problems for sophomore sophomores

1) f(a)=ln(a-1)=b, so, a-1=e b. i.e. a=1+e b.So (2+2e b,2b) on the image of g(x).

Let x=2+2e b, y=2b, and remove the parameter b, and get y=2ln[(x-2) 2].i.e. y=2ln(x-2)-2ln2Therefore, g(x)=2ln(x-2)-2ln2

2) When 0 a 2, cosa > 0, the function f(x) is a quadratic function, and its axis of symmetry is x=-2sina 2cosa=-tana

If f(x) is an increment on [-1, root number 3], only -tana<=-1, so 4 = a 2 (because 0 a).

When a=2, coea=0, f(x)=2x-1, the condition is satisfied; When 2 a, cosa<0, if f(x) is an increasing function on [-1, root number 3], just -tana>=root number 3, and the solution is 2 a = 2 3In summary, 4 = a = 2 3.

（abc）-（bc

d）=95x3-94x3=3

So a-d=3

Because A is the first place and E is the third place, 96 points are awarded.

Then A should be 98, 99 or 100

If A is 99, then D=96, because the data is not duplicated, Acan only be 98 or 100 If A is 100, then D=97BC

d=94x3=282

bc=185

bc=185, then there must be something greater than 97

It doesn't fit the meaning of the title.

If a=98

then d = 95b

cd=282

bc=187ab

c=285 also holds.

So d=95

75/3x7972

Let t=2 x then f(x)=t -2t-3=(t-1) -4

x∈【0,1】 ∴t-1）²∈0,1】 f(x）∈【4,-3】

The range of f(x) is [-4, -3].

Is that it? I don't understand what you mean)

Let t=1-x 1+x, then f(t)=t directly replace t with x, f(x)=x

The function relation is f(x)=x

By definition, the numerator of the fraction is the power of the base, and the denominator is the power of the base.

y=x numerator is 1, which is 1 to the power of x, and the denominator is 2, which is 1 power of x and then squared.

1.Let t=2 x then t:[1,2] then f(x)=t 2-2t-3

Actually, it's a simple parabolic evaluation domain, and the middle line is -2 2=1

It opens upwards, so the minimum value is 1-2-3=-4 and the maximum is 4-4-3=-3

1 is by t = 2 x x (0,1), then t (1,2)f(t) = t 2-2 t-3 = (t-1) 2-4 by t (1,2).

f(t) (4,-3).

2 by question 2f(t)=t 2, i.e., f(t)=t 2 2, i.e., f(x)=x, 2, 2

3. From y=x (1 2), the left and right sides are squared at the same time to get y 2=x, that is, y= x

ps.Just remember the formula a x*a y=a (x+y)(a m) n=a (mn).

1 Solution: Let t=2*x [1,2].

f(x)=t²-2t-3=（t-1）²-4（t-1）∈[0,1] （t-1）²∈0,1] （t-1）²-4∈[-4,-3]

2 Does your square mean square, or does it mean square?

Let (1-x) (1+x)=t f(t)=t 3 y=x * = x

From the condition -1 2x+1 2)(x-1 3)<0, the parentheses of the inequality are opened and simplified so that the inequality sign is the same as the inequality sign of the original inequality, and the constant term is the same. You can get the values of a and b. From this, we can calculate the solution set of x 2+bx+9<0 in the future.

I hope you can solve it by yourself.

The solution set for ax 2+bx+1>=0 is -1 2b a=-1 6,b=-1

The solution set of x 2-x+9<0 is an empty set.

Knowing the solution set, you can use Vedder's theorem, compare the magnitude of 2 values, and then sit with the sign to find the coefficient.

Then use Veda's theorem to compare the magnitude of 2 values to obtain the solution set.

1.Solution: f(9)=f(3*3)=f(3)+f(3)=2f(a)>f(a-1)+2=f(a-1)+f(9).

a>0, a-1>0 gets a>1

f(a)>f(9a-9), and f(x) is an additive function defined on (0,+).

a>9a-9, i.e. a<9 8

So, 1=0 (because of 0).

a>=3 2 or a"Oak cherry = -1;

Liang Zhencong (x-2a) 2=4a 2-2a-6 gives x=2a + root number (4a 2-2a-6) or x=2a - root number (4a 2-2a-6).

2a+ (4a, 2-2a-6), <0 or 2a-root (4a, 2-2a-6), <0

The solution is -3, so the value range of the actual number a is: a -1

<=-1) Looks like a mistake

f（-2）=0，f（f（-2））=f（0）=02.Straight point set p:

Take x=2,y=7, so (2,7) belongs to p (remember to use mathematical language, I can't type).

Remember to add points.