Senior 1 math problems. It s not hard. Ask for answers

1. Knowing a= , b= , a a, and a b, find aa a, and a b, and solve the simultaneous equations of y=2x-1, y=x+3 to obtain x=4, y=7a=(4,7).

2. Use the definition to prove that y=2x—4x+3 is a subtraction function in the interval (- 1).

y=2x²—4x+3=2(x^2-2x+1-1)+3=2(x-1)^2+1

Therefore y=2x—4x+3 is a subtraction function over the interval (- 1).

1.i.e. a a b, the solution is 2x-1=x+3 to get x=4, and substituting the original formula to get y=7, so a=(4,7).

2.Let x1, x2 (-1], and x1 x2

f(x1)-f(x2)=2x1²-4x1+3-2x2²+4x2-32(x1²-x2²)-4(x1-x2)

2(x1-x2)(x1+x2-2)

Because x1 x2

So x1-x2 0

Because x1, x2 (-1].

So x1+x2-2 0

So 2(x1-x2)(x1+x2-2) 0 is f(x1)-f(x2) 0

f(x1)＞f(x2)

Combine x1 x2, so f(x) is a subtractive function on (-1).

Solution: According to the title, the center of the circle c(1,1) has a radius of 2.

Connecting AC, the formula for the distance between the two points gives AC = root number 5 2.

So A is outside the circle C. There are two tangents to the circle C through A.

Let the tangent equation be y+1=k(x-2), and the distance from the circle c to the tangent d=r=2, we can get k=4 3 or k=0

That is, the linear equation is 4x-3y-11=0 or y+1=0

Offers you two methods:

1.Solution: The subset of the set sn can be divided into two categories:

contains a subset of 1; Does not contain a subset of 1. There are 2 (n-1) subsets of each of these two classes, and for any subset a of , there must be a unique subset of a corresponding to , and if a is an odd subset, then a is an even subset; If a is an even subset, then a is an odd subset. Therefore, if there are x odd subsets and y even subsets, then there must be x even subsets and y odd subsets.

Therefore, the number of odd and even subsets of SN is the same.

2.Solution: Let a be any odd subset of sn, and construct the map f as follows:

a a-, if 1 a

a a, if 1 a(a

Indicates the set obtained by subtracting 1 from set a).

So, the mapping f is a mapping that mirrors the odd subset as an even subset.

It is easy to know that if a1, a2 are two different odd subsets of sn. then f(a1) ≠ f(a2), i.e., f is monographic.

Wish you knew what a single shot is).

And for each even subset b of sn, if 1 b, then there is a=b (meaning b=, such that f(a)=b, so that f is full shot.

Knowing the full shot, ......）

Therefore, f is the one-to-one correspondence between the set of odd subsets of sn and the set of even subsets of s, so that the number of odd subsets and even subsets of sn is equal, both are 1 2 2 n=2 (n-1).

It involves the content of sets and functions, and typing hurts to the hand, hopefully.

n^2=1/3*n*3n=1/3*n*(3n+1-1)=1/3*n*(3n+1)-1/3*n

Then the accumulation, the things before the minus sign can be added up to the known conditions, and the things after the minus sign are too easy.

The second question looks at the picture.

1n(3n+1)=3n²+n

1×4＋2×7＋3×10＋..n(3n＋1)=3（1²+2²+.n²）+1+2+..

n）=3（1²+2²+.n²）+n（n+1）/2=n(n+1)²∴3（1²+2²+.n²）=n(n+1)²-n（n+1）/2=n(n+1)(2n+1)/2

1²+2²+.n = n(n+1)(2n+1) 62(a) to y.

Get dy dx=y =x *ln(2-x)+x*ln(2-x) =ln(2-x)-x (2-x).

b) The slope of the tangent at x=1 is ln(2-1)-1 (2-1)=-1, and the slope of the straight line x+ky+3=0 is -1 k

-1）*（1/k）=-1，k=-1

Note: y, x, ln(2-x) refer to the derivatives of y, x, ln(2-x), respectively.

Question 1:

Set 1 2+2 2+3 2+...n 2 = s compares the known equation of the problem with the expression of the left term of the equation to be proved n(3n+1)=3n 2+n and n 2

1x4+2x7+3x10+..n(3n+1)=3s+n(n+1)/2=n(n+1)^2

It is easy to get s=n(n+1)(2n+1) 6