Sophomore Science Mathematics Problems, Solving, Sophomore Science Math Problems

Solution 1, (let but not think) let a(x1,y1)b(x2,y2) be on the ellipse x +4y =36, and ab midpoint is (1,2).

x1+y1=2

x2+y2=4

and a(x1,y1)b(x2,y2) in the ellipse x +4y = 36, then above.

x1²+4y1²=36———

x2²+4y2²=36---

(x1 +4y1 )-x2 +4y2 )=0 (x1-x2)-4(y1-y2)=0, so that the slope of the linear equation where the string at the midpoint is located is: k=(y1-y2) (x1-x2)=linear equation: y-2=

That is: x+8y-17=0

2. Because x + (y 2) = 1

So x = 1-(y 2).

So x(1+y) = x(1+y) = (1-(y2))(1+y).

1-（y²/2）)(1+y²)

Let y = t(t>0) then (1-(y 2))(1+y )= (< = if and only if t=1, i.e., y=1, take the equal sign.

So when y=, x (1+y) is taken to the maximum.

1) Set a(x1,y1)b(x2,y2).

Then x1 +4y1 =36 x2 +4y2 =36 makes the difference.

x1+x2)(x1-x2)+4(y1+y2)(y1-y2)=0 where the midpoint (1,2).

Then x1+x2=2 y1+y2=4 substitution.

Get (x1-x2)+8(y1-y2)=0 known by symmetry x1≠x2 so k=y1-y2 x1-x2=-1 8

y=-1/8(x-1)+2

This method is called the point difference method. Take a good look at what is often used in analytic geometry in high school.

2) Since x +(y 2)=1, then 2x +y =2 so we recipe it this way.

x√(1+y²)=[√2x²（1+y²）]/√2≤[2x²+（1+y²）]/2√2=3√2/4

Think about the conditions for the establishment of the equal sign, and you can also use the triangular commutation method.

Let x=sin y=cos 2 0, 2).

It's an example ... It's good to look at a book, I haven't read a book for three years, I forgot.

I'll give you the way.

In the first question, let L, the distance from point A to L is equal to 1. The slope of l can be calculated. There should be two L's

Question 2. It can be known that the radius of the two circles is the same. The truncated strings are equal in length, and the distance from the center of the two circles to the corresponding straight line is equal. Set up the general formula of the straight line, and use the distance formula.

You can see if you can figure it out.

The first of the nth row is 2 (n-1), and the last one is 2 n-1;

2 10=1024,2 11=2048,,,So 2012 is the 11th row,The last one in the 10th row is 1023,j=2012-1023=989;;

an=2^(1-1)+2^(2-1)+1+2^(3-1)+2+2^(4-1)+3+..2^(n-1)+n-1；

2^0+2^1+2^2+..2^(n-1)+1+2+3+..n-1;

2^n-1+(n-1)*n/2;

n=1,an=1,n^2+n=2;;;n=2,an=4,n^2+n=6,;;n=3,an=10,n^2+n=12;;;

n=4,,an=21,n^2+n=20;;;When n>=4 an>n 2+n;;;

Proven by mathematical induction.

n=4.

Let n=k(k>=4) hold, i.e., ak=2 k-1+(k-1)k 2>k 2+k;;;

When n=k+1, an=2 (k+1)-1+k(k+1) 2>2 k-1+2 k+(k-1)k 2>k 2+k+2 k;

Because when n>4, 2 n>2n+2; Substituting the above formula an, >k 2+k+2 k>k 2+k+2k+2=k 2+3k+2=(k+1) 2+k+1;

So when n=k+1, it is true, and the original formula is proved.

In summary, when n<4 and an=4, an>n 2+n....

The first number in line i is 2 (i-1), 2 10=1024, 2012-1024=988, so 2012 is the 989th number in line 10.

The nth number in row nth is 2 (n-1)+(n-1), an=(2 0+0)+(2 1+1)+(2 2+2)+....2^(n-1)+(n-1)]

1+2+2^2+2^3+……2^(n-1)]+0+1+2+3+……n-1)]

2^n-1+n(n-1)/2

d(n)=an-[n²+n]=2^n-n(n+3)/2-1

d'(n)=2 n 2-n-3 2, d(n) decreases first and then increases.

d(1)=0、d(2)=-2、d(3)=-3、d(4)=0、d(5)=11

1.Observe the first number of each column in the table of numbers: 1, 2, 4, 8, 16....It can be seen that the first number in line i, is definitely 2 (i-1), then the first number in line 11 is 1024, because the tenth power of 2 is equal to 1024, and the first number in line 12 is 2048, obviously 2012 is in line 11, and now we know that the first number is 1024, then j=2012-1024+1=989

Hence i=11

j=989a22=2+1，a33=2^+2,..ann=2^(n-1)+(n-1)

an=2+4+8+16+..2^(n-1)+(1+2+3+..n-1)=2^n+1/2(n^2)-n/2-2

3.Let k=an-n -n=2 n-n 2-3n 2-2, and k=0 when n=4, i.e., an=n +n

When n<4, k<0 is an4, k>0 is an> n +n Say a word, you can check it yourself, and this is the method.

Drawing on scratch paper, because the bottom area is 2, the two bottom edges can be set to x and 2 x, and because the area of the diagonal plane perpendicular to the bottom is root number 5, so the square of x plus the square of 2 x + the square of 1 = the square of root number 5 gives x = root number 2, so the bottom surface is a square, and the side area is equal to 1 * root number 2 * 4 = 4 root number 2!!

1), value: 1, range: 2), let the imaginary number be: a+bi, and then simplify it by a 2 + b 2 = 1 in (1).

f=kx；So 10=k, so k=100;So to elongate 6cm, f=100; And the work w=fs, so w=6, 36j is needed

x+a+b=1

2ax+2ab+2bx=16/27

Find a, b (assuming x is a known number 0.)

Then v=abx

Do the rest for yourself, think carefully about such a simple question, don't take shortcuts, even if you want to get the solution steps, you can type out the question! Diligence is the last word.

This is a permutation and combination plus probability problem, which we also did at that time.

If you analyze it carefully, there are two roads on the first floor, four roads on the second floor, six roads on the third floor, and eight roads on the fourth floor.

But at every fork in the road, there are only two possible scenarios.

For example, if you want to fall to (4,2), you need to push up the count, analyze layer by layer, and multiply the probability of each layer.

The second question is simple, the same as the first question, just tremble out the distribution columns.

I don't understand, you can ask me again.