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If the line y = KX intersects the ellipse x ^ 2 / 4 + y ^ 2 = 1 at two points a and B, and ab ≥ √ 10, find the value range of K
Let x = 2sina, y = cosa, then k = Tana, get the relation from ab distance, 4sina ^ 2 cosa ^ 2 > = 5 / 2, simplify to get Sina > = root 2, get the value range of angle a as [pi / 4, PI / 2], thus get the value range of K as [1, positive infinity]
Let L: y = KX + 2 and the ellipse C: the square of x 2 / 2 plus the square of y equal to 1 intersect at two different points a, B, O as the origin of the coordinates, (1) find the straightening range of K;
x^2/2+y^2=1
Y = KX + 2 to get: x ^ 2 / 2 + K ^ 2x ^ 2 + 4kx + 4 = 1
(1+2k^2)x^2+8kx+6=0
If there are two intersections, the discriminant is more than 0
That is 64K ^ 2-4 * 6 [1 + 2K ^ 2] > 0
64k^2-24-48k^2>0
16k^2>24
k^2>3/2
k> Root 6 / 2 or K
The range of K can be obtained by comparing the straight line y = KX + 2 and ellipse x ^ 2 + 2Y ^ 2 = 2 with different two points
Y = KX + 2 Generation X ^ 2 + 2Y ^ 2 = 2
So (2k ^ 2 + 1) x ^ 2 + 8kx + 6 = 0,
There are two different roots in the above form
△=64k^2-24(2k^2+1)=8(2k^2-3)＞0
That is, K ^ 2 > 3 / 2,
K ＜ - √ 3 / 2 or K ＞ √ 3 / 2
Multiplication and summation of equal ratio sequence and equal difference sequence
a1=q
a2=q^2*2
a3=q^3*3
.
aN=q^N×N
Finding the sum of the first n terms
sn=a1+a2+a3+…… +a(n-1)+an=q+q^2*2+q^3*3+…… +q^(n-1)*(n-1)+q^(n+1)*nqsn=q^2+q^3*2+q^4*3+…… +Q ^ n * (n-1) + Q ^ (n + 1) * n: (1-Q) Sn = q + Q ^ 2 + Q ^ 3 + Q ^ 4 + +q^n-q^(n+1)*n =q[1-q^(n+1)]/(1-q)-q^(n+1)*n=[q-q^(n...
If the function f (x) = 2 √ 3sinxcosx + 2cos & # 178; the maximum value of X + m in the interval [0, Wu & # 189;] is 6, find the value of constant M and the minimum value of this function when x ∈ R, and find the set of corresponding values of X
F (x) = 2 √ 3sinxcosx + 2cos & # 178; X + M = √ 3sin2x + cos2x + 1 + M = 2Sin (2x + π / 6) + 1 + m in the interval x ∈ [0, Wu & # 189;] 2x + π / 6 ∈ [π / 6,7 π / 6] sin (2x + π / 6) maximum = 1, so 2 + 1 + M = 6m = 3f (x) = 2Sin (2x + π / 6) + 42x + π / 6 = 7 π / 6 sin (2x + π / 6) minimum
In △ ABC, ∠ a - C = 25 ° and ∠ B - a = 10 °, then ∠ B =?
Given that the rational number A.B.C satisfies the condition | a + C + 2 | + | 3a-6b + 7 | + (3b-3c + 4) & sup2; = 0, then ABC =?
Add two formulas ∠ a = ∠ C + 25 ∠ B - ∠ C = 35 ∠ a + ∠ B + ∠ C = 180 2 ∠ C + ∠ B + 25 = 180
From ∠ B - C = 35
2 ∠ C + B + 25 = 180
∠B=75
Q2
a+c+2=0
3a-6b+7=0
3b-3c + 4 = 0, a = - 3, B = - 1 / 3, C = 1
abc=1
The number of columns 2, 4, 8, 16, 32 ,
1) According to the arrangement rules of several numbers given in turn, continue to write the following three numbers;
2) What is the number 2014? What is the number n?
1) According to the arrangement rules of several numbers given in turn, continue to write the following three numbers;
64,128,236
2) What is the number 2014? What is the number n?
Is the 2014 power of 2, and the nth term is the nth power of 2
What is the binomial coefficient of the third power of X in the (x-1 / x) n power expansion?
How do I feel that I've given less conditions? N and R are unknown
Can calculate the trouble, the answer is more detailed, thank you
In the (x-1 / x) nth power expansion,
T=c(n,r)x^(n-r)*(-1/x)^r
=(-1)^r*c(n,r)x^(n-2r),
When n-2r = 3, r = (n-3) / 2 is a nonnegative integer,
The binomial coefficient of the cubic of X is C (n, (n-3) / 2)
When r = (n-3) / 2 is not a nonnegative integer, there is no solution
In the first year of junior high school, there should be the process and answer to the denominator in solving the equation of one variable. The more, the better. The best is 15 questions
1 2x-10.3x=15 2 0.52x-(1-0.52)x=80 3 x/2+3x/2=7 4 3x+7=32-2x 5 3x+5(138-x)=540 6 3x-7(x-1)=3-2(x+3) 7 18x+3x-3=18-2(2x-1) 8 3(20-y)=6y-4(y-11) 9 -(x/4-1)=5 10 3[4(5y-1)-8]=6 11 2(x-2)+2=x+112 2(x-2)-3...
There is a sequence {an} which is composed of the following rules: 11, 12, 21, 13, 22, 31, 14, 23, 32, 41, 15, 24, 33, 42, 51, 16 Question: (1) what is 2750 in the sequence? (2) Which score is item 200?
According to the meaning of the question, we can get 11, (12, 21), (13, 22, 31), (14, 23, 32, 41), (15, 24, 33, 42, 51), 16 If the numerator denominator is added to N, this group will have n-1 numbers, (1) ∵ 50 + 27 = 77, and the group of ∵ 2750 has 76 numbers, then the former group will have 75 numbers, and so on, the numbers of all the previous groups are: 1 + 2 + 3 + 4 + +75 = (1 + 75) × 752 = 2850, and 2850 + 27 = 2877, so 2750 is the 2877 item in the sequence; (2) 1 + 2 + 3 + 4 + ·· + n = (1 + n) × N2, when n = 19, 1 + 2 + 3 + +19 = (1 + 19) × 192 = 190, so the sum of numerator and denominator of item 200 is 21, and item 200 is 1011
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