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As shown in the figure, BD and CE are the two heights of triangle ABC, and m and N are the midpoint of BC and de respectively
Connecting DM, EM, ∵ m is the midpoint of BC, BD and CE are the two heights of △ ABC, ∵ EM = 12bc, DM = 12bc, ∵ EM = DM, ∵ n is the midpoint of De, ∵ Mn divides de vertically
Define the operation a * B as a * b = a (a > = b) or B (a > = b)
F (x) is a piecewise function
(1) When 2K π + π / 4 ≤ x ≤ 2K π + 5 π / 4 (K ∈ z), SiNx ≥ cosx
So at this time, f (x) = SiNx, the range is [- √ 2 / 2,1];
(2) When 2K π - 3 π / 4 ＜ x ＜ 2K π + π / 4 (K ∈ z), SiNx ＜ cosx
So at this time, f (x) = cosx, the range is (- √ 2 / 2,1];
The comprehensive value range is [- √ 2 / 2,1]
It is known that both {an} and {BN} are arithmetic sequences, CN = an * BN. Is that sequence {CN} arithmetic sequence
Only when one of the sequence {an} and the sequence {BN} is a constant sequence, can CN = an * BN be an arithmetic sequence, usually not an arithmetic sequence
For example:
{an} = n is an arithmetic sequence, {BN} = n is an arithmetic sequence
{CN} = an * BN = n ^ 2 is obviously not an arithmetic sequence
When the value of λ is, then the linear equations have the solution λ X1 + 11x2 + (λ + 1) X3 = 0 X1 - (λ - 8) x2 + 2x3 = 0 2x1 + 14x2 + (λ + 3) X3 = 0
λx1+11x2+(λ+1)x3=0
x1-（λ-8）x2+2x3=0
2x1+14x2+(λ+3）x3=0
First of all, the system of equations must have a zero solution (x1 = x2 = X3 = 0)
If you need a nonzero solution,
Take the sides AB and AC of triangle ABC as sides and make square AbfG and square ACDE inward. M is the midpoint of DF and N is the midpoint of BC. Connect Mn
Explore the relationship between line segment Mn and BC. This is different from the problem you have done before. The figure you drew is d on the left side of CM, while the title of my problem is d on the right side of CM. If you can't do that, it may be vertical, BC = 2Mn. I just don't know how to prove it
Extend cm to h, make cm = MH, connect FH, BH, cm, BM, extend CD, intersect BF at I
∵MF=MD CM=HM ∠CMD=∠HMF
∴△CMD≌△HMF
HF=CD=AC
∠HFJ=180°-∠JHF-∠HJF
∠HJF=∠IJC ∠JHF=∠DCM
∠BIC=∠IJC+∠DCM
In quadrilateral ABIC, ABI = ACI = RT ∠
∠BAC=360°-∠ABI-∠ACI-∠BIC=180°-∠BIC=180°-∠IJC-∠DCM=180°-∠JHF-∠HJF=∠HFB
∴△ABC≌△FBH
∠HBF=∠ABC
∠CBH=∠HBF+∠CBF=∠ABC+∠CBF=90°
BC⊥BH
N is the midpoint of BC and M is the midpoint of HC
MN‖BH
BC⊥MN
Well, if you can't, you can't
Can you attach your picture? I think we can solve it with vectors.
Extend cm to h, make cm = MH, connect FH, BH, cm, BM, extend CD, intersect BF at I
∵MF=MD CM=HM ∠CMD=∠HMF
∴△CMD≌△HMF
HF=CD=AC
∠HFJ=180°-∠JHF-∠HJF
∠HJF=∠IJC ∠JHF=∠DCM
∠BIC=∠IJC+∠DCM
In quadrilateral ABIC, ABI = ACI = RT ∠
The bac... Unfolds
Extend cm to h, make cm = MH, connect FH, BH, cm, BM, extend CD, intersect BF at I
∵MF=MD CM=HM ∠CMD=∠HMF
∴△CMD≌△HMF
HF=CD=AC
∠HFJ=180°-∠JHF-∠HJF
∠HJF=∠IJC ∠JHF=∠DCM
∠BIC=∠IJC+∠DCM
In quadrilateral ABIC, ABI = ACI = RT ∠
∠BAC=360°-∠ABI-∠ACI-∠BIC=180°-∠BIC=180°-∠IJC-∠DCM=180°-∠JHF-∠HJF=∠HFB
∴△ABC≌△FBH
∠HBF=∠ABC
∠CBH=∠HBF+∠CBF=∠ABC+∠CBF=90°
BC⊥BH
N is the midpoint of BC and M is the midpoint of HC
MN‖BH
BC ⊥ Mn
Define the operation a * B as: a * b = a (AB), for example, 1 * 2 = 1, then the range of function f (x) = SiNx * cosx is? For detailed explanation!
When SiNx ＞ cosx, π / 2 ＜ x ＜ 3 π / 2, then - 1 ≤ cosx ＜ 2 / 2
When SiNx ≤ cosx, 3 π / 2 ≤ x ≤ π / 2, then - 1 ≤ SiNx ≤ √ 2 / 2
So the range is [- 1, √ 2 / 2]
The operation a * B means to take the smaller one of AB, so the image of SiNx and cosx is drawn in a plane rectangular coordinate system, and the lower half of the image is needed, so the value range of F (x) is [- 1, two-thirds root sign 2]
It is known that both {an} and {BN} are arithmetical sequences, and CN = 2 * 3 (an + 2bn) times. It is proved that {CN} is an arithmetical sequence
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To prove that CN is an equal ratio sequence, we can find that CN / C (n-1) is a constant
Cn/C(n-1)＝{2×3^(An+2Bn)}/{2×3^(A(n-1)+2B(n-1))}
Reduction: ^ (2b3) ^ - 2b
=3^[An+2Bn-A(n-1)-2B(n-1)]
Because an and BN are arithmetic sequences, an-a (n-1) = constant D1
Bn-b (n-1) = constant D2
So the above formula is equal to 3 ^ (D1 + 2d2)
So it is an equal ratio sequence
It's too troublesome to write. It's about proving that an + 2bn is an arithmetic sequence. You can always calculate it by using that general term formula
Solving linear equations: X1 + x2 + X3 = 6 2x1-x2 + 3x3 = 9 2x1 + 3x2-2x3 = 2
Augmented matrix=
1 1 1 6
2 -1 3 9
2 3 -2 2
r2-2r1,r3-2r1
1 1 1 6
0 -3 1 -3
0 1 -4 -10
r1-r3,r2+3r3
1 0 5 16
0 0 -11 -33
0 1 -4 -10
r2*(-1/11),r1-5r2,r3+4r2
1 0 0 1
0 0 1 3
0 1 0 2
So, t = (1,3)
In the acute triangle ABC, take AB and AC as sides, make square abed and afgc. Connect DF, Mn passing through point A. Mn is perpendicular to n intersection, DF is at m.1: prove: DM = MF 2: prove: s △ DAF = s △ ABC
The last sentence of sensory condition should be "Mn is perpendicular to BC and N intersects DF with M". It is proved that: (1) FP and DQ are perpendicular to Mn and P, Q ∵ - fac = 90 °, m, a and N are in a straight line ∵ FAP + ≌ NAC = 90 ° and ∵ NAC + ≌ ACN = 90 ∩ FAP = ∩ ACN and ∩ FPA = ∩ ANC = 90 ° and FA = AC ≌ FAP ≌ ACN ≌ FP = an
Given that f (x) = SiNx + cosx has the domain of [a, b] and the range of [- 1, 2], then the range of B-A is___ .
∵ function f (x) = SiNx + cosx = 2Sin (x + π 4), ∵ and a ≤ x ≤ B, ∵ a + π 4 ≤ x + π 4 ≤ B + π 4, and - 1 ≤ 2Sin (x + π 4) ≤ 2, ∵ 22 ≤ sin (x + π 4) ≤ 1, in one period of sine function y = SiNx, in order to satisfy the above formula, then - π 4 ≤ x + π 4 ≤ 5 π 4, ∵ (B-A) max = 5 π 4 - (- π 4) = 3 π 2, (B-A) min = 5 π 4 - π 2 = 3 π 4
RELATED INFORMATIONS
- 1. It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence
- 2. In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC
- 3. Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n belongs to positive integer) is an arithmetic sequence Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n is a positive integer) is an arithmetic sequence, Sn is the sum of the first several terms of the sequence {BN}, and Sn = 2n-bn + 101) is the general term formula of an and BN (2) Does K belong to a positive integer, so that AK BK belongs to (0,1 / 2)? If so, find K; if not, explain the reason
- 4. In △ ABC, extend the midline BD on the AC side to F, so that DF = BD, extend the midline CE on the AB side to g, so that eg = CE
- 5. (1) In the plane rectangular coordinate system, move point a (- 3,4) to the right 5 units to point A1, and then rotate point A1 clockwise 90 ° around the origin of the coordinate to point A2. Directly write out the coordinates of points A1 and A2; (2) in the plane rectangular coordinate system, move point B (a, b) in the second quadrant to the right m units to point B1 in the first quadrant, and then rotate point B1 clockwise 90 ° around the origin of the coordinate to point B 2. Directly write out the coordinates of points B1 and B2; (3) in the plane rectangular coordinate system, translate point P (C, d) n units along the horizontal direction to point P1, and then rotate point P1 90 ° clockwise around the origin of the coordinate to point P2, directly write out the coordinates of point P2
- 6. [help] there is a point P in the triangle ABC, which satisfies "vector PA + vector Pb + vector PC = 0". Can you explain that P is the center of gravity of the triangle ABC?
- 7. The midpoint o (0,0) and a (2,0) B of the plane rectangular coordinate system are the midpoint of the line OA. Rotate OA clockwise about point O for 30 degrees. Mark the corresponding point of point B as C, and find the location of point C
- 8. As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connecting an and cm intersects at P, try to find the degree of ∠ APM
- 9. Given that the negative integer solution of inequality x-m > 0 is - 1, - 2, - 3, - 4, find the value range of M (and draw the number axis) It's OK not to draw, but it's not easy to write
- 10. If the value of the algebraic formula 2x + 33 is nonnegative, then the value range of X is () A. x≥32B. x≥-32C. x＞32D. x＞-32
- 11. In the arithmetic sequence {an}, if A1 = 3, A100 = 36, then A37 =?
- 12. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF
- 13. The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN
- 14. In the RT triangle ABC, the angle c = 90, BC = a, AC = B, a + B = 16, then the function analysis of the area s of the RT triangle with respect to the variable length a When s = 32, a=______
- 15. It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!
- 16. If AQ = 1 / mAb, AP = 1 / NAC, then M + n is equal to A.1 B.2 C.3 D.4
- 17. In the known sequence an, A1 = 2, the sum of the first n terms is SN. For any n ≥ 2, 3sn-4, an, 2-3 / 2S (n-1) is an arithmetic sequence (1) (2) if BN satisfies BN = 3Sn, find the first n terms and TN of BN
- 18. Ad is the height of the side BC of the triangle ABC. Take ad as the diameter to make a circle, intersect with AB and AC respectively, multiply AC by E and F, and prove that AE multiplied by ab equals AF
- 19. Let BN = an + n prove that {BN} is an equal ratio sequence
- 20. In the triangle ABC, D is the midpoint of BC, AE bisector angle BAC, make de perpendicular to AE at e, intersect AB at g, intersect AC extension line at h. proof: BG = ch = 1 / 2 (AB-AC)