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If the first term of the arithmetic sequence is 125, and it is larger than 1 from the 10th term, then the range of tolerance D is () A. d＞875B. d＜325C. 875＜d＜325D. 875＜d≤325
According to the meaning of the title, A10 ＞ 1A9 ≤ 1, 125 + 9D ＞ 1125 + 8D ≤ 1, 875 ＜ D ≤ 325
Let A1 = 125 and A10 be the first item larger than 1 in the arithmetic sequence {an}, then the tolerance D is ()
A. （875，+∞）B. （-∞，325）C. （875，325）D. （875，325]
According to the meaning of the question, the solution of A10 = 125 + 9D > 1, A9 = 125 + 8D ≤ 1 is 325 ≥ d > 875
If the first term is 125 and the 10th term is larger than 1, then the range of tolerance D of this arithmetic sequence is ()
A. d＞825B. d＜325C. 875＜d＜325D. 875＜d≤325
The number sequence is {an}. From the meaning of the question, we can get A9 = 125 + 8D ≤ 1A10 = 125 + 9D > 1, and the solution can get 875 < D ≤ 325, so we choose D
Given the function f (x) = a (COS ^ 2x + sinxcosx) + B, when a
The known function f (x) = cos (2x - π / 3) + 2Sin (x - π / 4) sin (x + π / 4) = sin (2x - π / 6),
Given the function f (x) = cos (2x - π / 3) + 2Sin (x - π / 4) sin (x + π / 4) = sin (2x - π / 6), let g (x) = f (x + a) be an even function and find the minimum positive number of A
F (x) = sin (2x - π / 6) = cos [π / 2 - (2x - π / 6)] = cos (2 π / 3-2x) = cos (2x-2 π / 3), so f (x) is even function, and it is symmetric with respect to x = π / 3, the period is π, so the minimum integer of a is π
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(2-cosx-2+4cosx)/(1-2cosx)(2-cosx)>0
3cosx/(1-2cosx)(2-cosx)>0
Because 2-cosx > 0
So Cos2 (1-2cosx) > 0
cosx(2cosx-1)
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X belongs to [- π / 4,3 π / 4]
As a result, we are going to be the 2-cosx ((x) as a result of the (1 + cos \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\sinx) > 0 is SiNx
Given n = (2cosx, 3sinx), M = (cosx, 2cosx), Let f (x) = n · m + A. (1) if x ∈ [0, π 2] and a = L, find the maximum and minimum of F (x), and the value of X when the maximum and minimum are obtained; (2) if x ∈ [0, π] and a = - 1, the equation f (x) = B has two unequal real roots X1 and X2, find the value range of B and the value of X1 + x2
F (x) = n · m + a = 2cos2x + 23sinxcosx + a = cos2x + 1 + 3sin2x + a = 2Sin (2x + π 6) + A + 1 (1) a = 1, f (x) = 2Sin (2x + π 6) + 2 ∵ 0 ≤ x ≤ π 2 ∵ π 6 ≤ 2x + π 6 ≤ & nbsp; 7 π 6 when 2x + π 6 = π 2, i.e. x = π 6, f (x) max = 4; X = π 2, f (x) min = 1. & nbsp; (2) A = - 1, f (x) = 2Sin (2x + π 6) ∵ 0 ≤ x ≤ π, ∪ π 6 ≤ 2x + π 6 ≤ 13 π 6 ∪ - 12 ≤ sin (2x + π 6) ≤ 1, ∪ - 1 ≤ f (x) ≤ 2 when f (x) = B has two unequal roots, we can get 1 ＜ B ＜ 2 or - 2 ＜ B ＜ 1b ∈ (- 2, 1) ∪ (1, 2); X1 + x2 = π 3, 4 π 3
Why does 4cosx-3sinx = - 5sin (x) have a maximum of 5 and a minimum of - 5
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φ = arctan 3 / 4, about = 37 degrees
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=-3sinx+4cosx
=5(-3/5sinx+4/5cosx)
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