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]

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

There are 20 items in an arithmetic sequence, and the sum of each item is 1050. The first item is 5. Find out the tolerance of the sequence and the 20th item

a1=5
S20=1050=(a1+a20)*20/2
5+a20=105
A20 = 100 item 20 100
Tolerance d = (a20-a1) / 19 = (100-5) / 19 = 5
The tolerance is 5

F (x) = the maximum of 3sinx + 5sin (x + 60 °)
F (x) = the minimum of 3sin (x + 120 °) + 5sin (x + 80 °)
The maximum value of F (x) = (2-cosx) (2-sinx)

f(x)=3sin(x+20°）＋5sin(x+80°)
=3sin(x+20°）+5sin（x+20+60）
=3sin (x + 20 °) + 5 / 2Sin (x + 20) + (2 / 5 root sign 3) cos (x + 20)
=Under the root sign [(11 / 2) square + 3 / 4] and multiply by sin (x + 20 + *)
The maximum value of sin is 1, so the maximum value of F (x) is under the root sign [(11 / 2) of the square + 3 / 4], which is 31

The reduced function f (x) = 2 cosx.cos (x - π / 6) - √ 3 (SiNx) ^ 2 + sinxcosx 0

f(x)=2 cosx.cos (x-π/6)-√3(sinx)^2+sinxcosx
=2cosx*(cosx*cosπ/6+sinx*sinπ/6)-√3(sinx)^2+sinx*cosx
=2cosx*(√3/2*cosx+1/2*sinx)-√3(sinx)^2+sinx*cosx
=√3(cosx)^2+sinx*cosx-√3(sinx)^2+sinx*cosx
=√3((cosx)^2-(sinx)^2)+2sinx*cosx
=√3cos2x+sin2x
=2sin(2x+π/3)

The reduced function f (x) = 2 cosx.cos (x-π/6)-√3(sinx)^2+sinxcosx

f(x)=2 cosx.cos (x-π/6)-√3(sinx)^2+sinxcosx =2cosx*(cosx*cosπ/6+sinx*sinπ/6)-√3(sinx)^2+sinx*cosx =2cosx*(√3/2*cosx+1/2*sinx)-√3(sinx)^2+sinx*cosx =√3(cosx)^2+sinx*cosx-√3(sinx)^2+sinx*cosx =√...

If 3sinx + 4cosx = 5cos (x + α), then sin α=

3sinx+4cisx=5[cosxcosa-sinxsina]=5cosacosx-5sinasinx
Then there must be 3sinx = - 5sinasinx, 4cosx = 5cosacosx
Then Sina = - 3 / 5, cosa = 4 / 5
So a = - 37 degrees

Y = (3sinx + 4cosx) ^ 2-3 / (3sinx + 4cosx + 6)

Let: M = 3sinx + 4cosx, then: m ∈ [- 5,5], then:
Y = (M & # 178; - 3) / (M + 6) [let t = m + 6, then t ∈ [1,11]]
=[(t－6)²－3]/(t)
=t＋(33/t)－12
If t ∈ [1,11], then: y ∈ [2 √ 33-12,22]

Find the maximum value of function f (x) = 2cosx square + 3sinx on (- Pie / 2, Pie / 2)

F (x) = 2cos & # 178; X + 3sinx = 2-2sin & # 178; X + 3sinx = - 2 (sinx-3 / 4) & # 178; + 25 / 8 because - π / 2 ＜ x ＜ π / 2, so - 1 ＜ SiNx ＜ 1, then - 7 / 4 ＜ sinx-3 / 4 ＜ 1 / 40 ≤ (sinx-3 / 4) & # 178; < 49 / 16-49 / 8 ＜ - 2 (sinx-3 / 4) & # 178; ≤ 0-3 ＜ - 2 (sinx-3 / 4) & # 178; + 2

Let f (x) = 3sinx + 2cosx + 1. If real numbers a, B and C make AF (x) + BF (x-C) = 1 hold for any real number x, then the value of bcosca is equal to ()
A. −12B. 12C. -1D. 1

Let C = π, then for any x ∈ R, f (x) + F (x - π) = 3sinx + 2cosx + 1 + 3sin (x - π) + 2cos (x - π) + 1 = 2, then a = b = 12, C = π, then for any x ∈ R, AF (x) + BF (x-C) = 1, thus bcosca = - 1

The maximum value of function f (x) = √ 3sinx + √ 3 / 2cosx is?

F (x) = √ 3sinx + √ 3 / 2cosx = √ 3 / 2 (2sinx + cosx) = √ 3 / 2 (asin α + bcos α) = √ 3 / 2 (√ (2 & sup2; + 1 & sup2;) sin (x + arctan (1 / 2)) = √ 15 / 2Sin (x + arctan (1 / 2)) function f (x) = √ 3sinx + √ 3 / 2cosx max = √ 15 / 2