The function f (x) is defined as a function with a period of 5 π 2 on R, and f (x) = {SiNx (0 ≤ π, less than or equal to 0) cosx (- π) It's not square. I wrote 5 π / 2 wrong

The function f (x) is defined as a function with a period of 5 π 2 on R, and f (x) = {SiNx (0 ≤ π, less than or equal to 0) cosx (- π) It's not square. I wrote 5 π / 2 wrong

The period is 5 π 2. Is that 2 not square?
My guess is 5 / 2 π
So the answer is 2 / 2
Let x = y = 1
So f (1 * 1) = f (1) + F (1) f (1) = 0
Let x = y = - 1
So f (- 1 * (- 1)) = f (- 1) + F (- 1) f (1) = 2F (- 1) f (- 1) = 0
Let y = - 1
So f (x * (- 1)) = f (x) + F (- 1)
f(-x)=f(x)+0
f(x)=f(-x)
F (x) is even function
The simple formula is 0.5 × 2.33 × 8
 
It is known that the minimum positive period of the function f (x) = sin (2 ω X - π / 6) + 1 / 2 (ω > 0) is π
1, find the value of Omega [is it 1? I figure it out to be 1]
2, find the value range of F (x) in [0,2 π / 3]
1. The minimum positive period T = 2 π / 2 ω = π, so ω = 1
2. In this case, f (x) = sin (2x - π / 6) + 1 / 2, when x is in the interval [0,2 π / 3], the range of 2x - π / 6 is [- π / 6,7 π / 6], and the function value in this range is determined by [- 1 / 2,1], so the value range of sin (2x - π / 6) is [- 1 / 2,1], while the value range of the original function f (x) = sin (2x - π / 6) + 1 / 2 is [0,3 / 2]
Please calculate and compare by yourself
33 and 3 / 45 + (2 / 45-4 / 27) - 5 / 27
33 and 3 / 45 + (2 / 45-4 / 27) - 5 / 27
=33 and 3 / 45 + 2 / 45 - (4 / 27 + 5 / 27)
=34-1/3
=33 and 1 / 3
Original formula = (33 + 3 / 45 + 2 / 45) - (4 / 27 + 5 / 27)
=33 1 / 9-1 / 3
=33 and 7 / 9
It is known that the minimum positive period of the function f (x) = sin (2wx - π / 6) plus 1 (W > 0, X belongs to R) is π
It is known that the minimum positive period of the function f (x) = sin (2wx - π / 6) plus 1 (W > 0, X belongs to R) is π
1. Find the analytic expression of FX and the monotone increasing interval of the function
2. Find the maximum and minimum of FX when x belongs to [π / 4, π / 2]
1) The minimum positive period of the function f (x) = sin (2wx - π / 6) plus 1 (W > 0, X belongs to R) is π, and the minimum positive period is 2 π / 2W = π / W, so w = 1F (x) = sin (2x - π / 6) + 1F (x) '= 2cos (2x - π / 6) > 02n π - π < 2x - π / 6 < 2n π n π - (5 / 12) π < x < n π + π / 122): X belongs to
How to calculate 9.9 times 5.5 + + in a simple way
The answer is 9.9 * 5.5 = 10 * 5.5-0.1 * 5.5 = 55-0.55 = 54.45
Given that the minimum positive period of the function f (x) = sin (2wx-u / 6) x 1 / 2 (W > 0) is u.1, find W
It is known that the minimum positive period of the function f (x) = sin (2wx-v / 6) x 1 / 2 (W > 0) is v?
It is known that the minimum positive period of the function f (x) = sin (2wx-v / 6) x 1 / 2 (W > 0) is v?
(1) Analysis: because the minimum positive period of the function f (x) = sin (2wx - Wu / 6) x 1 / 2 (W > 0) is Wu
So, 2W = 2 π / π = 2 = > W = 1
(2) Because f (x) = sin (2x - π / 6) + 1 / 2
Monotone increasing interval: 2K π - π / 2
15 and 3 / 3 * 14 and 1 / 3 calculated by simple method
15 and 2 / 3 * 14 and 1 / 3
=（15+2/3）×（15-2/3）
=225-4/9
=224 5 / 9
Given the function f (x) = radical 3 + acosx + B (a, B are constants, A.B belongs to R), we can find the minimum positive period of the function;
The period is only related to the period of the periodic function cosx and is independent of the constant
The minimum period is only related to cosx.
Tmin=2∏/1=2∏
When a = 0, there is no minimum positive period and the period is any positive real number
When a ≠ 0, the period is 2 π
2∏
Solution: F (x) = radical 3 + acosx + B is a function of X, and its period is independent of radical 3 + B.
From the definition of periodic function f (x + T) = f (x), we get
Root 3 + ACOS (x + T) + B = root 3 + acosx + B
Discussion: when a = 0, there is no minimum positive period, and the period is any positive real number
When a ≠ 0, the minimum positive period is 2 π
hey
1.49+499+4999+49999+499999；2.（1+11+21+31+41）＋（9+19+29+39+49）
1.49+499+4999+49999+499999； 2.（1+11+21+31+41）＋（9+19+29+39+49） 3.（9999+9997+1995+…… +9001）－（1+3+5+…… +999 ） 4.123+234+345+456+5657+678+789 5.1400÷25÷8+350÷4÷125 6.9÷13+13÷9+11÷13+14÷9+6÷13
6.（0.36×5.1×81）÷（0.17×8.1×0.4）
Write all six
1.49+499+4999+49999+499999；
=50+500+5000+50000+500000-5
=555550-5
=555545
2.（1+11+21+31+41）＋（9+19+29+39+49）
=（1+9）+（11+19）+21+29）+（31+39）+（41+49）
=10+30+50+70+90
=50×5
=250
1/ =500000+50000+5000+500+50-5=555545;
2/ =21*(2+2+1)+29*(2+2+1)=250
1.49+499+4999+49999+499999
=50-1+500-1+5000-1+50000-1+500000-1
=555550-5=555545
2.（1+11+21+31+41）＋（9+19+29+39+49）
=1+49+11+39+21+29+31+19+41+9
=5*50=250
1.49+499+4999+49999+499999
Original formula = 50 + 500 + 5000 + 50000 + 500000-5
=555550-5
=555545
2.（1+11+21+31+41）＋（9+19+29+39+49）
Original formula = (1 + 9) + (11 + 19) + 21 + 29) + (31 + 39) + (41 + 49)
=10+30+50+70+90
=50×5
=250
1 555550-5=555545
2 50X5=250
1.=49+1+499+1+4999+1+49999+1+499999+1-5
=555550-5
=555545
2.=1+9+11+19+21+29+31+39+41+49
=1+49+11+39+21+29+31+19+41+9
=50×5=250
3.=9000×999=8991000
Four
1.=50-1+500-1+5000-1+500000-1+500000-1=555550-5=555545
2.=1+49 +11+39 +21+29 +31+19+41+9=50*5=250
3.=9001-1+9003-3…… 9999-999=9000*500=4500000
4. The title should be 123 + 234 + 345 + 456 + 567 + 678 + 789.
100 is 1... Unfold
1.=50-1+500-1+5000-1+500000-1+500000-1=555550-5=555545
2.=1+49 +11+39 +21+29 +31+19+41+9=50*5=250
3.=9001-1+9003-3…… 9999-999=9000*500=4500000
4. The title should be 123 + 234 + 345 + 456 + 567 + 678 + 789.
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 * 100 = 2800
On the tenth, it is 2 + 3 + 4 + 5 + 6 + 7 + 8 = 35 * 10 = 350
3 + 4 + 5 + 6 + 7 + 8 + 9 = 42
It's 2800 + 350 + 42 = 3192
5.=1400/200+350/500=7.7
6.（9+11+6）/13+（13+14）/9+6=11
6. = (0.36 / 0.4) * (5.1 / 0.17) * (81 / 8.1) = 0.9 * 30 * 10 = 270