Given the absolute value of root a-5 + 5b-4a + (a + B-C) ^ 2 = 0, calculate the value of (B-A) ^ 2012 + radical C Don't delete Du Niang

Given the absolute value of root a-5 + 5b-4a + (a + B-C) ^ 2 = 0, calculate the value of (B-A) ^ 2012 + radical C Don't delete Du Niang

Radical a-5 + absolute value 5b-4a + (a + B-C) ^ 2 = 0
therefore
Radical a-5 = 0, absolute value 5b-4a = 0, (a + B-C) ^ 2 = 0
Namely
a-5=0,5b-4a=0,a+b-c=0
therefore
A=5
B=4
C=9
therefore
(B-A) ^ 2012 + radical C
=(4-5) ^ 2012 + radical 9
=1+3
=4

Find the maximum, minimum and period of the function y = root 3 (cosx) ^ 2 + SiNx * cosx

y=√3cos²x＋sinxcosx
=√3(1+cos2x)/2+1/2*2sinxcosx
=√3/2+√3/2*cos2x+1/*sin2x
=√3/2+sinπ/6cos2x+cosπ/6sin2x
=sin(2x+π/6)+√3/2
So the maximum value of the function is 1 + √ 3 / 2; the minimum value is - 1 + √ 3 / 2
The minimum positive period is 2 π / 2 = π

Given the function f (x) = sin (2x + π / 6) + sin2x - π / 6 + 2cos? X, find the minimum value and minimum positive period of the function

sinA+sinB=2sin(A+B)/2 *cos(A-B)/2
cos2a=2cosa*cosa-1
f(x)=2*sin2x*cos(π/6)+1+cos2x
=sin2x+cos2x
=√2*[sin2x*cos(π/4)+cos2x*sin(π/4)]
=√2*sin(2x+π/4)
Minimum - √ 2
Minimum positive period π

The minimum value of the function y = 4 / (COS? X) + 9 / (sin? X) is?

Sin? X + cos? X = 1, so y = (4 / cos? X + 9 / sin? X) (sin? X + cos? X) = 13 + (4sin? X / cos? X + 9cos? X / sin? X) 4sin? X / cos? X + 9cos? X / sin? X ≥ 2 √ [4sin? X / cos 

Let f (x) = sin 2 x + sin xcosx, X ∈ [0, π / 2]. (1) find the minimum positive period and range of F (x) 2) If f (α) = 5 / 6, find the value of sin2 α

(1)∵cos2x=cosxcosx-sinxsinx=cos²x-sin²x=(1-sin²x)-sin²x=1-2sin²x∴sin²x=(1-cos2x)/2∵sin2x=2sinxcosx∴sinxcosx=(sin2x)/2∴f(x)=(1-cos2x)/2+(sin2x)/2=1/2+(sin2x)/2-cos(2x...

Given the function f (x) = sinxcosx + sin? X, find the values of F (0) and f (π / 2)

f(x)=sinxcosx+sin²x
=1/2sin2x+(1-cos2x)/2
=1/2sin2x-1/2cos2x+1/2
=√2/2*(√2/2sin2x-√2/2cos2x)+1/2
=√2/2*sin(2x-π/4)+1/2
f(0)=√2/2*sin(2*0-π/4)+1/2
=√2/2*sin(-π/4)+1/2
=-√2/2*sinπ/4+1/2
=-√2/2*√2/2+1/2
=-1/2+1/2
=0
f(π/2)=√2/2*sin(2*π/2-π/4)+1/2
=√2/2*sin(π-π/4)+1/2
=√2/2*sinπ/4+1/2
=√2/2*√2/2+1/2
=1/2+1/2
=1

In order to make y = sin ω x (ω > 0) have at least 50 maxima in the interval [0, 1], then the minimum value of ω is () A. 98π B. 197π Two C. 199π Two D. 100π

∵ such that y = sin ω x (ω > 0) has at least 50 maxima in the interval [0, 1]
∴491
4 × t ≤ 1, i.e. 197
4×2π
ω≤1，
∴ω≥197π
2．
Therefore, B

Find the maximum and minimum values of the function y = - 2cos ^ 2x + 2sinx + 3, X ∈ [π / 6,5 π / 6]

Because x ∈ [π / 6,5 π / 6], so SiNx ∈ [1 / 2,1] so the maximum value of the function is 2 (1 + 1 / 2) 2 + 1 = 5, and the minimum value is 2 (1 / 2 + 1 / 2) 2 + 1 / 2

Find the maximum and minimum value of the function f (x) = 3 Λ sin Λ 2x + 2sinx + 5

       f(x)=3^(sin²x) + 2sinx +5
The maximum value of SiNx is 1, which appears at: x = π / 2, so the maximum value of F (x): Fmax = 3 + 2 + 5 = 10
f '(x)=cosx[2ln3 sinx 3^(sin²x)+2]=0         (1)   
The maximum value of F (x) corresponding to X1 = π / 2: 10
sinx 3^(sin²x)=-1/(ln3)                             (2) 
X 2 is the minimum point of F (x): it is difficult to get the analytical solution of (2), which seems to be only approximate,
By using Excel, we get the following results
X2 = - 0.63845 is the minimum point;
The minimum value of F (x 2) = 5.28534 is f (x 2)
The maximum value of F (x) is: 10
The minimum value of F (x) is approximately 5.28534
 

The maximum value of the function y = cos 2 x-3 / 2sinx + 2

y=cos²x -(3/2)sinx +2
=1-sin²x-(3/2)sinx+2
=-(sinx+3/4)²+57/16
When SiNx = - 3 / 4,
Y has a maximum value of 57 / 16