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It is known that the function f (x) defined on R is odd and the period of F (3x) is 3. If f (1) = 5, find the value of F (2010) + F (2011)
The period of F (3x) is 3
Then the f period of F (x) is 3
f(2010)+f(2011)=f(0)+f(1)
F (x) is an odd function
So f (0) = 0
So f (2010) + F (2011) = f (0) + F (1) = 5
Given the function f (x) = (x2-3x + 2) × g (x) + 3x-4 defined on R, where the image of function y = g (x) is a continuous curve, then the equation f (x) = 0 must have a solution in the following range? a.（0,1） b.（1,2）c.（2,3）d.（3,4）
Because f (1) = 0 * g (1) + 3-4 = - 1, f (2) = 0 * g (2) + 6-4 = 2
And the image of G (x) is a continuous curve
So f (1) f (2)
If f (x) satisfies 2F (x) + F (- x) = 3x + 4, then f (x)=______ .
∵2f（x）+f（-x）=3x+4，①
∴2f（-x）+f（x）=-3x+4，②
① The results show that f (x) = 3x + 4
3．
So the answer is: 3x + 4
3．
It is known that the function f (x) whose domain is R is a decreasing function on (2010, + ∞) and the function y = f (x + 2010) is even function f(2008)>f(2009) f(2008)>f(2011) f(2009)>f(2011) f(2009)>f(2012)
F (x + 2010) is even function, so f (x + 2010) = f (2010-x)
f(2008)=f(2010-2)=f(2010+2)=f(2012)
f(2009)=f(2010-1)=f(2010+1)=f(2011)
It's a minus function
The fourth answer
F (x) is an even function defined on R, which satisfies f (x + 2) = - f (x). When x ∈ [0,1], f (x) = x, then f (5.5)=
f(5.5)
=f(3.5+2)
=-f(3.5)
=-f(1.5+2)
=f(1.5)
=f(-0.5+2)
=-f(-0.5)
If f (x) is an even function, then f (x) = f (- x)
-f(-0.5)=-f(0.5)
When x ∈ [0,1], f (x) = X
So f (0.5) = 0.5
f(5.5)=-f(0.5)=-0.5
If sin2x and SiNx are the equidistant and proportional median of sin θ and cos θ respectively, the value of cos2x is: () A. 1+ Thirty-three Eight B. 1− Thirty-three Eight C. 1± Thirty-three Eight D. 1− Two Four
2 sin2x = sin θ + cos θ sin2x = sin θ + cos θ sin2x = sin θ + cos θ sin2x = sin θ + cos θ = (sin θ + cos θ) 2-2sin θ cos θ = 4sin22x-2sin22x = 1 ʎ 4 (1-cos22x) + cos2x-2 = 0, that is, 4ccos22x-cos2x-2 = 0, namely 4ccos22x-cos2x-2 = 0, get cos2x = 1 ± 338 sin2x = sin2x = sin2x = sin2x = 1 ± 338 sin2x = sin2x = sin2x = 1-1-1-1-1-1-1-cos2x = 1-338 = 1 ± 3382si
If f (SiNx) = cos2x, then f (COS x) =?
f(sinx)=cos2x
=1-2sin²x
Therefore, it can be concluded that:
f(cosx)=1-2cos²x
F (SiNx) = 3-cos2x find f (sin2x) + F (cos2x)
Six
f(sinx)=3-cos2x
f(sinx)=3-(1-2(sinx)∧2)
So f (x) = 3 - (1-2x Λ 2) = 2 + 2x Λ 2
So f (sin2x) + F (cos2x)
=2+2（sin2x）∧2+2+2（cos2x）∧2
=4+2（（sin2x）∧2+（cos2x）∧2）
=4+2
=6
Solve the equation cos 2x = cosx + SiNx and find the value of X
cos2x-sin2x=cosx+sinx，
（cosx+sinx）（cosx-sinx）-（cosx+sinx）=0，
（cosx+sinx）（cosx-sinx-1）=0．
If cosx + SiNx = 0, then 1 + TGX = 0, TGX = - 1,
∴x＝kπ−π
4. (k is an integer)
If cosx + sinx-1 = 0, then cosx SiNx = 1,
Qi
Two
2cosx−
Two
2sinx＝
Two
2，∴cos(x+π
4)＝
Two
2，
∴x+π
4＝2kπ±π
4，∴x＝
2kπ
2kπ−π
2 (k is an integer)
The equations are as follows: (1) SiNx = sin2x (2) SiNx = cos2x (3) SiNx = tan2x (4) SiNx = cot2x This is the problem of four trigonometric equations in our math paper this week
SiNx = sin2x, SiNx = 2sinxcosx, SiNx (1-2cosx) = 0, SiNx = 0 or cosx = 1 / 2, when SiNx = 0, x = k π, when cosx = 1 / 2, x = 2K π + π / 3 or x = 2K π - π / 3
SiNx = cos2x, SiNx = 1-2sin? X, 2Sin? X + sinx-1 = 0, (2sinx-1) (SiNx + 1) = 0, SiNx = - 1 or SiNx = 1 / 2, when SiNx = - 1, x = 2K π - π / 2, when SiNx = 1 / 2, x = 2K π + π / 6 or x = 2K π + 5 π / 6
SiNx = tan2x, SiNx = sin2x / cos2x, SiNx (cos2x-2cosx) = 0, SiNx (2cos? X-2cosx-1) = 0, SiNx = 0 or cosx = (1 + √ 3) / 2 (omitted) or cosx = (1 - √ 3) / 2, when SiNx = 0, x = k π, when cosx = (1 - √ 3) / 2, x = arccos [(1 - √ 3) / 2]
SiNx = cot2x, SiNx = cos2x / sin2x, SiNx = cos2x / sin2x, 2Sin? 2x = cos2x, 2 (1-cos? 2x) cosx = 2cos? X-1,2cosx-2cos? X = 2cos? X-1,2cos? X + 2cos? X-2cos X-1 = 0, make y = 2cos? X + 2cos? X-2cos X-1 = 0, make y = 2cos? X + 2cos? X-2cos X-1 = 0, let y = 2cos? X + 2cos? X-2cosx-1 = 0, use dichotomy to find points, its extreme value is when cos x = - 1 / 3, y = - 5 / 27, cos x = 1, cos x = 1, y = - 5 / 27, cos x let y = 1 always find a value between - 1 / 3 and 1 so that cosx ≈ 0
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