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It is known that the function f(x)=sinx/x, which of the following propositions is correct
1. f(x) is an odd function
②For any x in the definition domain, f(x)
③When x=3π/2, f(x) obtains the minimum value;
④f(2)>f(3)
5. When x>0, if the absolute value of the equation f(x)=k has and has only two different real solutions α, β (α>β), then β*cosα=-sinβ
Analysis: ∵ function f(x)=sinx/x, its domain is x≠0
f(-x)=-sinx/(-x)=f(x)==>even function;
∴(1)Wrong
∵When x tends to 0, the limit of function f(x) is 1
∴In the domain of definition f(x)
∴(2)Correct
When x>0, f'(x)=(xcosx-sinx)/x^2
f'(3π/2)=(0 1)/(3π/2)^2≠0
∴(3)Wrong
∵When x tends to 0, the limit of function f(x) is 1, f(π)=0
∴The function decreases monotonically on the interval (0, π]; ==>f(2)>f(3)
∴(4)Correct
When x>0,
When X∈(0, π), f(x)>0,
When X∈(π,2π), f(x)
takes the absolute value and becomes k
∵The absolute value of equation f(x)=k has and only two different real solutions α, β (α>β)
∴cosα=-k
f(β)=sinβ/β
∵k=f(β)=sinβ/β==>-cosα*(β)=sinβ
∴(5)Correct
To summarize: 2, 4, and 5 are correct
(Ⅰ) From the image, we know that A=2, the minimum positive period of f(x) is T=4*(
5π
12 -
π
6 )=π,∴ω=2
point (
π
6,2) Substitute to get sin(
π
3 φ)=1, and |φ|π
2, ∴φ=
π
6
So the analytical formula of function f(x) is f(x)=2sin(2x
π
6 )
(Ⅱ)g(x)=2sin(2x
π
6 )-2cos2x=
3 sin2x-cos2x=2sin(2x-
π
6 )
The transformation is as follows: translate the image of y=sinx to the right
π
6 Get y=sin(x-
π
6 ); then sin(x-
π
6 )
The abscissa coordinates of all points on the image are shortened to the original
1
2 If the vertical coordinate remains unchanged, we get y=sin(2x-
π
6) image;
Put y=sin(2x-
π
6) The vertical coordinates of all points on the image are expanded to twice the original size, and the horizontal coordinates remain unchanged to obtain y=2sin(2x-
π
6 ) image.
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