Answer:
The correct option is D.
Step-by-step explanation:
If function f(x) is called an even function if f(-x)=f(x).
The first function is
[tex]F(x)=2\sin(\frac{1}{2}x)[/tex]
Put x=-x,
[tex]F(-x)=2\sin(\frac{1}{2}(-x))=-2\sin(\frac{1}{2}x)=-F(x)\neq F(x)[/tex] [tex][\because \sin(-x)=-\sin x][/tex]
So, this function is not an even function.
The second function is
[tex]F(x)=2\cos(\frac{1}{2}x+\frac{\pi}{2})[/tex]
Put x=-x,
[tex]F(-x)=2\cos(\frac{1}{2}(-x)+\frac{\pi}{2})\neq F(x)[/tex]
So, this function is not an even function.
The third function is
[tex]F(x)=2\sin(\frac{1}{2}x+\pi)[/tex]
Put x=-x,
[tex]F(-x)=2\sin(\frac{1}{2}(-x)+\pi)\neq F(x)[/tex]
So, this function is not an even function.
The fourth function is
[tex]F(x)=2\cos(\frac{1}{2}x)[/tex]
Put x=-x,
[tex]F(-x)=2\cos(\frac{1}{2}(-x))=2\sin(\frac{1}{2}x)=F(x)[/tex] [tex][\because \cos(-x)=\cos x][/tex]
So, this function is an even function.
Hence the correct option is D.
