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lyudmila [28]
3 years ago
8

Which of the following is an even function need help quick

Mathematics
2 answers:
musickatia [10]3 years ago
8 0

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

F(x)=2\sin(\frac{1}{2}x)

Put x=-x,

F(-x)=2\sin(\frac{1}{2}(-x))=-2\sin(\frac{1}{2}x)=-F(x)\neq F(x)              [\because \sin(-x)=-\sin x]

So, this function is not an even function.

The second function is

F(x)=2\cos(\frac{1}{2}x+\frac{\pi}{2})

Put x=-x,

F(-x)=2\cos(\frac{1}{2}(-x)+\frac{\pi}{2})\neq F(x)

So, this function is not an even function.

The third function is

F(x)=2\sin(\frac{1}{2}x+\pi)

Put x=-x,

F(-x)=2\sin(\frac{1}{2}(-x)+\pi)\neq F(x)

So, this function is not an even function.

The fourth function is

F(x)=2\cos(\frac{1}{2}x)

Put x=-x,

F(-x)=2\cos(\frac{1}{2}(-x))=2\sin(\frac{1}{2}x)=F(x)       [\because \cos(-x)=\cos x]

So, this function is an even function.

Hence the correct option is D.

mrs_skeptik [129]3 years ago
4 0
Not sure but it is 3..

Thanks!
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