Question

Which of the following is an even function need help quick

Answer 1

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.

Answer 2

Not sure but it is 3..

Thanks!