Question

Which of the following options is an even function.

Answer 1

Answer:

Option C - $F(x)=\cos(x+\pi)$

Step-by-step explanation:

To find : Which of the following options is an even function?

Solution :

To determine the function is even,odd or neither we will find f(-x)

If f(-x) = - f(x) then function is odd.

If f(-x) = f(x) then function is even.

We know that,

$\sin(-x)=-\sin x$ is always a odd function.

$\cos(-x)=\cos x$ is always an even function.

Now, In the following options

A. $F(x)=\sin(x+\pi )$

$F(-x)=\sin(-(x+\pi))=-\sin (x+\pi)=-F(x)$

It is a odd function.

B. $F(x)=2sin(2x)$

$F(-x)=2\sin(-2x)=-2\sin (2x)=-F(x)$

It is a odd function.

C. $F(x)=\cos(x+\pi)$

$F(-x)=\cos(-(x+\pi))=\cos(x+\pi)=F(x)$

It is an even function.

D. $F(x)=2\sin(\pi+\pi)$

$F(-x)=2\sin(-2x)=-2\sin (2x)=-F(x)$

It is a odd function.

Therefore, Option C is correct.

Answer 2

 

D)  F(x) = 2sin(π+2π) = 2sin(3π) = 2 \cdot 0 = 0

==>  F(x) = 2sin(π+2π)    is an even function.

Correct answer:   (D)