Answer:
option b
Step-by-step explanation:
We are given that [tex]y=sin \frac{1}{2}(x-C)[/tex] be an even function
We have to find the value of C for which given function is even function
We know that sin x is odd function and cos is even function
Odd function : when f(x)[tex]\neqf(-x) [/tex] then the function is called an odd function.
Even function : When f(x)=f(-x) then the function is called an even function.
Sin(-x)=-Sin x
Cos (-x)= Cos x
When we take C=[tex]2\pi[/tex]
Then , y=Sin[tex]\frac{x}{2}-\frac{2\pi}{2}[/tex]
y=[tex]sin(\frac{x}{2}-\pi)[/tex]
[tex]y=-sin\frac{x}{2}[/tex] ( [tex]sin (x-\pi)=-sin x[/tex])
When x is replace by -x
Then, we get [tex]y=-sin(-\frac{x}{2})=sin\frac{x}{2}[/tex]
[tex]f(-x)\neq f( x)[/tex]
Hence, option a is false.
b.C=[tex]\pi[/tex]
[tex]y= sin (\frac{x}{2}-\frac{\pi}{2})[/tex]
[tex] y=-sin(\frac{\pi}{2}-\frac{x}{2})[/tex]
[tex]y=-cos \frac{x}{2}[/tex]
When x is replaced by -x then we get
[tex] y=-cos (-\frac{x}{2})=- cos \frac{x}{2}[/tex]
f(x)=f(-x) , Therefore, function is even,hence option b is true.
c.C=[tex]\frac{\pi}{2}[/tex]
[tex]y=sin (\frac{x}{2}-\frac{\pi}{4})[/tex]
[tex]Sin (A-B)=Sin A Cos B- Sin B Cos A[/tex]
[tex][y= sin \frac{x}{2} cos {\frac{\pi}{4}-cos\frac{x}{2} sin\frac{\pi}{4}[/tex]
[tex] sin\frac{\pi}{4}= cos \frac{\pi}{4}=\frac{1}{\sqrt2}[/tex]
[tex]y=\frac{1}{\sqrt2}(sin \frac{x}{2}- cos \frac{x}{2})[/tex]
When x is replaced by -x then we get
[tex]y=\frac{1}{\sqrt2}(-sin\frac{x}{2}-cos \frac{x}{2})[/tex]
[tex]f(x)\neq f(-x)[/tex]
Hence, function is odd .Therefore, option c is false.
