Find the value of k so that the function f defined by `f(x)={(kcosx)/(pi-2x),3\ \ \ ,\ \ \ \ \ \ \ \ \ \ "if"\ \ x\ !=pi/2"if"\ x=pi/2` is continuous at `x=pi/2`

Explanation

The given function `f` is `f(x)= frac{k cos x}{pi-2 x}, & text { if } x neq frac{pi}{2} 3, & text { if } x=frac{pi}{2}`


The given function `f` is continuous at , `x=frac{pi}{2}` if `f` is defined at `x=frac{pi}{2}` and if the value of the `f` at `x=frac{pi}{2}` equals the limit of `f` at `x=frac{pi}{2}`


It is evident that `f` is defined at `x=frac{pi}{2}` and `f(frac{pi}{2})=3`


`lim _{x -> frac{pi}{2}} f(x)=lim _{x -> frac{pi}{2}} frac{k cos x}{pi-2 x}`


Put `x=frac{pi}{2}+h`


Then, `x -> frac{pi}{2} Rightarrow h -> 0`


` therefore lim _{k -> frac{pi}{2}} f(x) &=lim _{x -> frac{pi}{2}} frac{k cos x}{pi-2 x}=lim _{h -> 0} frac{k cos (frac{pi}{2}+h)}{pi-2(frac{pi}{2}+h)} &=k lim _{h -> 0} frac{-sin h}{-2 h}=frac{k}{2} lim _{h -> 0} frac{sin h}{h}=frac{k}{2}, 1=frac{k}{2}`


`therefore lim _{x -> frac{pi}{2}} f(x)=f(frac{pi}{2})`


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