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Let f(x) =(kcosx)/(pi-2x) if x!=pi/2 and...

Let `f(x) =(kcosx)/(pi-2x)` if `x!=pi/2` and `f(x=pi/2)` if `x=pi/2`then find the value of `k` if `lim_(x->pi/2) f(x)=f(pi/2)`

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Given, `{{:((kcosx)/(pi-2x)"," , "When " xne (pi)/2),(3",", "When " x=(pi)/2):}`
`therefore` LHL =`lim_(xto(pi^(-))//2)(kcosx)/(pi-2x)= lim_(hto0)(kcos(pi/2-h))/(pi-2(pi/2-h))`
`k/2lim_(hto0)(sinh)/(h)=k/2.1=k/2` `[therefore lim_(hto0)(sinx)/(x)=1]`
RHL `=lim_(hto0)(kcosx)/(pi-2x)=underset(xto(pi//2))"lim"+(kcos(pi/2+h))/(pi-2(pi/2+h))`
`=lim_(hto0)(-ksinh)/(pi-pi-2h)=lim_(hto0)(ksinh)/(h)=k/2lim_(hto0)(sinh)/(2h)=k/2 "and" f(pi/2)=3`
It is given that, `therefore lim_(xto(pi//2))f(x)=f(pi/2)rArr k/2=3` k=6
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