1.

`f(x) = {{:((k cosx )/((pi - 2x)"," if x ne (pi)/(2))),(3"," if x = (pi)/(2)):}`` at x = (pi)/(2)` .

Answer» `f(pi//2) = 3 `
R.H.L = `underset(x to (pi^(+))/(2))("lim") f(x) = underset(x to 0)("lim")f((pi)/(2) + h)`
`= underset(h to 0)("lim") (k cos ((pi)/(2) + h))/(pi -2 ((pi)/(2) + h)) = underset(h to 0)("lim") (-k sin h)/(-2h) = (k)/(2)`
`(because underset(h to 0)("lim") (sinh)/(h) =1)`
and L.H.L =` underset(x to pi//2^(2))(lim)f(x)= underset(h to 0)(lim)f((pi)/(2)-h)`
`underset (h to 0)(lim)(k cos((pi)/(2)-h))/(pi-2((pi)/(2)-h))`
`= underset(h to 0 )(lim)(k sin h)/(2h)= (k)/(2)`
`because` Function is continuous at x = (`pi)/(2)`.
`:.` R.H.L. = `f ((pi)/(2)) = LI. `
`rArr (k)/(2) = 3 rArr k = 6`


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