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Question
Find the values of k so that the function f is continuous at the indicated point.
`f(x) = {((kcosx)/(pi-2x), "," if x != pi/2),(3, "," if x = pi/2):} " at x =" pi/2`
Solution 1
The given function f is `f(x) = {((kcosx)/(pi-2x), "," if x != pi/2),(3, "," if x = pi/2):} " at x " pi/2`
Solution 2
`f(x) = {(("k cos x")/(pi - 2"x")"," " if" "x" ne pi/2),(3"," " if" "x" = pi/2):}`
If f(x) is continuous at x = `pi/2`, it implies:
L.H.L. = `lim_(x -> pi^-/2) f(x) = lim_(h -> 0) (pi/2 - h)`
`= lim_(h -> 0) (k cos (pi/2 - h))/(pi - 2(pi/2 - h))`
`= lim_(h -> 0) (k sin h)/(pi - pi + 2h)`
`= lim_(h -> 0) k/2 (sin h)/h = k/2 ...(because lim_("h" -> 0) (sin h)/h = 1)`
R.H.L. = `lim_(x -> pi^+/2) f(x) = lim_(h -> 0) (pi/2 + h)`
`= lim_(h -> 0) (k cos (pi/2 + h))/(pi - 2(pi/2 + h))`
`= lim_(h -> 0) (- k sin h)/(- 2h)`
`= lim_(h -> 0) k/2 (sin h)/h = k/2`
`f(pi/2) = 3`
The function f will be continuous at x = `pi/2` if
L.H.L. = R.H.L. = `f(pi/2)`
`therefore k/2 = 3`
⇒ k = 6
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