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Find the value 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`
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f(x) = `{((kcosx)/(pi-2x)", if" x != 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`
Also `f(pi/2)` = 3
The function f will be continuous at x = `pi/2` if,
L.H.L. = R.H.L. = `f(pi/2)`
∴ `k/2 = 3`
⇒ k = 6
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