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Find the value of k so that the function f is continuous at the indicated point. f(x) = {ЁЭСШ тБвcos тБбЁЭСе/ЁЭЬЛ тИТ 2тБвЁЭСе, if ЁЭСе тЙа ЁЭЬЛ/2, 3, if ЁЭСе = ЁЭЬЛ/2, at x = ЁЭЬЛ/2

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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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рдкрд╛рда 5: Continuity and Differentiability - Exercise 5.1 [рдкреГрд╖реНрда резремрез]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдкрд╛рда 5 Continuity and Differentiability
Exercise 5.1 | Q 26 | рдкреГрд╖реНрда резремрез

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