मराठी

Find the values of k so that the function f is continuous at the indicated point. ,, at x =f(x)={kcosxπ-2x,ifx≠π23,ifx=π2 at x = π2 - Mathematics

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प्रश्न

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` 

बेरीज

उत्तर १

The given function f is `f(x) = {((kcosx)/(pi-2x), "," if x != pi/2),(3, "," if x = pi/2):}  " at x "  pi/2` 

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उत्तर २

`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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पाठ 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १६१]

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एनसीईआरटी Mathematics [English] Class 12
पाठ 5 Continuity and Differentiability
Exercise 5.1 | Q 26 | पृष्ठ १६१

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