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Discuss the continuity of the following function at the point(s) or on the interval indicated against them: f(x) =sin2πx3(1-x)2,for x≠1=π2sin2(πx2)3+4cos2(πx2),for x=1} at x = 1

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प्रश्न

Discuss the continuity of the following function at the point(s) or on the interval indicated against them:

f(x) `{:( = (sin^2pix)/(3(1 - x)^2) ",", "for"  x ≠ 1),(= (pi^2sin^2((pix)/2))/(3 + 4cos^2 ((pix)/2)) ",", "for"  x = 1):}}` at x = 1

योग
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उत्तर

f(1) = `(pi^2sin^2 (pi/2))/(3 + 4cos^2 (pi/2))`

= `(pi^2 xx 1^2)/(3 + 0)`

= `pi^2/3`    ...(1)

`lim_(x -> 1) "f"(x) =  lim_(x -> 1) (sin^2pix)/(3(1 - x)^2`

= `lim_(x -> 1) (sin^2(pi - pix))/(3(1 - x)^2`   ...[∵ sin (π – θ) = sin θ]

= `lim_(x -> 1) [sin{pi(1 - x)}]^2/(3pi^2(1 - x)^2) * pi^2`

= `pi^2/3 [lim_(x -> 1) (sin{pi(1 - x)})/[pi(1 - x))]^2`

= `pi^2/3 xx 1^2  ...[(because x -> 1","  (x - 1) -> 0 "," therefore pi(1 - x) -> 0),("and" lim_(theta -> 0)  sintheta/theta = 1)]`

= `pi^2/3`    ...(2)

From (1) and (2),

`lim_(x -> 1) "f"(x)` = f(1)

∴ f is continuous at x = 1

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अध्याय 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७७]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
MISCELLANEOUS EXERCISE-8 | Q (II) (4) | पृष्ठ १७७

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