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Select the correct answer from the given alternatives: If f(x) = 1-2sinxπ-4x,for x≠π4 is continuous at x = π4, then f(π4) =

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प्रश्न

Select the correct answer from the given alternatives:

If f(x) = `(1 - sqrt(2) sinx)/(pi - 4x), "for"  x ≠ pi/4` is continuous at x = `pi/4`, then `"f"(pi/4)` =

विकल्प

  • `1/sqrt(2)`

  • `-1/sqrt(2)`

  • `- 1/4`

  • `1/4`

MCQ
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उत्तर

`1/4`

Explanation:

f(x) is continuous at x = `pi/4`

`therefore "f"(pi/4) = lim_(x -> pi/4) "f"(x)`

`= lim_(x -> pi/4) (1 - sqrt2 sin x)/(pi - 4x)`

`= lim_(x -> pi/4) (sqrt2 (sin x - 1/sqrt2))/(4 (x - pi/4))`

`= sqrt2/4 lim_(x -> pi/4) (sin x - sin pi/4)/(x - pi/4)`

`= sqrt2/4 lim_(x -> pi/4) (2 cos ((x + pi/4)/2) * sin ((x - pi/4)/2))/(x - pi/4)`

`= sqrt2/4 * lim_(x -> pi/4) cos (x/2 + pi/8) * lim_(x -> pi/4) sin ((x - pi/4)/2)/((x - pi/4)/2)`

`= sqrt2/4 * cos (x/8 + pi/8) xx 1    ...[(because x -> pi/4 x - pi/4 -> 0),(therefore (x - pi/4)/2 -> and lim_(theta -> 0) (sin theta)/theta = 1)]`

`= sqrt2/4 xx cos pi/4`

`= 1/4`

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अध्याय 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७६]

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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 (I) (2) | पृष्ठ १७६

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