हिंदी

Select the correct answer from the given alternatives. limx→π2[3cosx+cos3x(2x-π)3] =

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प्रश्न

Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =

विकल्प

  • `3/2`

  • `1/2`

  • `-1/2`

  • `1/4`

MCQ
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उत्तर

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3] = underline (-1/2)`

Explanation:

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]`

= `lim_(x -> pi/2)(3cos x + 4cos^3x - 3cos x)/(2x - pi)^3`

= `lim_(x -> pi/2) (4cos^3x) /(8(x - pi/2)^3` 

Put `x = pi/2 + h,`

`x - pi/2 = h`

As `x -> pi/2, h -> 0`

= `lim_(x -> pi/2) (4 cos^3x) /(8(x - pi/2)^3` 

= `lim_(h-> 0) (4 cos^3(pi/2 + h))/(8h^3)`

= `lim_(h -> 0) (4(-sin h)^3)/(8h^3)`           ...`[∵ cos(pi/2 + θ) = -sinθ]`

= `-1/2(lim_(h->0)(sin h)/h)^3`

= `-1/2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.1 [पृष्ठ १५८]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.1 | Q I. (7) | पृष्ठ १५८

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