The value of limx→π/4 22−(cos⁡x+sin⁡x)31−sin⁡2x,is 

The value of limxπ/422(cosx+sinx)31sin2x,is 

  1. A

    32

  2. B

    23

  3. C

    12

  4. D

    2

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    Solution:

    We have,

    limxπ/422(cosx+sinx)31sin2x=limxπ/423/2(cosx+sinx)23/22(1+sin2x)=limxπ/423/2(1+sin2x)3/22(1+sin2x) 

    =limy2y3/223/2y2,where y=1+sin2x

    =32(2)3/21=32×2=32

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