The equation ksin⁡x+cos⁡2x=2k−7 possess a solution, if

The equation ksinx+cos2x=2k7 possess a solution, if

  1. A

    k>6

  2. B

    2k6

  3. C

    k>2

  4. D

    none of these

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    Solution:

    We have,

    ksinx+12sin2x=2k72sin2xksinx+2(k4)=0sinx=k±k216k+644=k±(k8)4=12(k4),2sinx=12(k4) [sinx2]

    Now, 1sinx11k4212k6

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