If ∫1a2sin2⁡x+b2cos2⁡x−dx=112tan−1⁡(3tan⁡x)+C, then the maximum value of a sin x + b cos x, is

If 1a2sin2x+b2cos2xdx=112tan1(3tanx)+Cthen the maximum value of a sin x + b cos x, is

  1. A

    41

  2. B

    40

  3. C

    39

  4. D

    38

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    Solution:

    We have,

             I=1a2sin2x+b2cos2xdx I=sec2xb2+a2tan2xdx I=1a1b2+(atanx)2d(atanx)         =1abtan1abtanx+C

     ab=12 and ab=3a2=36a=±6

     ab=12b=±2

    Thus, we have

           asinx+bcosx=±(6sinx+2cosx)

    We know that

                      a2+b2asinx+bcosxa2+b2 for all x

     40±6sinx±2cosx40 for all x.

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