If f′(x)=g(x) and g′(x)=−f(x) for all x and f(2)=4=f′(2),then (f(24))2+(g(24))2 is 

If f(x)=g(x) and g(x)=f(x) for all x and f(2)=4=f(2),then (f(24))2+(g(24))2 is 

  1. A

    32

  2. B

    24

  3. C

    64

  4. D

    48

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    Solution:

    We have,ddx(f(x))2+(g(x))2

    =2f(x)f(x)+2g(x)g(x)=2f(x)g(x)+2g(x)(f(x))=2f(x)g(x)2f(x)g(x)=0

    (f(x))2+(g(x))2 is constant.

    Again, g(x)=f(x)g(2)=f(2)=4(f(24))2+(g(24))2

    =(f(2))2+(g(2))2=(4)2+(4)2=16+16=32

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