If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2 and f(1)=2,f(p+q)=f(p)⋅f(q),∀p,q∈R then xf(1)+yf(2)+zf(3)−(x+y+z)xf(1)+yf(2)+zf(3) is equal to

If sin1x+sin1y+sin1z=3π2 and f(1)=2,f(p+q)=f(p)f(q),p,qR then xf(1)+yf(2)+zf(3)(x+y+z)xf(1)+yf(2)+zf(3) is equal to

  1. A

    0

  2. B

    1

  3. C

    2

  4. D

    3

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    Solution:

    π2sin1xπ2,π2sin1yπ2

    and  π2sin1zπ2

    Given that sin1x+sin1y+sin1z=3π2

    which is possible only when

    sin1x=sin1y=sin1z=π2x=y=z=1put  p=q=1then,   f(2)=f(1)f(1)=22=4and put  p=1,q=2then f(3)=f(1)f(2)=222=8

     xf(1)+yf(2)+zf(3)x+y+zxf(1)+yf(2)+zf(3)=1+1+131+1+1=31=2

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