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Question

If sin1x+sin1y+sin1z=3π2 and f(1)=2,f(a+b)=f(a)f(b) for all a,b, ϵ R, then
xf(1)+yf(2)+zf(3)x+y+zxf(1)+yf(2)+zf(3) is equal to :

A
10
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B
11
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C
2
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D
3
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Solution

The correct option is C 2
Since π2sin1x,sin1y,sin1zπ2
sin1x+sin1y+sin1z=3π2
sin1x=sin1y=sin1z=π2
x=y=z=1
xf(1)+yf(2)+zf(3)x+y+zxf(1)+yf(2)+zf(3)
=1+1+1(1+1+11+1+1)=31=2

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