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Question

If sin1a+sin1b+sin1c=3π2 and f(2)=2,f(x+y)=f(x)f(y)x,yϵR then af(2)+bf(4)+cf(6)3(af(2).bf(4).cf(6))af(2)+bf(4)+cf(6) equals

A
2
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B
4
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C
6
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D
8
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Solution

The correct option is A 2
Since
sin1(a)+sin1(b)+sin1(c)=3π2
Hence
a=b=c=1.
Now
f(x+y)=f(x).f(y) implies
f(x)=ax
Now
f(2)=2
Or
a2=2
Or
a=2.
Hence
f(x)=2x2.
Substituting the values of f(x) and a,b,c in the above expression, we get
12+122+1233(12.122.123)12+122+123
=33.(1)3
=2.

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