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Question

If 2f(xy)=(f(x))y+(f(y))x x,yR and f(1)=3, then the value of 10r=1f(r) is equal to

A
32(3101)
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B
32(391)
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C
12(3101)
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D
12(391)
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Solution

The correct option is A 32(3101)
Given, 2f(xy)=(f(x))y+(f(y))x and f(1)=3
Put y=1
2f(x)=f(x)+(f(1))x
f(x)=3x

10r=1f(r)
=10r=13r=3+32+33++310=32(3101)
( For G.P., Sn=a(rn1)r1)

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