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Question

If cos1(a)+cos1(b)+cos1(c)=3π and f be a function such that f(1)=2 and f(x+y)=f(x)f(y) for all x,yR, then the value of a2f(1)+b2f(2)+c2f(3)+a+b+ca2f(1)+b2f(2)+c2f(3) is

A
0
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B
2
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C
4
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D
3
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Solution

The correct option is B 2
We know that cos1x[0,π]
cos1(a)+cos1(b)+cos1(c)=3π is possible iff a=b=c=1

Now, f(1)=2 and f(x+y)=f(x)f(y)
Let, f(x)=akx
ak=2
Hence, f(x)=2x
f(2)=4, f(3)=8

a2f(1)+b2f(2)+c2f(3)+a+b+ca2f(1)+b2f(2)+c2f(3)
=1+1+131+1+1=2

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