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Question

Let f(x1+x2+...+xnn)=f(x1)+f(x2)+...+f(xn)n where all xiR are independent to each other and nN. if f(x) is differentiable and f(0)=a,f(0)=b and f(x) is equal to

A
a
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B
0
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C
b
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D
None of these
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Solution

The correct option is A a
Differentiating the given equation w.r.t. x1, we get
1nf(x1+x2+...+xnn)=f(x1)n
[Since all xis are independent to each other, dxidxj=0 if ij and dxidxj=1 if (i=j)]
On putting x1=x2=...=xn1=0 and xn=x, we get f(xn)=f(0)=a.
On integrating, we get nf(xn)=ax+c
Since f(0)=b, we have c=nb
nf(xn)=ax+nbnf(x)=nax+nbf(x)=ax+b.
f(x)=a,xR

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