Let f(x1+x2+...+xnn)=f(x1)+f(x2)+...+f(xn)n where all xi∈R are independent to each other and n∈N. 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 Aa
Differentiating the given equation w.r.t. x1, we get
1nf′(x1+x2+...+xnn)=f′(x1)n
[Since all x′is are independent to each other, ∴dxidxj=0 if i≠j and dxidxj=1 if (i=j)]
On putting x1=x2=...=xn−1=0 and xn=x, we get f′(xn)=f′(0)=a.