Question

# If f(x) satisfies the relation f(x+y)=f(x)+f(y) for all x,y∈R, and f(1)=5, then

A
f(x) is an odd function
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B
f(x) is an even function
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C
mr=1f(r)=5m+1C2
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D
mr=1f(r)=5m(m+2)3
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Solution

## The correct options are A f(x) is an odd function D m∑r=1f(r)=5m+1C2Let x=y=0, then the given equation is :f(0)=2×f(0)⇒f(0)=0Now for y=−x, we have 0=f(0)=f(x)+f(−x)∴f(x)=−f(−x)Hence the function is an odd function.Now, we will prove that f(k)=5k for any natural number kLet x=y=1, then f(2)=f(1)+f(1)=2×5 Let x=2,y=1, then f(3)=f(2)+f(1)=3×5In general,Let x=k−1,y=1, then f(k)=f(k−1)+f(1)=k×5Hence,m∑r=1f(r)=f(1)+f(2)+f(3)+⋯+f(m)=5+10+15+⋯5m=5(1+2+3+⋯+m)=5m(m+1)2=5(m+1C2)

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