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Question

If f(x) satisfies the relation f(x+y)=f(x)+f(y) for all x,yR, 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 mr=1f(r)=5m+1C2
Let x=y=0, then the given equation is :
f(0)=2×f(0)f(0)=0
Now 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 k
Let x=y=1, then f(2)=f(1)+f(1)=2×5
Let x=2,y=1, then f(3)=f(2)+f(1)=3×5
In general,
Let x=k1,y=1, then f(k)=f(k1)+f(1)=k×5
Hence,
mr=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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