f(x+y)=f(x)+f(y) where x,yεR
Prove thta nf(x-\frac{a}{n})+f(a) is an odd functionm wheere nεI-{0}
I am also givng ma soln...plz continue and finsh it off...
nf(x-a/n)=+f(a)
=>nf(x)-nf(a/n)+f(a)
=>nf(x)-n(f(a/n)-1/nf(a))
also 1/n f(an)=f(a)
what to do next ??
i did a couple of more steps..but they led nowhere...
any otehr method...then also reply..
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2 Answers
66Setting x=y=0, we have
f(0)=2f(0) => f(0)=0
Setting y=-x, we get f(0)=f(x)+f(-x) => f(-x)=f(x) since f(0)=0
Hence f is odd.
Also by induction, we have
f(x1+x2+....+xk)=f(x1)+f(x2)+.....+f(xk)
Setting x1=x2=...=xk=x, we get f(kx)=k f(x)
Let
g(x)=nf\left(x-\dfrac{a}{n}\right)+f(a)
Then,
g(x)=nf(x)+nf\left(-\dfrac{a}{n}\right)+f(a)=nf(x)-nf\left(\dfrac{a}{n}\right)+f(a)
On the other hand
f(a)=f\left(n\cdot \dfrac{a}{n}\right)=nf\left(\dfrac{a}{n}\right)
Hence, we get
g(x)=nf(x)
Obviously g(x) is odd.
