Why $f(t)= t+ 2t^2\ sin (\frac 1t) , t\neq 0$ & $f(t)=0 , t=0$. Prove that this function is not $1-1$ in any neighbourhood of zero. Why $f(t)= t+ 2t^2 \sin (\frac 1t) , t\neq 0$ & $f(t)=0 , t=0.\,$Prove that this function is not $1-1$ in any neighbourhood of zero. 
It is not possible with $\frac 1 {n \pi}$.

Hint: $f$ is continuously differentiable for all $t>0$; you can check that
$$f'\Bigl(\frac{1}{n\pi}\Bigr)$$
is positive if $n$ is odd and negative if $n$ is even.
