Let $A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$.
Then
-
$\sup (A)<\frac{2}{\pi}+\frac{1}{n\pi}$ for all $n\ge 1$.
-
$\inf (A)> \frac{-2}{3\pi}-\frac{1}{n\pi}$ for all $n\ge 1$.
- $\sup (A)=1$
- $\inf (A)=-1$
My answer was options $1$ and $2$ which matched with prelim answer key provided by organization, but the final answer key changed the answer to option $1$ only. Why is option $2$ not correct?
My attempt-
As $-t\le t\sin(1/t)\le t$. So $t\sin(1/t)$ will always be less than $\frac{2}{\pi}$ which is strictly less than $1$. So option $3$ is false and option $1$ is true. Also value of $t\sin(1/t)$ is always greater than or equal to $-2/\pi=-0.63$, so infimum cannot be $-1$.
Now at $t=2/3\pi$ , function has value $\frac{-2}{3\pi}$. So if option $2$ is false we must have some value of $t$ in $A$ for which the function attains value strictly less than $\frac{-2}{3\pi}$. Right?
Can you please help me with option $2$ now. Thanks in advance.
Best Answer
If the option $2$ is true, it means that $$\inf A\ge -{2\over 3\pi}$$ since $\inf A> -{2\over 3\pi}-{1\over n\pi}$ is true for all $n>1$. Since $$\large t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=-{2\over 3\pi}$$ then we must have $\inf A= -{2\over 3\pi}$ and $$\large {d\over dt}t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=0$$ because the infimum is equal to minimum now (since it is attained in $t={2\over 3\pi}\in A$) and as the function is differentiable in $A$. But this doesn't hold since $$\large {d\over dt}t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=\large \sin {1\over t}-{1\over t}\cos {1\over t}\Big|_{t={2\over 3\pi}}=-1\ne 0$$therefore option $2$ is incorrect.