Which of the following statements are true for $A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$?

Question

Let $A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$.

Then

1. $\sup (A)<\frac{2}{\pi}+\frac{1}{n\pi}$ for all $n\ge 1$.

2. $\inf (A)> \frac{-2}{3\pi}-\frac{1}{n\pi}$ for all $n\ge 1$.

3. $\sup (A)=1$
4. $\inf (A)=-1$

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.

Answer 1

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.

Answer 2

Let $x=1/t\implies A=\{\frac{\sin x}x:x>\pi/2\}$

Let $f(x)=\frac{\sin x}x, x>\pi/2$

$f(3\pi/2)=-\frac2{3\pi}<0, f'(3\pi/2)>0\implies f$ is strictly increasing at $3\pi/2$

This means you can find $a\in\varepsilon-$neighbourhood of $3\pi/2$ such that $f(a)<f(3\pi/2)\implies f(a)<-\frac2{3\pi}$

Answer 3

The graph of $\sin(1/t)$ turns at $t=2/3\pi$, but this is not true of $t\sin(1/t)$. Multiplying by $t$ has the effect of pushing the minimum point to the right and down slightly.

Since $a>b-\frac{1}{\pi n}$ for all $n\ge 1$ iff $a\ge b$, we are done.

To see this without a graph, lets start from your correct observation that for $f(t) = t\sin (1/t)$, $f(2/3\pi) = -2/3\pi$. The derivative at this point however is $$ f'(3\pi/2) = \sin(3\pi/2) + 2/3\pi \cos(3\pi/2) = -1 $$

Since $f$ is smooth on a small neighbourhood of $2/3\pi$, $f'\le -1/2<0$ on this neighbourhood. Thus $f$ is decreasing at $2/3\pi$.

Something slightly more general is true:

Suppose $f$ is smooth, $f(1)=0=f(2)$, $f\le 0$ and $f$ has a unique non degenerate minimum point $x_0$ in $[1,2]$. Then $g(x)=xf(x)$ has a minimum point to the right and below of $x_0$.

Proof- by a similar computation to the above, $g$ is decreasing at $x_0$. As $g(2)=2f(2)=0$, we conclude by Intermediate Value Theorem and Rolles' Theorem.

Answer 4

Like some other answers, but written differently:

Let $f(t)= t\sin(1/t).$ Then $f'(t)= \sin(1/t) - [\cos (1/t)]/t$ for $t>0.$ Thus $f'(2/(3\pi)) = -1.$ Hence for $t$ near and to the right of $2/(3\pi),$ we have

$$\frac{f(t)-f(2/(3\pi)}{t-2/(3\pi)}<-\frac{1}{2}.$$

For such $t$ we have $f(t)<f(2/(3\pi)=-2/(3\pi).$ Hence $\inf A <-2/(3\pi),$ which implies 2. does not hold.