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Let $A={tsin(frac{1}{t}) | tin (0,frac{2}{pi})}$.

Then

1. $sup (A)<frac{2}{pi}+frac{1}{npi}$ for all $nge 1$.




2. $inf (A)> frac{-2}{3pi}-frac{1}{npi}$ for all $nge 1$.



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 $-tle tsin(1/t)le t$. So $tsin(1/t)$ 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 $tsin(1/t)$ is always greater than or equal to $-2/pi=-0.63$, so infimum cannot be $-1$.

Now at $t=2/3pi$ , function has value $frac{-2}{3pi}$. So if option $2$ is false we must have some value of $t$ in $A$ for which the function attains value strictly less than $frac{-2}{3pi}$. Right?

Can you please help me with option $2$ now. Thanks in advance.

If the option $2$ is true, it means that $$inf Age -{2over 3pi}$$ since $inf A> -{2over 3pi}-{1over npi}$ is true for all $n>1$. Since $$large tsin {1over t}Big|_{t={2over 3pi}in A}=-{2over 3pi}$$ then we must have $inf A= -{2over 3pi}$ and $$large {dover dt}tsin {1over t}Big|_{t={2over 3pi}in A}=0$$ because the infimum is equal to minimum now (since it is attained in $t={2over 3pi}in A$) and as the function is differentiable in $A$. But this doesn't hold since $$large {dover dt}tsin {1over t}Big|_{t={2over 3pi}in A}=large sin {1over t}-{1over t}cos {1over t}Big|_{t={2over 3pi}}=-1ne 0$$therefore option $2$ is incorrect.

The graph of $sin(1/t)$ turns at $t=2/3pi$, but this is not true of $tsin(1/t)$. 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 $nge 1$ iff $age b$, we are done.

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

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

Something slightly more general is true:

Suppose $f$ is smooth, $f(1)=0=f(2)$, $fle 0$ 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.

Like some other answers, but written differently:

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

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

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

Let $x=1/timplies A={frac{sin x}x:x>pi/2}$

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

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

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