Bulgaria National Olympiad Problem 4

Let $y=3+\sin x$

Now $y\in [2,4]$ and takes all the values in this interval.

Consider $g(y)=y+\frac{2}{y}$

This function is increasing on $y\in [2,4]$

So $g(2) \leq g(y) \leq g(4) \implies 3 \leq g(y) \leq \frac{9}{2}$ .

So minimum of $f(b)=max(|g(y)+b-3|$ is $\frac{3}{4}$ which is attained when $b=-\frac{3}{4}$

Because if $b>-\frac{3}{4}$ when choose $x=\frac{\pi}{2}$ so $y=4$ and then $g(y) + b - 3 > \frac{3}{4}$ ,

while if $b < -\frac{3}{4}$ then choose $x =-\frac{\pi}{2}$ so $y=2$ and $g(y) + b - 3 = -\frac{3}{4}$ ; on the other hand, our range for $g(y)$ guarantees $-\frac{3}{4} ≤ g(y) + b - 3 ≤ \frac{3}{4}$ for $b = -\frac{3}{4}$ .

We conclude our final answer is $\boxed{\frac{3}{4}}$

If you differentiate it wrt x, you can see that it has a maximum at $\pi/2$ with a maximum value of $3/2$ (for b=0) as well as a minimum at $-\pi/2$ with a minimum value of 0 (again, for b=0). The modulus sign bounces back those values of y which are less than zero for some negative value of b in the function inside the modulus. Without the modulus, the maximum of the function would tend to negative infinity as b tends to an infinitely negative number. As the modulus sign bounces back those values, we can see that y would tend to a large, positive number whenever b goes to the extreme ends of the real line. Putting b=-3/2 gives back the same function, but with a phase difference of $\pi/2$ , i.e, the maximum points and the minimum points have swapped around. It becomes clearer that for reducing the maximum value, b needs to be between -3/2 and 0. If we put b= -3/4, i.e, the mean of 0 and -3/2, we can see that the maximum points have a value of 3/4, half the maximum value of the original cases with b=0,-3/2. This is as low as it's going to get because for lower values of b, the part that was bounced back would contain the maximum values (that are going to be greater than 3/4).