Question

For any real number b, let f(b) denotes the maximum of |$\sin x + \frac{2}{3 + \sin x} + b$| $\forall x \in R$. Then the minimum value of f(b) $\forall b \in R$ is:

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. 1

Answer 1

Answer: (B)

$\frac{3}{4}$

Answer 2

Let $y = 3 + \sin x$. Since $-1 \le \sin x \le 1$, the range of $y$ is $[2, 4]$. Let $g(y) = y + \frac{2}{y}$. The derivative $g'(y) = 1 - \frac{2}{y^2} > 0$ for $y \in [2, 4]$, so $g(y)$ is increasing. The range of $g(y)$ on $[2, 4]$ is $[g(2), g(4)] = [2 + \frac{2}{2}, 4 + \frac{2}{4}] = [3, \frac{9}{2}]$. The expression is $|\, g(y) - 3 + b \,|$. Let $k = g(y) - 3$. The range of $k$ is $[3-3, \frac{9}{2}-3] = [0, \frac{3}{2}]$. We want to find the minimum of $f(b) = \max_{k \in [0, 3/2]} |k+b|$. This is equivalent to finding the minimum of $\max_{A \in [b, b+3/2]} |A|$. This maximum is minimized when the interval $[b, b+3/2]$ is centered at 0. The midpoint is $b + \frac{3}{4}$. Setting $b + \frac{3}{4} = 0$, we get $b = -\frac{3}{4}$. The interval becomes $[-\frac{3}{4}, \frac{3}{4}]$, and the maximum value of $|A|$ is $\frac{3}{4}$.