Question

The function $f(x)=\cos^{-1}\left(\frac{2\lfloor |\sin x|+|\cos x| \rfloor}{\sin^2 x+2\sin x+\frac{11}{4}}\right)$ is defined if $x$ belongs to: (where $\lfloor . \rfloor$ represents the greatest integer function)

• A. [0, \frac{7\pi}{6}]
• B. (\frac{7\pi}{6}, \frac{11\pi}{6})
• C. [\frac{11\pi}{6}, 2\pi]
• D. [\pi, 2\pi]

Answer 1

Answer: (A)

[0, \frac{7\pi}{6}]

Answer 2

To find the domain of the function $f(x)=\cos^{-1}\left(\frac{2\lfloor |\sin x|+|\cos x| \rfloor}{\sin^2 x+2\sin x+\frac{11}{4}}\right)$, we need to ensure that the argument of the $\cos^{-1}$ function lies in the interval $[-1, 1]$.

Let the argument be $Y = \frac{2\lfloor |\sin x|+|\cos x| \rfloor}{\sin^2 x+2\sin x+\frac{11}{4}}$.

Step 1: Analyze the Numerator

Let $N = 2\lfloor |\sin x|+|\cos x| \rfloor$. We know that for any real $x$, $1 \le |\sin x|+|\cos x| \le \sqrt{2}$. To prove this, let $S = |\sin x|+|\cos x|$. Then $S^2 = (|\sin x|+|\cos x|)^2 = \sin^2 x + \cos^2 x + 2|\sin x \cos x| = 1 + |\sin(2x)|$. Since $0 \le |\sin(2x)| \le 1$, we have $1 \le S^2 \le 2$. Taking the square root, $1 \le S \le \sqrt{2}$. Since $\sqrt{2} \approx 1.414$, we have $1 \le |\sin x|+|\cos x| \le 1.414$. Therefore, the greatest integer less than or equal to $|\sin x|+|\cos x|$ is $\lfloor |\sin x|+|\cos x| \rfloor = 1$. So, the numerator $N = 2 \times 1 = 2$.

Step 2: Analyze the Denominator

Let $D = \sin^2 x+2\sin x+\frac{11}{4}$. Let $t = \sin x$. Since $-1 \le \sin x \le 1$, we have $-1 \le t \le 1$. The denominator can be rewritten by completing the square: $D = t^2+2t+1 + \frac{11}{4} - 1 = (t+1)^2 + \frac{7}{4}$. Now, we find the range of $D$ for $t \in [-1, 1]$. The minimum value of $(t+1)^2$ occurs at $t=-1$, where $(t+1)^2 = (-1+1)^2 = 0$. The maximum value of $(t+1)^2$ occurs at $t=1$, where $(t+1)^2 = (1+1)^2 = 2^2 = 4$. So, $0 \le (t+1)^2 \le 4$. Adding $\frac{7}{4}$ to all parts of the inequality: $\frac{7}{4} \le (t+1)^2 + \frac{7}{4} \le 4 + \frac{7}{4}$, which simplifies to $\frac{7}{4} \le D \le \frac{23}{4}$. Note that $D > 0$ for all $x$.

Step 3: Apply the condition for $\cos^{-1}$

The argument of $\cos^{-1}$ is $Y = \frac{N}{D} = \frac{2}{D}$. For $\cos^{-1}(Y)$ to be defined, we must have $-1 \le Y \le 1$. Since $D > 0$, $Y = \frac{2}{D}$ will always be positive. So, $Y > 0$. Therefore, we only need to satisfy the condition $Y \le 1$, which means $\frac{2}{D} \le 1$. Since $D > 0$, we can multiply by $D$ without changing the inequality direction: $2 \le D$.

Step 4: Find the condition on $\sin x$

Substitute the expression for $D$: $(\sin x+1)^2 + \frac{7}{4} \ge 2$. This simplifies to $(\sin x+1)^2 \ge \frac{1}{4}$. Taking the square root of both sides: $|\sin x+1| \ge \frac{1}{2}$. This inequality implies two cases:

Case 1: $\sin x+1 \ge \frac{1}{2}$, which gives $\sin x \ge -\frac{1}{2}$.

Case 2: $\sin x+1 \le -\frac{1}{2}$, which gives $\sin x \le -\frac{3}{2}$. Since the range of $\sin x$ is $[-1, 1]$, the condition $\sin x \le -\frac{3}{2}$ is impossible.

Thus, the only condition for $f(x)$ to be defined is $\sin x \ge -\frac{1}{2}$.

Step 5: Determine the interval for $x$

We need to find the values of $x$ for which $\sin x \ge -\frac{1}{2}$. In the interval $[0, 2\pi]$, $\sin x = -\frac{1}{2}$ at $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ and $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. The values of $x$ for which $\sin x \ge -\frac{1}{2}$ are $x \in \left[0, \frac{7\pi}{6}\right] \cup \left[\frac{11\pi}{6}, 2\pi\right]$ (and periodically).

Comparing with the options, $\left[0, \frac{7\pi}{6}\right]$ is a subset of the domain.