Question

Let $f(x) = [\sin x] + [2 \cos x]$ for $0 < x < \pi$, where $[t]$ denotes the greatest integer function. Then range of $f(x)$ contains number of elements equal to:

Answer 1

4

Answer 2

The function is given by $f(x) = [\sin x] + [2 \cos x]$ for $0 < x < \pi$. We need to find the range of this function. The range consists of all possible values that $f(x)$ can take for $x \in (0, \pi)$.

Let's analyze the possible values of $[\sin x]$ and $[2 \cos x]$ for $x \in (0, \pi)$.

For $x \in (0, \pi)$, the range of $\sin x$ is $(0, 1]$. So, $[\sin x]$ can take the value 0 (when $0 < \sin x < 1$, i.e., $x \in (0, \pi) \setminus \{\pi/2\}$) or 1 (when $\sin x = 1$, i.e., $x = \pi/2$).

For $x \in (0, \pi)$, the range of $\cos x$ is $(-1, 1)$. So, the range of $2 \cos x$ is $(-2, 2)$. The possible integer values for $[2 \cos x]$ are $-2, -1, 0, 1$.

Let's find the intervals of $x$ in $(0, \pi)$ for which $2 \cos x$ falls into the corresponding integer intervals.

1. $[2 \cos x] = -2 \iff -2 \le 2 \cos x < -1 \iff -1 \le \cos x < -1/2$. In $(0, \pi)$, this occurs when $x \in (2\pi/3, \pi)$. (Since $\cos(2\pi/3) = -1/2$ and $\cos(\pi) = -1$)
2. $[2 \cos x] = -1 \iff -1 \le 2 \cos x < 0 \iff -1/2 \le \cos x < 0$. In $(0, \pi)$, this occurs when $x \in (\pi/2, 2\pi/3]$. (Since $\cos(\pi/2) = 0$ and $\cos(2\pi/3) = -1/2$)
3. $[2 \cos x] = 0 \iff 0 \le 2 \cos x < 1 \iff 0 \le \cos x < 1/2$. In $(0, \pi)$, this occurs when $x \in (\pi/3, \pi/2]$. (Since $\cos(\pi/3) = 1/2$ and $\cos(\pi/2) = 0$)
4. $[2 \cos x] = 1 \iff 1 \le 2 \cos x < 2 \iff 1/2 \le \cos x < 1$. In $(0, \pi)$, this occurs when $x \in (0, \pi/3]$. (Since $\cos(0) = 1$ and $\cos(\pi/3) = 1/2$)

We need to consider the values of $f(x) = [\sin x] + [2 \cos x]$ in the intervals defined by the critical points $\pi/3, \pi/2, 2\pi/3$.

Case 1: $x \in (0, \pi/3)$ $[\sin x] = 0$ (since $0 < \sin x < \sin(\pi/3) = \sqrt{3}/2 < 1$) $[2 \cos x] = 1$ (from analysis above) $f(x) = 0 + 1 = 1$.

Case 2: $x = \pi/3$ $[\sin(\pi/3)] = [\sqrt{3}/2] = 0$ $[2 \cos(\pi/3)] = [2 \times 1/2] = [1] = 1$ $f(\pi/3) = 0 + 1 = 1$.

Case 3: $x \in (\pi/3, \pi/2)$ $[\sin x] = 0$ (since $\sin(\pi/3) = \sqrt{3}/2 < \sin x < \sin(\pi/2) = 1$) $[2 \cos x] = 0$ (from analysis above) $f(x) = 0 + 0 = 0$.

Case 4: $x = \pi/2$ $[\sin(\pi/2)] = [1] = 1$ $[2 \cos(\pi/2)] = [2 \times 0] = [0] = 0$ $f(\pi/2) = 1 + 0 = 1$.

Case 5: $x \in (\pi/2, 2\pi/3)$ $[\sin x] = 0$ (since $\sin(2\pi/3) = \sqrt{3}/2 < \sin x < \sin(\pi/2) = 1$) $[2 \cos x] = -1$ (from analysis above) $f(x) = 0 + (-1) = -1$.

Case 6: $x = 2\pi/3$ $[\sin(2\pi/3)] = [\sqrt{3}/2] = 0$ $[2 \cos(2\pi/3)] = [2 \times (-1/2)] = [-1] = -1$ $f(2\pi/3) = 0 + (-1) = -1$.

Case 7: $x \in (2\pi/3, \pi)$ $[\sin x] = 0$ (since $0 < \sin x < \sin(2\pi/3) = \sqrt{3}/2 < 1$) $[2 \cos x] = -2$ (from analysis above) $f(x) = 0 + (-2) = -2$.

The possible values of $f(x)$ obtained are $1, 0, -1, -2$. The range of $f(x)$ is the set $\{-2, -1, 0, 1\}$. The number of elements in the range is 4.