Question

Let $f: [0, 4\pi] \rightarrow [0, \pi], f(x) = \cos^{-1}(\cos x)$ then find number of solutions in $x \in [0, 4\pi]$ satisfying the equation $f(x) = \frac{10 - x}{10}$.

Answer 1

3

Answer 2

The function $f(x) = \cos^{-1}(\cos x)$ is defined piecewise on $[0, 4\pi]$ as:

$f(x) = x$ for $x \in [0, \pi]$

$f(x) = 2\pi - x$ for $x \in [\pi, 2\pi]$

$f(x) = x - 2\pi$ for $x \in [2\pi, 3\pi]$

$f(x) = 4\pi - x$ for $x \in [3\pi, 4\pi]$

The equation to solve is $f(x) = \frac{10 - x}{10}$. Let $g(x) = \frac{10 - x}{10} = 1 - \frac{x}{10}$.

We need to find the number of solutions to $f(x) = g(x)$ in the interval $[0, 4\pi]$.

Let's solve the equation in each interval:

Case 1: $x \in [0, \pi]$

$f(x) = x$. The equation is $x = \frac{10 - x}{10}$.

$10x = 10 - x \implies 11x = 10 \implies x = \frac{10}{11}$.

Since $0 \le \frac{10}{11} \le \pi$ (as $\frac{10}{11} \approx 0.909$ and $\pi \approx 3.14$), this solution is in the interval.

Also, $g(\frac{10}{11}) = \frac{10 - \frac{10}{11}}{10} = \frac{10}{11}$, which is in the range $[0, \pi]$ of $f(x)$.

So, $x_1 = \frac{10}{11}$ is a solution.

Case 2: $x \in (\pi, 2\pi]$

$f(x) = 2\pi - x$. The equation is $2\pi - x = \frac{10 - x}{10}$.

$10(2\pi - x) = 10 - x \implies 20\pi - 10x = 10 - x \implies 9x = 20\pi - 10 \implies x = \frac{20\pi - 10}{9}$.

Let's check if this solution is in $(\pi, 2\pi]$.

$\pi < \frac{20\pi - 10}{9} \le 2\pi$

$9\pi < 20\pi - 10 \le 18\pi$

$10 < 11\pi$ (which is true, $10 < 11 \times 3.14 = 34.54$)

$2\pi \le 10$ (which is false, $2\pi \approx 6.28 \le 10$)

$20\pi - 10 \le 18\pi \implies 2\pi \le 10$. This is false.

Let's recheck the inequality: $x = \frac{20\pi - 10}{9}$.

Is $x > \pi$? $\frac{20\pi - 10}{9} > \pi \implies 20\pi - 10 > 9\pi \implies 11\pi > 10$, which is true.

Is $x \le 2\pi$? $\frac{20\pi - 10}{9} \le 2\pi \implies 20\pi - 10 \le 18\pi \implies 2\pi \le 10$, which is true.

So $x = \frac{20\pi - 10}{9}$ is in $(\pi, 2\pi]$.

Let's check the value of $g(x)$ at this point: $g(\frac{20\pi - 10}{9}) = \frac{10 - \frac{20\pi - 10}{9}}{10} = \frac{90 - (20\pi - 10)}{90} = \frac{100 - 20\pi}{90} = \frac{10 - 2\pi}{9}$.

Is $\frac{10 - 2\pi}{9} \in [0, \pi]$?

$0 \le \frac{10 - 2\pi}{9} \le \pi$

$0 \le 10 - 2\pi \le 9\pi$

$10 \ge 2\pi$ (True, $10 \ge 6.28$)

$10 \le 11\pi$ (True, $10 \le 34.54$)

So the value of $g(x)$ is in the range $[0, 1]$ which is within $[0, \pi]$.

Thus, $x_2 = \frac{20\pi - 10}{9}$ is a solution.

Case 3: $x \in (2\pi, 3\pi]$

$f(x) = x - 2\pi$. The equation is $x - 2\pi = \frac{10 - x}{10}$.

$10(x - 2\pi) = 10 - x \implies 10x - 20\pi = 10 - x \implies 11x = 10 + 20\pi \implies x = \frac{10 + 20\pi}{11}$.

Let's check if this solution is in $(2\pi, 3\pi]$.

$2\pi < \frac{10 + 20\pi}{11} \le 3\pi$

$22\pi < 10 + 20\pi \le 33\pi$

$2\pi < 10$ (True, $6.28 < 10$)

$10 + 20\pi \le 33\pi \implies 10 \le 13\pi$ (True, $10 \le 13 \times 3.14 = 40.82$)

So $x = \frac{10 + 20\pi}{11}$ is in $(2\pi, 3\pi]$.

Let's check the value of $g(x)$ at this point: $g(\frac{10 + 20\pi}{11}) = \frac{10 - \frac{10 + 20\pi}{11}}{10} = \frac{110 - (10 + 20\pi)}{110} = \frac{100 - 20\pi}{110} = \frac{10 - 2\pi}{11}$.

Is $\frac{10 - 2\pi}{11} \in [0, \pi]$?

$0 \le \frac{10 - 2\pi}{11} \le \pi$

$0 \le 10 - 2\pi \le 11\pi$

$10 \ge 2\pi$ (True)

$10 \le 13\pi$ (True)

So the value of $g(x)$ is in the range $[0, 1]$ which is within $[0, \pi]$.

Thus, $x_3 = \frac{10 + 20\pi}{11}$ is a solution.

Case 4: $x \in (3\pi, 4\pi]$

$f(x) = 4\pi - x$. The equation is $4\pi - x = \frac{10 - x}{10}$.

$10(4\pi - x) = 10 - x \implies 40\pi - 10x = 10 - x \implies 9x = 40\pi - 10 \implies x = \frac{40\pi - 10}{9}$.

Let's check if this solution is in $(3\pi, 4\pi]$.

$3\pi < \frac{40\pi - 10}{9} \le 4\pi$

$27\pi < 40\pi - 10 \le 36\pi$

$10 < 13\pi$ (True)

$4\pi \le 10$ (False, $4\pi \approx 12.56 > 10$)

So $x = \frac{40\pi - 10}{9}$ is not in the interval $(3\pi, 4\pi]$.

We have found three solutions: $x_1 = \frac{10}{11}$, $x_2 = \frac{20\pi - 10}{9}$, and $x_3 = \frac{10 + 20\pi}{11}$.

All these solutions lie in the interval $[0, 4\pi]$.

Therefore, the total number of solutions is 3.

The final answer is $\boxed{3}$.