Let $f : [0, 4\pi] \to [0, \pi]$ be defined by $f(x) = \cos^{-1}(\cos x)$. The number of points $x$ in $[0, 4\pi]$ satisfying the equation $f(x) = \frac{10 - x}{10}$ is

Let f : [0, 4π] → [0, π] be defined by f(x) = cos⁻¹(cos x). The number of points x ∈ [0, 4π] satisfying the equation g(x) = (10 - x)/10 is

$f(x) = \cos^{-1}(\cos x),\qquad x \in [0, 4\pi]$

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

$f(x) = \dfrac{10-x}{10}$

$\text{Number of solutions of } f(x)=\dfrac{10-x}{10} \text{ in } [0,4\pi]$

$f(x)=\cos^{-1}(\cos x),\quad \text{period } 2\pi$

$0\le x\le\pi:\ f=x;\qquad \pi\le x\le 2\pi:\ f=2\pi-x$

$2\pi\le x\le 3\pi:\ f=x-2\pi;\qquad 3\pi\le x\le 4\pi:\ f=4\pi-x$

$L(x) = \dfrac{10-x}{10} = 1 - \dfrac{x}{10}$

$L(0)=1,\quad L(10)=0,\quad L(x)<0\ \text{for } x>10$

$\text{Solve } f(x)=1-\dfrac{x}{10} \text{ on each linear piece,}$

$\text{keep only solutions inside that piece's interval}$

$x = 1 - \dfrac{x}{10} \Rightarrow \dfrac{11x}{10}=1$

$x = \dfrac{10}{11} \approx 0.909 \in [0,\pi]\ \checkmark$

$2\pi - x = 1 - \dfrac{x}{10} \Rightarrow \dfrac{9x}{10} = 2\pi-1$

$x = \dfrac{10(2\pi-1)}{9} \approx 5.870 \in [\pi,2\pi]\ \checkmark$

$x - 2\pi = 1 - \dfrac{x}{10} \Rightarrow \dfrac{11x}{10} = 1+2\pi$

$x = \dfrac{10(1+2\pi)}{11} \approx 6.621 \in [2\pi,3\pi]\ \checkmark$

$4\pi - x = 1 - \dfrac{x}{10} \Rightarrow x = \dfrac{10(4\pi-1)}{9} \approx 12.85$

$12.85 > 4\pi \approx 12.566 \Rightarrow \text{reject}$

$\boxed{\text{Answer: } 3}$