Let $n \geq 2$ be a natural number and $f: [0, 1] \to \mathbb{R}$ be the function defined by $f(x) = \begin{cases} n(1-2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx-1) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1-nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx-1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}$. If $n$ is such that the area of the region bounded by the curves $x = 0$, $x = 1$, $y = 0$ and $y = f(x)$ is $4$, then the maximum value of the function $f$ is

Let n ≥ 2 be a natural number and f: [0, 1] → R be the function defined by f(x) = n(1-2nx) if 0 ≤ x ≤ 1/2n 2n(2nx-1) if 1/2n ≤ x ≤ 3/4n 4n(1-nx) if 3/4n ≤ x ≤ 1/n n/(n-1)(nx-1) if 1/n ≤ x ≤ 1 If n is such that the area of the region bounded by the curves x = 0, x = 1, y = 0 and y = f(x) is 4, then the maximum value of the function f is f(x)= n(1-2nx), 0≤ x≤ 1/2n 2n(2nx-1), 1/2n≤ x≤ 3/4n 4n(1-nx), 3/4n≤ x≤ 1/n n/(n-1)(nx-1), 1/n≤ x≤ 1

$\text{Area bounded by } x=0,\ x=1,\ y=0,\ y=f(x) = 4$

$\text{To find: maximum value of } f$

$f\ge 0 \Rightarrow \text{area} = \int_0^1 f(x)\,dx$

$1.\ \text{area of each of the } 4 \text{ pieces}$ $2.\ \text{total area} = g(n)$ $3.\ =4 \Rightarrow n \Rightarrow \max f$ $A_1 = \tfrac12\cdot\dfrac{1}{2n}\cdot(n+0) = \dfrac14$ $A_2 = \tfrac12\cdot\dfrac{1}{4n}\cdot(0+n) = \dfrac18$ $A_3 = \tfrac12\cdot\dfrac{1}{4n}\cdot(n+0) = \dfrac18$ $A_4 = \tfrac12\cdot\dfrac{n-1}{n}\cdot(0+n) = \dfrac{n-1}{2}$ $A = \dfrac14+\dfrac18+\dfrac18+\dfrac{n-1}{2} = \dfrac12+\dfrac{n-1}{2} = \dfrac{n}{2}$ $\dfrac{n}{2}=4 \Rightarrow n=8$ $f_{\max} = n = 8$ $\boxed{8}$