Let n $\ge$ 2 be a natural number and f :[0,1] $\to \mathrm{R}$ be the function defined by
$\mathrm{f}\left(\mathrm{x}\right)=$$\left\{\begin{array}{l}\mathrm{n}(1-2\mathrm{nx})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }0\le \mathrm{x}\le \frac{1}{2\mathrm{n}}\\ 2\mathrm{n}(2\mathrm{nx}-1)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }\frac{1}{2\mathrm{n}}\le \mathrm{x}\le \frac{3}{4\mathrm{n}}\\ 4\mathrm{n}(1-\mathrm{nx})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }\frac{3}{4\mathrm{n}}\le \mathrm{x}\le \frac{1}{\mathrm{n}}\\ \frac{\mathrm{n}}{\mathrm{n}-1}(\mathrm{nx}-1)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }\frac{1}{\mathrm{n}}\le \mathrm{x}\le 1\end{array}\right.$
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

$\mathrm{y}=\left\{\begin{array}{l}\mathrm{n}(1-2\mathrm{nx})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{, if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le \mathrm{x}\le \frac{1}{2\mathrm{n}}\\ 2\mathrm{n}(2\mathrm{nx}-1),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{2\mathrm{n}}\le \mathrm{x}\le \frac{3}{4\mathrm{n}}\\ 4\mathrm{n}(1-\mathrm{nx}),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{, if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{3}{4\mathrm{n}}\le \mathrm{x}\le \frac{1}{\mathrm{n}}\\ \frac{\mathrm{n}}{\mathrm{n}-1}(\mathrm{nx}-1),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{\mathrm{n}}\le \mathrm{x}\le 1\end{array}\right.$
$\mathrm{if} \mathrm{y}=\mathrm{n}(1-2\mathrm{nx})\Rightarrow \text{ if }\mathrm{y}=0\Rightarrow \mathrm{x}=\frac{1}{2\mathrm{n}}\text{ and if }\mathrm{x}=0\Rightarrow \mathrm{y}=\mathrm{n}$
if $\mathrm{y}=2\mathrm{n}(2\mathrm{nx}-1)\Rightarrow \text{ if }\mathrm{y}=0\Rightarrow \mathrm{x}=\frac{1}{2\mathrm{n}}\text{ and if }\mathrm{x}=\frac{3}{4\mathrm{n}}\Rightarrow \mathrm{y}=\mathrm{n}$
if $\mathrm{y}=4\mathrm{n}(1-\mathrm{nx})\Rightarrow \text{ if }\mathrm{y}=0\Rightarrow \mathrm{x}=\frac{1}{\mathrm{n}}\text{ and if }\mathrm{x}=\frac{1}{\mathrm{n}}\Rightarrow \mathrm{y}=\mathrm{n}$
$\mathrm{y}=\frac{\mathrm{n}}{\mathrm{n}-1}(\mathrm{nx}-1)\text{, if }\mathrm{y}=0\Rightarrow \mathrm{x}=\frac{1}{\mathrm{n}}\text{ and if }\mathrm{x}=1\Rightarrow \mathrm{y}=\mathrm{n}$

${\mathrm{A}}_1=\frac{1}{2}\cdot \frac{1}{2\mathrm{n}}\cdot \mathrm{n}{\mathrm{A}}_2=\frac{1}{2}\cdot \frac{1}{2\mathrm{n}}\cdot \mathrm{n}{\mathrm{A}}_3=\frac{1}{2}\left(1-\frac{1}{\mathrm{n}}\right)\cdot \mathrm{n}$
Total Area = ${\mathrm{A}}_1+{\mathrm{A}}_2+{\mathrm{A}}_3=4$
$=\frac{1}{2}\cdot \frac{1}{2\mathrm{n}}\cdot \mathrm{n}+\frac{1}{2}\cdot \frac{1}{2\mathrm{n}}\cdot \mathrm{n}+\frac{1}{2}\left(1-\frac{1}{\mathrm{n}}\right)\cdot \mathrm{n}=4\Rightarrow \frac{1}{2}(\mathrm{n}-1)=4-\frac{1}{2}=\frac{7}{2}\Rightarrow \mathrm{n}=8$
Maximum value of y = n = 8