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Mathematics Question on Methods of Integration

Let n ≥ 2 be a natural number and f:[0,1)]→\rightarrow R be the function defined by
f(x){n(1−2nx)         if 0≤x≤12n 2n((2nx−1)     if 12n≤x≤34n\4n(1−nx)         if 34n≤x≤1n nn−1(nx−1)       if 1n≤x≤1f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\\ 2n((2nx-1)\ \ \ \ \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \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

Answer

Given :
f(x){n(1−2nx)         if 0≤x≤12n 2n((2nx−1)     if 12n≤x≤34n\4n(1−nx)         if 34n≤x≤1n nn−1(nx−1)       if 1n≤x≤1f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\\ 2n((2nx-1)\ \ \ \ \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \end{cases}
x ∈ [0, 1]
f(x) is decreasing in [0,12n][0,\frac{1}{2n}]
Let's see the increase and decrease :
increasing in [12n,34n][\frac{1}{2n},\frac{3}{4n}]
decreasing in [34n,1n][\frac{3}{4n},\frac{1}{n}]
increasing in [1n,1][\frac{1}{n},1]
The graph is as follows :
Graph of the function
f(x) ∈ [0, n]
Area = 4
⇒ n = 8
f(x)max = n = 8
So, the correct answer is 8.