Let n∈N, n>1 and f:[0,1]→R be the function defined by f(x)={n(1−2nx) if 0≤x≤12n2n(2nx−1) if 12n≤x≤34n4n(1−nx) if 3 4n≤x≤1nnn−1(nx−1) if 1n≤x≤1If n is such that the area of the region bounded by the curves x = 0, x = 1, y = 0 and y = f (x) is 3 square units, then the maximum value of the function f is

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Let $n \in N, n > 1 a n d f : [0, 1] \to R$ be the function defined by
$f (x) = {\begin{cases} n (1 - 2 n x) i f 0 \leq x \leq \frac{1}{2 n} \\ 2 n (2 n x - 1) i f \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\ 4 n (1 - n x) i f \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n - 1} (n x - 1) i f \frac{1}{n} \leq x \leq 1 \end{cases}$
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 3 square units, then the maximum value of the function f is

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answer is 6.

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