Question
$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)$
- $\frac{\pi}{4}$
- $\frac{\pi}{3}$
- $\frac{\pi}{6}$
- $\frac{\pi}{2}$
$\frac{\pi}{4}$
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$f(x)= \begin{cases}n(1-2 n x) \text { if } 0 \leq x \leq \frac{1}{2 n}2 n(2 n x-1) \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n}4 n(1-n x) \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(n x-1) \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$
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