Question

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

Given :
$f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\ 2n((2nx-1)\ \ \ \  \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \end{cases}$
x ∈ [0, 1]
f(x) is decreasing in $[0,\frac{1}{2n}]$
Let's see the increase and decrease :
increasing in $[\frac{1}{2n},\frac{3}{4n}]$
decreasing in $[\frac{3}{4n},\frac{1}{n}]$
increasing in $[\frac{1}{n},1]$
The graph is as follows :

f(x) ∈ [0, n]
Area = 4
⇒ n = 8
f(x)_max = n = 8
So, the correct answer is 8.