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We are given a piecewise function $f : [0, 1] \to R$ defined for $n \geq 2$ as:

f (x) = {\begin{cases} n (1 - 2 n x) & if 0 \leq x \leq \frac{1}{2 n} \\ 2 n (2 n x - 1) & if \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\ 4 n (1 - n x) & if \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n - 1} (n x - 1) & if \frac{1}{n} \leq x \leq 1 \end{cases}

We need to find the maximum value of $f (x)$ given that the area bounded by $x = 0$ , $x = 1$ , $y = 0$ , and $y = f (x)$ is 4.

The function $f (x)$ is defined in four intervals:

First Interval ( $0 \leq x \leq \frac{1}{2 n}$ ): Linear decrease from $f (0) = n$ to $f (\frac{1}{2 n}) = 0$ .
Second Interval ( $\frac{1}{2 n} \leq x \leq \frac{3}{4 n}$ ): Linear increase from $f (\frac{1}{2 n}) = 0$ to $f (\frac{3}{4 n}) = n$ .
Third Interval ( $\frac{3}{4 n} \leq x \leq \frac{1}{n}$ ): Linear decrease from $f (\frac{3}{4 n}) = n$ to $f (\frac{1}{n}) = 0$ .
Fourth Interval ( $\frac{1}{n} \leq x \leq 1$ ): Linear increase from $f (\frac{1}{n}) = 0$ to $f (1) = \frac{n}{n - 1} (n - 1) = n$ .

The area $A$ under $f (x)$ from $x = 0$ to $x = 1$ is the sum of the areas in each interval:

First Interval: Triangle with base $\frac{1}{2 n}$ and height $n$ .
$A_{1} = \frac{1}{2} \times \frac{1}{2 n} \times n = \frac{1}{4}$
Second Interval: Triangle with base $\frac{3}{4 n} - \frac{1}{2 n} = \frac{1}{4 n}$ and height $n$ .
$A_{2} = \frac{1}{2} \times \frac{1}{4 n} \times n = \frac{1}{8}$
Third Interval: Triangle with base $\frac{1}{n} - \frac{3}{4 n} = \frac{1}{4 n}$ and height $n$ .
$A_{3} = \frac{1}{2} \times \frac{1}{4 n} \times n = \frac{1}{8}$
Fourth Interval: Triangle with base $1 - \frac{1}{n}$ and height $n$ .
$A_{4} = \frac{1}{2} \times (1 - \frac{1}{n}) \times n = \frac{n - 1}{2}$

Total Area:

A = A_{1} + A_{2} + A_{3} + A_{4} = \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{n - 1}{2} = \frac{1}{2} + \frac{n - 1}{2} = \frac{n}{2}

Given $A = 4$ :

\frac{n}{2} = 4 ⟹ n = 8

From the function definition, the maximum value occurs at $x = \frac{3}{4 n}$ and $x = 1$ :

f (\frac{3}{4 n}) = n and f (1) = n

For $n = 8$ :

f_{max} = 8

The maximum value of the function $f$ is $8$ .