Let n ≥ 2 be a natural number and f : [0, 1] → ℝ be the function defined by f(x) = n(1 - 2nx) if 0 ≤

Q: Wait, why do we split the area into triangles? Can't we just integrate the function directly?

You absolutely could integrate! However, because $f(x)$ is a piecewise linear function, the graph is composed of straight lines. Calculating the area of triangles is geometrically intuitive and much faster than performing four separate definite integrals. In JEE Advanced, recognizing these geometric shortcuts saves precious time.

Q: Is the function continuous at the boundaries like $x = \frac{1}{2n}$?

Great observation! Let's check: at $x = \frac{1}{2n}$, the first piece $n(1 - 2nx)$ gives $n(1 - 1) = 0$. The second piece $2n(2nx - 1)$ gives $2n(1 - 1) = 0$. Yes, the function is continuous. This confirms our geometric interpretation of the triangles is solid.

Q: Why is the maximum value of the function $n$?

If you look at the peaks of the triangles we plotted, the function reaches its highest point at $n$ in the first, second, and third intervals. Since $n$ is a natural number $\ge 2$, and the function values at the peaks are $n$, the maximum value is indeed $n$. Once you solve for $n=8$, the maximum value is simply 8.

Question

Let $n \ge 2$ be a natural number and $f : [0, 1] 	o \mathbb{R}$ be the function defined by $f(x) = \begin{cases} n(1 - 2nx) & 	ext{if } 0 \le x \le \frac{1}{2n} \ 2n(2nx - 1) & 	ext{if } \frac{1}{2n} \le x \le \frac{3}{4n} \ 4n(1 - nx) & 	ext{if } \frac{3}{4n} \le x \le \frac{1}{n} \ \frac{n}{n-1}(nx - 1) & 	ext{if } \frac{1}{n} \le x \le 1 \end{cases}$. 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

Sigma Percentile · Accepted Answer

### The Geometry of Piecewise Functions

Welcome, fellow explorer of the mathematical landscape! Today, we are going to dismantle a seemingly intimidating piecewise function and reveal the elegant geometry hiding beneath.

When you first look at a function defined in four different pieces like $f(x)$, it is natural to feel a bit overwhelmed. But remember, in the world of JEE Advanced, complexity is often just a mask for simplicity. Let us peel back that mask together.

### Phase 1: Mapping the Terrain

Our function $f(x)$ is defined on the interval $[0, 1]$. To understand it, we must walk through each interval one by one. Let us visualize the path:

1. **The First Descent:** On $[0, \frac{1}{2n}]$, we have $f(x) = n(1 - 2nx)$. At $x=0$, $f(0) = n$. At $x=\frac{1}{2n}$, $f(\frac{1}{2n}) = 0$. This is a straight line sloping downwards from $n$ to $0$.

2. **The First Ascent:** On $[\frac{1}{2n}, \frac{3}{4n}]$, we have $f(x) = 2n(2nx - 1)$. At $x=\frac{1}{2n}$, $f(\frac{1}{2n}) = 0$. At $x=\frac{3}{4n}$, $f(\frac{3}{4n}) = n$. This line climbs back up to $n$.

3. **The Second Descent:** On $[\frac{3}{4n}, \frac{1}{n}]$, we have $f(x) = 4n(1 - nx)$. At $x=\frac{3}{4n}$, $f(\frac{3}{4n}) = n$. At $x=\frac{1}{n}$, $f(\frac{1}{n}) = 0$. Another descent back to the x-axis.

4. **The Final Climb:** On $[\frac{1}{n}, 1]$, we have $f(x) = \frac{n}{n-1}(nx - 1)$. At $x=\frac{1}{n}$, $f(\frac{1}{n}) = 0$. At $x=1$, $f(1) = n$. A final climb to the peak.

### Phase 2: The Geometry of Area

Now that we have mapped the terrain, the area bounded by the curve and the x-axis becomes clear. It is not a chaotic mess; it is a series of three distinct triangles.

- **Triangle 1:** Formed by the first two intervals. The first interval $[0, \frac{1}{2n}]$ is a triangle with base $\frac{1}{2n}$ and height $n$. Its area is:
$$A_1 = \frac{1}{2} \cdot \frac{1}{2n} \cdot n = \frac{1}{4}$$

- **Triangle 2:** Formed by the next two intervals $[\frac{1}{2n}, \frac{1}{n}]$. The base is $\frac{1}{n} - \frac{1}{2n} = \frac{1}{2n}$. The height is $n$. Its area is:
$$A_2 = \frac{1}{2} \cdot \frac{1}{2n} \cdot n = \frac{1}{4}$$

- **Triangle 3:** Formed by the final interval $[\frac{1}{n}, 1]$. The base is $1 - \frac{1}{n} = \frac{n-1}{n}$. The height is $n$. Its area is:
$$A_3 = \frac{1}{2} \cdot \frac{n-1}{n} \cdot n = \frac{n-1}{2}$$

### Phase 3: The Grand Synthesis

We are almost at the finish line. The total area $A$ is the sum of these three triangles:

$$A = A_1 + A_2 + A_3 = \frac{1}{4} + \frac{1}{4} + \frac{n-1}{2}$$

Simplifying this, we get:

$$A = \frac{1}{2} + \frac{n-1}{2} = \frac{n}{2}$$

The problem tells us that this total area is exactly 4. So, we set our expression equal to 4:

$$\frac{n}{2} = 4 \Rightarrow n = 8$$

### Conclusion: The Final Victory

We have found $n=8$. The question asks for the maximum value of the function $f(x)$.

Looking back at our graph, the function peaks at $n$ in several places. Since $n=8$, the **maximum value of the function is 8**. You have successfully navigated the piecewise function, visualized the geometry, and solved for the unknown.

Enter Numerical Value:

Visualized Solution

Understanding the Function

First Interval $[0, \frac{1}{2 n}]$

Second Interval $[\frac{1}{2 n}, \frac{3}{4 n}]$

Third Interval $[\frac{3}{4 n}, \frac{1}{n}]$

Fourth Interval $[\frac{1}{n}, 1]$

Visualizing the Bounded Area

Area of First Triangle ( $A_{1}$ )

Area of Second Triangle ( $A_{2}$ )

Area of Third Triangle ( $A_{3}$ )

Summing the Areas

Solving for $n$

Maximum Value of $f (x)$

The Sigma Insight: Area Bounded by Curves

The Geometry of Piecewise Functions

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

Let $f (x) =$ Maximum ${x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ , where $0 \leq x \leq 1$ . Determine the area of the region bounded by the curves $y = f (x), x$ -axis, $x = 0$ and $x = 1$ .

Let $f (x)$ be a non-negative continuous function such that the area bounded by the curve $y = f (x)$ , x-axis and the ordinates $x = π /4$ and $x = β > π /4$ is $[β sin β + \frac{π}{4} cos β + 2 β]$ . Then $f (π /2)$ is

Find all maxima and minima of the function $y = x (x - 1)^{2}, 0 \leq x \leq 2$ . Also determine the area bounded by the curve $y = x (x - 1)^{2}$ , the $y$ -axis and the line $y = 2$ .

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f (x) = min {x^{2} + \frac{3}{4}, 1 + [x]}$ where $[x]$ denotes the greatest integer $\leq x$ , be $A$ . Then the value of $12 A$ is

If the area of the region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ is equal to $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ , then the natural number $n$ is equal to _______

One of the points of intersection of the curves $y = 1 + 3 x - 2 x^{2}$ and $y = \frac{1}{x}$ is $(\frac{1}{2}, 2)$ . Let the area of the region enclosed by these curves be $\frac{1}{24} (l 5 + m) - n lo g_{e} (1 + 5)$ , where $l, m, n \in N$ . Then $l + m + n$ is equal to

If the area of the bounded region $R = {(x, y) : max {0, lo g_{e} x} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2}$ is, $α (lo g_{e} 2)^{- 1} + β (lo g_{e} 2) + γ$ , then the value of $(α + β - 2 γ)^{2}$ is equal to :

The area of the region bounded by the curve $y = max {∣ x ∣, x ∣ x - 2∣}$ , then x-axis and the lines $x = - 2$ and $x = 4$ is equal to _______ .

The sum of squares of all possible values of $k$ , for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to :

If the area of the region bounded by the curves $y = 4 - \frac{x ^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals

Enter Numerical Value:

Visualized Solution

Understanding the Function

First Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,2n1​]

Second Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [2n1​,4n3​]

Third Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [4n3​,n1​]

Fourth Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [n1​,1]

Visualizing the Bounded Area

Area of First Triangle ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​)

Area of Second Triangle ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​)

Area of Third Triangle ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​)