Let f : [0, 1] → [0, 1] be the function defined by f(x) = x³/3 - x² + 5x/9 + 17/36. Consider the squ

Q: Wait, why do we calculate the total area of the Red region first? Can't we just jump straight to the line $L_h$?

That is a great question! While you could technically set up integrals for the sub-regions, it would be a nightmare of algebraic complexity. By calculating the total area $A_R = \int_0^1 f(x) dx$ first, we anchor our problem with a simple constant value (in this case, $1/2$). This turns the problem from a calculus exercise into a beautiful exercise in algebraic manipulation and geometric symmetry.

Q: Why do we reject the negative root or specific values in Option A?

In Option A, we are looking for a point where $G_{above} = G_{below}$. Since their sum is $1/2$, this implies $G_{above} = 1/4$. However, as $h$ increases, the area $G_{above}$ is strictly decreasing. By checking the boundaries of the interval $[1/4, 2/3]$, we can see that the value of $G_{above}$ never reaches $1/4$. It is not about rejecting a root, but about understanding the range of the function $G_{above}(h)$.

Q: Is there a trick to avoid calculating the integral of $f(x)$?

You cannot avoid the integral entirely, as you need to know the total area of the Red region to establish the relationships between the sub-regions. However, the 'trick' is to realize that once you have $A_R = 1/2$, you don't need to integrate again for the options. You can use the geometric relations $G_{above} + R_{above} = 1 - h$ and $R_{above} + R_{below} = 1/2$ to solve for the conditions in the options using simple algebra.

Question

Let $f : [0, 1] 	o [0, 1]$ be the function defined by $f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36}$. Consider the square region $S = [0, 1] 	imes [0, 1]$. Let $G = \{(x, y) \in S : y > f(x)\}$ be called the green region and $R = \{(x, y) \in S : y < f(x)\}$ be called the red region. Let $L_h = \{(x, h) \in S : x \in [0, 1]\}$ be the horizontal line drawn at a height $h \in [0, 1]$. Then which of the following statements is(are) true?
(A) There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the green region above the line $L_h$ equals the area of the green region below the line $L_h$
(B) There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the red region above the line $L_h$ equals the area of the red region below the line $L_h$
(C) There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the green region above the line $L_h$ equals the area of the red region below the line $L_h$
(D) There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the red region above the line $L_h$ equals the area of the green region below the line $L_h$

Options:

(A) (A)

(B) (B)

(C) (C)

(D) (D)

Sigma Percentile · Accepted Answer

### The Geometry of the Cubic Battlefield

Imagine you are standing on the unit square $S = [0, 1] 	imes [0, 1]$. It is a simple, elegant space.

Inside this square, a cubic curve $f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36}$ carves a path, dividing our square into two distinct territories: the Green region ($G$) above the curve and the Red region ($R$) below it.

At first glance, this looks like a problem that demands brute-force calculus. But as we will see, it is actually a symphony of geometric relationships waiting to be uncovered.

### Phase 1: The Foundation of Area

Before we introduce the horizontal line $L_h$, we must understand the total area of these regions. The Red region is the area under the curve, which we find using the definite integral:

$$A_R = \int_0^1 \left( \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36} ight) dx$$

Applying the power rule of integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get:

$$A_R = \left[ \frac{x^4}{12} - \frac{x^3}{3} + \frac{5x^2}{18} + \frac{17x}{36} ight]_0^1$$

Evaluating this at the limits, we find:

$$A_R = \frac{1}{12} - \frac{1}{3} + \frac{5}{18} + \frac{17}{36}$$

Finding a common denominator of $36$, this simplifies to:

$$\frac{3 - 12 + 10 + 17}{36} = \frac{18}{36} = \frac{1}{2}$$

Since the total area of the unit square is $1$, the Green region $A_G$ must also be $1 - \frac{1}{2} = \frac{1}{2}$. Both regions are perfectly balanced!

### Phase 2: The Slicing Line $L_h$

Now, we introduce the horizontal line $L_h$ at height $y = h$. This line slices our square into two rectangles: a top rectangle of height $1-h$ and a bottom rectangle of height $h$.

This line also partitions our Green and Red regions into four parts: $G_{above}$, $G_{below}$, $R_{above}$, and $R_{below}$. We know two fundamental truths:

$$G_{above} + G_{below} = \frac{1}{2} \quad 	ext{and} \quad R_{above} + R_{below} = \frac{1}{2}$$

Furthermore, the area of the top rectangle is $G_{above} + R_{above} = 1 - h$, and the area of the bottom rectangle is $G_{below} + R_{below} = h$. These are our master keys.

### Phase 3: The Algebraic Magic

Let us test the options. Option A claims $G_{above} = G_{below}$. Since their sum is $1/2$, this would require $G_{above} = 1/4$.

However, as $h$ increases from $1/4$ to $2/3$, the area $G_{above}$ decreases from $1/2$ to $1/3$. It never hits $1/4$, so Option A is incorrect.

Now, look at Option C: $G_{above} = R_{below}$. Instead of solving for these areas, let us look at their difference. From our relations:

$$G_{above} = 1 - h - R_{above} \quad 	ext{and} \quad R_{below} = \frac{1}{2} - R_{above}$$

Subtracting these, we get:

$$G_{above} - R_{below} = (1 - h - R_{above}) - \left( \frac{1}{2} - R_{above} ight) = \frac{1}{2} - h$$

For this to be zero, $h$ must be $1/2$. Since $1/2$ is in our interval, **Option C is correct!**

By using this exact same logic for Option D, we find that:

$$R_{above} - G_{below} = \left( \frac{1}{2} - R_{below} ight) - (h - R_{below}) = \frac{1}{2} - h$$

This also equals zero at $h = 1/2$. Finally, Option B works because at $h = 1/4$, the line is below the curve, making $G_{below} = 0$, which forces $R_{below} = h = 1/4$, satisfying the condition $R_{above} = R_{below} = 1/4$.

The beauty of this problem lies not in the integration, but in the elegant cancellation of terms.

Select Answer:

Visualized Solution

The Unit Square & Regions

Area of Red Region ( $A_{R}$ )

Setting up the Integral

Integrating the Polynomial

Evaluating the Limits

Area of Green Region ( $A_{G}$ )

Introducing the Line $L_{h}$

Fundamental Area Relations

Rectangle Area Relations

Testing Option A

Testing Option B

Testing Option C

Solving for Option C

Testing Option D

Solving for Option D

The Sigma Insight: Area Bounded by Curves

The Geometry of the Cubic Battlefield

Phase 1

Phase 2

Phase 3

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