If the line x = a divides the area of region R = {(x, y) ∈ ℝ² : x³ ≤ y ≤ x, 0 ≤ x ≤ 1} into two equa

Q: Wait, why is the upper curve $y=x$ and not $y=x^3$? How do I know for sure?

Great question! Always test a point in the interval $(0, 1)$. If you pick $x = 0.5$, then $x = 0.5$ and $x^3 = 0.125$. Since $0.5 > 0.125$, the line $y=x$ is clearly the 'ceiling' of your region.

Q: Why did we reject the root $a^2 = 1 + \frac{\sqrt{2}}{2}$?

Remember that our region is defined for $x \in [0, 1]$. If $a^2 > 1$, then $a$ would be greater than $1$, which falls outside the boundaries of our region. We must always check if our mathematical solutions respect the physical constraints of the problem.

Q: Can I just guess the value of $a$ by looking at the graph?

Visual estimation is a great way to check your work, but in JEE Advanced, precision is key. The algebra confirms that $a^2 \approx 0.293$, which is slightly larger than $0.25$, confirming $a > 0.5$.

Q: Is it possible to solve this using symmetry?

In this specific case, the curves $y=x$ and $y=x^3$ are not symmetric about a line that would make the area calculation trivial. Integration is the most robust and reliable path here.

Q: Why do we multiply the equation by 8 in the middle of the steps?

That is just a tactical move to clear the denominators! It makes the polynomial $2a^4 - 4a^2 + 1 = 0$ much easier to handle using the quadratic formula.

Question

If the line $x = a$ divides the area of region $R = \{(x, y) \in \mathbb{R}^2 : x^3 \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts, then

Options:

(A) $0 < a < \frac{1}{2}$

(B) $\frac{1}{2} < a < 1$

(C) $2a^4 – 4a^2 + 1 = 0$

(D) $a^4 + 4a^2 – 1 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the first quadrant of the Cartesian plane. You have two paths before you: a straight, honest line defined by $y = x$ and a more aggressive, curving path defined by $y = x^3$.

These two curves trap a small, leaf-like region $R$ between them as they travel from the origin $(0, 0)$ to the point $(1, 1)$. Our mission is to find the vertical 'knife' $x = a$ that slices this leaf into two pieces of exactly equal area.

### Measuring the Total Area

Before we can slice the region, we must know its total size. We define the area $A$ as the integral of the 'height' of the region at any point $x$.

Since the line $y = x$ sits above the curve $y = x^3$ for all $x$ in the interval $(0, 1)$, the height is simply $(x - x^3)$. We set up our integral:

$$
A = \int_{0}^{1} (x - x^3) \,dx
$$

Integrating this is a beautiful exercise in power rules. We calculate:

$$
A = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \left( \frac{1}{2} - \frac{1}{4} \right) - 0 = \frac{1}{4}
$$

So, the total area of our leaf is exactly $\frac{1}{4}$ square units. Now, we need to find the line $x = a$ such that the area from $0$ to $a$ is exactly half of this, which is $\frac{1}{8}$.

### The Slicing Equation

Now, let us place our vertical knife at $x = a$. The area of the left piece, $A_1$, is given by the integral from $0$ to $a$:

$$
A_1 = \int_{0}^{a} (x - x^3) \,dx = \frac{a^2}{2} - \frac{a^4}{4}
$$

We demand that $A_1 = \frac{1}{2} A$. Substituting our values, we get:

$$
\frac{a^2}{2} - \frac{a^4}{4} = \frac{1}{8}
$$

To make this equation look elegant, we multiply everything by $8$ to clear the denominators, yielding the biquadratic equation:

$$
2a^4 - 4a^2 + 1 = 0
$$

This is a moment of triumph! We have successfully translated a geometric requirement into a clean algebraic form. This matches option **[C]** perfectly.

### Finding the Value of $a$

To solve for $a$, we treat this as a quadratic in terms of $u = a^2$. The equation becomes $2u^2 - 4u + 1 = 0$.

Using the quadratic formula, we find:

$$
u = \frac{4 \pm \sqrt{16 - 8}}{4} = 1 \pm \frac{\sqrt{2}}{2}
$$

We have two potential values for $a^2$. However, we must be physically grounded. Since our region exists only between $x=0$ and $x=1$, $a^2$ must be less than $1$.

The value $1 + \frac{\sqrt{2}}{2}$ is clearly greater than $1$, so we reject it. We are left with $a^2 = 1 - \frac{\sqrt{2}}{2} \approx 0.293$.

Since $0.293 > 0.25$ (which is $(\frac{1}{2})^2$), it follows that $a > \frac{1}{2}$. Thus, our dividing line $x=a$ must lie in the interval $(\frac{1}{2}, 1)$.

We have navigated the calculus, mastered the algebra, and respected the geometric constraints. You have successfully sliced the leaf!

Animated Solution for Mathematics - Definite Integration: If the line $x = a$ divides the area of region $R = {(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1}$ into two equal parts, then

Select Answer:

Visualized Solution

Visualizing Region $R$

Identifying Upper and Lower Curves

Setting up the Integral

Integrating the Function

Evaluating Total Area

The Dividing Line $x = a$

Setting up Area of First Half

Evaluating First Half Area

Equating the Areas

Simplifying the Equation

Quadratic Formula Setup

Calculating the Roots

Checking Domain Constraints

Locating $a$ in Intervals

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Measuring the Total Area

The Slicing Equation

Finding the Value of $a$

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line x=a divides the area of region R={(x,y)∈R2:x3≤y≤x,0≤x≤1} into two equal parts, then