One of the points of intersection of the curves y = 1 + 3x - 2x² and y = 1/x is (1/2, 2). Let the ar

Q: Wait, why do we get a cubic equation? I thought we were just finding intersection points.

That is a great observation! When you equate $1 + 3x - 2x^2 = \frac{1}{x}$, you are essentially looking for the points where the heights of both curves are identical. To solve this, we multiply both sides by $x$ to clear the denominator, which naturally elevates the degree of the equation from a quadratic to a cubic. It is a standard maneuver in coordinate geometry—don't let the degree scare you; it just means there are potentially three intersection points, and we are hunting for the ones that define our region.

Q: How do I handle the $\ln \alpha$ term? It looks messy when $\alpha = \frac{1+\sqrt{5}}{2}$.

I understand the hesitation! When you evaluate the integral, you will encounter $\ln(\frac{1+\sqrt{5}}{2})$. Remember your logarithmic properties: $\ln(\frac{a}{b}) = \ln a - \ln b$. So, $\ln(\frac{1+\sqrt{5}}{2}) = \ln(1+\sqrt{5}) - \ln 2$. When you subtract the lower limit evaluation, which also contains a $\ln 2$ term, you will find that the $\ln 2$ terms cancel out beautifully. Trust the process!

Q: Is there a shortcut for the polynomial evaluation at $\alpha$? Plugging in $\frac{1+\sqrt{5}}{2}$ seems like a nightmare.

You are absolutely right to look for a shortcut! Since $\alpha$ is a root of $x^2 - x - 1 = 0$, we know that $\alpha^2 = \alpha + 1$. You can use this to reduce any power of $\alpha$. For instance, $\alpha^3 = \alpha \cdot \alpha^2 = \alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1$. By substituting these linear expressions, you avoid the messy arithmetic of squaring and cubing the radical. It is the mark of a seasoned JEE aspirant to use the properties of the root rather than brute force.

Question

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

Options:

(A) 29

(B) 31

(C) 30

(D) 32

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not merely solving a calculus problem; we are embarking on a journey to find the hidden geometry between two distinct mathematical entities: the parabola $y = 1 + 3x - 2x^2$ and the hyperbola $y = \frac{1}{x}$.

In the world of JEE Advanced, problems like this are not just about integration; they are about recognizing patterns and simplifying complexity. Let us break this down step by step.

### The Algebraic Hunt

Our first task is to define the boundaries of our region. We are given one intersection point, $(\frac{1}{2}, 2)$, but to find the area, we need the full extent of the enclosed region. We equate the two functions:

$$1 + 3x - 2x^2 = \frac{1}{x}$$

Multiplying by $x$ transforms this into the cubic equation $2x^3 - 3x^2 - x + 1 = 0$. We already know that $x = \frac{1}{2}$ is a root.

By the Factor Theorem, $(2x - 1)$ must be a factor. Performing polynomial division, we find:

$$(2x - 1)(x^2 - x - 1) = 0$$

The quadratic part, $x^2 - x - 1 = 0$, gives us the other roots: $x = \frac{1 \pm \sqrt{5}}{2}$. We are interested in the positive root, the famous Golden Ratio, which we will call $\alpha = \frac{1 + \sqrt{5}}{2}$.

### The Geometry of Integration

Now that we have our limits, $x = \frac{1}{2}$ and $x = \alpha$, we set up the integral for the area $A$. The area is the integral of the upper curve minus the lower curve:

$$A = \int_{1/2}^{\alpha} \left( 1 + 3x - 2x^2 - \frac{1}{x} \right) dx$$

We integrate term by term: $\int 1 dx = x$, $\int 3x dx = \frac{3x^2}{2}$, $\int -2x^2 dx = -\frac{2x^3}{3}$, and $\int -\frac{1}{x} dx = -\ln x$.

Our expression becomes:

$$\left[ x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x \right]_{1/2}^{\alpha}$$

### The Golden Ratio Trick

Instead of plugging in $\alpha = \frac{1+\sqrt{5}}{2}$ directly into the polynomial, we use the property of the root: $\alpha^2 - \alpha - 1 = 0$, which implies $\alpha^2 = \alpha + 1$.

Consequently, $\alpha^3 = \alpha(\alpha^2) = \alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1$. By substituting these into our evaluated expression, the polynomial part simplifies beautifully to $\frac{7}{6}\alpha + \frac{5}{6} - \ln \alpha$.

When we evaluate at the lower limit $x = \frac{1}{2}$, we get $\frac{19}{24} + \ln 2$. Subtracting the lower limit from the upper limit, and remembering that $\ln \alpha = \ln(\frac{1+\sqrt{5}}{2}) = \ln(1+\sqrt{5}) - \ln 2$, we see the $\ln 2$ terms vanish.

### Final Calculation

After careful algebraic simplification, we arrive at the final form:

$$A = \frac{14\sqrt{5} + 15}{24} - \ln(1 + \sqrt{5})$$

Comparing this to the required form $\frac{1}{24}(l\sqrt{5} + m) - n \ln(1 + \sqrt{5})$, we identify $l = 14$, $m = 15$, and $n = 1$.

The sum is:

**$l + m + n = 14 + 15 + 1 = 30$**

See how the complexity collapsed into a simple integer? That is the beauty of mathematics. You didn't just calculate an area; you navigated a path through cubic roots and logarithmic identities to find a singular, elegant truth.

Select Answer:

Visualized Solution

Visualizing the Curves

Finding Intersection Points

Forming the Cubic Equation

Factorizing the Cubic Equation

Solving for Other Roots

Setting up the Area Integral

Performing the Integration

Simplifying with the Golden Ratio

Evaluating at Upper Limit

Evaluating at Lower Limit

Combining and Final Calculation

The Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Algebraic Hunt

The Geometry of Integration

The Golden Ratio Trick

Final Calculation

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