Let the domain of the function f(x) = log₂ log₄ log₆ (3 + 4x - x²) be (a, b). If ∫₀^{b-a} [x²] dx =

Q: Wait, why do we use the power rule $A > b^C$ instead of just setting the argument greater than zero for every log?

That is a fantastic question! Setting the argument greater than zero is the *first* step to define the domain, but solving the inequality requires us to respect the *value* of the logarithm. When we have $\log_b(A) > C$, we are not just saying $A > 0$; we are saying $A$ must be large enough to produce a value greater than $C$. Since the log function is strictly increasing (for base $> 1$), we exponentiate both sides: $b^{\log_b(A)} > b^C$, which simplifies to $A > b^C$. It is the only way to capture the full constraint imposed by the nested structure.

Q: Why do we split the integral at $\sqrt{2}$ and $\sqrt{3}$? Can't we just integrate $[x^2]$ as a constant?

I understand the temptation! However, the greatest integer function $[x^2]$ is not a constant; it is a step function. It remains constant only between the points where $x^2$ hits an integer. Since $x^2$ changes continuously as $x$ increases, the value of $[x^2]$ 'jumps' at specific values of $x$. We must find where $x^2 = 1, 2, 3$ to identify these jump points. If we don't split the integral at these exact points, we would be treating a staircase as a flat line, which would lead to a completely incorrect area calculation.

Q: Is it possible to have a negative value for the integral? My result has negative signs.

Do not panic! The presence of negative signs in your final expression $p - \sqrt{q} - \sqrt{r}$ does not mean the integral itself is negative. The integral represents an area, which must be positive. If you evaluate $5 - \sqrt{2} - \sqrt{3}$, you will find it is approximately $5 - 1.414 - 1.732 \approx 1.854$, which is positive. The form $p - \sqrt{q} - \sqrt{r}$ is just a specific algebraic representation required by the problem. Trust your algebra; the structure is designed to fit this form perfectly.

Question

Let the domain of the function $f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2)$ be $(a, b)$. If $\int_0^{b-a} [x^2] dx = p - \sqrt{q} - \sqrt{r}, p, q, r \in \mathbb{N}, 	ext{gcd}(p, q, r) = 1$, where [.] is the greatest integer function, then $p + q + r$ is equal to

Options:

(A) 10

(B) 8

(C) 11

(D) 9

Sigma Percentile · Accepted Answer

### The Architecture of Nested Logs: Peeling the Onion Welcome, fellow traveler on the path to JEE Advanced excellence. Today, we are not just solving a problem; we are dissecting a mathematical structure. We are faced with a function that looks like a fortress: $$f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2)$$ It is a nested logarithmic function, a classic test of your ability to maintain composure when faced with complexity. Our mission is to find the domain $(a, b)$ and then evaluate a definite integral involving the greatest integer function. Let us begin. ### Phase 1: The Logarithmic Onion Imagine this function as a set of nested Russian dolls. To reach the core—the quadratic expression $3 + 4x - x^2$—we must peel away the layers one by one. The fundamental rule of logarithms is that for $\log_b(y)$ to be defined, $y$ must be strictly greater than zero. We are looking for the domain, which implies we need the function to be defined at every step. We start with the outermost layer: $\log_2( ext{something}) > 0$. For this to be defined, the argument must be positive. We peel the first layer: $$\log_4 \log_6 (3 + 4x - x^2) > 0$$ Using the property that if $\log_b(A) > C$, then $A > b^C$, we transform this into $\log_6 (3 + 4x - x^2) > 4^0$, which simplifies to $\log_6 (3 + 4x - x^2) > 1$. Now, we repeat the process for the second layer. Again, we apply the exponentiation rule: $3 + 4x - x^2 > 6^1$. This is the moment where the abstract logarithmic problem collapses into a concrete, familiar quadratic inequality: $$3 + 4x - x^2 > 6$$ ### Phase 2: The Quadratic Threshold Now, let us bring this quadratic to life. Rearranging the terms to keep the $x^2$ coefficient positive, we get $x^2 - 4x + 3 < 0$. This is a beautiful, factorable quadratic. We are looking for the values of $x$ where the parabola $y = x^2 - 4x + 3$ dips below the x-axis. Factoring gives us: $$(x - 1)(x - 3) < 0$$ Visualize the parabola. It opens upward, crossing the x-axis at $x = 1$ and $x = 3$. The inequality tells us we are interested in the region *between* these roots. Thus, our domain is $(1, 3)$. We have found our $a = 1$ and $b = 3$. The fortress has been breached, and the domain is revealed. ### Phase 3: The Step Function Dance With $a = 1$ and $b = 3$, the upper limit of our integral becomes $b - a = 2$. We are tasked with evaluating: $$I = \int_0^2 [x^2] \, dx$$ This is where many students stumble, but you will not. The greatest integer function $[x^2]$ is a step function; it remains constant only as long as $x^2$ stays between two consecutive integers. We are integrating from $x = 0$ to $x = 2$. In this range, $x^2$ ranges from $0$ to $4$. The integers in this range are $1, 2,$ and $3$. Therefore, the function $[x^2]$ will jump at the values of $x$ where $x^2 = 1, x^2 = 2,$ and $x^2 = 3$. These critical points are $x = 1, x = \sqrt{2},$ and $x = \sqrt{3}$. We must break our integral into four distinct segments based on these points: 1. From $0$ to $1$: $[x^2] = 0$ 2. From $1$ to $\sqrt{2}$: $[x^2] = 1$ 3. From $\sqrt{2}$ to $\sqrt{3}$: $[x^2] = 2$ 4. From $\sqrt{3}$ to $2$: $[x^2] = 3$ ### Phase 4: The Final Calculation Now, we simply calculate the area of these rectangles. The integral becomes: $$I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx$$ Evaluating these is straightforward: $$I = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$$ Let us expand this carefully. Precision is the hallmark of a master: $$I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$$ Grouping the terms: - Integers: $6 - 1 = 5$ - $\sqrt{2}$ terms: $\sqrt{2} - 2\sqrt{2} = -\sqrt{2}$ - $\sqrt{3}$ terms: $2\sqrt{3} - 3\sqrt{3} = -\sqrt{3}$ We arrive at $I = 5 - \sqrt{2} - \sqrt{3}$. Comparing this to the form $p - \sqrt{q} - \sqrt{r}$, we identify $p = 5, q = 2, r = 3$. The condition $ ext{gcd}(5, 2, 3) = 1$ is satisfied. Finally, the sum **$p + q + r = 5 + 2 + 3 = 10$**.

Select Answer:

Visualized Solution

Analyze the Function $f (x)$

Outer Log Condition

Middle Log Condition

Solving the Quadratic Inequality

Identify $a$ and $b$

Setup the Integral

Critical Points for $[x^{2}]$

Break the Integral

Evaluate the Parts

Simplify the Expression

Compare and Find $p, q, r$

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Architecture of Nested Logs

Phase 1

Phase 2

Phase 3

Phase 4

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