Let the function f(x) = x/3 + 3/x + 3, x ≠ 0 be strictly increasing in (-∞, α1) ∪ (α2, ∞) and strict

Q: Wait, why can't I just say the decreasing interval is $(-3, 3)$?

That is a classic trap! While it is true that $f'(x) < 0$ for all $x$ in the interval $(-3, 3)$, the function itself is not defined at $x = 0$. Because the function has a vertical asymptote at $x = 0$, you cannot include it in the interval. You must split the interval at the point of discontinuity, resulting in $(-3, 0) \cup (0, 3)$.

Q: How do I know if the function is increasing or decreasing without plotting?

The derivative $f'(x)$ tells you the slope of the tangent line at any point. If $f'(x) > 0$, the slope is positive, meaning the function is climbing (strictly increasing). If $f'(x) < 0$, the slope is negative, meaning the function is falling (strictly decreasing). By finding the critical points where $f'(x) = 0$ and checking the sign of $f'(x)$ in the intervals between them, you can map the behavior perfectly.

Q: Does the vertical asymptote at $x=0$ affect the derivative?

Absolutely. The derivative is only defined where the function itself is defined. Since $f(x)$ is undefined at $x=0$, $f'(x)$ is also undefined there. This is why $x=0$ acts as a boundary for our intervals. Always check the domain of the original function before you start differentiating!

Question

Let the function $f(x) = \frac{x}{3} + \frac{3}{x} + 3, x 
eq 0$ be strictly increasing in $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $\sum_{i=1}^5 \alpha_i^2$ is equal to :-

Options:

(A) 48

(B) 28

(C) 40

(D) 36

Sigma Percentile · Accepted Answer

### The Geometry of Change: Unlocking the Monotonicity of $f(x)$ Welcome, fellow traveler on the JEE journey! Today, we are going to dissect a function that, at first glance, seems simple, but hides a beautiful, subtle trap. We are looking at the function: $$f(x) = \frac{x}{3} + \frac{3}{x} + 3$$ Our mission is to find where this function climbs and where it falls. ### Phase 1: The Anatomy of the Function Before we dive into the calculus, let us pause and look at the function's structure. The term $\frac{3}{x}$ is our red flag. In the world of real numbers, we cannot divide by zero. Therefore, the domain of our function is $x \in \mathbb{R} \setminus \{0\}$. This means there is a vertical asymptote at $x = 0$. Imagine the graph: it is split into two distinct branches. This is the first piece of our puzzle. ### Phase 2: The Power of the Derivative To understand the monotonicity—the increasing and decreasing behavior—we need to know the slope of the tangent line at any point. We turn to the derivative, $f'(x)$. Using the power rule, we differentiate term by term: $$f'(x) = \frac{d}{dx}\left(\frac{x}{3} ight) + \frac{d}{dx}\left(\frac{3}{x} ight) + \frac{d}{dx}(3)$$ The derivative of $\frac{x}{3}$ is $\frac{1}{3}$. The derivative of $\frac{3}{x}$ (which is $3x^{-1}$) is $-3x^{-2}$, or $-\frac{3}{x^2}$. The constant $3$ vanishes. Thus, we have: $$f'(x) = \frac{1}{3} - \frac{3}{x^2}$$ ### Phase 3: Simplifying the Expression Let us make this expression easier to handle by finding a common denominator: $$f'(x) = \frac{x^2 - 9}{3x^2}$$ Now, look at this expression. The denominator $3x^2$ is always positive for any $x eq 0$. This is a powerful realization! It means the sign of the derivative depends entirely on the numerator, $x^2 - 9$. If $x^2 - 9 > 0$, the function is increasing; if $x^2 - 9 < 0$, the function is decreasing. ### Phase 4: The Critical Points and the Trap We find the critical points by setting $f'(x) = 0$, which means $x^2 - 9 = 0$. This gives us $x = 3$ and $x = -3$. These are the points where the function turns. For $f'(x) > 0$, we need $x^2 > 9$, which happens when $x < -3$ or $x > 3$. So, the function is strictly increasing in $(-\infty, -3) \cup (3, \infty)$. This matches the form $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$, so $\alpha_1 = -3$ and $\alpha_2 = 3$. Now, for the decreasing part, we need $x^2 < 9$, which means $-3 < x < 3$. But wait! We must respect the domain. We cannot include $x = 0$. So, the interval is $(-3, 0) \cup (0, 3)$. This matches the form $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$, giving us $\alpha_3 = -3$, $\alpha_4 = 0$, and $\alpha_5 = 3$. ### Phase 5: The Final Calculation We have our values: $\alpha_1 = -3$, $\alpha_2 = 3$, $\alpha_3 = -3$, $\alpha_4 = 0$, and $\alpha_5 = 3$. The problem asks for the sum of their squares: $$\sum_{i=1}^5 \alpha_i^2 = (-3)^2 + 3^2 + (-3)^2 + 0^2 + 3^2$$ Calculating this, we get: $$9 + 9 + 9 + 0 + 9 = 36$$ And there it is! The elegance of the result lies in how the discontinuity at $x=0$ was not just a hurdle, but a fundamental part of the function's definition. The final answer is **36**.

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Visualized Solution

Analyze the Function $f (x)$

Differentiate $f (x)$

Simplify $f^{'} (x)$

Find Critical Points

Strictly Increasing Intervals

Find $α_{1}$ and $α_{2}$

Strictly Decreasing Intervals

Find $α_{3}, α_{4}, α_{5}$

Calculate the Final Sum

The Sigma Insight: Monotonicity

The Geometry of Change

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Visualized Solution

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