Let S = (-1, ∞) and f : S → R be defined as f(x) = ∫(-1 to x) (e^t - 1)¹¹ (2t - 1)⁵ (t - 2)⁷ (t - 3)

Q: Wait, why can't we just integrate the function first? It looks like a standard polynomial integral.

That is the classic trap! If you try to expand that expression, you will be stuck for hours. The problem is designed to test your understanding of the Fundamental Theorem of Calculus. We only care about the *slope* of the function, not its explicit form. Newton-Leibniz is the elegant shortcut that saves you from an impossible calculation.

Q: Why does the sign not change at $x=3$?

Great observation! Look at the exponent of the factor $(x-3)$. It is $12$, which is an even number. In the Wavy Curve method, an even power acts like a mirror; the function touches the x-axis but doesn't cross it. Since the sign was negative before $x=3$, it remains negative after $x=3$.

Q: How do I handle the $(e^x - 1)^{11}$ term? Is it different from a polynomial?

It behaves exactly like a polynomial near the root! Since $e^x - 1 = 0$ only at $x=0$, and the function $e^x - 1$ is strictly increasing, it changes sign from negative to positive at $x=0$ just like $(x-0)$ would. The power $11$ is odd, so the sign change is guaranteed.

Question

Let $S = (-1, \infty)$ and $f : S 	o \mathbb{R}$ be defined as $f(x) = \int_{-1}^{x} (e^t - 1)^{11} (2t - 1)^5 (t - 2)^7 (t - 3)^{12} (2t - 10)^{61} dt$. Let $p = $ Sum of square of the values of $x$, where $f(x)$ attains local maxima on $S$. and $q = $ Sum of the values of $x$, where $f(x)$ attains local minima on $S$. Then, the value of $p^2 + 2q$ is _______.

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: A Lesson in Perspective

My dear student, take a deep breath. When you first looked at this problem, I know exactly what went through your mind. You saw that massive integral, that terrifying product of powers, and perhaps a moment of panic set in.

You thought, "How on earth am I supposed to integrate this?"

Here is the first secret of JEE Advanced: **The problem is not asking you to solve the integral; it is asking you to understand the behavior of the function.**

In mathematics, as in life, we often get bogged down by the sheer volume of information. We see a complex expression like

$$f(x) = \int_{-1}^{x} (e^t - 1)^{11} (2t - 1)^5 (t - 2)^7 (t - 3)^{12} (2t - 10)^{61} dt$$

and we want to simplify it. But we don't need to simplify it. We need to differentiate it.

### The Newton-Leibniz Revelation

We invoke the Newton-Leibniz rule, the bridge between the integral and the derivative. It tells us that if we have a function defined as an integral with a variable upper limit, its derivative is simply the integrand itself, evaluated at that variable.

$$f'(x) = (e^x - 1)^{11} (2x - 1)^5 (x - 2)^7 (x - 3)^{12} (2x - 10)^{61}$$

Look at that! In one stroke, the integral has vanished. We are left with a product of factors.

This is the moment where the "impossible" problem becomes a beautiful, structured puzzle. We are no longer doing calculus; we are doing logic.

### The Wavy Curve Odyssey

To find the local maxima and minima, we need to know where the slope, $f'(x)$, changes sign. We need to find the critical points where $f'(x) = 0$.

By setting each factor to zero, we find our key coordinates:

1. $e^x - 1 = 0 \Rightarrow x = 0$
2. $2x - 1 = 0 \Rightarrow x = 0.5$
3. $x - 2 = 0 \Rightarrow x = 2$
4. $x - 3 = 0 \Rightarrow x = 3$
5. $2x - 10 = 0 \Rightarrow x = 5$

Now, imagine these points laid out on a number line. We start from the far right, where $x > 5$. If you pick a number like $100$, every single bracket in our expression for $f'(x)$ is positive. Thus, $f'(x) > 0$. The curve is above the axis.

### The Parity Trap: Why Exponents Matter

This is where the true test of your conceptual clarity begins. As we move from right to left, we cross these critical points. The rule is simple: if the exponent of a factor is **odd**, the sign flips. If the exponent is **even**, the sign stays the same.

Let's walk through this together:

*   **At $x = 5$:** The factor is $(2x - 10)^{61}$. The power is $61$ (odd). The sign flips from positive to negative. We are now in the valley.

*   **At $x = 3$:** The factor is $(x - 3)^{12}$. The power is $12$ (even). **Stop!** Do not flip the sign. The curve touches the axis and bounces back. We remain in the negative territory. This is the trap that catches the unwary.

*   **At $x = 2$:** The factor is $(x - 2)^7$. The power is $7$ (odd). The sign flips from negative to positive. We are climbing a hill.

*   **At $x = 0.5$:** The factor is $(2x - 1)^5$. The power is $5$ (odd). The sign flips from positive to negative. We are descending again.

*   **At $x = 0$:** The factor is $(e^x - 1)^{11}$. The power is $11$ (odd). The sign flips from negative to positive. We are climbing once more.

### The Final Tally

Now, we identify our peaks and valleys. A local maximum occurs when the slope changes from positive to negative (the peak of a hill). Looking at our analysis, this happens at $x = 0$ and $x = 2$.

We are asked for $p$, the sum of the squares of these values:

$$p = 0^2 + 2^2 = 4$$

A local minimum occurs when the slope changes from negative to positive (the bottom of a valley). This happens at $x = 0.5$ and $x = 5$. We are asked for $q$, the sum of these values:

$$q = 0.5 + 5 = 5.5 = \frac{11}{2}$$

Finally, we calculate the value of $p^2 + 2q$:

$$p^2 + 2q = (4)^2 + 2\left(\frac{11}{2}\right) = 16 + 11 = 27$$

There it is. The answer is **27**.

Do you see the elegance? We didn't need to perform a single complex integration. We didn't need to expand a polynomial of degree $100+$. We only needed to understand the nature of the function's derivative and the behavior of its roots.

Keep this perspective, my friend. In JEE Advanced, the most complex-looking problems often yield to the simplest, most fundamental principles. You have the tools; you just need to trust them.

Enter Numerical Value:

Visualized Solution

The Objective and The Tool

Differentiating the Integral

Finding Critical Points

Wavy Curve: $x > 5$

Crossing $x = 5$

The Trap at $x = 3$

Crossing $x = 2$ and $x = 0.5$

Crossing $x = 0$

Identifying Local Maxima ( $p$ )

Identifying Local Minima ( $q$ )

Final Calculation

The Sigma Insight: Maxima and Minima

The Illusion of Complexity

The Newton-Leibniz Revelation

The Wavy Curve Odyssey

The Parity Trap

The Final Tally

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Enter Numerical Value:

Visualized Solution

The Objective and The Tool

Differentiating the Integral

Finding Critical Points

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