If the absolute maximum value of the function f(x) = (x² - 2x + 7) e^(4x³ - 12x² - 180x + 31) in the

Q: Wait, why do we use the substitution $t = x^2 - 2x$ instead of just differentiating directly?

That is a great question! While you *could* differentiate directly, the expression would become incredibly messy and prone to algebraic errors. In JEE Advanced problems, we often look for patterns. Notice how $x^2 - 2x$ appears in the derivative of the exponent and the original function structure. By substituting $t = x^2 - 2x$, we transform a complex expression into a much simpler polynomial form, allowing us to analyze its sign without getting lost in the algebra.

Q: How do I know for sure that $f'(x) < 0$ for the entire interval $[-3, 0]$?

Excellent skepticism! We broke $f'(x)$ into two parts: $12(t + 7)(t - 15)$ and $2(x - 1)$. For $x \in [-3, 0]$, $t = x^2 - 2x$ ranges from $0$ to $15$. In this range, $(t + 7)$ is always positive, and $(t - 15)$ is non-positive. Thus, the first term is $\le 0$. The second term, $2(x - 1)$, is strictly negative because $x \le 0$. The sum of a non-positive and a strictly negative number is always strictly negative. Therefore, $f'(x) < 0$ is guaranteed.

Q: Why does $f'(x) < 0$ imply the maximum is at $x = -3$?

Think of the derivative as the 'slope' of the function. If the slope is negative ($f'(x) < 0$) everywhere in an interval, it means the function is 'sliding downhill' or strictly decreasing. If you are walking downhill from $x = -3$ to $x = 0$, where would you be at your highest point? Exactly, at the very beginning—the leftmost endpoint, $x = -3$.

Question

If the absolute maximum value of the function $f(x) = (x^2 - 2x + 7) e^{(4x^3 - 12x^2 - 180x + 31)}$ in the interval $[-3, 0]$ is $f(\alpha)$, then :

Options:

(A) $\alpha = 0$

(B) $\alpha = -3$

(C) $\alpha \in (-1, 0)$

(D) $\alpha \in (-3, -1)$

Sigma Percentile · Accepted Answer

### The Monster Function: A Lesson in Perspective

Imagine you are standing before a massive, intimidating mountain. The function $f(x) = (x^2 - 2x + 7) e^{(4x^3 - 12x^2 - 180x + 31)}$ is that mountain.

At first glance, with its quadratic front and that terrifying cubic exponent, it looks like a nightmare. But in the world of JEE Advanced, we do not climb the mountain by brute force; we find the path of least resistance.

Our goal is to find the absolute maximum of this function on the interval $[-3, 0]$. Let us embark on this journey together.

### Phase 1: The Power of Differentiation

When we seek the maximum or minimum of a function, our most reliable compass is the derivative. The derivative, $f'(x)$, tells us the slope.

If the slope is positive, the function climbs; if it is negative, it descends. If we can prove the function is strictly decreasing, the maximum is simply the starting point.

We define $g(x) = 4x^3 - 12x^2 - 180x + 31$, turning our function into $f(x) = (x^2 - 2x + 7) e^{g(x)}$. Applying the product rule, we get:

$$f'(x) = e^{g(x)} [ (x^2 - 2x + 7) g'(x) + (2x - 2) ]$$

### Phase 2: The Elegance of Substitution

We calculate $g'(x) = 12x^2 - 24x - 180$, which factors beautifully into $12(x^2 - 2x - 15)$. Now, look at the structure of our derivative.

We see $x^2 - 2x$ appearing repeatedly. This is a classic JEE trap—if you try to expand everything, you will drown in terms.

Instead, let us use a substitution: $t = x^2 - 2x$. Suddenly, the derivative transforms into:

$$f'(x) = e^{g(x)} [ 12(t + 7)(t - 15) + 2(x - 1) ]$$

The mountain just became a hill.

### Phase 3: The Sign Analysis

Now, we must determine the sign of $f'(x)$ for $x \in [-3, 0]$. First, let us find the range of $t$.

Since $t = x^2 - 2x$, its derivative is $2x - 2$, which is negative for $x \le 0$. This means $t$ is strictly decreasing.

At $x = -3$, $t = 15$. At $x = 0$, $t = 0$. So, $t \in [0, 15]$.

Now, look at our derivative expression again. The term $12(t + 7)(t - 15)$ involves $t \in [0, 15]$.

Since $t+7$ is always positive and $t-15$ is non-positive, this entire term is $\le 0$. The second term, $2(x - 1)$, is strictly negative for $x \in [-3, 0]$.

When you add a non-positive number to a strictly negative number, the result is always strictly negative.

### The Conclusion: A Victory of Logic

We have proven that $f'(x) < 0$ for the entire interval $[-3, 0]$. This means our function is strictly decreasing.

Like a ball rolling down a hill, the highest point must be where it started. Therefore, the absolute maximum occurs at the leftmost endpoint, **$x = -3$**.

We have conquered the mountain not by brute force, but by understanding its shape. Keep this mindset—look for the structure, simplify with substitution, and let the logic guide you to the answer.

Animated Solution for Mathematics - Differentiation: If the absolute maximum value of the function $f (x) = (x^{2} - 2 x + 7) e^{(4 x^{3} - 12 x^{2} - 180 x + 31)}$ in the interval $[- 3, 0]$ is $f (α)$ , then :

Select Answer:

Visualized Solution

Defining the Problem Space

The Strategy: Derivative for Extrema

Applying the Product Rule

Differentiating the Exponent

Assembling the Derivative

Smart Substitution

Analyzing the Range of $t$

Sign Analysis: First Term

Sign Analysis: Second Term

Concluding the Sign of $f^{'} (x)$

Finding the Absolute Maximum

The Sigma Insight: Maxima and Minima

The Monster Function

Phase 1

Phase 2

Phase 3

The Conclusion

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