The function f(x) = ∫_{-1}^x t(e^t - 1)(t - 1)(t - 2)³(t - 3)⁵ dt has a local minimum at x = (A) 0 (

Q: Wait, why do we ignore the integral sign entirely? Shouldn't we try to integrate this monster first?

That is a classic trap! If you try to integrate that expression, you will be stuck in a nightmare of integration by parts for hours. The question asks for the local minimum of $f(x)$, which depends on $f'(x)$. By the Leibniz Rule, $f'(x)$ is simply the integrand itself. Always look for the shortcut that bypasses the brute-force calculation!

Q: Why does the term $x(e^x - 1)$ not cause a sign change at $x=0$?

Great observation! If $x > 0$, then $e^x > 1$, so $x(e^x - 1)$ is positive. If $x < 0$, then $e^x < 1$, so $(e^x - 1)$ is negative. Since $x$ is also negative, the product of two negatives is positive. Because the expression is positive on both sides of $x=0$, it acts like a squared term—it doesn't flip the sign of the derivative.

Q: How do I know if a point is a local minimum or a local maximum?

Think of the slope. If the derivative $f'(x)$ changes from negative to positive, the function was decreasing and then starts increasing—that is a valley, or a local minimum. If it changes from positive to negative, the function was increasing and then starts decreasing—that is a peak, or a local maximum.

Q: Does the power of the factors matter in the Wavy Curve Method?

Absolutely! If a factor $(x-a)^n$ has an odd power $n$, the sign of the expression will flip as you cross $x=a$. If the power $n$ is even, the sign remains the same on both sides. Here, the powers 1, 3, and 5 are all odd, so they all trigger sign changes.

Q: Is it possible for $f'(x)$ to be zero but not be a local extremum?

Yes, exactly! This happens at $x=0$ in this problem. Even though $f'(0) = 0$, the function doesn't change direction there because the sign of $f'(x)$ remains negative on both sides. This is called a point of inflection, not a local extremum.

Question

The function $f(x) = \int_{-1}^x t(e^t - 1)(t - 1)(t - 2)^3(t - 3)^5 dt$ has a local minimum at $x =$

Options:

(A) 0

(B) 1

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing before a massive, intimidating wall of symbols. You see an integral, a product of polynomials, and an exponential function all tangled together: $$f(x) = \int_{-1}^x t(e^t - 1)(t - 1)(t - 2)^3(t - 3)^5 dt$$ Your instinct might be to panic, to reach for a pen and try to integrate this expression. But stop. Take a deep breath. In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions. We don't need to solve the integral; we only need to understand its behavior. ### The Leibniz Key To find the local minima of $f(x)$, we need to know where the function stops decreasing and starts increasing. This is the domain of the derivative, $f'(x)$. How do we differentiate an integral with a variable upper limit? We use the **Leibniz Rule**. It tells us that the derivative of the integral is simply the integrand itself, evaluated at the upper limit. It is as if the integral sign and the derivative operator cancel each other out, leaving us with the core of the problem: $$f'(x) = x(e^x - 1)(x - 1)(x - 2)^3(x - 3)^5$$ Suddenly, the "monster" has been tamed. We are no longer looking at an integral; we are looking at a polynomial-like function whose roots are easy to identify: $x = 0, 1, 2, 3$. ### The Hidden Trap: The Sign of $x(e^x - 1)$ Now, we must determine where $f'(x)$ changes sign. This is where most students stumble. Look closely at the term $x(e^x - 1)$. If you test a positive value, say $x=1$, you get $1(e^1 - 1) > 0$. If you test a negative value, say $x=-1$, you get $(-1)(e^{-1} - 1)$. Since $e^{-1} \approx 0.36$, the term $(e^{-1} - 1)$ is negative. A negative times a negative is a positive! This means $x(e^x - 1)$ is always non-negative. It never crosses the x-axis to become negative; it just touches it at $x=0$ and bounces back. Therefore, $x=0$ is not a point where the sign of $f'(x)$ changes. It is a point of inflection, not a local extremum. ### The Wavy Curve Dance With that trap neutralized, we focus on the remaining factors: $(x - 1)(x - 2)^3(x - 3)^5$. We use the Wavy Curve Method, starting from the rightmost interval ($x > 3$). Here, all factors are positive, so $f'(x) > 0$. As we move left and cross each root, we check the power of the factor. Since the powers of $(x-1)$, $(x-2)$, and $(x-3)$ are all odd (1, 3, and 5 respectively), the sign of $f'(x)$ will flip at each of these points: * For $x > 3$, $f'(x) > 0$. * At $x = 3$, we cross into the interval $(2, 3)$, and the sign flips to $f'(x) < 0$. * At $x = 2$, we cross into $(1, 2)$, and the sign flips to $f'(x) > 0$. * At $x = 1$, we cross into $(0, 1)$, and the sign flips to $f'(x) < 0$. ### The Moment of Truth A local minimum occurs when the slope changes from negative to positive—the function goes down, hits a bottom, and starts climbing back up. Looking at our sign analysis: 1. At $x = 1$, the sign changes from negative to positive. This is a **local minimum**. 2. At $x = 2$, the sign changes from positive to negative. This is a **local maximum**. 3. At $x = 3$, the sign changes from negative to positive. This is another **local minimum**. And there you have it. By refusing to be intimidated by the integral and carefully analyzing the sign changes of the derivative, we have navigated the trap and arrived at the solution: **$x = 1$ and $x = 3$**. Physics and math are not about brute force; they are about seeing the structure beneath the surface.

Animated Solution for Mathematics - Differentiation: The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

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Visualized Solution

Analyzing the Integral Function

Newton-Leibniz Rule

Finding $f^{'} (x)$

Identifying Critical Points

Analyzing the Trap: $x (e^{x} - 1)$

Wavy Curve Method

Marking the Signs

Finding Local Minima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Leibniz Key

The Hidden Trap

The Wavy Curve Dance

The Moment of Truth

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=∫−1x​t(et−1)(t−1)(t−2)3(t−3)5dt has a local minimum at x=