Let f(x) = ∫_0^{x²} (t² - 8t + 15)/e^t dt, x ∈ R Then the numbers of local maximum and local minimum

Q: Wait, why can't we just integrate the function first?

Great question! If you try to integrate $\frac{t^{2}-8t+15}{e^{t}}$, you'll be stuck in a loop of integration by parts for hours. The beauty of the Leibniz Rule is that it lets us bypass the integration entirely. In JEE, recognizing when *not* to integrate is just as important as knowing how to integrate.

Q: Why can we ignore the $e^{x^{2}}$ in the denominator?

Think about the nature of exponential functions. $e^{u}$ is always positive for any real $u$. Since $x^{2}$ is always non-negative, $e^{x^{2}}$ is always at least $1$. Because it's strictly positive, it cannot change the sign of our derivative. It's just a bystander!

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

Visualize the slope. If the derivative changes from positive to negative, the function was climbing and then started falling—that's a peak, a local maximum. If it changes from negative to positive, it was falling and then started climbing—that's a valley, a local minimum. It's all about the direction of the slope!

Question

Let $f(x)=\int_{0}^{x^{2}}\frac{t^{2}-8t+15}{e^{t}}$ dt, $x\in R$ Then the numbers of local maximum and local minimum points of f, respectively, are:

Options:

(A) 2 and 3

(B) 2 and 2

(C) 3 and 2

(D) 1 and 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex function, $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-8t+15}{e^{t}} dt$. At first glance, it looks like a nightmare.

You might be tempted to reach for your integration toolkit, trying to find the antiderivative of $\frac{t^{2}-8t+15}{e^{t}}$. But stop!

In the world of JEE Advanced, the most powerful tool is often the one that lets you avoid the grunt work. We don't need to integrate; we need to understand how the function *changes*.

### The Secret Weapon: The Leibniz Rule

When you see a function defined as an integral with a variable in the limit, your mind should immediately jump to the Leibniz Rule. This rule is the bridge between the integral and the derivative.

It tells us that to find $f'(x)$, we don't need to solve the integral; we just need to evaluate the integrand at the limits and multiply by the derivative of those limits. Specifically:

$$ \frac{d}{dx} \int_{a(x)}^{b(x)} g(t) dt = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x) $$

In our case, $b(x) = x^{2}$ and $a(x) = 0$. Substituting these into the formula, we get:

$$ f'(x) = \frac{(x^{2})^{2} - 8(x^{2}) + 15}{e^{x^{2}}} \cdot \frac{d}{dx}(x^{2}) - 0 $$

Suddenly, the complexity vanishes, and we are left with a clear, manageable expression:

$$ f'(x) = \frac{x^{4} - 8x^{2} + 15}{e^{x^{2}}} \cdot (2x) $$

### Factorization and the Critical Points

Now, let's look at the numerator. It is a polynomial in $x^{2}$.

If we let $y = x^{2}$, we have $y^{2} - 8y + 15$, which factors beautifully into $(y-3)(y-5)$. Translating this back to $x$, our derivative becomes:

$$ f'(x) = \frac{2x(x^{2}-3)(x^{2}-5)}{e^{x^{2}}} $$

To find the local maxima and minima, we set $f'(x) = 0$. Since $e^{x^{2}}$ is always positive, it doesn't affect the sign or the roots.

We only care about $2x(x^{2}-3)(x^{2}-5) = 0$. This gives us five critical points: $x = 0$, $x = \sqrt{3}$, $x = -\sqrt{3}$, $x = \sqrt{5}$, and $x = -\sqrt{5}$.

### The Wavy Curve: Visualizing the Peaks and Valleys

Now, we plot these points on the number line: $-\sqrt{5}, -\sqrt{3}, 0, \sqrt{3}, \sqrt{5}$. These points divide the real line into six intervals.

Using the Wavy Curve method, we check the sign of $f'(x)$ in each interval. For $x > \sqrt{5}$, all factors are positive, so $f'(x) > 0$.

As we cross each root (all of which have odd multiplicity), the sign alternates: $+,-,+,-,+,-$.

A local minimum occurs where the slope changes from negative to positive (a valley), and a local maximum occurs where it changes from positive to negative (a peak).

Looking at our signs, we find **local minima at $x = -\sqrt{5}, 0, \sqrt{5}$ (total of 3)** and **local maxima at $x = -\sqrt{3}, \sqrt{3}$ (total of 2)**.

We have successfully navigated the function's landscape without ever performing a single integration! The answer is **2 and 3**.

Animated Solution for Mathematics - Differentiation: Let $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}}$ dt, $x \in R$ Then the numbers of local maximum and local minimum points of f, respectively, are:

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Applying the Leibniz Rule

Substituting the Values

Simplifying $f^{'} (x)$

The Final Derivative Expression

Finding the Critical Points

Plotting on the Number Line

The Wavy Curve Method

Identifying Local Minima

Identifying Local Maxima

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Secret Weapon

Factorization and the Critical Points

The Wavy Curve

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