Let F : [3, 5] → R be a twice differentiable function on (3, 5) such that F(x) = e⁻ˣ ∫(3 to x) (3t²

Q: Wait, why do we reject the negative root here? Or rather, why can't we just use energy conservation?

My dear student, this is not a mechanics problem, so energy conservation doesn't apply here! We are dealing with a pure calculus problem. There are no roots to reject because this is a linear differential equation, not a quadratic one. We are solving for a function $F(x)$, not a variable $x$.

Q: How do we know $F(3) = 0$?

That is a brilliant question! Look at the original definition: $F(x) = e^{-x} \int_3^x (3t^2 + 2t + 4F'(t)) dt$. If you substitute $x = 3$, the integral becomes $\int_3^3 (…) dt$. By the fundamental properties of definite integrals, the integral of any function over an interval of zero length is always zero. Thus, $F(3) = 0$ is our golden key to finding the constant of integration $C$.

Q: Is the integrating factor always this simple?

In JEE Advanced problems, the integrating factor is often designed to be elegant. Whenever you see a fraction like $\frac{e^x}{e^x - 4}$, always check if the numerator is the derivative of the denominator. If it is, the integral is just the natural logarithm of the denominator, and the $e^{\ln(…)}$ structure will simplify beautifully. Always look for that hidden symmetry!

Question

Let $F : [3, 5] 	o \mathbf{R}$ be a twice differentiable function on $(3, 5)$ such that $F(x) = e^{-x} \int_3^x (3t^2 + 2t + 4F'(t)) dt$. If $F'(4) = \frac{\alpha e^\beta - 224}{(e^\beta - 4)^2}$, then $\alpha + \beta$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of calculus.

We have been presented with an integral equation:

$$F(x) = e^{-x} \int_3^x (3t^2 + 2t + 4F'(t)) dt$$

At first glance, it looks like a tangled mess of functions, but I want you to see the beauty beneath the surface. This is a classic setup designed to test your ability to transform a complex integral into a manageable differential equation.

### The Strategic Rearrangement

Our first instinct, when we see that $e^{-x}$ sitting outside the integral, should be to liberate the integral. It is a barrier, a weight holding us back.

By multiplying both sides by $e^x$, we transform the equation into:

$$e^x F(x) = \int_3^x (3t^2 + 2t + 4F'(t)) dt$$

Now, the integral stands alone on the right, and the left side is a product of two functions. We have prepared the battlefield for the next step.

### The Newton-Leibniz Revelation

Now, we differentiate both sides with respect to $x$. On the left, we apply the Product Rule:

$$\frac{d}{dx}[e^x F(x)] = e^x F(x) + e^x F'(x)$$

On the right, we invoke the Newton-Leibniz rule. It is a powerful tool that allows us to differentiate an integral with respect to its upper limit by replacing the dummy variable $t$ with $x$.

Equating the two sides, we get:

$$e^x F(x) + e^x F'(x) = 3x^2 + 2x + 4F'(x)$$

### The LDE Architecture

We are now in the realm of Linear Differential Equations. Let us group the $F'(x)$ terms.

Rearranging, we get:

$$F'(x)(e^x - 4) + e^x F(x) = 3x^2 + 2x$$

To reach the standard form $\frac{dF}{dx} + P(x)F = Q(x)$, we divide by $(e^x - 4)$. This gives us:

$$\frac{dF}{dx} + \left(\frac{e^x}{e^x - 4}\right) F = \frac{3x^2 + 2x}{e^x - 4}$$

### The Integrating Factor Magic

Every LDE has a secret key: the Integrating Factor. We calculate it as $I.F. = e^{\int P(x) dx}$.

Here, $P(x) = \frac{e^x}{e^x - 4}$. Notice the symmetry; the numerator is the derivative of the denominator.

This makes the integral $\int P(x) dx = \ln|e^x - 4|$. Thus, the Integrating Factor is:

$$I.F. = e^{\ln(e^x - 4)} = e^x - 4$$

It is a moment of pure mathematical joy when the terms align so perfectly.

### The Final Stretch

Multiplying the LDE by our $I.F.$, we get:

$$F(x)(e^x - 4) = \int (3x^2 + 2x) dx = x^3 + x^2 + C$$

To find $C$, we use our initial condition $F(3) = 0$, which yields $C = -36$. Our function is:

$$F(x) = \frac{x^3 + x^2 - 36}{e^x - 4}$$

Finally, we apply the Quotient Rule to find $F'(4)$. After careful calculation, we find:

$$F'(4) = \frac{12e^4 - 224}{(e^4 - 4)^2}$$

Comparing this to the given form, we identify $\alpha = 12$ and $\beta = 4$. The sum **$\alpha + \beta = 16$**. You have conquered the problem!

Animated Solution for Mathematics - Differential Equations: Let $F : [3, 5] \to R$ be a twice differentiable function on $(3, 5)$ such that $F (x) = e^{- x} \int_{3}^{x} (3 t^{2} + 2 t + 4 F^{'} (t)) d t$ . If $F^{'} (4) = \frac{α e ^{β} - 224}{( e ^{β} - 4 ) ^{2}}$ , then $α + β$ is equal to

Enter Numerical Value:

Visualized Solution

Rearranging the Integral Equation

Differentiating with Respect to $x$

Identifying the Linear Differential Equation

Calculating the Integrating Factor

Finding the General Solution

Determining the Initial Condition

Expressing $F (x)$ Explicitly

Applying the Quotient Rule

Evaluating $F^{'} (4)$

Comparing and Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Strategic Rearrangement

The Newton-Leibniz Revelation

The LDE Architecture

The Integrating Factor Magic

The Final Stretch

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