Let f : [1, ∞) → ℝ be a differentiable function. If 6 ∫₁ˣ f(t)dt = 3xf(x) + x³ - 4 for all x ≥ 1, th

Q: Why do we differentiate? Can't we just solve for $f(t)$ directly?

You cannot solve for $f(t)$ directly because it is trapped inside the integral. Differentiation is the operator that 'unlocks' the function from the integral sign. By applying the Leibniz Rule, we convert the integral equation into a first-order linear differential equation, which is much easier to solve.

Q: How did you know to divide by $x^2$ in the middle of the problem?

It is about pattern recognition. We arrived at $xf'(x) - f(x) = -x^2$. The quotient rule for $\frac{d}{dx}(\frac{f(x)}{x})$ is $\frac{xf'(x) - f(x)}{x^2}$. Seeing the numerator $xf'(x) - f(x)$, the denominator $x^2$ is the natural missing piece to complete the derivative. Always look for these standard derivative forms!

Q: Why is $f(1)=1$ so important? Can't we just find the general solution?

Without the boundary condition, the constant of integration $C$ remains unknown, and you cannot find the specific value of $f(2) - f(3)$. The integral equation gives us a free point at $x=1$ because the integral from $1$ to $1$ is zero, allowing us to solve for $C$ immediately.

Question

Let $f : [1, \infty) 	o \mathbb{R}$ be a differentiable function. If $6 \int_1^x f(t)dt = 3xf(x) + x^3 - 4$ for all $x \ge 1$, then the value of $f(2) - f(3)$ is

Options:

(A) -4

(B) 3

(C) 4

(D) -3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex integral equation:

$$6 \int_1^x f(t)dt = 3xf(x) + x^3 - 4$$

It looks intimidating, but in the world of JEE Advanced, intimidation is just a mask for elegance. This problem is a beautiful dance between integration and differentiation.

### The Leibniz Revelation

The first thing that should catch your eye is the variable $x$ in the upper limit of the integral. This is a classic signal.

Whenever you see $\int_a^x f(t)dt$, you should immediately think of the **Leibniz Rule**. It is our magic wand. By differentiating both sides with respect to $x$, we strip away the integral.

The left-hand side, $6 \int_1^x f(t)dt$, becomes $6f(x)$.

### The Differential Dance

Now, we turn our attention to the right-hand side: $3xf(x) + x^3 - 4$. We must apply the product rule to $3xf(x)$, which gives us $3f(x) + 3xf'(x)$.

The derivative of $x^3$ is $3x^2$, and the constant $-4$ vanishes into zero. Equating the two sides, we get:

$$6f(x) = 3f(x) + 3xf'(x) + 3x^2$$

Rearranging this, we find $3f(x) - 3xf'(x) = 3x^2$. Dividing by $3$, we arrive at the differential equation:

$$f(x) - xf'(x) = x^2$$

### The Quotient Pattern

Here is the 'Aha!' moment. We have $f(x) - xf'(x) = x^2$. If we multiply by $-1$, we get $xf'(x) - f(x) = -x^2$.

Does this look familiar? It is the numerator of the quotient rule for $\frac{f(x)}{x}$. If we divide both sides by $x^2$, we get:

$$\frac{xf'(x) - f(x)}{x^2} = -1$$

This is exactly the derivative of a quotient:

$$\frac{d}{dx} \left( \frac{f(x)}{x} \right) = -1$$

We have transformed a complex equation into a simple integration problem. Integrating both sides gives us $\frac{f(x)}{x} = -x + C$, or:

$$f(x) = -x^2 + Cx$$

### The Final Anchor

We are almost there, but we have an unknown constant $C$. To find it, we go back to the original equation.

By substituting $x=1$, the integral $\int_1^1 f(t)dt$ becomes zero. This gives us:

$$0 = 3(1)f(1) + 1^3 - 4$$

This simplifies to $3f(1) = 3$, so $f(1) = 1$. Plugging this into our general form $f(x) = -x^2 + Cx$, we get $1 = -1 + C$, meaning **$C = 2$**.

Our function is **$f(x) = -x^2 + 2x$**.

### Final Calculation

Now, calculating $f(2) - f(3)$ is just arithmetic:

$$f(2) = -(2)^2 + 2(2) = 0$$

$$f(3) = -(3)^2 + 2(3) = -3$$

Thus, $f(2) - f(3) = 0 - (-3) = 3$. We have conquered the problem, and the final answer is **$3$**.

Animated Solution for Mathematics - Differential Equations: Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is

Select Answer:

Visualized Solution

Given Equation

Differentiating using Leibniz Rule

Differentiating the Right Hand Side

Equating and Simplifying

Recognizing the Quotient Rule Pattern

Integrating to find $f (x)$

Finding the Initial Condition $f (1)$

Solving for the Constant $C$

Calculating $f (2)$

Calculating $f (3)$

Final Result: $f (2) - f (3)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Leibniz Revelation

The Differential Dance

The Quotient Pattern

The Final Anchor

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Given Equation

Differentiating using Leibniz Rule

Differentiating the Right Hand Side

Equating and Simplifying

Recognizing the Quotient Rule Pattern

Integrating to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding the Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)

Solving for the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3)

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)−f(3)