Let $$f : [1, \infty) \to [2, \infty)$$ be a differentiable function. If $$10\int_{1}^{x} f(t)\,dt = 5xf(x) - x^5 - 9$$ for all $$x \geq 1$$, then the value of $$f(3)$$ is :

The functional equation is given for all $$x \ge 1$$:

$$10\int_{1}^{x} f(t)\,dt = 5x\,f(x) \;-\; x^{5} \;-\; 9 \qquad -(1)$$

Step 1: Differentiate both sides of $$(1)$$ with respect to $$x$$.
• By the Fundamental Theorem of Calculus, $$\dfrac{d}{dx}\Bigl[10\int_{1}^{x} f(t)\,dt\Bigr] = 10\,f(x)$$.
• For the right‐hand side use the product rule on $$5x\,f(x)$$:

$$\dfrac{d}{dx}\!\left[5x\,f(x) - x^{5} - 9\right] = 5\Bigl[f(x) + x\,f'(x)\Bigr] \;-\; 5x^{4}$$

Equating the two derivatives:

$$10f(x) = 5f(x) + 5x\,f'(x) - 5x^{4}$$

Simplify by dividing every term by $$5$$:

$$2f(x) = f(x) + x\,f'(x) - x^{4}$$

Rearrange to obtain the first-order linear differential equation:

$$x\,f'(x) - f(x) = x^{4} \qquad -(2)$$

Step 2: Write $$(2)$$ in standard linear form.

Divide by $$x$$ (valid for $$x \gt 0$$):

$$f'(x) \;-\; \dfrac{1}{x}\,f(x) = x^{3} \qquad -(3)$$

Step 3: Solve $$(3)$$ using an integrating factor.
For a linear ODE $$f' + P(x)f = Q(x)$$, the integrating factor is $$\mu(x)=e^{\int P(x)\,dx}$$.
Here $$P(x)= -\dfrac{1}{x}$$, so

$$\mu(x)= e^{\int -\frac{1}{x}\,dx}=e^{-\ln x}=x^{-1}$$

Multiply $$(3)$$ by $$\mu(x)=x^{-1}$$:

$$\dfrac{f'(x)}{x} \;-\; \dfrac{f(x)}{x^{2}} = x^{2}$$

The left side is the derivative of $$\dfrac{f(x)}{x}$$ because

$$\dfrac{d}{dx}\Bigl[\dfrac{f(x)}{x}\Bigr] = \dfrac{x\,f'(x) - f(x)}{x^{2}}$$

Comparing with $$(2)$$ confirms this fact. Therefore

$$\dfrac{d}{dx}\Bigl[\dfrac{f(x)}{x}\Bigr] = x^{2}$$

Integrate both sides from $$1$$ to $$x$$:

$$\dfrac{f(x)}{x} - \dfrac{f(1)}{1} = \int_{1}^{x} t^{2}\,dt = \left[\dfrac{t^{3}}{3}\right]_{1}^{x} = \dfrac{x^{3}-1}{3}$$

Hence

$$\dfrac{f(x)}{x} = \dfrac{f(1)}{1} + \dfrac{x^{3}-1}{3}$$

Multiply by $$x$$:

$$f(x) = \dfrac{x^{4}}{3} \;+\; Cx \qquad\text{where}\quad C = f(1) - \dfrac{1}{3}$$

Step 4: Find the constant $$C$$ using the condition at $$x=1$$.
Put $$x=1$$ in $$(1)$$:

Left side: $$10\int_{1}^{1} f(t)\,dt = 0$$.
Right side: $$5(1)f(1) - 1^{5} - 9 = 5f(1) - 10$$.

Equate: $$0 = 5f(1) - 10 \;\Longrightarrow\; f(1) = 2$$.

Therefore

$$C = 2 - \dfrac{1}{3} = \dfrac{5}{3}$$

So the explicit form of $$f$$ is

$$f(x) = \dfrac{x^{4}}{3} + \dfrac{5x}{3} = \dfrac{x^{4} + 5x}{3}$$

Step 5: Evaluate $$f(3)$$.

$$f(3) = \dfrac{3^{4} + 5\cdot 3}{3} = \dfrac{81 + 15}{3} = \dfrac{96}{3} = 32$$

Hence $$f(3) = 32$$.

The correct option is Option B.