Let f:[1,∞)→R Be a Differentiable Function Such That f(1) = 1/3 and 3∫₁ˣ f(t)dt = x f(x) − x³/3. Then the Value of f(e) Is

JEE Advanced 2023, Paper 2. Previous Year Question.

Problem

Let $f:[1,\infty)\to\mathbb R$ be a differentiable function such that $f(1)=\dfrac13$ and $3\displaystyle\int_1^xf(t)\,dt = xf(x)-\dfrac{x^3}{3}$, $x\in[1,\infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is:

Key insight. An equation with an integral on one side and $f(x)$ tangled up on the other is a strong hint to differentiate both sides — this turns the integral equation into an ordinary (if slightly disguised) linear differential equation, solvable with the standard integrating-factor method.

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Approach

Differentiate both sides of the given equation with respect to $x$, using the Fundamental Theorem of Calculus on the left and the product rule on $xf(x)$ on the right. Rearrange into a standard linear differential equation, find the integrating factor, and integrate. Apply the initial condition $f(1)=\dfrac13$ to fix the constant, then evaluate at $x=e$.

Solution

Step 1 — Differentiate both sides

$$\frac{d}{dx}\left[3\int_1^xf(t)\,dt\right] = \frac{d}{dx}\left[xf(x)-\frac{x^3}{3}\right]$$

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

Step 2 — Simplify into a linear ODE

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

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

Dividing by $x$:

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

Step 3 — Find the integrating factor

$$\text{I.F.} = e^{\int -\frac{2}{x}\,dx} = e^{-2\ln x} = x^{-2}$$

Step 4 — Multiply through and integrate

$$\frac{d}{dx}\left[\frac{f(x)}{x^2}\right] = \frac{x}{x^2} = \frac{1}{x}$$

$$\frac{f(x)}{x^2} = \ln x + C$$

Step 5 — Solve for f(x)

$$f(x) = x^2(\ln x+C)$$

Step 6 — Apply the initial condition f(1) = 1/3

$$f(1) = 1^2(\ln1+C) = 0+C = C$$

$$C = \frac13$$

Step 7 — Write the full solution and evaluate at x = e

$$f(x) = x^2\left(\ln x+\frac13\right)$$

$$f(e) = e^2\left(\ln e+\frac13\right) = e^2\left(1+\frac13\right) = e^2\times\frac43 = \frac{4e^2}{3}$$

Answer

Common mistakes

• Forgetting the product rule when differentiating $xf(x)$ on the right-hand side. This term differentiates to $f(x)+xf'(x)$, not just $xf'(x)$ or just $f'(x)$ — dropping the $f(x)$ term derails the entire subsequent simplification.
• Losing track of which variable of integration is which. The integral $\int_1^xf(t)\,dt$ uses a dummy variable $t$ while the upper limit and everything outside the integral is in terms of $x$ — the Fundamental Theorem of Calculus differentiates cleanly with respect to $x$ precisely because $t$ is just a placeholder.

Practise next

• Let $g:[1,\infty)\to\mathbb R$ be differentiable with $g(1)=\dfrac12$ and $2\displaystyle\int_1^xg(t)\,dt=xg(x)-\dfrac{x^2}{2}$. Find $g(e)$, using the same differentiate-and-solve approach.