Question:

<p><strong>Step 1: Analyze the limit form</strong></p><p>As t → x, the numerator: t³f(x) - x³f(t) → x³f(x) - x³f(x) = 0</p><p>The denominator: t² - x² → 0</p><p>This is a 0/0 indeterminate form, so we apply L'Hôpital's rule (differentiate with respect to t).</p><p><strong>Step 2: Apply L'Hôpital's rule</strong></p><p>$$\lim_{t \to x} \frac{t^3 f(x) - x^3 f(t)}{t^2 - x^2} = \lim_{t \to x} \frac{3t^2 f(x) - x^3 f'(t)}{2t}$$</p><p>Substituting t = x:</p><p>$$\frac{3x^2 f(x) - x^3 f'(x)}{2x} = \frac{1}{2}$$</p><p><strong>Step 3: Simplify to get a differential equation</strong></p><p>$$3x^2 f(x) - x^3 f'(x) = x$$</p><p>Dividing by x³:</p><p>$$\frac{3f(x)}{x} - f'(x) = \frac{1}{x^2}$$</p><p>Rearranging:</p><p>$$f'(x) - \frac{3f(x)}{x} = -\frac{1}{x^2}$$</p><p><strong>Step 4: Solve the linear differential equation</strong></p><p>This is a first-order linear ODE. The integrating factor is:</p><p>$$\mu(x) = e^{\int -3/x \, dx} = e^{-3\ln x} = x^{-3}$$</p><p>Multiplying both sides by x⁻³:</p><p>$$x^{-3}f'(x) - 3x^{-4}f(x) = -x^{-5}$$</p><p>The left side is the derivative of x⁻³f(x):</p><p>$$\frac{d}{dx}[x^{-3}f(x)] = -x^{-5}$$</p><p><strong>Step 5: Integrate both sides</strong></p><p>$$x^{-3}f(x) = \int -x^{-5} \, dx = \frac{x^{-4}}{4} + C = \frac{1}{4x^4} + C$$</p><p>$$f(x) = \frac{1}{4x} + Cx^3$$</p><p><strong>Step 6: Apply initial condition f(1) = 1</strong></p><p>$$f(1) = \frac{1}{4} + C = 1$$</p><p>$$C = \frac{3}{4}$$</p><p><strong>Step 7: Write the final function</strong></p><p>$$f(x) = \frac{1}{4x} + \frac{3x^3}{4}$$</p><p><strong>∴ Answer: c</strong></p>