Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \(f(x)\) is a differentiable function in the interval \((0, \infty)\) such that \(f(1) = 1\) and \(\lim_{t \to x} \dfrac{t^2 f(x) - x^2 f(t)}{t - x} = 1\), for each \(x > 0\), then \(f(3/2)\) is equal to</p>
<p>\(\dfrac{23}{18}\)</p>
<p>\(\dfrac{13}{6}\)</p>
<p>\(\dfrac{25}{9}\)</p>
<p>\(\dfrac{31}{18}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the given limit is the derivative of x²f(x) at point t=x. By expanding and simplifying the limit expression, we can write it as d/dx[x²f(x)] = 1, which gives us a differential equation to solve.
<p><strong>Step 1:</strong> Recognize the limit form. Rewrite the limit:</p><p>$$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = \lim_{t \to x} \frac{t^2 f(x) - x^2 f(x) + x^2 f(x) - x^2 f(t)}{t - x}$$</p><p><strong>Step 2:</strong> Split into two parts:</p><p>$$= \lim_{t \to x} \frac{f(x)(t^2 - x^2)}{t - x} + \lim_{t \to x} \frac{x^2(f(x) - f(t))}{t - x}$$</p><p>$$= f(x) \lim_{t \to x} \frac{(t-x)(t+x)}{t-x} - x^2 \lim_{t \to x} \frac{f(t) - f(x)}{t - x}$$</p><p><strong>Step 3:</strong> Evaluate the limits:</p><p>$$= f(x) \cdot 2x - x^2 f'(x) = 1$$</p><p>This is equivalent to: $\frac{d}{dx}[x^2 f(x)] = 1$</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>$$x^2 f(x) = x + C$$</p><p><strong>Step 5:</strong> Use the condition f(1) = 1:</p><p>$$1^2 \cdot f(1) = 1 + C \implies 1 = 1 + C \implies C = 0$$</p><p><strong>Step 6:</strong> Therefore $x^2 f(x) = x$, so $f(x) = \frac{1}{x}$</p><p><strong>Step 7:</strong> Calculate f(3/2):</p><p>$$f(3/2) = \frac{1}{3/2} = \frac{2}{3}$$</p><p>∴ Answer: D</p>
Correct Answer: D