Question

Let \(f:(0,\infty) \to (0,\infty)\) be a differentiable function such that \(f(1) = e\) and
\(\lim_{t \to x} \frac{t^2 f^2(x) - x^2 f^2(t)}{t - x} = 0\).
If \(f(x) = 1\), then x is equal to :

• A. \(\frac1e\)
• B. e
• C. \(\frac{1}{2e}\)
• D. 2e

Hint:

The limit is a 0/0 form; apply L'Hospital's rule in t to get a differential equation for f.

Answer 1

The Correct Option is A

The limit is of the form 0/0 at \(t = x\). Differentiate numerator and denominator w.r.t. t:
\(\lim_{t\to x}\frac{2t f^2(x) - 2x^2 f(t)f'(t)}{1} = 2x f^2(x) - 2x^2 f(x)f'(x) = 0\).
Since \(f(x) \gt 0\): \(f(x) = x f'(x) \Rightarrow \frac{f'}{f} = \frac1x\).
Integrate: \(\ln f = \ln x + \ln c \Rightarrow f(x) = cx\).
\(f(1) = e \Rightarrow c = e\), so \(f(x) = ex\).
\(f(x) = 1 \Rightarrow x = \frac1e\).
Answer: (1) \(\frac1e\)