JEE PYQ: Limits Question 14
Question 14 - 2020 (04 Sep Shift 2)
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:
(1) $\frac{1}{e}$
(2) $2e$
(3) $\frac{1}{2e}$
(4) $e$
Show Answer
Answer: (1)
Solution
$\lim_{t\to x}\frac{2tf^2(x) - 2x^2 f(t)f’(t)}{1} = 0 \Rightarrow f(x) = xf’(x)$. So $\frac{f’(x)}{f(x)} = \frac{1}{x}$, giving $\ln f(x) = \ln x + C$. Since $f(1) = e$, $C = e$, so $f(x) = ex$. When $f(x) = 1$: $ex = 1 \Rightarrow x = \frac{1}{e}$.
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