JEE PYQ: Differentiation Question 1
Question 1 - 2021 (16 Mar Shift 2)
Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f(x+1) = xf(x)$. If $g : S \to \mathbb{R}$ be defined as $g(x) = \log_e f(x)$, then the value of $|g’’(5) - g’’(1)|$ is equal to:
(1) $\frac{205}{144}$ (2) $\frac{197}{144}$ (3) $\frac{187}{144}$ (4) $1$
Show Answer
Answer: (1) $\frac{205}{144}$
Solution
From $f(x+1) = xf(x)$, taking log: $g(x+1) - g(x) = \ln x$, so $g’’(x+1) - g’’(x) = -\frac{1}{x^2}$. Summing for $x = 1,2,3,4$: $g’’(5) - g’’(1) = -1 - \frac{1}{4} - \frac{1}{9} - \frac{1}{16} = -\frac{205}{144}$. Hence $|g’’(5) - g’’(1)| = \frac{205}{144}$.
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