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 :
Show Hint
- 1
- \(\frac{187}{144}\)
- \(\frac{197}{144}\)
- \(\frac{205}{144}\)
The Correct Option is D
Solution and Explanation
Differentiate twice: \(g''(x+1) = -\frac{1}{x^2} + g''(x)\), i.e. \(g''(x+1) - g''(x) = -\frac{1}{x^2}\).
Put \(x = 1,2,3,4\) and add: \(g''(5) - g''(1) = -\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}\right) = -\frac{144 + 36 + 16 + 9}{144} = -\frac{205}{144}\).
\(|g''(5) - g''(1)| = \frac{205}{144}\).
Answer: (4) \(\frac{205}{144}\)
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