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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Let g(x)=log⁡ef(x)g(x)=\log _e f(x), where f(x)f(x) is twice differentiable positive function on (0,∞)(0, \infty) such that f(x+1)=xf(x)f(x+1)=x f(x), for N=1,2,3,…N=1,2,3, \ldots, then g′′(N+12)−g′′(12)equalsg''\left( {N + \frac{1}{2}} \right) - g''\left( {\frac{1}{2}} \right)\mathrm{ equals }
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Correct option: A
−4{1+19+125+…+1(2N−1)2}-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}
Option A (−4{1+19+125+…+1(2N−1)2}-4\left\{1+\frac{1}{9}+ \frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}) is correct because it is the result obtained after differentiating the given logarithmic function and summing the series.

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