Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $g(x) = \ln f(x)$ where $f(x)$ is a twice differentiable positive function on $(0,\infty)$ such that $f(x+1) = xf(x)$. Then, for $N = 1, 2, 3, \ldots$: $g''\\!\left(N+\frac{1}{2}\right) - g''\\!\left(\frac{1}{2}\right) = -4\left(1 + \frac{1}{9} + \frac{1}{25} + \cdots + \frac{1}{(2N-1)^2}\right)$</p>
<p>Statement 1 true, Statement 2 true and explanation</p>
<p>Statement 1 true, Statement 2 true but NOT explanation</p>
<p>Statement 1 true, Statement 2 false</p>
<p>Statement 1 false</p>
Step-by-Step Solution
Key Concept: General
<b>nth Derivative + Functional Equation [JEE Advanced 2008]</b><br>$f(x+1) = xf(x) \Rightarrow$ this is the Gamma function property.<br>$g(x) = \ln f(x)$, so $g(x+1) = \ln f(x+1) = \ln(xf(x)) = \ln x + g(x)$<br>$g'(x+1) = g'(x) + 1/x$<br>$g''(x+1) = g''(x) - 1/x^2$<br>$g''\\!\left(n+\frac{1}{2}\right) = g''\\!\left(\frac{1}{2}\right) - \sum_{k=0}^{n-1}\frac{1}{(k+\frac{1}{2})^2}$<br>$= g''\\!\left(\frac{1}{2}\right) - 4\sum_{k=0}^{n-1}\frac{1}{(2k+1)^2} = g''\\!\left(\frac{1}{2}\right) - 4\left(1+\frac{1}{9}+\cdots+\frac{1}{(2N-1)^2}\right)$<br>✓ Statement 1 is proven. Statement 2 follows from the recursion.<br><b>Answer: A</b> — both true, S2 explains S1.
Correct Answer: A