IIT-JEE 2008 · Paper 2 · Mathematics Q9

← IIT-JEE 2008 · Paper 2

MathematicsSingle correct+3 / -1Question 9 of 66

Let $g(x)=\log f(x)$ where $f(x)$ is a twice differentiable positive function on $(0,\infty)$ such that $f(x+1)=xf(x)$ . Then, for $N=1,2,3,\ldots$ ,
$g''\left(N+\frac12\right)-g''\left(\frac12\right)=$

-4\left\{1+\frac19+\frac1{25}+\cdots+\frac1{(2N-1)^2}\right\}

4\left\{1+\frac19+\frac1{25}+\cdots+\frac1{(2N-1)^2}\right\}

-4\left\{1+\frac19+\frac1{25}+\cdots+\frac1{(2N+1)^2}\right\}

4\left\{1+\frac19+\frac1{25}+\cdots+\frac1{(2N+1)^2}\right\}

Answer and worked solution

Official answer: A

Functional equation. $g(x+1)=\log x+g(x)$ , so $g''(x+1)=g''(x)-\dfrac1{x^2}$ .

Telescope from $x=\frac12$ :
$g''\left(N+\tfrac12\right)-g''\left(\tfrac12\right)=-\sum_{k=0}^{N-1}\frac1{(k+\frac12)^2}=-4\sum_{k=0}^{N-1}\frac1{(2k+1)^2}.$

(A) $-4\left\{1+\frac19+\cdots+\frac1{(2N-1)^2}\right\}$

Topics:Higher-order derivatives Special series sums

← Previous question Next question →

IIT-JEE 2008 · Paper 2, Mathematics question 9