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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Mathematics
Chapter
Functions

JEE Advanced 2008 · Paper 2 · Mathematics · Functions

Let g(x)= f(x) , where f(x) is a twice differentiable positive...

Let g(x)=log⁡f(x)g(x)=\log f(x), where f(x)f(x) is a twice differentiable positive function on (0,∞)(0, \infty) such that f(x+1)=xf(x)f(x+1)=x f(x). Then, for N=1,2,3,….g′′N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}(12)\left(\frac{1}{2}\right) is equal to

Options

  1. 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\}
  2. B4{1+19+125+…+1(2N−1)2}4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}
  3. C−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\}
  4. D4{1+19+125+…+1(2N+1)2}4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}

Correct answer

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\}

Step-by-step solution

Since, f(x)=eg(x)\quad f(x)=e^{g(x)}⇒eg(x+1)=f(x+1)=xf(x)=xeg(x)\begin{aligned} \Rightarrow \quad e^{g(x+1)} & =f(x+1) \\ & =x f(x) \\ & =x e^{g(x)} \end{aligned}andg(x+1)=log⁡x+gand g(x+1)=\log x+g(x) \Rightarrow \quad g(x+1)-g(x)=log⁡x\log x Replacing xx by x−12x-\frac{1}{2}, we get g(x+12)−g(x−12)=log⁡(x−12)=log⁡(2x−1)−log⁡2∴g′′(x+12)−g′′(x−12)=−4(2x−1)2\begin{aligned} g\left( x+\frac{1}{2}\right)- g\left( x-\frac{1}{2}\right)& =\log\left( x-\frac{1}{2}\right)=\log(2 x-1)-\log 2 \\ \therefore g^{\prime \prime}\left( x+\frac{1}{2}\right)- g^{\prime \prime}\left( x-\frac{1}{2}\right)& =-\frac{4}{(2 x-1)^2} \end{aligned} Substituting, x=1,2,3,…,Nx=1,2,3, \ldots, N in Eq. (ii) and adding, we get g^{\prime \prime}\left( N+\frac{1}{2}\right)- g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
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