Search any question & find its solution
Question:
Answered & Verified by Expert
Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)$ is equal to
Options:
Solution:
2467 Upvotes
Verified Answer
The correct answer is:
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$
Since, $\quad f(x)=e^{g(x)}$
$$
\begin{aligned}
\Rightarrow \quad e^{g(x+1)} & =f(x+1) \\
& =x f(x) \\
& =x e^{g(x)}
\end{aligned}
$$
and
$$
g(x+1)=\log x+g(x)
$$
$$
\Rightarrow \quad g(x+1)-g(x)=\log x
$$
Replacing $x$ by $x-\frac{1}{2}$, we get
$$
\begin{aligned}
g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\
\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =-\frac{4}{(2 x-1)^2}
\end{aligned}
$$
Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$$
g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
$$
$$
\begin{aligned}
\Rightarrow \quad e^{g(x+1)} & =f(x+1) \\
& =x f(x) \\
& =x e^{g(x)}
\end{aligned}
$$
and
$$
g(x+1)=\log x+g(x)
$$
$$
\Rightarrow \quad g(x+1)-g(x)=\log x
$$
Replacing $x$ by $x-\frac{1}{2}$, we get
$$
\begin{aligned}
g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\
\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =-\frac{4}{(2 x-1)^2}
\end{aligned}
$$
Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$$
g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.