Question

For $\left|x\right|>1,$ if $\underset{n\to \infty }{lim}\prod _{k=0}^n(1+\frac{2}{x^{2^k}+x^{-2^k}})=f\left(x\right),$ then

• A. $\underset{2}{\overset{5}{\int }}f\left(x\right)dx=3+\ln 16$
• B. $\underset{x\to \infty }{lim}f\left(x\right)=1$
• C. $f\left(x\right)=0$ has exactly one solution
• D. $f\left(x\right)$ is a decreasing function

Answer 1

Answer: (A), (B), (D)

$f\left(x\right)$ is a decreasing function

$a_n=\prod _{k=0}^n(1+\frac{2}{x^{2^k}+x^{-2^k}}),\left|x\right|>1=\prod _{k=0}^n\frac{{(x^{2^k}+1)}^2}{x^{2^{k+1}}+1}=\frac{(x+1{)}^2}{x^2+1}\cdot \frac{(x^2+1{)}^2}{x^4+1}\cdot \frac{(x^4+1{)}^2}{x^8+1}\dots \frac{(x^{2^n}+1{)}^2}{x^{2^{n+1}}+1}=(x+1{)}^2\cdot (x^2+1)\cdot (x^4+1)\dots \frac{(x^{2^n}+1)}{x^{2^{n+1}}+1}$

Now, multiply and divide by $(x-1),$ we get
$a_n=\frac{x-1}{x-1}(x+1{)}^2\cdot (x^2+1)\cdot (x^4+1)\dots \frac{(x^{2^n}+1)}{x^{2^{n+1}}+1}$
Now apply $(a-b)(a+b)=a^2-b^2$ formula consecutively
$a_n=\frac{x+1}{x-1}\cdot \frac{x^{2^{n+1}}-1}{x^{2^{n+1}}+1}$

$\underset{n\to \infty }{lim}{a}_n=\left(\frac{x+1}{x-1}\right)\left(\frac{1-0}{1+0}\right)=\frac{x+1}{x-1}=f\left(x\right)$
$\therefore f\left(x\right)=\frac{x+1}{x-1}$

$\underset{2}{\overset{5}{\int }}f\left(x\right)dx=\underset{2}{\overset{5}{\int }}\frac{x+1}{x-1}dx$
$=\underset{2}{\overset{5}{\int }}(1+\frac{2}{x-1})dx=[x+2\ln (x-1){]}_2^5=3+\ln 16$

$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to \infty }{lim}\frac{x+1}{x-1}=1$

$f\left(x\right)=\frac{x+1}{x-1}=0$
$\Rightarrow x=-1$ $($not possible as $\left|x\right|>1)$

$f\left(x\right)=\frac{x+1}{x-1}$
$f^{\prime }\left(x\right)=\frac{-2}{(x-1{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)<0$ for $|x|>1$
$\therefore f\left(x\right)$ is decreasing.