Question

For any positive integer $n$, define $f_n:(0,\infty )\to R$ as
$f_n\left(x\right)=\sum _{j=1}^n{\tan }^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right)$ for all $x\in (0,\infty )$
$($Here, the inverse trigonometric function ${\tan }^{-1}x$ assumes values in $(\frac{-\pi }{2},\frac{\pi }{2})$ $)$
Then, which of the following statement(s) is (are) TRUE?

• A. $\sum _{j=1}^5{\tan }^2\left(f_j\right(0\left)\right)=55$
• B. $\sum _{j=1}^{10}(1+f_j^{\prime }(0\left)\right){\sec }^2\left(f_j\right(0\left)\right)=10$
• C. For any fixed positive integer $n,\underset{x\to \infty }{lim}\tan \left(f_n\right(x\left)\right)=\frac{1}{n}$
• D. For any fixed positive integer $n,\underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=1$

Answer 1

The correct option is D For any fixed positive integer $n,\underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=1$

$f_n\left(x\right)=\sum _{j=1}^n{\tan }^{-1}\left(\frac{(x+j)-(x+j-1)}{1+(x+j)(x+j-1)}\right)=\sum _{j=1}^n\left[{\tan }^{-1}\right(x+j)-{\tan }^{-1}(x+j-1\left)\right]={\tan }^{-1}(x+n)-{\tan }^{-1}x{f}_n\left(x\right)={\tan }^{-1}\left[\frac{n}{1+(x+n)x}\right]\Rightarrow \tan \left[f_n\right(x\left)\right]=\left[\frac{n}{1+(x+n)x}\right]\underset{x\to \infty }{lim}\tan \left(f_n\right(x\left)\right)=0\therefore \left(C\right)\text{is wrong.}\underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=\underset{x\to \infty }{lim}(1+{\tan }^2(f_n\left(x\right))=1+\underset{x\to \infty }{lim}{\left[\frac{n}{1+(x+n)x}\right]}^2=1\therefore (D\left)\text{is correct.}\sum _{j=1}^5{\tan }^2\right(f_j\left(0\right))={\left[\frac{5}{1+0}\right]}^2=25\therefore (A\left)\text{is wrong.}{f}_j^{\prime }\right(x)=\frac{1}{1+{\left(\frac{n}{1+(x+n)x}\right)}^2}\times (2x+n\left)f_j^{\prime }\right(0)=\frac{n}{1+n^2}\sum _{j=1}^{10}(1+f_j^{\prime }\left(0\right)\left){\sec }^2\right(f_j\left(0\right))=\sum _{j=1}^{10}(1+\frac{n}{1+n^2})(1+{\left[\frac{n}{1+0}\right]}^2)=\sum _{j=1}^{10}\left(\frac{1+n+n^2}{1+n^2}\right)(1+n^2)=1+10+100=111\therefore (B)\text{is wrong.}$