For any positive integer n, define ${{f}_{n}}:\left( 0,\infty \right)\to \mathbb{R}$ as
\[{{f}_{n}}\left( x \right)=\sum\limits_{j=1}^{n}{{{\tan }^{-1}}}\left( \dfrac{1}{1+\left( x+j \right)\left( x+j-1 \right)} \right)\] for all $x\in \left( 0,\infty \right)$ .
(Here, the inverse trigonometric function $ta{{n}^{-1}}x$ assume values in $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ )
Then, which of the following statement(s) is (are) TRUE?
This question has multiple correct options
A)$\sum\limits_{j=1}^{5}{{{\tan }^{2}}}\left( {{f}_{j}}\left( 0 \right) \right)=55$
B)$\sum\limits_{j=1}^{10}{\left( 1+{{f}_{j}}\left( 0 \right) \right){{\sec }^{2}}\left( {{f}_{j}}\left( 0 \right) \right)}=10$
C)For any fixed positive integer n, $\underset{x\to \infty }{\mathop{\lim }}\,\tan \left( {{f}_{n}}\left( x \right) \right)=\dfrac{1}{n}$
D)For any fixed positive integer n, $\underset{x\to \infty }{\mathop{\lim }}\,{{\sec }^{2}}\left( {{f}_{n}}\left( x \right) \right)=1$

Hint: Replace ‘1’ in numerator by difference of $x+j$ and $x+j-1$ . And now observe the trigonometric identity \[{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)\] to write the given series in difference form. Verify all the options for getting answers.

Complete step-by-step answer:

Here, function ${{f}_{n}}\left( x \right)$ is given as

\[{{f}_{n}}\left( x \right)=\sum\limits_{j=1}^{n}{{{\tan }^{-1}}}\left( \dfrac{1}{1+\left( x+j \right)\left( x+j-1 \right)} \right)\] for all $x\in \left( 0,\infty \right)$.

Let us replace ‘1’ in the numerator by relation \[\left( x+j \right)-\left( x+j-1 \right)\] so that we can convert the given function ${{f}_{n}}\left( x \right)$ in \[{{\tan }^{-1}}A-{{\tan }^{-1}}B\] in the following way:-

\[{{f}_{n}}\left( x \right)=\sum\limits_{j=1}^{n}{{{\tan }^{-1}}}\left( \dfrac{\left( x+j \right)-\left( x+j-1 \right)}{1+\left( x+j \right)\left( x+j-1 \right)} \right)\] (i)

Now compare the above relation with the trigonometric identity of inverse trigonometry, given as

\[{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)\] (ii)
Hence, we can re-write the equation (i) by comparing equation (ii), hence we get \[{{f}_{n}}\left( x \right)=\sum\limits_{j=1}^{n}{\left( {{\tan }^{-1}}\left( x+j \right)-{{\tan }^{-1}}\left( x+j-1 \right) \right)}\]

Now, put values of j from ‘1’ to ‘n’ in the above expression, we get \[\begin{align}

  & {{f}_{n}}\left( x \right)=\left( {{\tan }^{-1}}\left( x+1 \right)-{{\tan }^{-1}}x \right)+\left( {{\tan }^{-1}}\left( x+2 \right)-{{\tan }^{-1}}\left( x+1 \right) \right)+\left( {{\tan }^{-1}}\left( x+3 \right)-{{\tan }^{-1}}\left( x+2 \right) \right)+\ldots \ldots \\

 & \left( {{\tan }^{-1}}\left( x+n \right)-{{\tan }^{-1}}\left( x+n+1 \right) \right) \\

\end{align}\]

Above expression can be added in the following manner as well:-

\[\begin{align}

  & {{\tan }^{-1}}\left( x+1 \right)-{{\tan }^{-1}}x \\

 & {{\tan }^{-1}}\left( x+2 \right)-{{\tan }^{-1}}\left( x+1 \right) \\

 & {{\tan }^{-1}}\left( x+3 \right)-{{\tan }^{-1}}\left( x+2 \right) \\

\end{align}\]

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\[\dfrac{+{{\tan }^{-1}}\left( x+n \right)-{{\tan }^{-1}}\left( x+n+1 \right)}{{{f}_{n}}\left( x \right)={{\tan }^{-1}}\left( x+n \right)-{{\tan }^{-1}}\left( x \right)}\]

Now, apply the identity given in equation (ii) with the above equation. Hence, we get \[{{f}_{n}}\left( x \right)={{\tan }^{-1}}\left( x+n \right)-{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{x+n-x}{1+x\left( x+n \right)} \right)\]

\[{{f}_{n}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{n}{1+x\left( x+n \right)} \right)\]

     (iii)
Now, Let us verify the given options with the help of equation (iii).

We cannot put $n=0$ in ${{f}_{n}}\left( x \right)$ as it is not included in the domain of the given function i.e. $\left( 0,\infty \right)$ .

Hence, we can ignore option (A)

Similarly, we can ignore option (B)

Option (C) is given as $\underset{x\to \infty }{\mathop{\lim }}\,\tan \left( {{f}_{n}}\left( x \right) \right)=\dfrac{1}{n}$,

LHS $=\underset{x\to \infty }{\mathop{\lim }}\,\tan \left( {{\tan }^{-1}}\left( \dfrac{n}{1+x\left( x+n \right)} \right) \right)$

LHS $=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{n}{1+x\left( x+n \right)}$

Hence, LHS=0 if $x\to \infty $.

So, option (C) is not correct.

Option (D) is given as

$\underset{x\to \infty }{\mathop{\lim }}\,{{\sec }^{2}}\left( {{f}_{n}}\left( x \right) \right)=1$ (iv)

Use ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $

Hence, LHS of equation (iv) can be given as LHS $=\underset{x\to \infty }{\mathop{\lim }}\,1+{{\left( \tan \left( {{\tan }^{-1}}\dfrac{n}{1+x\left( x+n \right)} \right) \right)}^{2}}$

LHS $=1+\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{n}{1+x\left( x+n \right)} \right)}^{2}}$

Hence, we get LHS

LHS=1 as $n\to \infty $ for $\dfrac{n}{1+x\left( x+n \right)}$ will give zero.

Hence, option (D) is the correct answer.


Note: As ‘0’ is not included in the domain but one can miss and calculate $\sum\limits_{j=1}^{5}{{{\tan }^{2}}fj\left( 0 \right)}$ and will get answer 55 of it as well. So, one can tick the option (A) as well. But always keep in mind the domain and range of the functions while solving these kinds of problems.

Writing ‘1’ as $\left( x+j \right)-\left( x+j-1 \right)$ was the key point of the solution. We can always break any series of ${{\tan }^{-1}}$ in the same way expressed in the solution.