Question

If $n=4\lambda +3$, $\lambda ϵI$ & $y={\tan }^{-1}x$ then $y_n\left(0\right)$ is

• A. $0$
• B. $n!$
• C. $-(n-1)!$
• D. $(n-1)!$

Answer 1

Answer: (C)

$-(n-1)!$

$\because$ $y={\tan }^{-1}x$

$\therefore$ $y_1=\frac{1}{1+x^2}=\frac{1}{x^2-i^2}$

$\therefore$ $y_1=\frac{1}{(x+i)(x-i)}=\frac{1}{2i}[\frac{1}{x-i}-\frac{1}{x+i}]$

$y_1=\frac{1}{2i}[{(x-i)}^{-1}-{(x+i)}^{-1}]$

$\therefore$ $y_2=\frac{1}{2i}[(-1){(x-i)}^{-1}-(-1){(x+i)}^{-2}]=\frac{1}{2i}{(-1)}^{2-1}[{(x-i)}^{-2}-{(x+i)}^{-2}]$

$\therefore$ $y_n\left(x\right)=\frac{1}{2i}{(-1)}^{n-1}[{(x-i)}^{-n}-{(x+i)}^{-n}](n-1)!$

$=\frac{1}{2i}{(-1)}^{n-1}[\frac{1}{{(x-i)}^n}-\frac{1}{{(x+i)}^n}](n-1)!$

$\therefore$ $y_n\left(0\right)=\frac{1}{2i}{(-1)}^{n-1}[\frac{1}{{(-i)}^n}-\frac{1}{{\left(i\right)}^n}](n-1)!$

Putting $n=4\lambda +3$

$y_n\left(0\right)=\frac{1}{2i}{(-1)}^{4\lambda +2}(4\lambda +2)![\frac{1}{{(-i)}^{4\lambda +3}}-\frac{1}{{\left(i\right)}^{4\lambda +3}}]$

$=\frac{1}{2i}(4\lambda +3)![\frac{1}{i}+\frac{1}{i}]=-(4\lambda +2)!$

$=-(4\lambda +3-1)!=-(n-1)!$