Question

If $f\left(x\right)={\cos }^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)$, then the value of $f\left(0\right)+f^{\prime }\left(0\right)+\frac{f^{\prime \prime }\left(0\right)}{2!}+...+\frac{f^n\left(0\right)}{n!}$ is

• A. $n$
• B. $0$
• C. $2^n$
• D. $2^n-1$

Answer 1

Answer: (B)

$0$

Given :

$f\left(x\right)={\cos }^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)$

$={\cos }^{-1}\left(\frac{\frac{1}{x}-x}{\frac{1}{x}+x}\right)$

$={\cos }^{-1}\left(\frac{\frac{1{-x}^2}{x}}{\frac{1{+x}^2}{x}}\right)$

$f\left(x\right)={\cos }^{-1}\left(\frac{1{-x}^2}{1{+x}^2}\right)\Rightarrow f\left(0\right)={\cos }^{-1}\left(\frac{1-0}{1+0}\right)={\cos }^{-1}\left(1\right)=0$

Differentiate w.r.t 'x'

i) $f^1\left(x\right)=\frac{-1}{\sqrt{1-\left(\frac{1-x^2}{1+x^2}\right)}}\times \frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)$

$=\frac{-1}{\sqrt{\frac{1+x^2-1+x^2}{1+x^2}}}\times \frac{-2x(1+x^2)-2x(1-x^2)}{{(1+x^2)}^2}$

$=\frac{-1}{\sqrt{\frac{2x^2}{1{+x}^2}}}\times \frac{-2x{-2x}^3-2x+{2x}^3}{\left(1{+x}^2\right)\left(1{+x}^2\right)}$

$=\frac{-\sqrt{1{+x}^2}}{\sqrt{2}x}\times \frac{-4x}{\left(1{+x}^2\right)\left(1{+x}^2\right)}$

$=\frac{2\sqrt{2}}{\left(1{+x}^2\right)\left(1{+x}^2\right)}$

$f^1\left(x\right)=\frac{2\sqrt{2}}{{\left(1{+x}^2\right)}^{\frac{3}{2}}}=2\sqrt{2}{\left(1{+x}^2\right)}^{\frac{-3}{2}}$

$f^1\left(0\right)=\frac{2\sqrt{2}}{{(1+0)}^{\frac{3}{2}}}$
$f^1\left(0\right)=2\sqrt{2}x=0$
$f^{11}\left(x\right)=2\sqrt{2}[\frac{-3}{2}{(1+x^2)}^{\frac{-3}{2}-1}\times 2x]=2\sqrt{2}[\frac{-3}{2}{(1+x^2)}^{\frac{-5}{2}}\times 2x]=-6\sqrt{2}x{(1+x^2)}^{\frac{-5}{2}}{f}^{11}\left(0\right)=0$

$f^{111}\left(x\right)=-6\sqrt{2}x\times \frac{-5}{2}{(1+x^2)}^{\frac{-5}{2}-1}\times 2x=-30\sqrt{2}x{(1+x^2)}^{\frac{-7}{2}}=0$
Hence the further derivatives would also sum upto Zero ,since x=0
$f\left(0\right)+f^1\left(0\right)+\frac{f^{11}\left(0\right)}{2!}+.......\frac{f^n\left(0\right)}{n!}=0$