Question

Let $f\left(x\right)=\{\begin{array}{c}xta{n}^{-1}x+se{c}^{-1}\frac{1}{x},xϵ(-1,1)-\left\{0\right\}\\ \frac{\pi }{2},x=0\end{array}$ If $f^{\prime }\left(0^{+}\right)=l$ and $f^{\prime }\left(0^{-}\right)=m$ then find the value of $(l^2+m^2).$

Answer 1

$f\left(x\right)=\{\begin{array}{c}xta{n}^{-1}x+se{c}^{-1}\frac{1}{x},xϵ(-1,1)-\left\{0\right\}\\ \frac{\pi }{2},x=0\end{array}$
$l=f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{h{\tan }^{-1}h+{\cos }^{-1}h-\frac{\pi }{2}}{h}$
It is of the form $\frac{0}{0}$, so applying L-Hopital's rule
$l=f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{\frac{h}{1+h^2}+{\tan }^{-1}h-\frac{1}{\sqrt{1-h^2}}}{1}$
$\Rightarrow l=-1$
$m=f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{-h{\tan }^{-1}(-h)+{\cos }^{-1}(-h)-\frac{\pi }{2}}{(-h)}$
It is of the form $\frac{0}{0}$, so applying L-Hopital's rule
$m=\underset{h\to 0}{lim}\frac{h{\tan }^{-1}\left(h\right)+\pi -{\cos }^{-1}(h)-\frac{\pi }{2}}{(-h)}$
$m=f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{\frac{h}{1+h^2}+{\tan }^{-1}h+\frac{1}{\sqrt{1-h^2}}}{-1}$
$\Rightarrow m=-1$
Hence, $l^2+m^2=2$