A) \[{{\tan }^{-1}}{{k}^{2}}\]
B) \[{{\cot }^{-1}}({{k}^{2}})\]
C) \[2{{\sin }^{-1}}\left( \frac{1}{\sqrt{1+{{k}^{4}}}} \right)\]
D) \[{{\sec }^{-1}}(\sqrt{1+{{k}^{4}}})\]
Correct Answer: B
Solution :
| \[{{\left. \frac{dy}{dx} \right]}_{x=0}}={{k}^{2}}\] |
| \[\Rightarrow \tan \,\,\psi ={{k}^{2}}\]\[\Rightarrow \cot \left( \frac{\pi }{2}-\Psi \right)={{k}^{2}}\]\[\Rightarrow \left( \frac{\pi }{2}-\Psi \right)={{\cot }^{-1}}{{k}^{2}}={{\sin }^{-1}}\frac{1}{\sqrt{1+{{k}^{4}}}}\]\[\Rightarrow B\] |
You need to login to perform this action.
You will be redirected in
3 sec
