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
