A) \[{{\cot }^{-1}}(-1/\pi )\]
B) \[{{\cot }^{-1}}\pi \]
C) \[{{\tan }^{-1}}(-\pi )\]
D) \[{{\cot }^{-1}}\left( \frac{1}{\pi } \right)\]
Correct Answer: B
Solution :
Given equation of curves are \[{{y}^{2}}=2x/\pi \]and\[y=\sin x\] \[\therefore \]The point of intersections are \[(0,\,\,0)\] and\[(\pi /2,\,\,1)\]. \[\therefore \]The slope of curve\[{{y}^{2}}=2x/\pi \]at point\[(\pi /2,\,\,1)\]is \[{{m}_{1}}=\frac{dy}{dx}=\frac{1}{y\pi }=\frac{1}{\pi }\] And slope of curve \[y=\sin x\] at point\[(\pi /2,\,\,1)\]is \[{{m}_{2}}=\frac{dy}{dx}=\cos x=0\] \[\therefore \] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\] Note: Whenever we find the angle of intersection of two curves, we have to find the angle between the tangents at that point.
You need to login to perform this action.
You will be redirected in
3 sec
