A) \[{{\cot }^{-1}}(-1/\pi )\]
B) \[{{\cot }^{-1}}\pi \]
C) \[{{\cot }^{-1}}(-\pi )\]
D) \[{{\cot }^{-1}}(1/\pi )\]
Correct Answer: B
Solution :
The curves \[{{y}^{2}}=2x/\pi \] and \[y=\sin x\] intersect at \[(0,0)\] and \[(\pi /2,\,1)\]. Let the gradients of the tangents to the curves be \[{{m}_{1}}\] and \[{{m}_{2}}\]respectively. Then \[{{m}_{1}}=\frac{dy}{dx}=\frac{1}{\pi y}\] and \[{{m}_{2}}=\frac{dy}{dx}=\cos x\] At \[(\pi /2\,,1)\], \[{{m}_{1}}=\frac{1}{\pi }\], \[{{m}_{2}}=\cos \frac{\pi }{2}=0\] Thus \[\tan \theta =\frac{(1/\pi )-0}{1+(1/\pi )(0)}=\frac{1}{\pi }\]Þ\[\theta ={{\cot }^{-1}}\pi \].You need to login to perform this action.
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