A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[2\pi \]
D) \[3\pi \]
Correct Answer: D
Solution :
[d] \[{{y}^{2}}(2-x)={{x}^{3}}\] \[\therefore {{y}^{2}}=\frac{{{x}^{3}}}{2-x}\] \[\therefore y=\pm \frac{{{x}^{3/2}}}{\sqrt{2-x}}\] Clearly, \[x\in \left[ 0,2 \right)\] Consider \[y=\frac{{{x}^{3/2}}}{\sqrt{2-x}}\] When x=0, then y=0. Also, when x increases from 0 to 2, y increases from 0 to\[\infty \]. Hence, graph of given relation is as shown in the following figure:You need to login to perform this action.
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