JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    The area of the region described by \[A=\{(x,\text{ }y):{{x}^{2}}+{{y}^{2}}\le 1\]and \[{{y}^{2}}\le 1-x\}\]is   JEE Main  Solved  Paper-2014

    A) \[\frac{\pi }{2}+\frac{4}{3}\]       

    B) \[\frac{\pi }{2}-\frac{4}{3}\]

    C) \[\frac{\pi }{2}-\frac{2}{3}\]                        

    D) \[\frac{\pi }{2}+\frac{2}{3}\]

    Correct Answer: A

    Solution :

    The required region is The area of region A is \[A=\int\limits_{-1}^{1}{(1-{{y}^{2}})dy=\frac{4}{3}}\] and area of region B is\[B=\frac{\pi }{2}{{(1)}^{2}}=\frac{\pi }{2}\] \[\therefore \]required area\[A+B=\frac{\pi }{2}+\frac{4}{3}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner