JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer
    The area (in sq. units) of the region \[A=\{(x,y):{{x}^{2}}\le y\le x+2\}\] is                         [JEE Main 9-4-2019 Morning]

    A) \[\frac{10}{3}\]                                  

    B) \[\frac{9}{2}\]

    C) \[\frac{31}{6}\]                                  

    D) \[\frac{13}{6}\]

    Correct Answer: B

    Solution :

    \[{{x}^{2}}\le y\le x+2\]           \[{{x}^{2}}=y;y=x+2\]           \[{{x}^{2}}=x+2\]           \[{{x}^{2}}-x-2=0\]           \[(x-2)(x-1)=0\]           \[x=2,-1\]           Area,\[=\int\limits_{-1}^{2}{(x+2)}-{{x}^{2}}dx=\frac{9}{2}\]          

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