JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The area (in sq. units) of the region \[\{(x,\text{ }y)\in {{R}^{2}}|4{{x}^{2}}\le y\le 8x+12\}\] is [JEE MAIN Held on 07-01-2020 Evening]

    A) \[\frac{128}{3}\]        

    B) \[\frac{125}{3}\]

    C) \[\frac{127}{3}\]        

    D) \[\frac{124}{3}\]

    Correct Answer: A

    Solution :

    [a] For point of intersections \[4{{x}^{2}}=8x+12\] \[{{x}^{2}}2x3=0\] \[\therefore \,\,\,\,\,x=\,1,3\] The required area =\[\int_{-1}^{3}{(8x+12-4{{x}^{2}})}\,dx\] \[=\,4\left( 2\cdot \frac{{{x}^{2}}}{2}+3x-\frac{{{x}^{3}}}{3} \right)_{-\,1}^{3}\] \[=\,4\,\left\{ (9+9-9)-\left( 1-3+\frac{1}{3} \right) \right\}\] \[=\frac{128}{3}\,square\,units\]

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