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JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    The ends A and B of a rod of length \[\sqrt{5}\] are sliding along the curve \[y=2{{x}^{2}}.\] Let \[{{x}_{A}}\] and \[{{x}_{B}}\] be the x-coordinates of the ends. At the moment when A is at (0,0) and B is at (1,2), the derivative \[\frac{d{{x}_{B}}}{d{{x}_{A}}}\] has the value equal to

    A)  1/3                       

    B)  1/5     

    C)  1/8                       

    D)  1/9

    Correct Answer: D

    Solution :

    Given that \[y=2{{x}^{2}}.\] \[|A{{B}^{2}}|={{({{x}_{B}}-{{x}_{A}})}^{2}}+(2x_{B}^{2}-2x_{A}^{2})=5\] \[\Rightarrow \]\[{{({{x}_{B}}-{{x}_{A}})}^{2}}+4{{(x_{B}^{2}-x_{A}^{2})}^{2}}=5\] On differentiating \[w.r.t\,{{x}_{A}}\] and denoting \[\frac{d{{x}_{B}}}{d{{x}_{A}}}=D\] \[2({{x}_{B}}-{{x}_{A}})(D-1)+8(x_{B}^{2}-x_{A}^{2})(2{{x}_{B}}D-2{{x}_{A}})=0\] On putting \[{{x}_{A}}=0\]and \[{{x}_{B}}=1\] \[2(1-0)(D-1)+8(1-0)(2D-0)=0\] \[2D-2+16D=0\]\[\Rightarrow \]\[D=\frac{1}{9}\]


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