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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 40

  • question_answer
    The equation of motion of a particle is given by \[\frac{dx}{dt}=t(t+1),\] \[\frac{dy}{dt}=\frac{1}{t+1},\] where particle is at \[(x(t),y(t))\]at time t. If the particle is at the origin at \[t=0\]and passes through the point \[(d,2),\] then the value of d is

    A) \[\frac{{{({{e}^{2}}+1)}^{3}}}{3}+\frac{{{({{e}^{2}}+1)}^{2}}}{2}\]

    B) \[\frac{{{({{e}^{2}}-1)}^{3}}}{3}+\frac{{{({{e}^{2}}-1)}^{2}}}{2}\]

    C) \[\frac{{{({{e}^{2}}+1)}^{3}}}{3}-\frac{{{({{e}^{2}}+1)}^{2}}}{2}\]

    D) None of these

    Correct Answer: B

    Solution :

    [b] \[\frac{dx}{dt}=t(t+1)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{{t}^{3}}}{3}+\frac{{{t}^{2}}}{2}+c\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{{t}^{3}}}{3}+\frac{{{t}^{2}}}{2}\] (\[\because \,\,\,\,\,\,x=0\]at \[t=0\])      ?(1)             \[\frac{dy}{dt}=\frac{1}{t+1}\] \[\Rightarrow \,\,\,\,\,\,\,\,y={{\log }_{e}}(t+1)+c\] \[\Rightarrow \,\,\,\,\,\,\,\,y={{\log }_{e}}(t+1)\]     (\[\because \,\,\,\,\,\,\,y=0\] at \[t=0\]) Since the particle passes through the point \[(d,2),\] \[2={{\log }_{e}}(t+1)\] \[\Rightarrow \,\,\,\,\,\,t={{e}^{2}}-1\] Putting the value of t into (1), we get \[d=\frac{{{({{e}^{2}}-1)}^{3}}}{3}+\frac{{{({{e}^{2}}-1)}^{2}}}{2}\]


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