2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    The continuously differentiable function \[y=f(x),\,\,x\in \,(0,\,\pi )\] satisfying \[y'=1+{{y}^{2}},y(0)=0=y(\pi )\] is

    A)  \[\tan \,x\]                   

    B)  \[x(x-\pi )\]

    C)  \[(x-\pi )\,(1-{{e}^{x}})\]    

    D)  not possible

    Correct Answer: A

    Solution :

     Given,             \[y'=1+{{y}^{2}}\] \[\Rightarrow \]            \[\frac{dy}{1+{{y}^{2}}}=dx\] \[\Rightarrow \]            \[{{\tan }^{-1}}y=x+C\]        (on integrating) At \[x=0,\,ta{{n}^{-1}}y=0+C\] \[\Rightarrow \]            \[0=0+C\] \[\Rightarrow \]            \[C=0\] \[\therefore \]    \[{{\tan }^{-1}}y=x\] \[\Rightarrow \]            \[y=\tan \,x\]


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