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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    A continuously differentiable function \[\varphi (x)\,\text{in}\,(0,\,\pi )\] satisfying \[{y}'=1+{{y}^{2}},\,\,y(0)=0=y(\pi )\] is [MP PET 2000]

    A) \[\tan x\]                                    

    B) \[x(x-\pi )\]

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

    D) Not possible

    Correct Answer: D

    Solution :

    • \[\frac{dy}{dx}=1+{{y}^{2}}\] Þ \[\frac{dy}{1+{{y}^{2}}}=dx\]        
    • Integrating both sides,        
    • \[\int{\frac{dy}{1+{{y}^{2}}}=\int{dx}}\] Þ  \[{{\tan }^{-1}}y=x+c\]        
    • At \[x=0,\]\[y=0,\] then \[c=0\]        
    • At \[x=\pi ,\]\[y=0,\] then \[{{\tan }^{-1}}0=\pi +c\] Þ \[c=-\pi \]        
    • \[\therefore {{\tan }^{-1}}y=x\]Þ\[y=\tan x=\varphi (x)\]        
    • Therefore, solution is \[y=\tan x\]        
    • But \[\tan x\] is not continuous function in \[(0,\,\pi )\]           
    • Hence, \[\varphi \,(x)\] is not possible in \[(0,\,\pi )\].


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