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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    The curve represented by the differential equation \[xdy-ydx=ydy\], intersects the y-axis at A(0, 1) and the line \[y=e\] at (a, b), then the value of \[(a+b)\] is equal to (Here e denotes napier's constant)

    A) 0              

    B)                                    1

    C) 2                             

    D) 3

    Correct Answer: A

    Solution :

    \[\frac{ydx-xdy}{{{y}^{2}}}=\frac{-ydy}{{{y}^{2}}}\Rightarrow \int{{}}d\left( \frac{x}{y} \right)\] \[=\int{{}}\frac{-dy}{y}\Rightarrow \frac{x}{y}+\ell ny=C\]                             As, A(0, 1) lies on it, so \[0+0=C\Rightarrow C=0\]                 \[\therefore \frac{x}{y}+\ell ny=0\]                                        ?(i)                 Put \[y=e\] in (1), we get                 \[\frac{x}{e}\ell ne=0\Rightarrow x=-e\] \[\therefore (a=-e,\,\,b=e)\]                 Hence, \[(a+b)=0\]


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