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

  • question_answer
    A curve y = f(x) passes through the point P( 1,1). The normal to the curve at P is a (y - 1) + (x - 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is

    A) \[y={{e}^{a}}^{(x-1)}\]               

    B) \[y={{e}^{a}}^{(1-x)}\]

    C) \[y={{e}^{a/2}}^{(x-1)}\]            

    D) \[{{e}^{a/2}}^{(x+1)}\]

    Correct Answer: A

    Solution :

    Slope of the normal at (1,1) is - 1 /a \[\Rightarrow \]Slope of the tangent at (1, 1) is 'a' i. e., \[{{\left. \frac{dy}{dx} \right]}_{(1,1)}}=a\]                             ......(1) We are given that \[\frac{dy}{dx}\propto y;\frac{dy}{dx}=ky,\] where k is some constant = k dx \[\log |y|=kx+c,\]where c is a constant\[|y|={{e}^{kx+c}}\]\[y=\pm {{e}^{c}}{{e}^{kx}}=\]\[A{{e}^{kx}},\]where A is a constant. Since the curve passes through (I,I), therefore, \[1=A{{e}^{k}}\Rightarrow A={{e}^{-k}}\] Therefore, \[y={{e}^{-k}}.{{e}^{kx}}={{e}^{k}}(x-1)\] \[\Rightarrow \]\[\frac{dy}{dx}=k{{e}^{k}}(x-1)\] \[\Rightarrow \]\[{{\left. \frac{dy}{dx} \right]}_{(1,1)}}=k\Rightarrow a=k\]               [Using (1)] Thus, the required curve is \[y={{e}^{a}}\] (x - 1).


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