JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    If curve passes through the point                 \[\left( 2,\frac{7}{2} \right)\]  and has slope                 \[\left( 1-\frac{1}{{{x}^{2}}} \right)\]  at any point                 \[(x,y)\]  on it, then the ordinate of the point on the curve whose abscissa is -2 is:     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                                   \[-\frac{3}{2}\]                

    B)                                                          \[\frac{3}{2}\]

    C)                                                          \[\frac{5}{2}\]                                                

    D)                                                          \[-\frac{5}{2}\]

    Correct Answer: A

    Solution :

                     Slope                 \[=\frac{dy}{dx}=1-\frac{1}{{{x}^{2}}}\]                 \[\Rightarrow\]                 \[\int_{{}}^{{}}{dy}=\int_{{}}^{{}}{\left( 1-\frac{1}{{{x}^{2}}} \right)dx}\]                 \[\Rightarrow\]                 \[y=x+\frac{1}{x}+C,\] which is the equation of the curve since curve passes through the point                 \[\left( 2,\frac{7}{2} \right)\]                 \[\therefore\]                 \[\frac{7}{2}=2+\frac{1}{2}+C\Rightarrow C=1\]                 \[\therefore\]                 \[y=x+\frac{1}{x}+1\] when                 \[x=-2,\] then                 \[y=-2+\frac{1}{-2}+1=\frac{-3}{2}\]                                


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