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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer
    If the tangent to the curve, \[y={{x}^{3}}+axb\] at the point (1, -5) is perpendicular to the line, \[x+y+4=0,\]then which one of the following points lies on the curve? [JEE Main 9-4-2019 Morning]

    A) (-2, 2)             

    B) (2, -2)

    C) (2, -1)             

    D) (-2, 1)

    Correct Answer: B

    Solution :

    \[y={{x}^{3}}+ax-b\] (1, -5) lies on the curve \[\Rightarrow -5=1+a-b\Rightarrow a-b=-6\]                         ... (i)             Also,\[y'=3{{x}^{2}}+a\] \[y{{'}_{\left( 1,5 \right)}}=3+a\]               (slope of tangent) \[\because \]this tangent is \[\bot \]to \[x+y+4=0\] \[\Rightarrow (3+a)\](1) = -1 \[\Rightarrow a=-4\]                                        ....(ii) By (i) and (ii) : \[a=-4,b=2\] \[\therefore \]\[y={{x}^{3}}-4x-2.\] (2,-2) lies on this curve.


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