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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Tangent and Normal

  • question_answer
    The point at which the tangent to the curve \[y=2{{x}^{2}}-x+1\] is parallel to \[y\text{ }=\text{ 3}x+\text{9 }\]will be                      [Karnataka CET 2001]

    A)            (2, 1)

    B)            (1, 2)

    C)            (3, 9)

    D)            (?2, 1)

    Correct Answer: B

    Solution :

               \[y=2{{x}^{2}}-x+1\] Þ \[\frac{dy}{dx}=4x-1\].                    We know that this equation gives the slope of tangent to the curve. Since this tangent is parallel to \[y=3x+9,\] therefore slope of the tangent is 3, so \[4x-1=3\] or \[x=1.\] Therefore \[y=2{{x}^{2}}-x+1=2-1+1=2.\] Thus the point \[(x,\,y)\] is (1, 2).


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