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J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    The tangent to the curve \[y=2{{x}^{2}}-x+1\]is parallel to the line \[y=3x+9\]at the point

    A)  \[(3,\,\,9)\]         

    B)  \[(2,\,\,-1)\]

    C)  \[(2,\,1)\]

    D)  \[(1,\,2)\]

    Correct Answer: D

    Solution :

    Given curve is \[y=2{{x}^{2}}-x+1\] ?..(i) On differentiating w.r.t. x, we get \[\frac{dy}{dx}=4x-1\] Since, tangent to the curve is parallel to the  given line y = 3x + 9. Then, slopes will be  equal. \[\therefore \] \[4x-1=3\] \[\Rightarrow \] \[y=2\] Hence, required point is \[(1,2)\].


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