2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2006

  • question_answer
    The equation of the tangent at \[(1,5)\] to the curve \[y=5{{x}^{4}}\] is

    A)  \[20x-y=15\] 

    B)  \[x+20y=101\]

    C)  \[20x+y=15\]  

    D)  \[x-20y=101\]

    Correct Answer: A

    Solution :

    Given curve is \[y=5{{x}^{4}}\] \[\therefore \] \[\frac{dy}{dx}=20{{x}^{3}}\] At   \[(1,\,5),\frac{dy}{dx}=20\] Now,  equation of tangent at \[(1,5)\] is \[y-5=20(x-1)\] \[\Rightarrow \] \[20x-y-15=0\] or \[20x-y=15\]


You need to login to perform this action.
You will be redirected in 3 sec spinner