2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Applications of Derivatives

JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Self Evaluation Test - Application of Derivatives

  • question_answer
    The equation of the normal to the curve \[y=\left| {{x}^{2}}-\left| x \right| \right|\] at\[x=-2\].

    A) \[3y=x+8\]

    B) \[x=3y+4\]

    C) \[y=2x+8\]

    D) \[y=3x\]

    Correct Answer: A

    Solution :

    [a] In the neighborhood of \[x=-2,y={{x}^{2}}+x\]. Hence, the point on curve is \[(-2,2)\]. \[\frac{dy}{dx}=2x+1\] or \[{{\left. \frac{dy}{dx} \right|}_{x=-2}}=-3\] So, the slope of the normal at \[(-2,2)\] is \[\frac{1}{3}\]. Hence, the equation of the normal is \[\frac{1}{3}(x+2)=y-2\] or \[3y=x+8\].


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