2. Solved Papers
  3. CET Karnataka Engineering

CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    If \[y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4\,!}....\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to :

    A)  \[-x\]                                   

    B)  \[x\]

    C)  \[y\]                                    

    D)  \[-y\]

    Correct Answer: C

    Solution :

    Here we have                 \[y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}....\]                 \[y={{e}^{-x}}\left[ 1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}....={{e}^{-x}} \right]\] \[\therefore \]  \[\frac{dy}{dx}=-{{e}^{-x}}\] \[\therefore \]  \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-x}}=y\]


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