2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    If \[y=\sqrt{x+\sqrt{x+\sqrt{x+....\infty ,}}}\]then \[\frac{dy}{dx}\]is equal to:

    A) \[\frac{1}{2y+1}\]

    B)         \[\frac{1}{2y-1}\]                

    C)  \[\frac{1}{xy}\]                    

    D)         1

    Correct Answer: B

    Solution :

    Let     \[y=\sqrt{x+\sqrt{x+\sqrt{x+....\infty }}}\] \[\Rightarrow \]               \[y=\sqrt{x+y}\] \[\Rightarrow \]               \[{{y}^{2}}=x+y\] \[\Rightarrow \]               \[{{y}^{2}}-y-x=0\] On differentiating w.r.t. \[x,\]we get \[2y\frac{dy}{dx}-\frac{dy}{dx}-1=0\]                 \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{1}{(2y-1)}\]


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