2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Differential Equations

JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    If \[y\cos x+x\cos y=\pi \], then \[{{y}'}'(0)\] is [IIT Screening 2005]

    A) 1

    B) \[\pi \]

    C) 0    

    D) \[-\pi \]

    Correct Answer: B

    Solution :

    • \[y\cos x+x\cos y=\pi \] Differentiate both sides with respect to x, we get \[-y\sin x+\cos x.{y}'+x(-\sin y){y}'+\cos y\] Again differentiate with respect to x \[-{{y}'}'\sin x-y\cos x+\cos x.{{y}'}'+\sin x.{y}'-\sin y.{y}'\]                             
    • \[-x[\cos y.{{({y}')}^{2}}+\sin y.{{y}'}']-\sin y.{y}'\] Putting \[x=0\], we get \[-y+{{y}'}'-2\sin y\,{y}'=0\] \[{{y}'}'=y+2{y}'\sin y\] Since at \[x=0\], \[y=\pi \]; \[{{({{y}'}')}_{0}}=\pi \].


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