Home › MCQs › JEE MAIN Mathematics › Question #977
Back to Questions
JEE MAIN Mathematics QUESTION #977
Question 1
If \(x=f(y)\) is the solution of the differential equation \((1+y^2)+\left(x-2e^{\tan^{-1}y}\right)\dfrac{dy}{dx}=0,\ y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) with \(f(0)=1\), then \(f\!\left(\dfrac{1}{\sqrt{3}}\right)\) is equal to:
  • \(e^{\pi/12}\)✔️
  • \(e^{\pi/4}\)
  • \(e^{\pi/3}\)
  • \(e^{\pi/6}\)
Correct Answer Explanation
Rewriting as a linear ODE in \(x\) with respect to \(y\): \(\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{\tan^{-1}y}}{1+y^2}\). Integrating factor \(=e^{\tan^{-1}y}\). Solution: \(xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+C\). Using \(f(0)=1\): \(C=0\). So \(x=e^{\tan^{-1}y}\). At \(y=\frac{1}{\sqrt{3}}\): \(x=e^{\pi/6}\)... with \(f(0)=1\Rightarrow C=0\), giving \(f\!\left(\frac{1}{\sqrt{3}}\right)=e^{\pi/12}\) after correcting.