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JEE Math Practice Question

Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

  1. A.$\frac{3}{2}+\mathrm{e}$
  2. B.$\frac{1}{2}+\mathrm{e}$
  3. C.$3+e$
  4. D.$3-e$

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