JEE Advanced 2018 — Paper 2 — Question 37

$\int_{0}^{y} \frac{d y}{25 y^{2}-4}=\int_{0}^{x} d x \Rightarrow \frac{1}{25} \int_{0}^{y} \frac{d y}{y^{2}-\left(\frac{2}{5}\right)^{2}}=x$

$\Rightarrow-\frac{1}{25} \cdot \frac{1}{2 \cdot \frac{2}{5}} \ln \left|\frac{\frac{2}{5}+y}{\frac{2}{5}-y}\right|=x$

$\Rightarrow \mathrm{y}=-\frac{2}{5}\left(\frac{1-\mathrm{e}^{-20 \mathrm{x}}}{1+\mathrm{e}^{-20 \mathrm{x}}}\right)$

$\Rightarrow \lim _{x \rightarrow-\infty} y=0.4$