Answer:
Option B
Explanation:
We have , $\frac{\text{d}y}{\text{d}x}=(2+5y)(5y-2)$
$\Rightarrow $ $\frac{dy}{25y^{2}-4}=dx$
$\Rightarrow $ $\frac{1}{25}(\frac{dy}{y^{2}-\frac{4}{25}})=dx$
On integrating both sides, we get ,$\frac{1}{25}\int_{}^{} \frac{dy}{y^{2}-(\frac{2}{5})^{2}}=\int_{}^{} dx$
$\Rightarrow $ $\frac{1}{25}\times\frac{1}{2\times 2/5}\log_{}{\mid\frac{y-2/5}{y+2/5}}\mid =x+C$
$\Rightarrow $ $log_{}{\mid\frac{5y-2}{5y+2}}\mid =20(x+C)$
$\Rightarrow $ ${\mid\frac{5y-2}{5y+2}}\mid =Ae^{20x}$ [ $\therefore$ $e^{20C}=A$ ]
when x=o $\Rightarrow $ y =0 then A =1
$\therefore$ $\mid \frac{5y-2}{5y+2}\mid =e^{20x}$
$\lim_{x \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =\lim_{x \rightarrow -\infty}e^{20x}$
$\Rightarrow $ $\Rightarrow \lim_{n \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =0$
$\Rightarrow \lim_{n \rightarrow -\infty} 5 f(x) -2=0$
$\Rightarrow \lim_{n \rightarrow -\infty} f(x) =\frac{2}{5}=0.4$
