Q28 / 54 · JEE Advanced 2018 (Paper 2) · Mathematics

Let $f : ℝ \to ℝ$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (5 y - 2)$ , then the value of $\lim_{x \to - \infty} f (x)$ is

Correct answer: 0.4

Solution

\frac{d y}{d x} = 25 y^{2} - 4

\frac{d y}{25 y^{2} - 4} = d x

Integrating

\frac{1}{25} \times \frac{1}{2 \times \frac{2}{5}} \ln |\frac{y - \frac{2}{5}}{y + \frac{2}{5}}| = x + c

\Rightarrow \ln |\frac{5 y - 2}{5 y + 2}| = 20 (x + c)

Now,

c = 0

f (0) = 0

Hence,

|\frac{5 y - 2}{5 y + 2}| = e^{(20 x)}

\lim_{x \to - \infty} |\frac{5 f (x) - 2}{5 f (x) + 2}| = \lim_{x \to - \infty} e^{(20 x)}

Now,

R H S = 0 \Rightarrow \lim_{x \to - \infty} (5 f (x) - 2) = 0

\Rightarrow \lim_{x \to - \infty} f (x) = \frac{2}{5}

Hence, the correct answer is 0.4.