Let f : ℝ → ℝ be a differentiable function with f 0 = 0 . If y = f x satisfies the…

Let f:ℝ→ℝ be a differentiable function with f0=0. If y=fx satisfies the differential equation dydx=2+5y5y-2, then the value of  limx→-∞fx is 

Solution

dydx=25y2-4
So  dy25y2-4=dx
Integrating  125×12×25ln⁡y-25y+25=x+c
⇒   ln⁡5y-25y+2=20x+c
Now,  c=0 as f0=0
Hence, 5y-25y+2=e20x
limx→-∞⁡5fx-2 5fx+2= limx→-∞⁡e20x
Now,  
RHS=0⇒ lim x→−∞ ( 5f( x )−2 )=0
⇒ lim x→−∞ f( x )= 2 5

Asked in: JEE Advanced 2018 (Paper 2)

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