Let f x be a differentiable function such that f ' x = 7 - 3 4 f x x , x 0 and f 1 ≠ 4 . Then lim x → 0 + ⁡ x f 1 x

Question

Let f x be a differentiable function such that f ' x = 7 - 3 4 f x x , x > 0 and f 1 ≠ 4 . Then lim x → 0 + ⁡ x f 1 x, does not exist., exists and equals 4 ., exists and equals 4 7 ., exists and equals 0 .

MARKS App · Accepted Answer

We have, d y d x + 3 y 4 x = 7 ( where y = f ( x ) ) This is a linear differential equation. Integrating factor, I . F . = e ∫ 3 d x 4 x = e 3 4 l n x = x 3 4 So, integrating the given differential equation using I . F . ⇒ y . x 3 4 = ∫ 7 . x 3 4 d x ⇒ y . x 3 4 = 7 . 4 x 7 4 7 + C ⇒ y x 3 4 = 4 x 7 4 + C ⇒ f ( x ) = 4 x + C x - 3 4 ⇒ l i m x → 0 + x f 1 x ⇒ l i m x → 0 + x 4 x + C x 3 4 = 4

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