JEE PYQ: Differential Equation Question 45
Question 45 - 2019 (10 Jan Shift 2)
Let $f$ be a differentiable function such that $f’(x) = 7 - \frac{3}{4}\cdot\frac{f(x)}{x}$, $(x > 0)$ and $f(1) \neq 4$. Then $\lim_{x \to 0^+} xf\left(\frac{1}{x}\right)$:
(1) exists and equals $\frac{4}{7}$
(2) exists and equals 4
(3) does not exist
(4) exists and equals 0
Show Answer
Answer: (2)
Solution
$f’(x) + \frac{3}{4x}f(x) = 7$. I.F. $= x^{3/4}$. Solution: $f(x)x^{3/4} = 7\cdot\frac{x^{7/4}}{7/4} + C = 4x^{7/4} + C$. So $f(x) = 4x + Cx^{-3/4}$. Then $f\left(\frac{1}{x}\right) = \frac{4}{x} + Cx^{3/4}$, so $xf\left(\frac{1}{x}\right) = 4 + Cx^{7/4} \to 4$ as $x \to 0^+$.
BETA
