The condition on a and b, such that for $y=\frac{a}{x}-\frac{b}{x^2}, \frac{d y}{d x}=0$ at x = 1 is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → a = 2b
$\frac{dy}{dx}=-\frac{a}{x^2}+\frac{2b}{x^3}$
$\left.\frac{dy}{dx}\right]_{x=1}=0=-a+2b$
$⇒a = 2b$