CUET Preparation Today
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: |
b = 2a b = -2b a = 2b b = -2a |
a = 2b |
The correct answer is Option (3) → a = 2b $\frac{dy}{dx}=-\frac{a}{x^2}+\frac{2b}{x^3}$ $\left.\frac{dy}{dx}\right]_{x=1}=0=-a+2b$ $⇒a = 2b$ |
