Question

The condition on a and b, such that for \(y = \frac{a}{x}-\frac{b}{x²}\), \(\frac{dy}{dx} =0\) at x=1 is:

• A. b=2a
• B. b=-2b
• C. a=2b
• D. b=-2a

Answer 1

The Correct Option is C

To solve for the condition on \(a\) and \(b\) such that for the given function \(y = \frac{a}{x} - \frac{b}{x^2}\), the derivative \(\frac{dy}{dx} = 0\) at \(x = 1\), follow these steps: 
Calculate the derivative \(\frac{dy}{dx}\) using standard differentiation rules. For \(y = \frac{a}{x} - \frac{b}{x^2}\), we recognize the terms as \(\frac{a}{x} = ax^{-1}\) and \(\frac{b}{x^2} = bx^{-2}\). 
The derivative is: \[ \frac{dy}{dx} = \frac{d}{dx}(ax^{-1}) - \frac{d}{dx}(bx^{-2}) \] Using the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\), we find: \[ \frac{d}{dx}(ax^{-1}) = -ax^{-2} \quad \text{and} \quad \frac{d}{dx}(bx^{-2}) = -2bx^{-3} \] Thus, \[ \frac{dy}{dx} = -ax^{-2} + 2bx^{-3} \] Evaluating at \(x = 1\): \[ \frac{dy}{dx} = -a(1)^{-2} + 2b(1)^{-3} = -a + 2b \] Set this expression equal to zero for the condition: \[ -a + 2b = 0 \] Solving for a gives: \[ a = 2b \] Therefore, the condition on \(a\) and \(b\) is \(a = 2b\).