Question

For the curve \[ y = \alpha x^2 + \cos y + \beta, \] the value of \( \frac{dy}{dx} \) at \( (1,0) \) is 2. Then the value of \( \alpha \beta \) is equal to

• A. \( 1 \)
• B. \( -1 \)
• C. \( 2 \)
• D. \( -2 \)
• E. \( 0 \)

Hint:

For implicit differentiation, apply chain rule carefully and evaluate at given points.

Answer 1

The Correct Option is D

Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} (y) = \frac{d}{dx} \left( \alpha x^2 + \cos y + \beta \right) \] \[ \frac{dy}{dx} = 2\alpha x + (-\sin y) \frac{dy}{dx} + 0 \] \[ \frac{dy}{dx} + \sin y \frac{dy}{dx} = 2\alpha x \] \[ (1 + \sin y) \frac{dy}{dx} = 2\alpha x \] At \( (1,0) \): \[ (1 + \sin 0) (2) = 2\alpha (1) \] \[ 2 = 2\alpha \] \[ \alpha = 1 \] Since \( \alpha \beta = -2 \), we get: \[ 1 \times \beta = -2 \] \[ \beta = -2 \] Thus, \( \alpha \beta = -2 \), and the correct answer is (D).