Question

Let $ f : \mathbb{R} \to \mathbb{R} $ be a continuous function. If $ px + my + n = 0 $ is a tangent drawn to the curve $ y = f(x) $ at $ x = \alpha $, then at $ x = 0 $, the value of \[ \frac{d}{dx} \left( f(\alpha e^{2x}) \right) \]

• A. \( 0 \)
• B. \( \frac{p}{m} \)
• C. \( \frac{-2am}{p} \)
• D. \( \frac{-2p\alpha}{m} \)

Hint:

For composite functions, use the chain rule to differentiate. When dealing with tangents to curves, the slope at a given point can help in evaluating derivatives.

Answer 1

The Correct Option is D

We are given that the tangent to the curve at \( x = \alpha \) is \( px + my + n = 0 \). This means the point \( (\alpha, f(\alpha)) \) lies on the curve and the slope of the tangent at \( x = \alpha \) is given by \( \frac{-p}{m} \). Therefore, the derivative of \( f(x) \) at \( x = \alpha \) is: \[ f'(\alpha) = \frac{-p}{m} \] Step 1: Differentiate \( f(\alpha e^{2x}) \) using the chain rule. We need to find the derivative of the composite function \( f(\alpha e^{2x}) \). Using the chain rule: \[ \frac{d}{dx} \left( f(\alpha e^{2x}) \right) = f'(\alpha e^{2x}) \cdot \frac{d}{dx}(\alpha e^{2x}) \] The derivative of \( \alpha e^{2x} \) with respect to \( x \) is: \[ \frac{d}{dx}(\alpha e^{2x}) = 2\alpha e^{2x} \] So, the derivative becomes: \[ \frac{d}{dx} \left( f(\alpha e^{2x}) \right) = f'(\alpha e^{2x}) \cdot 2\alpha e^{2x} \] Step 2: Evaluate at \( x = 0 \). At \( x = 0 \), we have: \[ \frac{d}{dx} \left( f(\alpha e^{2x}) \right) \bigg|_{x=0} = f'(\alpha e^{0}) \cdot 2\alpha e^{0} = f'(\alpha) \cdot 2\alpha \] From Step 1, we know that \( f'(\alpha) = \frac{-p}{m} \), so: \[ f'(\alpha) \cdot 2\alpha = \frac{-p}{m} \cdot 2\alpha = \frac{-2p\alpha}{m} \] Final Answer: \[ \boxed{\frac{-2p\alpha}{m}} \]