Question

Let $y = 4e^{-x} - 2e^{-2x} - e^{-3x}, x \in \mathbb{R}$. If $\frac{d^2y}{dx^2} = e^{\alpha x}(4e^{2x} - 8e^x - 9)$ for all $x$, then the value of the constant $\alpha$ is

• A. -3
• B. -2
• C. 3
• D. 2

Hint:

When matching a factor like $e^{\alpha x}$ to a polynomial-like expression in $e^x$, look at the term with the smallest exponent in your second derivative. Factoring out that smallest term (here $e^{-3x}$) will naturally create the $e^{2x}$ and $e^x$ terms inside the parentheses.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a function \(y\) and the structured format of its second derivative. We need to calculate the second derivative explicitly and match it to the given format to find \(\alpha\).
Step 2: Key Formula or Approach:
Calculate the first derivative \(\frac{dy}{dx}\) using \(\frac{d}{dx}(e^{kx}) = ke^{kx}\).
Calculate the second derivative \(\frac{d^2y}{dx^2}\).
Factor out the highest negative exponential to match the polynomial-like structure in the parentheses.
Step 3: Detailed Explanation:
Given \(y = 4e^{-x} - 2e^{-2x} - e^{-3x}\).
Find the first derivative:
\[ \frac{dy}{dx} = -4e^{-x} - 2(-2)e^{-2x} - (-3)e^{-3x} \] \[ \frac{dy}{dx} = -4e^{-x} + 4e^{-2x} + 3e^{-3x} \] Find the second derivative:
\[ \frac{d^2y}{dx^2} = -4(-1)e^{-x} + 4(-2)e^{-2x} + 3(-3)e^{-3x} \] \[ \frac{d^2y}{dx^2} = 4e^{-x} - 8e^{-2x} - 9e^{-3x} \] To match the given form \(e^{\alpha x}(4e^{2x} - 8e^x - 9)\), factor out \(e^{-3x}\):
\[ \frac{d^2y}{dx^2} = e^{-3x} (4e^{2x} - 8e^x - 9) \] Compare this with the given expression:
\[ e^{-3x} (4e^{2x} - 8e^x - 9) = e^{\alpha x} (4e^{2x} - 8e^x - 9) \] Thus, \(\alpha = -3\).
Step 4: Final Answer:
The value of the constant \(\alpha\) is -3.