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:
Differentiate the function $y$ twice with respect to $x$ and rearrange the result to match the given form $e^{\alpha x}(\dots)$.

Step 2: Detailed Explanation:
1. First derivative $\frac{dy}{dx}$:
\[ \frac{dy}{dx} = 4(-1)e^{-x} - 2(-2)e^{-2x} - (-3)e^{-3x} = -4e^{-x} + 4e^{-2x} + 3e^{-3x} \]
2. Second derivative $\frac{d^2y}{dx^2}$:
\[ \frac{d^2y}{dx^2} = -4(-1)e^{-x} + 4(-2)e^{-2x} + 3(-3)e^{-3x} = 4e^{-x} - 8e^{-2x} - 9e^{-3x} \]
3. Factor out $e^{-3x}$:
\[ \frac{d^2y}{dx^2} = e^{-3x} (4e^{2x} - 8e^x - 9) \]
4. Comparing this to the given form $e^{\alpha x}(4e^{2x} - 8e^x - 9)$, we find:
\[ \alpha = -3 \]

Step 3: Final Answer:
The constant $\alpha$ is -3.