Let \(y = x\,e^{-x}\). Consider the following statements:
(A) \(\displaystyle\frac{dy}{dx} = 0\) at x = 0.
(B) \(\displaystyle\frac{d^2y}{dx^2} = e^{-x}(x - 2)\).
(C) y is strictly increasing in \((1, \infty)\).
(D) \(\displaystyle\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
By the product rule, using \(\frac{d}{dx}e^{-x} = -e^{-x}\):
\( \displaystyle \frac{dy}{dx} = e^{-x} - x\,e^{-x} = e^{-x}(1 - x) \)
\( \displaystyle \frac{d^2y}{dx^2} = -e^{-x}(1 - x) + e^{-x}(-1) \)
\( \displaystyle \frac{d^2y}{dx^2} = e^{-x}(x - 2) \)
Since \(e^{-x} \gt 0\), the sign of \(\frac{dy}{dx}\) is the sign of \(1 - x\).
So \(\frac{dy}{dx} = e^{-x}(1 - x)\), \(\frac{d^2y}{dx^2} = e^{-x}(x - 2)\), and y is increasing for x < 1 and decreasing for x > 1.
Checking the statements
(A) \(\frac{dy}{dx} = 0\) at x = 0. — Incorrect. At x = 0, \(\frac{dy}{dx} = e^{0}(1 - 0) = 1\).
(B) \(\frac{d^2y}{dx^2} = e^{-x}(x - 2)\). — Correct. \(-e^{-x}(1 - x) - e^{-x} = e^{-x}(x - 2)\).
(C) y is strictly increasing in \((1, \infty)\). — Incorrect. For x > 1, \(1 - x \lt 0\), so y is decreasing.
(D) \(\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\). — Correct. \(e^{-x}[(x - 2) + 2(1 - x) + x] = e^{-x} \times 0 = 0\).