Question

Let \( y \) be a twice differentiable function on \( \mathbb{R} \) satisfying \[ y''(x) = 2 + e^{-|x|}, \quad x \in \mathbb{R}, \quad y(0) = -1, \quad y'(0) = 0. \] Then

• A. \(y = 0\) has exactly one root.
• B. \(y = 0\) has exactly two roots.
• C. \(y = 0\) has more than two roots.
• D. There exists an \(x_0 \in \mathbb{R}\) such that \(y(x_0) \ge y(x)\) for all \(x \in \mathbb{R}.\)

Hint:

If \(y''(x)>0\) everywhere, the function is convex, and any local minimum is global. Such graphs typically have at most two roots.

Answer 1

The Correct Option is B

Step 1: Analyze \(y''(x).\)
Since \(y''(x) = 2 + e^{-|x|}>0\) for all \(x \in \mathbb{R}\), \(y'(x)\) is strictly increasing.
Step 2: Integrate for \(y'(x).\)
For \(x>0\): \[ y'(x) = \int_0^x (2 + e^{-t}) \, dt = 2x + (1 - e^{-x}). \] For \(x<0\): \[ y'(x) = \int_0^x (2 + e^{t}) \, dt = 2x + (e^{x} - 1). \] Thus \(y'(x)<0\) for \(x<0\) and \(y'(x)>0\) for \(x>0\).
Step 3: Behavior of \(y(x)\).
Since \(y'(x)\) changes sign from negative to positive at \(x=0\), \(y(x)\) has a minimum at \(x=0\) where \(y(0) = -1.\) As \(x \to \infty\), \(y(x) \to \infty\); as \(x \to -\infty\), \(y(x) \to \infty\).
Step 4: Conclusion.
Therefore, \(y(x)\) decreases to \(-1\) at \(x=0\) and increases on both sides, crossing \(y=0\) exactly twice. Hence, (B) is correct.