Question

Let y : \(\R → \R\) be a twice differentiable function such that y" is continuous on [0, 1] and y(0) = y(1) = 0. Suppose y"(x) + x² < 0 for all x ∈ [0, 1]. Then

• A. y(x) > 0 for all x ∈ (0, 1)
• B. y(x) < 0 for all x ∈ (0, 1)
• C. y(x) = 0 has exactly one solution in (0, 1)
• D. y(x) = 0 has more than one solution in (0, 1)

Answer 1

The Correct Option is A

To solve this problem, we start by analyzing the information given about the function \( y(x) \). We know that:

• \( y : \R \to \R \) is a twice differentiable function.
• \( y''(x) \) is continuous on the interval \([0, 1]\), and \( y(0) = y(1) = 0 \).
• It is given that \( y''(x) + x^2 < 0 \) for all \( x \in [0, 1] \). 

Let's analyze what \( y''(x) + x^2 < 0 \) implies:

• Rewriting, we have \( y''(x) < -x^2 \). Since \( x^2 \) is non-negative for \( x \in [0, 1] \), this implies \( y''(x) \) is strictly negative.

Since \( y''(x) < 0 \), this implies that \( y'(x) \) is a decreasing function over the interval \([0, 1]\).

Additionally, since \( y(0) = 0 \) and \( y(1) = 0 \), it means that as per the boundary conditions, the function value returns to zero at both ends of the interval:

• At \( x = 0 \), the function starts from 0.
• At \( x = 1 \), the function ends at 0.

To understand the behavior of \( y(x) \), consider the implications of \( y''(x) < 0 \):

• The function \( y'(x) \) is decreasing on \([0, 1]\), which means it reaches a maximum value at some point \( c \in (0, 1) \). Since the values return to zero, this maximum implies positivity in the middle of the interval.
• The condition given and combined continuity suggest that \( y(x) \) does not dip below the x-axis within this range.

Given that \( y(x) \) is continuous and differentiable, and \( y''(x) < 0 \) ensures a pure downward concavity, it ensures \( y(x) \) would be positive for all \( x \in (0,1) \).

Therefore, the correct statement regarding the function \( y(x) \) is:

• \( y(x) > 0 \) for all \( x \in (0, 1) \).