Question

Let 𝑓: ℝ→ ℝ be a function defined by 𝑓(𝑥)=𝑥²−𝑥, 𝑥 ∈ ℝ. Let 𝑔: ℝ→ℝ be a twice differentiable function such that 𝑔(𝑥)=0 has exactly three distinct roots in the open interval (0, 1). Let ℎ(𝑥) = 𝑓(𝑥)𝑔(𝑥), 𝑥 ∈ ℝ, and ℎ′′ be the second order derivative of the function ℎ. If 𝑛 is the number of roots of ℎ′′(𝑥)=0 in (0, 1), then the minimum possible value of 𝑛 equals ________

Answer 1

Correct Answer: 3

To solve the problem, we need to consider the function \( h(x) = f(x)g(x) \) where \( f(x) = x^2 - x \) and \( g(x) \) has three distinct roots in the interval (0, 1). We are tasked with finding the number of roots of \( h''(x) = 0 \) in the interval (0, 1), and ensuring this value matches the minimum required, which is 3.

1. Start by computing \( h'(x) \) using the product rule:
\( h'(x) = f'(x)g(x) + f(x)g'(x) \).
Here, \( f'(x) = 2x - 1 \). Thus, \( h'(x) = (2x - 1)g(x) + (x^2 - x)g'(x) \).

2. Now compute \( h''(x) \):
Apply the product rule again:
\( h''(x) = ((2x - 1)g(x) + (x^2 - x)g'(x))' \).
Break it down:
\((2x - 1)g(x)\) gives \((2)g(x) + (2x - 1)g'(x)\).
\((x^2 - x)g'(x)\) gives \((2x - 1)g'(x) + (x^2 - x)g''(x)\).
Thus,
\( h''(x) = 2g(x) + 2(2x - 1)g'(x) + (x^2 - x)g''(x) \).

3. Consider the roots of \( g(x) = 0 \) in (0, 1): say \( r_1, r_2, r_3 \). In-between these roots, by Rolle's Theorem, there must exist at least two roots of \( g'(x) = 0 \) because \( g(x) \) is twice differentiable.

4. To find the roots of \( h''(x) = 0 \), note that \( h''(x) \) involves \( g'(x) \) and \( g''(x) \), and by step 3, \( g'(x) \) will have roots between \( r_1, r_2, \) and \( r_3 \), giving \( n = 3 \) as the minimum number of roots (one between each pair of consecutive roots of \( g(x) \)).

5. Verify the minimum possible value of \( n \) in the open interval (0, 1). As derived, \( n = 3 \), which matches our expected range and confirms the solution.

Therefore, the minimum possible value of \( n \) is 3.