Question

Define f: ℝ → ℝ and g: ℝ → ℝ as follows
\(f(x)=\sum\limits^{\infin}_{m=0}\frac{(-1)^mx^{2m}}{2^{2m}(m!)^2}\) and \(g(x)=\frac{x}{2}\sum\limits^{\infin}_{m=0}\frac{(-1)^mx^{2m}}{2^{2m}(m+1)!m!}\) for x ∈ \(\R\).
Let x₁, x₂, x₃, x₄ ∈ ℝ be such that 0 < x₁ < x₂ , 0 < x₃ < x₄,
f(x₁) = f(x₂) = 0,    f(x) ≠ 0 when x₁ < x < x₂,
g(x₃) = g(x₄) = 0 and g(x) ≠ 0 when x₃ < x < x₄.
Then, which of the following statements is/are TRUE ?

• A. The function f does not vanish anywhere in the interval (x₃, x₄)
• B. The function f vanishes exactly once in the interval (x₃, x₄)
• C. The function g does not vanish anywhere in the interval (x₁, x₂)
• D. The function g vanishes exactly once in the interval (x₁, x₂)

Answer 1

The Correct Option is B, D

- (A) is false because \( f(x) \) vanishes exactly at \( x_1 \) and \( x_2 \), so there is a possibility that it vanishes at some point in the interval \( (x_3, x_4) \). - (B) is true because \( f(x) \) is defined as a series that has zeros at \( x_1 \) and \( x_2 \), and based on the functional form, we can conclude that \( f(x) \) vanishes exactly once in the interval \( (x_3, x_4) \). - (C) is false. The function \( g(x) \) is a sum of series that vanish at \( x_3 \) and \( x_4 \), so it is not necessarily non-zero in the interval \( (x_1, x_2) \). - (D) is true. The function \( g(x) \) vanishes at \( x_3 \) and \( x_4 \), so it vanishes exactly once in the interval \( (x_1, x_2) \). Thus, the correct answers are (B) and (D).