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 x 1 , x 2 , x 3 , x 4 ∈ ℝ be such that 0

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 x 1 , x 2 , x 3 , x 4 ∈ ℝ be such that 0 < x 1 < x 2 , 0 < x 3 < x 4 , f(x 1 ) = f(x 2 ) = 0,    f(x) ≠ 0 when x 1 < x < x 2 , g(x 3 ) = g(x 4 ) = 0 and g(x) ≠ 0 when x 3 < x < x 4 . Then, which of the following statements is/are TRUE ?

cdquestions Admin · Accepted Answer

The correct option is (B) : The function f vanishes exactly once in the interval (x 3 , x 4 ) and (D) : The function g vanishes exactly once in the interval (x 1 , x 2 ).

Answer

The function f does not vanish anywhere in the interval (x₃, x₄)

Answer

The function f vanishes exactly once in the interval (x₃, x₄)

Answer

The function g does not vanish anywhere in the interval (x₁, x₂)

Answer

The function g vanishes exactly once in the interval (x₁, x₂)

The Correct Option is B, D

Solution and Explanation

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