For a positive integer n, let $$f^{(n)}$$​ denote the n-th derivative of f.

Suppose anentire function f satisfies: $$f^{(2)}+f=0$$​​

Which of the following is correct?

Correct option is D

​$$\text{The given differential equation is:} \\[10pt]f^{(2)} + f = 0. \\[10pt]\textbf{Step 1: General solution of the differential equation} \\[10pt]\text{The characteristic equation is:} \\[10pt]m^2 + 1 = 0 \implies m = \pm i. \\[10pt]\text{Thus, the general solution of } f(z) \text{ is:} \\[10pt]f(z) = C_1 \cos z + C_2 \sin z, \\[10pt]\text{where } C_1 \text{ and } C_2 \text{ are constants.} \\[10pt]\textbf{Step 2: Compute } f^{(n)}(0) \\[10pt]\text{The derivatives of } f(z) \text{ are:} \\[10pt]f^{(n)}(z) = C_1 \cos^{(n)} z + C_2 \sin^{(n)} z. \\[10pt]\text{At } z = 0: \\[10pt]f^{(n)}(0) = C_1 \cos^{(n)}(0) + C_2 \sin^{(n)}(0). \\[10pt]\textbf{Step 3: Behavior of derivatives at } z = 0 \\[10pt]\text{Using the periodicity of } \cos z \text{ and } \sin z, \text{ we observe:} \\[10pt]\text{If } n \text{ is even:} \\[10pt]f^{(n)}(0) = C_1 \cdot (-1)^{n/2} \quad \text{(only cosine contributes)}. \\[10pt]\text{If } n \text{ is odd:} \\[10pt]f^{(n)}(0) = C_2 \cdot (-1)^{(n-1)/2} \quad \text{(only sine contributes)}. \\[10pt]\textbf{Step 4: Convergence of subsequences} \\[10pt]\text{From the above:} \\[10pt]f^{(n)}(0) \text{ alternates based on whether } n \text{ is odd or even.} \\[10pt]\text{The subsequences } f^{(\text{even})}(0) \text{ and } f^{(\text{odd})}(0) \text{ are constant (depending on } C_1 \text{ and } C_2), \\[10pt]\text{and hence each subsequence is convergent.} \\[10pt]\textbf{Final Answer:} \\[10pt]\textbf{Option D is correct.}$$​​