Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∈ N. The function f(x) is defined asf(x)=\(\begin{cases}2x+g(2x)& :x\in [0,\tau] \\ 2x+g(4x)& :x\in (\tau,2\tau] \\2x+g(6x)& :x\in (2\tau,3\tau]\\.\\.\\2x+g(2nx)& :x\in [(n-1)\tau,n\tau)\\.\\.\end{cases}\) How many points c ∈ [0, 18τ] exist such that f'(c) = tan-1(2)

Let g(x) be periodic and non-constant, with period τ. Also we have g

1.	Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∈ N. The function f(x) is defined asf(x)=\(\begin{cases}2x+g(2x)& :x\in [0,\tau] \\ 2x+g(4x)& :x\in (\tau,2\tau] \\2x+g(6x)& :x\in (2\tau,3\tau]\\.\\.\\2x+g(2nx)& :x\in [(n-1)\tau,n\tau)\\.\\.\end{cases}\) How many points c ∈ [0, 18τ] exist such that f'(c) = tan-1(2)
A.	325
B.	323
C.	= tan-1(2)a) 325b) 323c) 324
D.	162view answer
Answer» D. 162view answer

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Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∈ N. The function f(x) is defined asf(x)=\(\begin{cases}2x+g(2x)& :x\in [0,\tau] \\ 2x+g(4x)& :x\in (\tau,2\tau] \\2x+g(6x)& :x\in (2\tau,3\tau]\\.\\.\\2x+g(2nx)& :x\in [(n-1)\tau,n\tau)\\.\\.\end{cases}\) How many points c ∈ [0, 18τ] exist such that f'(c) = tan-1(2)
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