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Question: If \((1,\infty)\) be two fixed positive integers such that \(\lbrack 1,\infty)\) for all real \(y =...

If (1,∞)(1,\infty) be two fixed positive integers such that

[1,∞)\lbrack 1,\infty) for all real y=12−sin⁡3xy = \frac{1}{2 - \sin 3x} then 13≤y≤1\frac{1}{3} \leq y \leq 1 is a periodic function with period

A

a

B

2 a

C

b

D

2 b

Answer

2 a

Explanation

Solution

f(a+x)=b+(1+{b−f(x)}3)1/3f ( a + x ) = b + \left( 1 + \{ b - f ( x ) \} ^ { 3 } \right) ^ { 1 / 3 }

⇒f(a+x)−b={1−{f(x)−b}3}1/3f ( a + x ) - b = \left\{ 1 - \{ f ( x ) - b \} ^ { 3 } \right\} ^ { 1 / 3 }

⇒ϕ(a+x)={1−{ϕ(x)}3}1/3\phi ( a + x ) = \left\{ 1 - \{ \phi ( x ) \} ^ { 3 } \right\} ^ { 1 / 3 } [ϕ(x)=f(x)−b\phi ( x ) = f ( x ) - b]

⇒ ϕ(x+2a)={1−{ϕ(x+a)}3}1/3=ϕ(x)\phi ( x + 2 a ) = \left\{ 1 - \{ \phi ( x + a ) \} ^ { 3 } \right\} ^ { 1 / 3 } = \phi ( x )

⇒f(x+2a)−b=f(x)−b⇒f(x+2a)=f(x)f ( x + 2 a ) - b = f ( x ) - b \Rightarrow f ( x + 2 a ) = f ( x )

∴f(x)f ( x ) is periodic with period 2a2 a.