91Ó°ÊÓ

Let \(x = \sin^2{\frac{\theta}{n}}\). Then the function becomes: $$f(\theta) = \frac{x^2 - x + 1}{x^2 + x + 1}$$

We can find the domain of the function by finding the range of \(x = \sin^2{\frac{\theta}{n}}\). Since \(\sin\) function ranges from -1 to 1, and we are squaring the function, x can be any value between 0 and 1, inclusive. So the domain of x is: \(0 \le x \le 1\)

Take the derivative of the function with respect to x and set it to zero to find critical points: $$\frac{d}{dx} \frac{x^2 - x + 1}{x^2 + x + 1} = 0$$ To find the derivative, we will use the quotient rule: \(\frac{d}{dx}\frac{u}{v} = \frac{u'v - uv'}{v^2}\) Where \(u = x^2 - x + 1\) and \(v = x^2 + x + 1\). Then: $$u' = 2x - 1$$ $$v' = 2x + 1$$ Using quotient rule: $$\frac{d}{dx}\frac{x^2 - x + 1}{x^2 + x + 1} = \frac{(2x - 1)(x^2 + x + 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}$$ Solving for the numerator: Numerator = \((2x^3 + 2x^2 + 2x) - (2x^3 - x^2 - x) = 3x^2 + 3x\) Setting the derivative to zero: \(3x^2 + 3x = 0\) \(x * (3x + 3) = 0\) We have 2 critical points: \(x = 0\) and \(x = -1\). Since x is in the domain of [0, 1], we only consider x = 0 as a critical point.

Evaluate the function \(f(\theta)\) at \(x = 0\) and at the endpoints of the domain (0 and 1) to find the minimum and maximum values: $$f(0) = \frac{(-0)^2 - 0 + 1}{0^2 + 0 + 1} = 1$$ $$f(1) = \frac{1^2 -1 + 1}{1^2 + 1 + 1} = \frac{1}{3}$$ Thus, the function's minimum value is \(\frac{1}{3}\) and its maximum value is 1.

Since the minimum value is \(\frac{1}{3}\) and the maximum value is 1, the range of \(f(\theta)\) is [1/3, 1]. Hence the correct option is: (B) \([1 / 3,1]\)