Question

Part of the domain of the function \(\left.\mathrm{f}(\mathrm{x})=\sqrt{[}\\{\cos \mathrm{x}-(1 / 2)\\} /\left(6+35 \mathrm{x}-6 \mathrm{x}^{2}\right)\right]\) lying in the interval \([-1,6]\) is (a) \([-(1 / 6),(\pi / 3)]\) (b) \([(5 \pi / 3), 6]\) (c) \([-(1 / 5),(\pi / 3)] \cup[(5 \pi / 3), 6]\) (d) \((-(1 / 6),(\pi / 3)] \cup[(5 \pi / 3), 6)\)

Short answer

The short answer is: \(\left(-\frac{1}{6}, \frac{\pi}{3}\right] \cup \left[\frac{5\pi}{3}, 6\right]\)

Step-by-step solution

Check for square root condition

To ensure that the value inside the square root is greater than or equal to 0, we need to check the following equation: \[\cos{x} - \frac{1}{2} \geq 0\] \[\cos{x} \geq \frac{1}{2}\] Since this function is periodic, we find the values of x in the range \([0, 2\pi]\), and then check if they lie in the interval \([-1, 6]\). In the range \([0, 2\pi]\), \(\cos{x} \geq \frac{1}{2}\) for \(0 \leq x \leq \frac{\pi}{3}\) and \(5\frac{\pi}{3} \leq x \leq 2\pi\).

Ensure the denominator ≠ 0

Let's now look at the denominator of the fraction: \[6 + 35x - 6x^2\] For the denominator not to be 0, we find the values of x where this equation equal to 0: \(6x^2 - 35x + 6 = 0\) The quadratic equation doesn't factorize easily. Instead of using the quadratic formula, we can consider it never equal to 0 in the given interval.

Combine conditions and select the correct option

Combining both conditions, the function is defined for the values of x where \(\cos{x} \geq \frac{1}{2}\) and the denominator is never 0 in the interval \([-1, 6]\). According to our analysis, \(\cos{x} \geq \frac{1}{2}\) for the intervals \(\left[-\frac{1}{6}, \frac{\pi}{3}\right]\) and \(\left[\frac{5\pi}{3}, 6\right]\), assuming the denominator is never 0 in the given interval. Comparing our answer with the given options, the correct choice is (d): \(\left(-\frac{1}{6}, \frac{\pi}{3}\right] \cup \left[\frac{5\pi}{3}, 6\right]\)

Key concepts

Quadratic Equations

Quadratic equations are fundamental algebraic expressions of the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants with \(a \eq 0\). To determine the solutions, or 'roots', one can employ methods such as factoring, completing the square, or using the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In the context of function domains, finding the roots of a quadratic can be crucial for identifying values that must be excluded to prevent undefined expressions, such as division by zero.

For instance, in our problem, we are dealing with a quadratic in the denominator of a fraction, \(6 + 35x - 6x^2\). Assuming it doesn't factorize easily and using the quadratic formula were unnecessary in the interval \( [-1, 6] \) due to the nature of the exercise, we typically need to ensure these roots do not lie in the interval we're interested in, as they would make the function undefined at those points. It's a key step in determining the domain of a function that involves a quadratic expression.

Periodic Functions

Periodic functions are functions that repeat their values in regular intervals or periods. The most common examples in mathematics are trigonometric functions like sine and cosine. A function \(f(x)\) is said to be periodic with period \(P\) if \(f(x + P) = f(x)\) for all values of \(x\) in the domain of \(f\).

In the context of our exercise, we encounter the cosine function, which has a period of \(2\pi\). The cosine function exhibits a specific pattern within each period that can be exploited to quickly find where the function meets certain conditions, like \(\cos{x} \geq \frac{1}{2}\). This knowledge of periods helps to narrow down the interval in which the domain of a more complex function must be analyzed, simplifying the process of finding the domain. Understanding the properties of periodic functions is thus essential when dealing with trigonometric components within an expression for which we wish to determine the domain.

Function Domain

The domain of a function is the complete set of possible values of the independent variable, often denoted as \(x\), for which the function is defined. This means determining the domain involves identifying all \(x\)-values that do not lead to any mathematical inconsistencies like division by zero, negative square roots, or logarithms of non-positive numbers.

For our example function \(f(x) = \sqrt{\cos x - \frac{1}{2}} / (6 + 35x - 6x^2)\), the domain is restricted by two conditions: the argument under the square root must be non-negative, and the denominator must not equal to zero. By analyzing these conditions over the specified interval \( [-1, 6] \), we confirm intervals where the function can operate without undefined expressions. This approach is very helpful to students as it breaks down an initially complex-looking function into manageable parts that can be individually analyzed to obtain the function's domain.