Problem 80 - A Textbook of Differential Calculus

Question

Match the statements of Column I with values of Column II. Column I (A) \(\sqrt{\sin (\cos x)}\), has domain (B) \((\sqrt{\cos (\sin x)})^{-1}\), has domain (C) \(\tan (\pi \sin x)\), has domain (D) \(\ln (\tan x)\), has domain Column II (p) \(\quad x \in R\) (q) \(R-\left\\{n \pi \pm \frac{\pi}{6}\right\\}\) (r) \(\quad x \in\left(n \pi, n \pi+\frac{\pi}{2}\right)\) (s) \(\quad x \in\left[2 n \pi-\frac{\pi}{2}, 2 n \pi+\frac{\pi}{2}\right]\)

Step-by-Step Solution

Answer

(A) matches (s), (B) matches (p), (C) matches (q), (D) matches (r).

Step 1

Analyze the expression \( \sqrt{\sin (\cos x)} \). The domain of \( \sin \theta \) is all real numbers, giving \( \cos x \) the domain of all real numbers. Since we need \( \sin (\cos x) \geq 0 \) for the square root to be defined, and \( \sin \theta \geq 0 \) when \( \theta \) is between \( 2n\pi \) and \( 2n\pi + \pi \), we substitute \( \theta = \cos x \) and solve for \( x \). Therefore, \( x \) falls in the intervals \( [2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}] \). So, (A) matches (s).

Step 2

Analyze the expression \( \left(\sqrt{\cos (\sin x)}\right)^{-1} \). First, ensure that \( \cos (\sin x) \geq 0 \). Since \( -1 \leq \sin x \leq 1 \), \( \sin x \) is never outside this interval. We know \( \cos \theta \geq 0 \) when \( \theta \) lies between \( 2n\pi \) and \( 2n\pi + \pi \). Hence, the domain for \( x \) is all real numbers, as there are no additional restrictions. (B) matches (p).

Step 3

Analyze the expression \( \tan (\pi \sin x) \). The tangent function has a domain with exclusions where the argument equals \( (2n+1)\frac{\pi}{2} \). For \( \pi \sin x eq (2n+1)\frac{\pi}{2} \), solve for \( \sin x \), which gives \( \sin x eq \pm \frac{1}{2} \). Thus, \( x eq n\pi \pm \frac{\pi}{6} \). Therefore, (C) matches (q).

Step 4

Analyze the expression \( \ln (\tan x) \). The logarithm function requires its argument, \( \tan x \), to be greater than 0. Consider that \( \tan x > 0 \) within the intervals \( n\pi, n\pi+\frac{\pi}{2} \). This means \( x \) must lie within these intervals as well. Hence, (D) matches (r).

Key Concepts

Domain of Functions

Understanding the domain of functions is crucial in calculus and algebra. The domain represents all the possible inputs (or values of x) for which the function is defined. When we look at expressions involving trigonometric or inverse functions, identifying their domain is often the first step in analysis. For instance, in the expression \( \sqrt{\sin(\cos x)} \), we have to ensure the inner function \( \cos x \) is defined for all real numbers and that \( \sin(\cos x) \geq 0 \). This limitation leads us to specific intervals where \( \cos x \) results in non-negative values for \( \sin(\cos x) \). Thus, the domain of this function corresponds to the set where \( x \) belongs to \([2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}]\).

Always remember, the function must operate correctly within its domain, ensuring output values are real and valid under the given conditions.

In summary, identifying the domain is all about finding which values of x keep the function "working without breaking the rules." Make sure to analyze all transformations and compositions within the function.

Trigonometric Functions

Trigonometric functions, like \(\sin\), \(\cos\), and \(\tan\), play a fundamental role in mathematics, particularly in calculus. Each of these functions has specific properties and periodic behaviors. The function \(\tan(\pi \sin x)\) from the given exercise highlights these features.

The tangent function has vertical asymptotes and is undefined where its input equals \((2n+1)\frac{\pi}{2}\). Thus, to find its domain, we avoid these values, solving inequalities like \(\pi \sin x eq (2n+1)\frac{\pi}{2}\). This requires understanding when \(\sin x \) equals specific values, leading us to exclusions like \(x eq n\pi \pm \frac{\pi}{6}\).

Another notable point is the range of sine, which is confined between -1 and 1, impacting how it interacts within other trigonometric expressions. This can affect the resulting intervals and exclusions, essential for correctly defining the domain.

Inverse Functions

Inverse functions allow us to reverse operations, solving equations that include functions by 'undoing' them. The idea is that if you have a function that takes you from A to B, its inverse will take you from B back to A.

The function \((\sqrt{\cos(\sin x)})^{-1}\) from the exercise asks us to consider the inverse square root. Knowing that the inverse function swaps the domain and range of the original function provides significant insight. Here, \( \cos(\sin x) \) was already conditioned to be non-negative, making the inverse of the square root function defined across all real x-values without additional restrictions, unlike regular scenarios involving inverse functions.

In essence, when dealing with inverse functions, focus on the necessary preconditions for both the function and its proposed inverse to ensure they are properly defined. Always ensure the output of the original function falls within the inputs possible for its inverse.