Problem 46 - Calculus for JEE ADVANCED

Question

Since \(\\{x\\} \in[0,1), \sin \\{x\\} \in(0, \sin 1)\) as \(f(x)\) is defined if \(\sin \\{x\\} \neq 0\), i.e. $$ \frac{1}{\sin \\{x\\}} \in\left(\frac{1}{\sin 1}, \infty\right) \text { or }\left[\frac{1}{\sin \\{x\\}}\right] \in\\{1,2,3, \ldots\\} $$ Note that \(1<\frac{\pi}{3}\) or \(\sin 1<\sin \frac{\pi}{3}=0.866\) or \(\frac{1}{\sin 1}>1.155\)

Step-by-Step Solution

Answer

The function is defined for integer values \([\frac{1}{\sin \{x\}}] \geq 2\).

Step 1: Understand the Given Range for \( \{x\} \)

We are given that \( \{x\} \in [0,1) \), which indicates that we are dealing with the fractional part of \( x \). This means the value of \( \{x\} \) is between 0 (inclusive) and 1 (exclusive).

Step 2: Determine the Range of \( \sin \{x\} \)

Since \( \{x\} \in [0,1) \), it follows that \( \sin \{x\} \) is in the range \((0, \sin 1)\) because \( \sin x \) increases from \(0\) to approximately \(0.84147\) as \(x\) goes from \(0\) to \(1\) radian.

Step 3: Analyze the Condition \( \frac{1}{\sin \{x\}} \)

For the function \( f(x) \) to be defined, \( \sin \{x\} eq 0 \). Thus, \( \frac{1}{\sin \{x\}} \) must be in the interval \( \left( \frac{1}{\sin 1}, \infty \right) \). Here, \( \frac{1}{\sin 1} \approxeq 1.188395 \).

Step 4: Simplify Using \( \frac{1}{\sin 1} > 1.155 \)

Since \( \sin 1 < \sin \frac{\pi}{3} \), we have that \( \frac{1}{\sin 1} > 1.155 \). It implies that the minimum value for \( \left[ \frac{1}{\sin \{x\}} \right] \) can start from 2.

Step 5: Conclusion Using Integer Values

Given \( \left[ \frac{1}{\sin \{x\}} \right] \in \{1, 2, 3, \ldots\} \), and knowing that \( \frac{1}{\sin x} > 1.155 \), the integer values satisfying this are \( \left[ \frac{1}{\sin \{x\}} \right] \geq 2 \). Therefore, the domain is defined where this condition holds.

Key Concepts

Understanding the Sine Function

The sine function, one of the fundamental functions in trigonometry, relates the angle of a right triangle to the length of its opposite side and its hypotenuse. It is expressed mathematically as \( \sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} \). This function oscillates between -1 and 1 as the angle changes from \( 0 \) to \( 2\pi \).

• The sine function is periodic with a period of \( 2\pi \), meaning \( \sin(x) = \sin(x + 2\pi) \).
• For angles in radians, as in our exercise, \( \sin(x) \) gradually rises from 0 to 1 between 0 and \(\frac{\pi}{2}\), and then returns to 0 by \(\pi\).
• The graph of sine function shows a smooth wave, known as a sine wave, representing these oscillations.

The value of \( \sin(x) \) around very small fractions of \( x \), such as seen in \( \sin \{x\} \) (where \( \{x\} \) describes fractional parts), involves values between 0 and \( \sin(1) \). Since \( \sin(1) \approx 0.84147 \), this gives insight into the smallest interval for \( \sin \{x\} \) in our problem.

Exploring the Domain of a Function

The domain of a function concerns itself with all the possible input values, or "x-values," for which a function is defined. For practical solutions, it is vital to ensure that we do not encounter undefined mathematical operations such as division by zero.

• In our specific case, for the function \( \frac{1}{\sin \{x\}} \) to be defined, it's crucial that \( \sin \{x\} eq 0 \) because division by zero is undefined and problematic in mathematics.
• Thus, we derive the condition \( \{x\} \in [0,1) \) to mean the fractional part \( \{x\} \) cannot result in a sine value of 0.
• An additional step involves ensuring that the fraction \( \frac{1}{\sin \{x\}} \) results in an integer. We establish a specific interval \( (\frac{1}{\sin(1)}, \infty) \) where these calculated values exist, helping further define our function's domain.

By defining these critical intervals, we find integer values starting from 2 that keep our function well-defined and functioning, ensuring proper operation within this spectral range.

Delving into Inverse Trigonometric Functions

Inverse trigonometric functions enable the calculation of angles from known trigonometric ratios. These functions help bridge the gap when you know the sine, cosine, or tangent and want to find the respective angle, different from the usual trigonometric functions that start with angles.

• The conventional inverse sine function, \( \text{asin} \), is expressed as \( \sin^{-1}(y) \), where \( -1 \leq y \leq 1 \), producing angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• Important in problems involving inverse relationships, the inverse sine can answer queries like "What angle has a sine of 0.5?"
• This function and its limitations, such as the range of outputs, are necessary for complex calculations associated with both geometric and real-world problems.
• Understanding these relationships, particularly when working backwards from a fractional or known sine, can aid greatly in solving advanced trigonometry problems.

Commonly, inverse trigonometric functions are represented as \( \sin^{-1}, \cos^{-1}, \tan^{-1} \) and are vital in gathering insight, allowing the tracing back from trigonometric values to their respective angles, as utilized in a myriad of scientific and engineering applications.