Problem 151 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The domain of \(\sin ^{-1}\left[\log _{3}(x / 3)\right]\) is : (A) \([1,9]\) (B) \([-1,9]\) (C) \([-9,1]\) (D) \([-9,-1]\)

Step-by-Step Solution

Answer

The domain is \([1, 9]\).

Step 1: Understand the domain of arcsine function

The function \( \sin^{-1} \) is defined only for inputs within the range \([-1, 1]\). This means that for \( \sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right] \) to be defined, \( \log_3 \left( \frac{x}{3} \right) \) must lie between \(-1\) and \(1\).

Step 2: Set inequalities for logarithmic condition

We need the expression \( \log_3 \left( \frac{x}{3} \right) \) to satisfy: \(-1 \leq \log_3 \left( \frac{x}{3} \right) \leq 1\)

Step 3: Solve inequality \(-1 \leq \log_3(\frac{x}{3})\)

This inequality can be solved by converting it to exponential form: \(-1 \leq \log_3(\frac{x}{3})\) implies \( \left( \frac{x}{3} \right) \geq 3^{-1} = \frac{1}{3}\). So, multiplying through by 3, we get \( x \geq 1 \).

Step 4: Solve inequality \(\log_3(\frac{x}{3}) \leq 1\)

Similarly, this inequality can be solved by rewriting: \(\log_3(\frac{x}{3}) \leq 1\) implies \( \frac{x}{3} \leq 3^1 = 3 \). Multiplying through by 3, we get \( x \leq 9 \).

Step 5: Find the domain considering both inequalities

Combining both conditions from Steps 3 and 4, we find that \( 1 \leq x \leq 9 \). Hence, the domain of the function is \([1, 9]\).

Key Concepts

Arcsine Function

The arcsine function, denoted as \(\sin^{-1}(x)\), is the inverse of the sine function. It only handles inputs in the restricted range of \([-1, 1]\). This restriction ensures the output, or angle, falls within the principal range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

• When dealing with problems involving arcsine, always remember: the value you calculate inside must lie between \(-1\) and \(1\).
• If it falls outside this range, the arcsine function will be undefined for those values.

This requirement means that any expression passed to \(\sin^{-1}(x)\), must firstly be logically checked against these boundaries, as is seen in this textbook problem. In this particular exercise, \(\sin^{-1}\left[\log_3\left(\frac{x}{3}\right)\right]\), asserts that the result of \(\log_3\left(\frac{x}{3}\right)\) must also lie between \(-1\) and \(1\), before the arcsine function can accurately process it.

Logarithmic Inequality

Logarithmic inequalities are inequalities that involve logarithmic expressions. Solving them often requires converting the logarithmic inequality to an exponential one. Let's consider a general equation, like \(\log_b(y) = x\):

• Raising the base \(b\) to the power of \(x\) gives us \(y\), which results in the equation \(y = b^x\).
• This conversion principle helps us solve the inequalities effectively.

For the exercise at hand, we have the inequality \(-1 \leq \log_3(\frac{x}{3}) \leq 1\). These two constraints can be tackled separately:

• To solve \(-1 \leq \log_3(\frac{x}{3})\), we convert it to \(\frac{x}{3} \geq 3^{-1}\), leading us to \(x \geq 1\) after multiplying both sides by 3.
• Similarly, solving \(\log_3(\frac{x}{3}) \leq 1\) results in \(\frac{x}{3} \leq 3\), leading to \(x \leq 9\).

These transformations help pinpoint the range of acceptable \(x\) values by checking them against the logarithmic boundaries dictated by the base 3.

Function Domain

Understanding a function's domain is crucial, as it defines all possible input values for which the function is valid. For composite functions like \(\sin^{-1}\left[\log_3\left(\frac{x}{3}\right)\right]\), determining the domain involves:

• Evaluating each separate function involved.
• Analyzing how the results interplay to satisfy the conditions of each part.

In our problem, understanding the domain revolves around acknowledging that the output of the logarithm must satisfy the input conditions of the arcsine function. After understanding both the arcsine restriction of \([-1, 1]\) and the logarithmic inequalities, we combine inequalities to find a shared range where all conditions are satisfied.

• This range, coming from \(x \geq 1\) and \(x \leq 9\), establishes the valid domain as \([1, 9]\).
• Hence, every acceptable input within this range meets both the requirements of the arcsine function and the logarithmic part.

This method ensures that every possible \(x\) produces a well-defined and computable result for the entire function.