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

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: Understanding the inverse sine function

The function \( \sin^{-1}(y) \) is defined only when \( y \) lies in the interval \([-1, 1]\). Therefore, 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: Setting up the inequalities

We need to establish two inequalities based on the condition from the inverse sine function. Thus, we have: 1. \( \log_{3}\left( \frac{x}{3} \right) \geq -1 \)2. \( \log_{3}\left( \frac{x}{3} \right) \leq 1 \)

Step 3: Solving the first inequality

Consider the inequality \( \log_{3}\left( \frac{x}{3} \right) \geq -1 \). This implies:\[\frac{x}{3} \geq 3^{-1} = \frac{1}{3}\]Therefore, multiplying both sides by 3 gives:\[x \geq 1\]

Step 4: Solving the second inequality

Now, consider the inequality \( \log_{3}\left( \frac{x}{3} \right) \leq 1 \). This implies:\[\frac{x}{3} \leq 3^{1} = 3\]Therefore, multiplying both sides by 3 gives:\[x \leq 9\]

Step 5: Combining the inequalities

By combining the results from the first and second inequalities, we find the domain of the function is:\[x \in [1, 9]\]

Step 6: Verifying the endpoint inclusion

Both endpoints 1 and 9 satisfy the inequalities, as substituting them into the logarithmic function results in values \(-1\) and \(1\) respectively, which lie exactly at the boundary of the definition for \( \sin^{-1}(y) \).

Key Concepts

Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1}(y) \), is fundamental in trigonometry. It allows us to find the angle whose sine is a given number, \( y \). However, it's only defined when \( y \) is between -1 and 1, inclusive. This is crucial because it restricts which values you can plug into the function. For example, you can compute \( \sin^{-1}(0.5) \), but not \( \sin^{-1}(1.5) \) because 1.5 is outside the acceptable range.
When dealing with composite functions like \( \sin^{-1}\left[ \log_{3}(x/3) \right] \), the inner function, \( \log_{3}(x/3) \), must produce results that are within the range of -1 to 1, so the inverse sine function is valid.

Logarithmic Inequalities

Logarithmic inequalities involve expressions like \( \log_{b}(x) \leq c \), where we determine the values of \( x \) that make the inequality true. These inequalities often need to be solved in stages, first by addressing the logarithm and then proceeding with algebraic manipulation.
For example, solving \( \log_{3}\left( \frac{x}{3} \right) \leq 1 \) begins by rewriting it as \( \frac{x}{3} \leq 3^{1} \), which simplifies to \( x \leq 9 \). The property of logarithms that \( \log_{b}(y) = c \) implies \( y = b^{c} \) is a key tool here.
Combining solutions from multiple inequalities gives you the range of \( x \) values that satisfy the original problem.

Function Inequalities

Function inequalities occur when we set conditions for composite functions to be defined, like ensuring \( \sin^{-1}\left[ \log_{3}(x/3) \right] \) is valid by checking the domain of each individual part. This involves setting up and solving separate inequalities for each part.

• Start by focusing on the range of the inner function (here, \( \log_{3}(x/3) \)). This value must satisfy the inverse sine function's requirements, hence it needs to be between -1 and 1.
• Next, convert these requirements into inequalities, and solve them to find the domain of the variable \( x \).

This approach of breaking down and recombining inequalities can be applied to tackle various function constraints in mathematics.