Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The domain of the function \(f(x)=\log _{2}\left(-\log _{1 / 2}\left(1+\frac{1}{\sqrt[4]{x}}\right)-1\right)\) is (A) \(01\)

Step-by-Step Solution

Answer

The domain of the function is (A) \(0 < x < 1\).

Step 1: Identify the log conditions

Given the function \( f(x) = \log_{2}(-\log_{1 / 2}(1+\frac{1}{\sqrt[4]{x}})-1) \), we start by identifying the conditions under which the logarithms are defined. The inside expression of any logarithm must be greater than zero. Start with the innermost log: \(-\log_{1 / 2}(1+\frac{1}{\sqrt[4]{x}})-1>0\).

Step 2: Solve for the base condition

Rearranging the inequality from Step 1: \(-\log_{1 / 2}(1+\frac{1}{\sqrt[4]{x}}) > 1\). Now, since \( \log_{1/2}a = -\log_{2}a \), we can rewrite this as \(\log_{2}(1+\frac{1}{\sqrt[4]{x}}) > 1\).

Step 3: Simplify and solve the inequality

Since \(\log_{2}(1+\frac{1}{\sqrt[4]{x}}) > 1\), we convert this to an exponential form: \(1+\frac{1}{\sqrt[4]{x}} > 2^1 = 2\). Thus, \(\frac{1}{\sqrt[4]{x}} > 1\).

Step 4: Solve the fractional inequality

The inequality \(\frac{1}{\sqrt[4]{x}} > 1\) can be converted to an expression of \(x\): \(1 > \sqrt[4]{x}\), which implies raising both sides to the fourth power gives \(1 > x\). Thus, \(0 < x < 1\) is our domain.

Key Concepts

Logarithmic Functions

Logarithmic functions play a crucial role in this exercise, and understanding their properties is essential for solving it. The logarithm function is an inverse of the exponential function. For the function \( y = \log_b(x) \), \( b \) is the base and must be a positive number not equal to 1, and \( x \) is the value we're taking the logarithm of.

Here are some key points to remember about logarithms:

• The domain of a logarithmic function \( \log_b(x) \) is \( x > 0 \), meaning the function is only defined for positive values of \( x \).
• Specific bases, like the natural logarithm (base \( e \)) or the common logarithm (base 10), have their particular uses, but the rules for defining a logarithm remain the same across all bases.

In the given function, we have a nested logarithm, which means we must ensure the argument of each logarithm is positive. This condition is the starting point for determining the domain of the entire function.

Inequality Solving

Solving inequalities is a fundamental skill in working with functions, especially when determining domains. In the context of our exercise, once we find the inequalities from the logarithmic conditions, we need to solve them to pinpoint valid values of \( x \).

The critical inequality identified here is \(-\log_{\frac{1}{2}}(1+\frac{1}{\sqrt[4]{x}}) - 1 > 0\). Rearranging and solving inequalities involves several steps:

• First, rearrange the inequality to isolate the term involving \( x \).
• Use properties of logarithms to manipulate terms effectively, such as converting negative logarithms to positive ones by changing the base.

In our example, converting the logarithmic inequality to its exponential form allowed us to solve for \( x \) more directly. Always ensure to maintain the inequality direction unless you multiply or divide by a negative number, which will flip the inequality sign.

Function Conditions

Function conditions are about finding valid inputs for a function, ensuring all internal and external conditions are met. We already know the main function condition is ensuring the input to a log function is positive. However, it's essential to dig deeper into other conditions that could influence the function's domain.

For our given function, consider these extra checks:

• Investigate any additional functions involved, like square roots or fractional powers, that may further restrict the domain.
• Combine all conditions from multiple operations within the function (like the boundaries set by roots or fractions) to achieve a final, accurate domain.

Ensuring all function conditions are met is crucial for the accurate determination of the domain. Missing a condition could lead to an incorrect assumption about the input values that the function can take. In our case, after considering all conditions, we found the domain to be \( 0 < x < 1 \), where every part of the function will remain defined.