Problem 25 - Calculus for JEE ADVANCED

Question

It is given that \(2^{x}+2^{y}=2 \forall x, y \in R\) or \(\quad 2^{y}=2-2^{x}\) or \(y=\log _{2}\left(2-2^{x}\right)\) Therefore, function is defined only when \(2-2^{x}>0\) or \(2^{x}<2\) or \(x<1\)

Step-by-Step Solution

Answer

The function is defined for \(x < 1\).

Step 1: Understand the Given Equation

The problem involves the equation \(2^{x} + 2^{y} = 2\), which implies that the exponential terms must add to 2.

Step 2: Rearrange the Equation

Rearrange the given equation to isolate one of the exponential terms. From \(2^{x} + 2^{y} = 2\), we can express \(2^{y}\) as \(2^{y} = 2 - 2^{x}\).

Step 3: Apply Logarithmic Transformation

Convert the equation \(2^{y} = 2 - 2^{x}\) to a logarithmic form to solve for \(y\). This gives us \(y = \log_{2}(2 - 2^{x})\).

Step 4: Determine Domain Restrictions

For the logarithmic function \(y = \log_{2}(2 - 2^{x})\) to be defined, the argument \(2 - 2^{x}\) must be greater than 0. Thus, \(2^{x} < 2\), which simplifies to \(x < 1\).

Key Concepts

Logarithmic Transformation

Logarithmic transformation is a powerful mathematical tool used to solve exponential equations. In our exercise, we encounter the equation \(2^{y} = 2 - 2^{x}\). To express \(y\) in a format that's easier to interpret, we apply a logarithmic transformation. By using the definition of logarithms, which states that \(y = \log_{b}(a)\) whenever \(b^{y} = a\), we can transform \(2^{y} = 2 - 2^{x}\) into a logarithmic form. This gives us \(y = \log_{2}(2 - 2^{x})\).

• The base of the logarithm is 2, corresponding to the base of the exponent in the equation.
• This transformation helps us rearrange and simplify the problem, making it manageable to solve for \(y\).

The logarithmic transformation provides a neat way to handle exponential components, especially when solving problems that involve finding values for unknown variables.

Domain Restrictions

In mathematics, the concept of "domain restrictions" refers to the set of permissible input values for which a function is defined. When dealing with logarithmic functions, it's crucial to ensure that the argument inside the logarithm is positive, as logarithms of non-positive numbers are undefined in the real number system.
For the function \(y = \log_{2}(2 - 2^{x})\), we derive domain restrictions based on the requirement that the argument \(2 - 2^{x}\) must be greater than zero:

• This results in the inequality \(2 - 2^{x} > 0\), which ensures that the logarithmic expression is valid.
• Solving this inequality gives \(2^{x} < 2\), which simplifies to \(x < 1\).

These domain conditions are vital, as they define the range of \(x\) values that make the function valid and solvable.

Rearranging Equations

Rearranging equations is a fundamental skill in algebra that involves manipulating an equation to isolate a particular variable on one side of the equation. In our exercise, this technique is crucial in handling the equation \(2^{x} + 2^{y} = 2\).
By rearranging, we aim to express one of the variables, such as \(2^{y}\), on one side:

• Starting with \(2^{x} + 2^{y} = 2\), we isolate \(2^{y}\) by subtracting \(2^{x}\) from both sides, obtaining \(2^{y} = 2 - 2^{x}\).
• This step simplifies the equation, setting the path toward solving it using logarithms.

Rearranging allows you to adjust the form of an equation to make further operations, like logarithmic transformations, more straightforward. It streamlines the problem-solving process by breaking down complex equations into simpler, more accessible forms.