Problem 60 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the equation \(x 2^{x}=1\) has at least one positive root not exceeding 1 .

Step-by-Step Solution

Answer

By evaluating the function \(f(x) = x2^x\) at the endpoints of the interval \([0, 1]\), we find that \(f(0) = 0 < 1\) and \(f(1) = 2 > 1\). Since the function \(f(x)\) is continuous over the interval, we can apply the Intermediate Value Theorem, which guarantees that there exists at least one point within the interval \([0, 1]\) where \(f(x) = x2^x = 1\). Thus, the equation \(x2^x = 1\) has at least one positive root not exceeding 1.

Step 1: Find the value of f(x) at x=0

Plug x=0 into the function: \(f(0) = 0 \cdot 2^0 = 0\)

Step 2: Find the value of f(x) at x=1

Plug x=1 into the function: \(f(1) = 1 \cdot 2^1 = 2\)

Step 3: Apply the Intermediate Value Theorem

Since \(f(0) = 0 < 1\) and \(f(1) = 2 > 1\), we can see that the function assumes values both below and above 1 within the interval \([0, 1]\). Since the function \(f(x) = x2^x\) is continuous over the interval, we can apply the Intermediate Value Theorem. This guarantees that there must be at least one point within the interval \([0, 1]\) where \(f(x) = x2^x = 1\). Therefore, the equation \(x2^x = 1\) has at least one positive root not exceeding 1.

Key Concepts

Continuous Functions

In mathematics, continuous functions play a crucial role in connecting points in a manner that there are no breaks or gaps in the graph of the function. It's like drawing a line without lifting the pencil, ensuring that every movement from one point to another is smooth and connected. A function is deemed continuous over an interval if it is continuous at every point within that interval. This means, for every point 'c' in that interval, the limit of the function as it approaches 'c' from both directions is equal to its value at 'c'.
For our problem, the function in question, \( f(x) = x \cdot 2^x \), is continuous because both multiplication and exponential functions are inherently continuous operations. Thus, throughout the interval \([0, 1]\), this function does not jump or miss any values, providing a foundational requirement for further analysis like the Intermediate Value Theorem.

Positive Root

A positive root of an equation is essentially a solution where the variable, when substituted, results in a non-negative value. For the equation \( x2^x = 1 \), a positive root would be a value of \( x \) within the acceptable domain that satisfies this equation.
Since the original problem statement requests confirmation of at least one positive root not exceeding 1, it implies probing within the interval \([0, 1]\). Hence, finding such a root confirms that the equation crosses the horizontal line (where the function equals 1) within this range.
Within the context of the solution provided, the Intermediate Value Theorem gives us assurance that such a crossing point—representing the positive root—indeed exists. The focus on continuity and changing values within the specified interval aids in identifying the presence of such a positive root.

Function Evaluation

To understand the behavior of the function \( f(x) = x \cdot 2^x \), an important step is evaluating the function at specific points. This involves calculating \( f(x) \) for given \( x \) values, which can sometimes reveal critical information about intersections or behavior of the curve.
In the exercise, we evaluate \( f(x) \) at \( x=0 \) giving \( f(0) = 0 \) and \( x=1 \) giving \( f(1) = 2 \). These evaluations are fundamental for applying the Intermediate Value Theorem. It shows that the output of the function transitions from below 1 to above 1 within the interval. Such transitions indicate that at some point between \( x=0 \) and \( x=1 \), \( f(x) \) should equal 1. Evaluations at key points help outline these scenarios and set the stage for making broader mathematical conclusions.

Equation Solving

Equation solving is the process of finding the values of variables that satisfy a given equation. In our exercise, solving \( x \cdot 2^x = 1 \) involves looking for those particular \( x \) values where the function equals 1.
Direct solving for \( x \) in this equation can be complex because it intertwines both linear and exponential components. However, by employing methods like the Intermediate Value Theorem, we're not explicitly calculating exact solutions but instead proving their existence within a given range.
This theorem assures us that if the function moves smoothly from one side of a target value to the other, as it does from \( f(0) = 0 \) to \( f(1) = 2 \), then a solution exists. This approach often makes equation solving feasible for complex functions that resist traditional algebraic manipulation.