Problem 30 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the equation \(x e^{x}=2\) has only one solution which lies in the interval \((0,1)\).

Step-by-Step Solution

Answer

The function \(f(x) = xe^{x} - 2\) is continuous and strictly increasing in the interval (0, 1) as shown by its derivative \(f'(x) = e^{x} + xe^{x} > 0\). Since it takes different signs at the endpoints, \(f(0) < 0\) and \(f(1) > 0\), the Intermediate Value Theorem implies the existence of at least one solution in the interval. As the function is strictly increasing, there can be only one solution in the interval (0, 1).

Step 1: 1. Check the continuity of the function

The function is given by \(y = xe^{x} - 2\). We will first check if it is continuous in the interval (0, 1). Since \(xe^{x}\) is the product of two continuous functions, \(x\) and \(e^{x}\), it is also continuous. Therefore, \(xe^{x} - 2\) is continuous in the interval (0, 1).

Step 2: 2. Determine the signs at the endpoints

Next, we'll determine the values of the function at the endpoints of the interval (0, 1). \(f(0) = 0\cdot e^{0} - 2 = -2\) \(f(1) = 1\cdot e^{1} - 2 = e - 2\) Since \(e > 2\), we have \(f(1) > 0\). Therefore, the function takes different signs at the endpoints of the interval (0, 1): \(f(0) < 0\) and \(f(1) > 0\).

Step 3: 3. Apply the Intermediate Value Theorem

Now, we can apply the Intermediate Value Theorem (IVT). Since \(f(x) = xe^{x} - 2\) is continuous in the interval (0, 1) and takes different signs at the endpoints, there must be a point \(c \in (0, 1)\) where \(f(c) = 0\). This means at least one solution exists in (0, 1).

Step 4: 4. Check if the function is strictly increasing

To prove that there is only one solution, we'll check if the function is strictly increasing in the interval (0,1). To do this, we'll find the derivative of the function. \(f'(x) = \frac{d}{dx}(xe^{x} - 2) = e^{x} + xe^{x}\) Since both terms (\(e^{x}\) and \(xe^{x}\)) are positive in the interval (0, 1), the derivative is strictly positive, which implies that the function is strictly increasing in the interval (0,1).

Step 5: 5. Conclude that there is only one solution

Since \(f(x)=xe^{x}-2\) is strictly increasing in the interval (0, 1), it can have at most one root in this interval. We've previously shown the existence of a solution using the Intermediate Value Theorem. Therefore, there is only one solution in the interval (0, 1).

Key Concepts

Continuity

Continuity in mathematics refers to a property of functions. A continuous function is one that does not have any gaps, jumps, or sudden changes in value. In simpler terms, if you can draw the graph of the function without lifting your pencil, the function is continuous. For example, in the function from our exercise, \( y = xe^{x} - 2 \), we see that it is continuous. This is because it is the product of two continuous functions: the linear function \( x \) and the exponential function \( e^{x} \).

• The linear function \( x \) is continuous everywhere.
• The exponential function \( e^{x} \) is also continuous everywhere.

Therefore, the product \( xe^{x} \) is continuous. When the constant \( -2 \) is subtracted, the function remains continuous, particularly in the domain we're examining, which is the interval \((0,1)\). This property of continuity is fundamental for applying concepts like the Intermediate Value Theorem.

Exponential Functions

Exponential functions are a special class of mathematical functions characterized by the form \( f(x) = a^{x} \), where \( a \) is a positive constant. In our exercise, we are dealing with the exponential function \( e^{x} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Exponential functions have unique properties:

• The function \( e^{x} \) is always positive.
• It is continuous and differentiable everywhere.
• It increases rapidly as \( x \) increases, exhibiting what's known as exponential growth.

In the context of our exercise, \( e^{x} \) is part of the expression \( xe^{x} \). It's important not only because it contributes to the continuity of the function but also because it affects the behavior and growth of the function. Exponential functions often appear in problems involving growth processes, finance, or natural phenomena, making them fundamental in both pure and applied mathematics.

Derivative of a Product

The derivative is a foundational tool in calculus used to determine the rate of change of a function. When you have a function that is the product of two functions, like \( g(x) = xe^{x} \), you apply the product rule to find its derivative. The product rule states that the derivative of a product \( uv \) is \( u'v + uv' \). Here's how it works in this context:

• Let \( u(x) = x \) and \( v(x) = e^{x} \).
• The derivative of \( u(x) \), \( u'(x) \), is 1.
• The derivative of \( v(x) \), \( v'(x) \), is \( e^{x} \).

Applying the product rule, the derivative \( g'(x) = u'(x)v(x) + u(x)v'(x) \) becomes:
\[ g'(x) = 1 \, e^{x} + x \, e^{x} = e^{x} + xe^{x} \] This result shows that the derivative, \( e^{x} + xe^{x} \), is positive for all \( x \) in the interval \((0, 1)\). A positive derivative indicates that the original function \( g(x) = xe^{x} \) is increasing. Understanding the behavior of derivatives helps in sketching the graph of functions and explaining their growth or decay trends.