Problem 66 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the equation \(\frac{a_{1}}{x-\lambda_{1}}+\frac{a_{2}}{x-\lambda_{2}}+\frac{a_{3}}{x-\lambda_{3}}=0\), where \(a_{1}>0, a_{2}>0, a_{3}>0\) and \(\lambda_{1}<\lambda_{2}<\lambda_{3}\), has two real roots lying in the intervals \(\left(\lambda_{1}, \lambda_{2}\right)\) and \(\left(\lambda_{2}, \lambda_{3}\right)\).

Step-by-Step Solution

Answer

The given equation is \(f(x) = \frac{a_{1}}{x-\lambda_{1}} + \frac{a_{2}}{x-\lambda_{2}} + \frac{a_{3}}{x-\lambda_{3}}\). By analyzing the signs of the function at the endpoints of the intervals \((\lambda_{1}, \lambda_{2})\) and \((\lambda_{2}, \lambda_{3})\) and applying the Intermediate Value Theorem, we find that there must be one root in each interval, resulting in two real roots lying in the given intervals.

Step 1: Define the given function

Let's define the function as follows: \(f(x) = \frac{a_{1}}{x-\lambda_{1}} + \frac{a_{2}}{x-\lambda_{2}} + \frac{a_{3}}{x-\lambda_{3}}\)

Step 2: Analyze the function in the interval (\(\lambda_{1}, \lambda_{2}\))

Let's check the signs of \(f(x)\) at the endpoints of this interval: \(f(\lambda_{1}) = \frac{a_{1}}{\lambda_{1}-\lambda_{1}} + \frac{a_{2}}{\lambda_{1}-\lambda_{2}} + \frac{a_{3}}{\lambda_{1}-\lambda_{3}} = -\infty\) \(f(\lambda_{2}) = \frac{a_{1}}{\lambda_{2}-\lambda_{1}} + \frac{a_{2}}{\lambda_{2}-\lambda_{2}} + \frac{a_{3}}{\lambda_{2}-\lambda_{3}} = \infty\) Since \(f(\lambda_{1}) < 0\) and \(f(\lambda_{2}) > 0\), by the Intermediate Value Theorem, there must be a root in the interval \((\lambda_{1}, \lambda_{2})\).

Step 3: Analyze the function in the interval (\(\lambda_{2}, \lambda_{3}\))

Again, check the signs of \(f(x)\) at the endpoints of this interval: \(f(\lambda_{2}) = \frac{a_{1}}{\lambda_{2}-\lambda_{1}} + \frac{a_{2}}{\lambda_{2}-\lambda_{2}} + \frac{a_{3}}{\lambda_{2}-\lambda_{3}} = \infty\) \(f(\lambda_{3}) = \frac{a_{1}}{\lambda_{3}-\lambda_{1}} + \frac{a_{2}}{\lambda_{3}-\lambda_{2}} + \frac{a_{3}}{\lambda_{3}-\lambda_{3}} = -\infty\) Since \(f(\lambda_{2}) > 0\) and \(f(\lambda_{3}) < 0\), by the Intermediate Value Theorem, there must be a root in the interval \((\lambda_{2}, \lambda_{3})\). Therefore, the equation \(\frac{a_{1}}{x-\lambda_{1}}+\frac{a_{2}}{x-\lambda_{2}}+\frac{a_{3}}{x-\lambda_{3}}=0\) has two real roots lying in the intervals \((\lambda_{1}, \lambda_{2})\) and \((\lambda_{2}, \lambda_{3})\).

Key Concepts

Real Roots

Real roots are the values of a variable that make an equation equal to zero. Solving for real roots often involves finding intersections of a function with the x-axis. In mathematical terms, if you have a function \( f(x) \), a real root is a point where \( f(x) = 0 \). In the given exercise, the function is \( f(x) = \frac{a_{1}}{x-\lambda_{1}} + \frac{a_{2}}{x-\lambda_{2}} + \frac{a_{3}}{x-\lambda_{3}} \), and our task is to find the values of \(x\) for which the function outputs zero.

• These real root values are significant because they represent the solutions to the equation.
• Determining their existence and location involves evaluating the behavior of the function over specific intervals.

By using the Intermediate Value Theorem, we confirmed that there are real roots in the intervals \( (\lambda_{1}, \lambda_{2}) \) and \( (\lambda_{2}, \lambda_{3}) \) because the function changes from negative to positive, ensuring it crosses the x-axis.

Interval Analysis

Interval analysis is a mathematical approach used to examine the behavior of a function over a certain range. This is done to identify the potential presence of roots or solutions.
In our context, we are looking at the intervals \((\lambda_{1}, \lambda_{2})\) and \((\lambda_{2}, \lambda_{3})\). We first check the function's value at the boundaries of these intervals.

• If \( f(\lambda_{1}) = -\infty \) and \( f(\lambda_{2}) = \infty \), then the function is negative at one endpoint and positive at the other.
• This indicates a root exists between these points.
• Similarly, if \( f(\lambda_{2}) = \infty \) and \( f(\lambda_{3}) = -\infty \), then another root is present in the second interval.

The Intermediate Value Theorem is crucial here—it states that if a function is continuous on a closed interval and changes signs, then it must have at least one root within that interval.

Function Behavior

Understanding function behavior involves recognizing how a function acts over different sections of its domain. For a rational function like \( f(x) = \frac{a_{1}}{x-\lambda_{1}} + \frac{a_{2}}{x-\lambda_{2}} + \frac{a_{3}}{x-\lambda_{3}} \), this means exploring how it behaves as \( x \) approaches different values, particularly those that make the denominator zero.

• In our function, \( f(x) \) has vertical asymptotes at \( x = \lambda_{1}, \lambda_{2}, \lambda_{3} \), where the function goes to positive or negative infinity.
• It's crucial to know these points of discontinuity to understand how the function might cross the x-axis between them.
• The change in signs between these points, as deduced from the Intermediate Value Theorem, helps pinpoint where the function hits zero.

Recognizing these behaviors assists in predicting where solutions may exist without solving the equation algebraically by clarifying how the function navigates its domain.