Problem 70 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f(x)\) be continuous in \([a, b]\) and the equation \(f(x)=0\) has only two roots \(\alpha\) and \(\beta(\alpha<\beta)\) in the interval \([a, b]\), then prove that \(f(x)\) retains the same sign in the interval \([\alpha, \beta]\).

Step-by-Step Solution

Answer

Since there is no point in the interval \((\alpha, \beta)\) with a sign change without having an additional root, it means that the function \(f(x)\) does not have any point in \((\alpha, \beta)\) where its sign changes. Thus, \(f(x)\) retains the same sign in the interval \([\alpha, \beta]\).

Step 1: Show that there cannot exist a point with a sign change without an additional root

Assume that there exists a point \(c\) in the interval \([\alpha, \beta]\) such that the function changes its sign at \(c\). Then \(f(c) \neq 0\) and either \(f(x) > 0\) for all \(x\) in \((\alpha, c)\) and \(f(x) < 0\) for all \(x\) in \((c, \beta)\) or vice versa. Since \(\alpha\) and \(\beta\) are the roots of the function, we have: \(f(\alpha) = 0\), and \(f(\beta) = 0\). Now, if there exists a point \(c\) in the interval \((\alpha, \beta)\) with a sign change, it means that \(f(x)\) is continuous and goes from positive to negative or negative to positive in between \(\alpha\) and \(\beta\). Since \(f(x)\) is continuous, according to the Intermediate Value Theorem (IVT), for any value \(k\) between \(f(\alpha)\) and \(f(c)\) or \(f(c)\) and \(f(\beta)\), there exists a point \(x \in [\alpha, c]\) or \(x \in [c, \beta]\) such that \(f(x) = k\). In this case, there must exist an \(x \in (\alpha, \beta)\) such that \(f(x) = 0\). But this contradicts the given that there are only two roots \(\alpha\) and \(\beta\) in the interval \([a, b]\). Therefore, there cannot exist any point in the interval \((\alpha, \beta)\) with a sign change without having an additional root.

Step 2: Use the continuity of \(f(x)\) to prove that it retains the same sign in interval \([\alpha, \beta]\).

Since there is no point in the interval \((\alpha, \beta)\) with a sign change without having an additional root, it means that the function \(f(x)\) does not have any point in \((\alpha, \beta)\) where its sign changes. Let's say the function is positive in the interval \((\alpha, \beta)\). Then, we have \(f(x) > 0\) for all \(x\) in \((\alpha, \beta)\). Since \(f(x)\) is continuous and \(f(\alpha) = 0\), then \(f(x) \geq 0\) for all \(x\) in \([\alpha, \beta]\). Similarly, if the function is negative in the interval \((\alpha, \beta)\), then we have \(f(x) < 0\) for all \(x\) in \((\alpha, \beta)\). Hence, \(f(x) \leq 0\) for all \(x\) in \([\alpha, \beta]\). So, we have proved that \(f(x)\) retains the same sign in the interval \([\alpha, \beta]\).

Key Concepts

Intermediate Value Theorem

In the world of calculus, the Intermediate Value Theorem (IVT) is an essential concept that helps us understand the behavior of continuous functions. This theorem states that if a function \( f(x) \) is continuous over an interval \([a, b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), there exists at least one number \( c \) in the interval \([a, b]\) such that \( f(c) = N \).
This theorem is crucial when analyzing whether a function crosses the x-axis (i.e., has a root) within a given interval:

• If \( f(a) \) and \( f(b) \) have different signs, then the IVT guarantees that there is a root between \( a \) and \( b \).
• It ensures that a continuous function cannot just "jump over" a value without passing through it, making it a valuable tool in proving the existence of roots.

In our exercise, applying the IVT helps us understand why the function \( f(x) \), continuous over \([\alpha, \beta]\), cannot change its sign without encountering an additional root, which contradicts the given conditions.

Roots of Equations

A root of an equation \( f(x) = 0 \) is a value of \( x \) where the function equals zero, indicating that the graph of the function intersects the x-axis. Finding roots is essential in solving equations, and roots tell us important information about the function's behavior.
In the provided exercise, we know:

• \(\alpha\) and \(\beta\) are the only roots of the function \( f(x) = 0 \) within the interval \([a, b]\).
• Being the only roots ensures no other places in the interval where the function crosses the x-axis, except potentially at \( \alpha \) and \( \beta \).

Understanding roots enable us to effectively use the sign of the function to establish other critical properties, such as ensuring that \( f(x) \) does not change signs in the interval \((\alpha, \beta)\), based on the Intermediate Value Theorem.

Sign of Function

The sign of a function within an interval indicates whether the function's outputs are positive or negative. When dealing with continuous functions, understanding the sign can provide valuable insights into the behavior and characteristics of the function over certain intervals.
In our exercise, the continuity and the absence of additional roots between \( \alpha \) and \( \beta \) ensure:

• If \( f(x) \) is positive at the beginning and end, it remains positive throughout \((\alpha, \beta)\).
• Similarly, if \( f(x) \) starts and ends negatively, it stays negative across \((\alpha, \beta)\).

This principle stems from the fact that a change of sign would imply another root within the interval, contradicting the premise that only \( \alpha \) and \( \beta \) are roots.
Hence, analyzing the sign of the function helps in proving that \( f(x) \) retains a consistent sign over \( [\alpha, \beta] \), leveraging its continuity and the absence of additional roots.