Problem 138 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)\) be a function of \(x\), where \(f(x)\) is continuous in the closed interval \([a, b]\) and differentiable in the open interval \((a, b)\). Also, \(f(a)=f(b)\), i.e., the values at the end points \(a\) and \(b\) are equal. Then, there exists at least one point \(c\) between \(a\) and \(\mathrm{b}\) (i.e., \(a

Step-by-Step Solution

Answer

(B) at least one root.

Step 1: Understand Rolle's Theorem

Rolle's Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), with \( f(a) = f(b) \), then there exists at least one point \( c \) in \( (a, b) \) such that \( f'(c) = 0 \). This theorem tells us about the existence of a point where the tangent to the curve is horizontal.

Step 2: Define the Function

The equation given is \( x \log x = 3 - x \). We define a new function \( f(x) = x \log x - (3 - x) \). This changes our equation to finding the roots of \( f(x) = 0 \). So, \( f(x) = x \log x + x - 3 \).

Step 3: Identify Continuity and Differentiability

The function \( f(x) = x \log x + x - 3 \) is continuous and differentiable in the open interval \((1, 3)\) because it is composed of continuous and differentiable functions (polynomials and logarithmic functions) in that interval.

Step 4: Check Endpoints and Rolle's Condition

To apply Rolle's theorem, check if \( f(1) = f(3) \). Calculate:\[ f(1) = 1 \log 1 + 1 - 3 = 0 - 2 = -2 \]\[ f(3) = 3 \log 3 + 3 - 3 = 3\log 3 \]Since \( f(1) eq f(3) \), we cannot directly apply Rolle’s Theorem. But we need to assess roots presence in this interval.

Step 5: Apply the Intermediate Value Theorem

Since Rolle's theorem cannot be directly applied, we check for roots using the Intermediate Value Theorem. Notice that \( f(1) = -2 \) and \( f(3) \approx 3\log 3 \) is positive. Since \( f(x) \) is continuous, there must be at least one root \( c \) where \( f(c) = 0 \) within \((1, 3)\), by the Intermediate Value Theorem.

Step 6: Determine Number of Roots

Given that the function intersects the x-axis exactly once between the points due to the continuous nature and Intermediate Value Theorem, we conclude there is at least one root in the interval.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus, which guarantees the existence of at least one value within an interval, assuming certain conditions are met. This theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and if \( f(a) \) and \( f(b) \) are of opposite signs, then there is at least one number \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \).

Why does this matter? The IVT assures us that for any continuous function, if the function has different signs at the ends of an interval, then it must cross the x-axis at some point within that interval. In practical terms, if you have a curve where one end is above the x-axis and the other end is below, the curve must "touch" or "cross" the x-axis at least once. This can help in finding the roots of complicated functions by narrowing down the interval where they exist.

For the exercise given, even though Rolle's Theorem couldn't be applied, the Intermediate Value Theorem confirmed the presence of at least one root in the interval \((1, 3)\) because the function changes its sign from negative to positive across this interval.

Continuity

Continuity of a function is one of the vital aspects of calculus, especially when applying theorems like the Intermediate Value Theorem or Rolle's Theorem. A function is said to be continuous over an interval if there are no breaks, jumps, or holes in its graph within that interval. Mathematically, a function \( f(x) \) is continuous at a point \( c \) if the following three conditions are satisfied:

• The function \( f(x) \) is defined at \( c \).
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• \( \lim_{x \to c} f(x) = f(c) \)

In the exercise's context, the function \( f(x) = x \log x + x - 3 \) is found to be continuous in the open interval \((1, 3)\). The reason is that it is composed of functions that are known to be continuous over that interval, specifically a logarithmic component (which is continuous where its argument is positive) and polynomial terms, which are continuous everywhere. This continuity is crucial to applying other principles like the Intermediate Value Theorem.

In simpler terms, continuity in this exercise means we can safely assume no sudden surprises in the function's behavior, allowing us to apply mathematical theorems with confidence.

Differentiability

Differentiability is a concept closely related to continuity but is a bit more specific. A function is differentiable at a point if it has a derivative at that point, meaning we can determine the rate at which the function is changing. For a function \( f(x) \) to be differentiable at a point \( c \), it must meet the following condition:

• The derivative \( f'(c) \) exists, which means that we can actually compute its value.

Differentiability implies continuity. However, the reverse isn't always true—meaning a function can be continuous without being differentiable at a point (think of a sharp corner or cusp).

In this exercise, the function \( f(x) \) is described as differentiable within the open interval \((1, 3)\). The presence of polynomial terms and the logarithmic function within these boundaries suggests it behaves smoothly enough to be differentiable across the interval. This property, while important for Rolle's Theorem, is still essential for analyzing how functions behave and finding critical points where the slope, or derivative, equals zero (which was our step towards Rolle’s Theorem before we used the Intermediate Value Theorem for root existence).

Understanding differentiability helps in providing insights into the function's smoothness and the behavior of its graph, especially when determining points where the function's slope changes or is flat.