Problem 22 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\left\\{\begin{array}{cl}\frac{1+\cos x}{(\pi-x)^{2}} \cdot \frac{\sin ^{2} x}{\log \left(1+\pi^{2}-2 \pi x+x^{2}\right)} & , x \neq \pi \\\ k & , x=\pi\end{array}\right.\) If \(f(x)\) is continuous at \(x=\pi\), then \(k\) is equal to (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) \(\frac{-1}{2}\) (D) \(\frac{-1}{4}\)

Step-by-Step Solution

Answer

The correct value of \(k\) is 1, but none match the options given.

Step 1: Understand the Problem

We need to find the limit of the function \(f(x)\) as \(x\) approaches \(\pi\) and compare it to the value of \(f(x)\) at \(x=\pi\). The function is defined differently at \(x=\pi\), using the value \(k\), so we'll find the limit first to determine what \(k\) must be for continuity.

Step 2: Calculate the limit of \(f(x)\) as \(x\to \pi\)

The continuous function is the limit of the piecewise function as \(x\) approaches \(\pi\). For continuity, \(\lim_{{x \to \pi}} f(x) = k\). Substitute \(x = \pi - h\) and take \(h \to 0^+\). Substitute the expression into \(f(x)\) and simplify.

Step 3: Simplify the Function

Substitute \(x = \pi - h\) in the function terms: \(1 + \cos x \rightarrow 1 - \cos h\), \((\pi - x)^2 \rightarrow h^2\), \(\sin^2 x \rightarrow \sin^2 h\), and the log term to simplify. Use the approximations \(\cos h \approx 1\) and \(\sin h \approx h\) since \(h \to 0^+\).

Step 4: Calculate Individual Limits

Calculate the limits: \(1 + \cos x = 2\), \((\pi - x)^2 = h^2\), \(\sin^2 x = h^2\), and \(\log(1 + h^2) \approx h^2\) as \(h \to 0\). The expression simplifies to \(\frac{1}{1}\).

Step 5: Evaluate Final Limit

Since \(f(x)\) simplifies to \(\frac{1}{1} = 1\), continue this approximation over \(h \to 0^+\) direction. Therefore, the limit \(\lim_{{x \to \pi}} f(x) = 1\).

Step 6: Solve for Continuity

For \(f(x)\) to be continuous at \(x = \pi\), it must be that \(k = 1\), since continuous functions must have \(\lim_{{x \to \pi}} f(x) = f(\pi) = k\). Hence, verify through limit calculation and matching to options.

Key Concepts

Limit of a function

When exploring functions in calculus, one fundamental concept is the "limit of a function." This represents the value that a function approaches as the input (or variable) comes infinitely close to a particular point. It's crucial in understanding how functions behave near specific values, even if the function is not explicitly defined at that point.

In the given exercise, the function \(f(x)\) is defined differently at \(x = \pi\) compared to values close to \(\pi\). We use limits to analyze the behavior of the function when \(x\) is near \(\pi\). Calculating the limit involves substituting \(x = \pi - h\) and letting \(h\) approach zero, thereby transforming the problem into examining how the function behaves as \(x\) gravitates closer to \(\pi\).

By understanding the limit, we determine the value \(k\) must take so that \(f(x)\) maintains its continuity at \(x = \pi\). Thus, the idea of limits helps bridge the gap between function behavior at and around specific points, establishing a foundation for continuous functions.

Piecewise functions

Piecewise functions are defined by different expressions over various intervals. They are prevalent in real-world applications where conditions or rules change based on different scenarios or inputs.

In this exercise, \(f(x)\) is a piecewise function. It dictates that for any input \(x eq \pi\), the function adopts a complex expression involving trigonometric and logarithmic operations. However, at \(x = \pi\), the value of the function is the constant \(k\).

Understanding piecewise functions helps to analyze scenarios where the function’s definition or behavior changes dramatically at specific points. It is crucial to determine how such functions coordinate between different pieces to ensure properties like continuity are preserved. In this instance, finding the correct \(k\) ensures the function blends smoothly across the pieces, showing the importance of calculating limits for continuity.

Calculus continuity

Continuity is a key concept in calculus that ensures a function is predictable and smooth without any breaks or jumps at a specific point. A function is continuous at a point \(a\) if the following conditions are satisfied:

• The function \(f(x)\) is defined at \(x = a\).
• The limit of the function as \(x\) approaches \(a\), \(\lim_{{x \to a}} f(x)\), exists.
• The value of the limit equals the function’s value at that point: \(\lim_{{x \to a}} f(x) = f(a)\).

In the discussed problem, continuity at \(x = \pi\) necessitates that the limit of \(f(x)\) as \(x\) approaches \(\pi\) matches the function's value at \(x = \pi\), which is \(k\). By simplifying the expression for \(f(x)\) and calculating the limit, we ensure that the curve of the function does not have any abrupt changes at \(x = \pi\).

Continuity ties together the concepts of limits and piecewise functions to provide a seamless transition across different function behaviors. It's essential in maintaining mathematical consistency and predictability in calculus-based analysis.