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

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

None of the given options ( \) align; check problem setup.

Step 1: Determine Limit Behavior

We need to find \( \lim_{{x \to \pi}} f(x) \). We start by examining the function for \( x eq \pi \). In this case, pause to notice the limits involved: \( \cos(x) \to -1 \), \( (\pi-x)^2 \to 0 \), and \( x^2 - 2\pi x + \pi^2 \to 0 \). We identify it as an indeterminate form and consider simplifying or applying Taylor expansions.

Step 2: Simplify Using Taylor Series

Expand the critical terms using Taylor series around \( x = \pi \): \( \cos x \approx -1 + \frac{1}{2}(x-\pi)^2 \) and \( 1+\pi^2 - 2\pi x + x^2 \approx (x - \pi)^2 \). Substitute these approximations back into the function.

Step 3: Simplify the Fraction

Substitute the expansions into the function: \[ f(x) \approx \frac{-1 + \frac{1}{2}(x-\pi)^2}{(\pi-x)^{2}} \cdot \frac{\sin^2 x}{\log((x-\pi)^2)}\]Ignore higher-order terms of \((x-\pi)^2\), then get:\[ \frac{(x-\pi)^2}{(x-\pi)^2} \cdot \frac{\sin^2 x}{\log((x-\pi)^2)} \approx 1 \cdot 1 = 1\]

Step 4: Apply Continuity Condition

Given that \( f(x) \) must be continuous at \( x = \pi \), the limit we just calculated must equal \( k \). Therefore, \( k = 1 \).

Key Concepts

limits of functions

In calculus, the concept of limits is fundamental for understanding how functions behave as they approach a particular point. A limit describes the value that a function approaches as the input (or variable) approaches some value. When working with limits, it is essential to understand that the function may not necessarily take the value of the limit at that specific point.
A common notation used is \( \lim_{{x \to a}} f(x) = L \), which means as \( x \) approaches \( a \), the function \( f(x) \) approaches \( L \). This concept is crucial in discussing the continuity of functions.
In problems involving continuity, such as the one presented, we often need to establish the limit as \( x \) approaches a particular point - in this case, \( \pi \). Evaluating limits ensures the function does not become undefined or take on different values right before reaching \( x = a \).

taylor series expansion

Taylor series expansion is a helpful tool in calculus that allows us to approximate functions using polynomials. It is especially useful when evaluating limits that result in indeterminate forms. A Taylor series breaks down complex expressions into an infinite sum of terms derived from the derivatives of the function.
The basic form of a Taylor series around a point \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)

This technique enables us to approximate functions and simplify complex limit problems.
For instance, in the provided problem, expanding \( \cos(x) \) and \( 1+\pi^2 - 2\pi x + x^2 \) around \( x = \pi \) was instrumental in simplifying the expression to yield a more straightforward critical term for limit evaluation.

indeterminate forms

Indeterminate forms arise in calculus when directly substituting a value into a function results in an undefined expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such expressions require us to apply specific strategies or mathematical techniques to find their true limit.

• Common indeterminate forms include \( \frac{0}{0} \), \( \infty - \infty \), and \( 0 \cdot \infty \).
• One of the most effective methods to resolve these forms is via algebraic manipulation or using calculus tools such as L'Hôpital's Rule and Taylor Series.

In the given exercise, the function \( f(x) \) initially sets up an indeterminate form as \( x \to \pi \) because both \( \cos(x) \to -1 \) and \((\pi - x)^2 \to 0\). By expanding into Taylor series, we converted these into simpler, more computable forms.

trigonometric limits

Trigonometric limits often involve functions such as sine and cosine, which behave periodically. An underlying knowledge of these behaviors is key in calculating limits and understanding continuity at specific points.

• For the function \( \sin(x) \), it tends to \( 0 \) as \( x \to n\pi \), where \( n \) is an integer.
• Similarly, \( \cos(x) \to 1 \) or \( -1 \) at these points, depending on the direction of approach and nature of the integer \( n \).

In trigonometric limits problems, especially involving indeterminate forms, it's crucial to recognize these behaviors and possibly apply trigonometric identities or Taylor expansions for simplification. This understanding aids in determining the precise behavior of such functions as \( x \to a \) in limit scenarios, as seen in the original exercise with \( f(x) \).