Problem 73 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f\) be a continuous function on \(R\) such that \(f(1 / 4 n)=\left(\sin e^{n}\right) e^{-n^{2}}+\frac{n^{2}}{n^{2}+1}\). Then, the value of \(f(0)\) is (A) 1 (B) \(\frac{1}{2}\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The value of \(f(0)\) is 1.

Step 1: Understanding the Problem

We are given a function \( f(x) \) that is continuous for all real numbers \( x \). The specific form of the function is given for \( f(\frac{1}{4n}) \) as \( (\sin e^{n}) e^{-n^{2}} + \frac{n^{2}}{n^{2}+1} \). We need to evaluate \( f(0) \). Due to the continuity of \( f(x) \), \( f(0) \) can be found as \( \lim_{{n \to \infty}} f(\frac{1}{4n}) \).

Step 2: Analyzing the Sine Exponent Term

First, consider the term \((\sin e^{n}) e^{-n^{2}}\). It involves \(\sin e^{n}\), which is bounded between -1 and 1, and \(e^{-n^{2}}\), which approaches 0 as \(n\) goes to infinity. Therefore, the entire expression \((\sin e^{n}) e^{-n^{2}}\) approaches 0 as \(n\) becomes very large.

Step 3: Analyzing the Rational Term

Consider the term \(\frac{n^{2}}{n^{2}+1}\). As \(n\) approaches infinity, both the numerator and the denominator approach \(n^{2}\), thus the expression approaches \(\frac{n^2}{n^2} = 1\). Thus, \(\lim_{{n \to \infty}} \frac{n^{2}}{n^{2}+1} = 1\).

Step 4: Calculating the Limit of the Entire Expression

We combine the results of the previous steps to find the limit of \(f(\frac{1}{4n})\) as \(n \to \infty\). So, we have:\[\lim_{{n \to \infty}} \left((\sin e^{n}) e^{-n^{2}} + \frac{n^{2}}{n^{2}+1}\right) = 0 + 1 = 1\]Thus, \(f(0) = 1\) since \(f(x)\) is continuous.

Step 5: Concluding the Value of f(0)

From the calculations, the continuity of \(f(x)\), and the computed limit, we conclude that the value of \(f(0)\) is 1.

Key Concepts

Limit of a Function

The concept of the limit of a function is fundamental in understanding continuous functions. It helps determine the behavior of a function as it approaches a certain point. In our exercise, we evaluate the limit of the function as the variable approaches 0, specifically focusing on a sequence \(\frac{1}{4n}\) where \(n\) tends to infinity.

A limit formally tells us the value that a function approaches as the input approaches some value. For continuous functions, the limit and the function value at that point are the same. This is why we focus on finding \(\lim_{{n \to \infty}} f(\frac{1}{4n})\) to find \(f(0)\).

By understanding limits, we can analyze parts of the function separately and then combine their effects, as was done in the detailed step-by-step solution.

Sine Function

The sine function, denoted as \(\sin(x)\), is a periodic function with a range of values between -1 and 1. This bounded behavior is crucial when it's used within more complex expressions.

In our function example, \(\sin e^{n}\) plays an important role. Since the sine function's output remains between -1 and 1 regardless of the input, the product \((\sin e^{n}) e^{-n^{2}}\) simplifies because \(e^{-n^{2}}\) tends to zero as \(n\) grows. This makes the entire expression approaching zero straightforward.

Using basic properties of the sine function aids in simplifying complex expressions and clarifying what the result would yield at the limit.

Exponential Function

The exponential function, typically represented as \(e^x\), is a crucial function characterized by growing very fast with increasing values of \(x\). Conversely, \(e^{-x}\) represents exponential decay, rapidly approaching zero as \(x\) becomes very large.

In the given exercise, understanding \(e^{-n^{2}}\) is pivotal. It tells us that as \(n\) approaches infinity, \(e^{-n^{2}}\) diminishes to zero. This component hugely impacts the term \((\sin e^{n}) e^{-n^{2}}\), pulling it closer to zero as \(n\) increases.

Recognizing exponential decay's impact helps in grasping how one part of the function leads to zero, which later simplifies the calculation of the overall function's limit.

Rational Functions

Rational functions are functions expressed as the ratio of two polynomials. An example is \( \frac{n^{2}}{n^{2}+1} \), as seen in our problem. Such expressions often include simplifications, especially as one those variables grows indefinitely.

The concept of horizontal asymptotes applies here, which is critical for students to understand. As \(n\) gets extremely large, the \(+1\) on the denominator becomes negligible, causing \( \frac{n^{2}}{n^{2}+1} \) to essentially approach \(1\).

Understanding the behavior of rational functions at infinity is important for computing limits and observing function tendencies over a wide range. This insight consolidates the part of the function that determines \(f(0)\) to be \(1\), after considering the zero contribution from the sine-exponential term.