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

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, so the correct option is (A).

Step 1: Understanding the Problem

We need to evaluate the function \( f(x) \) at \( x = 0 \). The given function is \( f(\frac{1}{4n}) = (\sin e^n) e^{-n^2} + \frac{n^2}{n^2 + 1} \). Since \( f \) is continuous everywhere, we can use a limit approach to find \( f(0) \) by considering the limit of \( f(\frac{1}{4n}) \) as \( n \) approaches infinity.

Step 2: Express Function Terms for Large n

Observe that as \( n \to \infty \), the term \( (\sin e^n)e^{-n^2} \) approaches zero because \( e^{-n^2} \) tends towards zero very quickly due to the exponent \(-n^2\), which is a rapidly decreasing exponential function.

Step 3: Evaluate the Second Term's Limit

Consider the term \( \frac{n^2}{n^2 + 1} \). As \( n \to \infty \), this expression approaches 1, since the higher power term \( n^2 \) in both the numerator and the denominator dominate the constant 1.

Step 4: Calculate the Limit of the Entire Function

Calculate the limit of the entire function expression as \( n \to \infty \): \[ \lim_{n \to \infty} \left((\sin e^n)e^{-n^2} + \frac{n^2}{n^2 + 1}\right) = \lim_{n \to \infty} (0 + 1) = 1. \] Thus, \( f(\frac{1}{4n}) \to 1 \) as \( n \to \infty \), which implies \( f(0) = 1 \) due to the continuity of \( f \).

Step 5: Final Conclusion

Since \( f(\frac{1}{4n}) \to 1 \) as \( n \to \infty \) and \( f \) is continuous, it follows that \( f(0) = 1 \). The correct option is (A) 1.

Key Concepts

Limits

Limits are a fundamental concept in calculus that allow us to evaluate the behavior of a function as its input approaches a particular value. This is especially useful when the function might not be directly defined at that point. In this exercise, we used limits to determine the value of the function \( f \) at \( x = 0 \).

The idea is to look at what happens to \( f(x) \) as \( x \) gets arbitrarily close to zero. This is done by examining \( f(1/4n) \) as \( n \) approaches infinity, effectively making \( 1/4n \) approach zero. The limit allows us to bypass potential undefined situations by focusing on the trend of the function inputs.

• The first term, \( (\sin e^n)e^{-n^2} \), contributes zero to the limit as \( n \to \infty \) due to the rapid decay of the exponential function.
• The second term, \( \frac{n^2}{n^2 + 1} \), approaches one, providing a stable contribution to the limit.

Thus, the limit \[ \lim_{n \to \infty} \left((\sin e^n)e^{-n^2} + \frac{n^2}{n^2 + 1}\right) = 1, \] solidifies our evaluation of \( f(0) = 1 \). Limits help bridge the gap between what we observe in the behavior of functions and the values at specific points.

Continuity

Continuity refers to a function's property where it does not have sudden jumps or breaks. It intuitively means that you can draw the graph of the function without lifting your pencil. For the function \( f \) given in the exercise, continuity is crucial to justify using the limit to find \( f(0) \).

Continuity ensures that:

• If \( f(x) \to L \) as \( x \to c \), then \( f(c) = L \).
• In this exercise, as \( n \to \infty \), the function at \( 1/4n \to 0 \). Therefore, due to continuity, \( f(0) = \lim_{n \to \infty} f(1/4n). \)

Because \( f \) is continuous on \( \mathbb{R} \), the calculated limit reflects the exact value of the function at \( x = 0 \). Thus, continuity assures us that our limit evaluation is not just an approximation; it's the precise value of the function at that point.

Exponential function

Exponential functions are a key type of function in mathematics, particularly identified by their form \( a^x \) or \( e^x \) where \( e \) is the base of natural logarithms. They are known for their rapid growth or decay properties. In the given exercise, the term \( e^{-n^2} \) is a classic example of an exponentially decaying function.

When \( n \) gets large, the value of \( e^{-n^2} \) becomes very small very quickly. This explains why the term \( (\sin e^n)e^{-n^2} \) approaches zero as \( n \to \infty \). The exponential decay overpowers the oscillating \( \sin e^n \) term, which does not grow exponentially and instead remains bounded between \(-1\) and \(1\).

• Exponential growth or decay is critical in many fields like finance, natural sciences, and engineering because they model processes that involve continual growth or breakdown.
• Understanding these properties helps us in predicting how certain terms in functions behave as the input becomes large or small.

Recognizing and understanding exponential functions assures us that the exponential decay is significant in controlling the behavior of complex functions and thus, aids us in reliably determining limits.