Problem 100 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a real-valued function defined on an interval \(I\). If \(f\) be derivable at a point \(x_{0} \in I\), then it is continuous at \(x_{0}\). The converse of the above statement does not hold. That is, a function may be continuous at a point but may fail to be derivable at that point. Thus, derivability is a more restrictive property than continuity. In fact, there are functions which are continuous everywhere but differentiable nowhere. If \(R f^{\prime}\left(x_{0}\right)\) and \(L f^{\prime}\left(x_{0}\right)\) are finite (they may or may not be equal), then \(f(x)\) is continuous at \(x=x_{0}\). If \(f(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}(\log a)^{n}\), then at \(x=0, f(x)\) (A) has no limit (B) is continuous (C) is continuous but not differentiable (D) is differentiable

Step-by-Step Solution

Answer

(D) is differentiable.

Step 1: Identify the Function

The function given is \( f(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n !}(\log a)^{n} \). This series resembles a power series expression.

Step 2: Evaluate the Function at x=0

Evaluate \( f(x) \) at \( x = 0 \). Notice that each term with \( x^n \) where \( n > 0 \) becomes 0. Thus, \( f(0) = \log a \cdot \frac{0^0}{0!} \) which simplifies to \( 1 \) since the \( 0^0 \) term is considered as 1 when \( n = 0 \).

Step 3: Check Continuity at x=0

For \( f(x) \) to be continuous at \( x = 0 \), the limit as \( x \) approaches 0 must equal \( f(0) \). Consider the limit \( \lim_{x \to 0} f(x) = \sum_{n=0}^{\infty} \frac{0^n}{n!} (\log a)^{n} = 1 \), which is precisely \( f(0) \). This shows the function is continuous at \( x = 0 \).

Step 4: Evaluate Differentiability at x=0

To check for differentiability, examine the derivative. Assume \( f'(x) \) is obtained by term-by-term differentiation. The derivative at \( x = 0 \) boils down to \( f'(0) = \sum_{n=1}^{\infty} \frac{n \, 0^{n-1}}{n!} (\log a)^n \), which sums to zero. This implies that \( f'(0) \) exists.

Step 5: Conclusion

Since \( f(x) \) at \( x = 0 \) is both continuous and has a well-defined derivative, \( f(x) \) is differentiable at \( x = 0 \). Thus, the answer is (D).

Key Concepts

Continuity

Continuity of a function at a given point is a fundamental concept in calculus. It implies that as we zoom in closer and closer to that point, the function's behavior doesn't abruptly change. For a function to be continuous at a particular point, essentially three conditions must be satisfied:

• The function is defined at that point.
• The limit of the function as it approaches that point from either side exists.
• The value of the function at that point equals the limit of the function as it approaches that point.

If these conditions hold, the function is continuous at that point. For example, consider the function from the exercise: it is confirmed to be continuous at \( x = 0 \) because the limit as \( x \) approaches 0 is equal to the value of the function at that point, which is 1.

Differentiability vs Continuity

Differentiability and continuity are closely related, yet distinct concepts. While all differentiable functions are continuous, not all continuous functions are differentiable. Differentiability at a point means that the function has a defined derivative there. This implies smoothness - no sharp corners or cusps at that point.
On the other hand, continuity merely requires a function to not have breaks or jumps. You can think of differentiability as a stronger condition, which includes continuity but goes further.
An illustrative example is the absolute value function \( f(x) = |x| \) at \( x = 0 \). This function is continuous across its domain; however, it is not differentiable at \( x = 0 \) due to the sharp corner at that point. In our provided exercise, the function \( f(x) \) not only is continuous but also differentiable at \( x = 0 \), because it meets both criteria - having a limit and a derivative.

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are coefficients and \( x \) is the variable. Power series are immensely powerful tools in calculus because they allow functions to be expressed as an infinite sum of terms. This makes them incredibly useful for analyzing functions, especially around a certain point, often called the center of the series.
In the given exercise, the function is expressed in terms of a power series. Evaluating it at \( x = 0 \), we see all terms in the series involving \( x^n \) for \( n > 0 \) evaluate to zero, significantly simplifying the function to just a constant term. More broadly, power series can converge within a certain radius, allowing precise function approximation in this interval.
Working with power series involves checking for convergence and ensuring that they appropriately fit the mathematical scenario being analyzed. In our case, the series converges and shows that the function is both continuous and differentiable at a specific point.