Problem 64 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(f(x)=\left\\{\begin{array}{ll}x e^{-\left(\frac{1}{|x|} \mid \frac{1}{x}\right)}, x \neq 0 \\ 0 & , x=0\end{array}\right.\) then \(\mathrm{f}(\mathrm{x})\) is (a) discontinuous every where (b) continuous as well as differentiable for all \(x\) (c) continuous for all \(x\) but not differentiable at \(x=0\) (d) neither differentiable nor continuous at \(x=0\)

Step-by-Step Solution

Answer

f(x) is continuous for all \(x\) but not differentiable at \(x=0\). Answer is (c).

Step 1: Analyze the function for points where \(x \neq 0\)

For \(x eq 0\), the function is given as \(f(x) = x e^{-\left(\frac{1}{|x|} \mid \frac{1}{x}\right)}\). As \(e^{-x}\) is a well-defined, continuous, and differentiable function for all real numbers, the product \(x e^{-\left(\frac{1}{|x|} \mid \frac{1}{x}\right)}\) is continuous and differentiable for \(x eq 0\).

Step 2: Analyze continuity at \(x = 0\)

To determine continuity at \(x = 0\), we need to check \(\lim_{x \to 0} f(x)\) and compare it with \(f(0) = 0\). The limit is expressed as:\[\lim_{x \to 0} x e^{-\left(\frac{1}{|x|} \mid \frac{1}{x}\right)} = \lim_{x \to 0} x e^{-\left(\frac{1}{x}\right)}\]Let's take \( \left|x\right| = t \), then as \( x \to 0 \), \( t \to 0 \). Thus, \( x = \pm t \), and the expression becomes:\[\lim_{t \to 0} t e^{-\left(\frac{1}{t}\right)} = 0\]This evaluates to \(0\), matching \(f(0)\), showing the function is continuous at \(x=0\).

Step 3: Analyze differentiability at \(x = 0\)

A function is differentiable at a point if its derivative exists at that point. The derivative \[ f'(x) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]needs to be evaluated. The expression\[ \lim_{h \to 0} \frac{h e^{-\left(\frac{1}{h}\right)} - 0}{h} = e^{-\left(\frac{1}{h}\right)} \to \text{does not exist as } h \to 0\] The limit does not exist because \(e^{-\left(\frac{1}{h}\right)}\) does not tend to any finite value. So, \(f\) is not differentiable at \(x=0\).

Step 4: Synthesize the findings

The analysis indicates that \(f(x)\) is continuous for all \(x\) since the limit at \(x = 0\) is the same as the function value, confirming continuity. However, we have shown that the function is not differentiable at \(x = 0\) due to the non-existence of the derivative at that point.

Key Concepts

Differentiability at a Point

Differentiability at a point essentially means that the function has a defined slope, or rate of change, at that specific point. For a function to be differentiable at a point, say at \(x = a\), it requires that the derivative exists at that point.

• This is checked using the formula for the derivative:

\[f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\]In the given function, we analyze differentiability at \(x=0\). We attempted to compute the derivative and found that the expression involves \(e^{-\left(\frac{1}{h}\right)}\). As \(h\) approaches zero, \(\frac{1}{h}\) becomes infinitely large, which causes \(e^{-\left(\frac{1}{h}\right)}\) to approach zero at an undefined rate. Without a definite value for the limit, we conclude that \(f(x)\) is not differentiable at \(x=0\). This highlights how the exponential component affects derivatives at boundary points.

Limit of a Function

The concept of the limit is foundational to understanding calculus and continuity. It refers to the value that a function approaches as its input approaches a given point. For a function \(f(x)\) to be continuous at some point \(x = c\), the limit \(\lim_{x \to c} f(x)\) must exist and be equal to \(f(c)\). In our exercise, we analyzed \(\lim_{x \to 0} f(x)\) to determine if the function is continuous at \(x=0\).

• We substituted the function form given by \(x e^{-\left(\frac{1}{x}\right)}\) and then evaluated it as \(x\) approaches zero.

Switching to a variable \(t = |x|\), we took the limit and found that it equaled zero, matching \(f(0)\). Thus, the limit exists and matches the function's value at that point, indicating \(f(x)\) is continuous at \(x=0\). This emphasizes the importance of limits in determining continuity.

Discontinuity

Discontinuity in a function occurs when there is a sudden break or jump at a point in the function's graph. This means that the function does not smoothly continue at that point. For a function to be continuous at a point, it cannot be discontinuous there. In our problem, we checked for discontinuity at \(x=0\).

• The limit \(\lim_{x \to 0} f(x)\) method was used to ascertain if there were breaks in continuity.
• Since \(\lim_{x \to 0} f(x) = f(0)\), this confirmed no discontinuity at \(x=0\). This ensures a smooth pass through that point on the graph.

By understanding continuity and limits, we can confirm that \(f(x)\) does not have any jumps, gaps, or interruptions at zero, illustrating a vital aspect of how functions behave.

Exponential Function

Exponential functions, typically expressed in the form \(e^x\), play a critical role in the behavior of many functions in calculus due to their continuous and differentiable nature across all real numbers. The exponential function \(e^{-x}\) decreases rapidly as \(x\) increases. In the provided exercise, the component \(e^{-\left(\frac{1}{|x|} \mid \frac{1}{x}\right)}\) enters the analysis.

• This particular exponential expression significantly influences the behavior of the function even at boundary conditions, such as when \(x \to 0\).
• Its rapid decay means it approaches zero much faster than any polynomial or linear component might.

This exponential function contributes to \(f(x)\) being continuous everywhere but not differentiable at \(x=0\). Understanding how exponentials behave is key to analyzing the limits and differentiability in more complex functions.