Problem 112 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\left\\{\begin{array}{cc}\frac{x\left(3 e^{1 / x}+4\right)}{2-e^{1 / x}}, x \neq 0 \\ 0 & , x=0\end{array}\right.\), then \(f(x)\) is (A) continuous as well as differentiable at \(x=0\) (B) continuous but not differentiable at \(x=0\) (C) differentiable but not continuous at \(x=0\) (D) None of these

Step-by-Step Solution

Answer

(D) None of these

Step 1: Identify Continuity at x=0

To determine continuity at a point, the limit as x approaches the point must equal the function value at that point. Thus, we need to check if \( \lim_{{x \to 0}} f(x) = f(0) \). Given that \( f(0) = 0 \), we calculate the limit:\[ \lim_{{x \to 0}} \frac{x(3e^{1/x} + 4)}{2 - e^{1/x}}. \]Because the term \( e^{1/x} \) approaches \( \infty \) as \( x \to 0^+ \) and approaches \( 0 \) as \( x \to 0^- \), this function is not defined closely around \( x=0 \), making the limit impossible to find directly either positively or negatively. Thus, \( \lim_{{x \to 0}} f(x) \) does not exist, and \( f(x) \) is not continuous at \( x=0 \).

Step 2: Evaluate Differentiability at x=0

Since we have determined the function is not continuous at \( x=0 \), it cannot be differentiable at that point either. Differentiability requires continuity first; without continuity, differentiability cannot be achieved.

Key Concepts

Continuity

Continuity in a function is an essential idea in calculus. It describes a function that, if plotted, has no gaps, breaks, or jumps at any point within its domain. To determine if a function is continuous at a certain point, you check if the limit of the function as it approaches that point is equal to the function's value at that point. The mathematical expression for continuity at a point \( c \) is given as:\[ \lim_{{x \to c}} f(x) = f(c) \]For the specific function in question, it turns out that the limit as \( x \to 0 \) doesn't match the function value at \( x=0 \), implying a lack of continuity.

• Continuity requires \( \lim_{{x \to c}} f(x) = f(c) \)
• If the limit isn't equal to the function’s value, like in this exercise, the function is not continuous at that point.

Understanding continuity helps us predict the behavior of functions without interruptions. If a function isn't continuous, we also know differentiability fails.

Differentiability

Differentiability is about whether a function has a derivative at a certain point. For a function to be differentiable at a point, it must also be continuous there. This means the graph has to be smooth and not have sharp corners or discontinuities. The derivative represents the slope of the tangent line to the curve at a point.

• The main condition is: if a function is differentiable, it must be continuous.
• If a function isn't continuous at a point, it can't have a derivative at that point.

In the context of the exercise, since the function is not continuous at \( x = 0 \), it cannot be differentiable there either. Always remember: differentiation is a more stringent condition than continuity. Without meeting the continuity requirement, differentiability is out of the question.

Limits

Limits are at the heart of continuity and differentiability. They help us understand the behavior of a function as the input approaches a particular value. A limit takes the form:\[ \lim_{{x \to c}} f(x) \]This expression asks the question: "What value does \( f(x) \) approach when \( x \) gets really close to \( c \)?"

• Limits allow examination of function behavior without requiring specific values.
• They can help detect if a function is about to jump, like in the case when \( x \to 0 \) for the problem function.

For many functions, limits can be determined using algebraic manipulation or known limit properties. However, in more complex cases, like when \( e^{1/x} \) behaves differently from \( x \to 0^+ \) and \( x \to 0^- \), limits might not exist or become very difficult to assess. Recognizing the subtle nuances that limits reveal in a function’s behavior is crucial for deeper calculus understanding.