Problem 90 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\left\\{\begin{array}{cc}x-3, & x<0 \\ x^{2}-3 x+2, & x \geq 0\end{array}\right.\) and \(g(x)=f(|x|)+|f(x)|\), then \(g(x)\) is (A) continuous in \(R-\\{0\\}\) (B) continuous in \(R\) (C) differentiable in \(R-\\{0,1,2\\}\) (D) differentiable in \(R-\\{1,2\\}\)

Step-by-Step Solution

Answer

Answer is (D): differentiable in \( R-\{1,2\} \).

Step 1: Analyze Function f(x)

The function \( f(x) \) is piecewise defined as \( f(x) = x - 3 \) for \( x < 0 \) and \( f(x) = x^2 - 3x + 2 \) for \( x \geq 0 \). We will analyze the continuity and differentiability of \( f(x) \) across these domains.

Step 2: Check Continuity of f(x) at x = 0

To check for continuity at \( x = 0 \), find \( \lim_{x \to 0^-} f(x) \) which equals \( -3 \), and \( \lim_{x \to 0^+} f(x) \) which equals \( 2 \). Since \( \lim_{x \to 0^-} f(x) eq \lim_{x \to 0^+} f(x) \), \( f(x) \) is not continuous at \( x = 0 \).

Step 3: Check Differentiability of f(x)

For \( x < 0 \), \( f(x) = x - 3 \) is differentiable, and for \( x \geq 0 \), \( f(x) = x^2 - 3x + 2 \) is differentiable. However, \( f(x) \) cannot be differentiable at \( x = 0 \) due to discontinuity.

Step 4: Analyze g(x) = f(|x|) + |f(x)|

For \( x < 0 \), \( f(|x|) = f(-x) = x - 3 \) and \( |f(x)| = |x-3| \). For \( x \geq 0 \), \( f(|x|) = f(x) = x^2 - 3x + 2 \) and \( |f(x)| = |x^2 - 3x + 2| \).

Step 5: Continuity and Differentiability of g(x) at Various Points

To check continuity of \( g(x) \) away from \( 0 \), analyze \( x = 1 \) and \( x = 2 \). At \( x = 1 \), \( f(x) = 0 \), so \( f(|x|) + |f(x)| = 0 + 0 = 0 \). At \( x = 2 \), \( f(x) = 0 \), so \( g(x) = 0 \). Both points show continuity. However, since the derivative depends on the absolute value function generating cusps at transitions, \( g(x) \) is not differentiable at \( x = 0, 1, 2 \).

Step 6: Select the Correct Answer from Options

Based on the analysis, \( g(x) \) is continuous in \( \mathbb{R} \) but not differentiable due to the problematic points at \( x = 0, 1, 2 \). The correct answer for the differentiability condition is option D, indicating differentiability in \( \mathbb{R} - \{1, 2\} \).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus, and it describes whether a function has any sudden jumps or breaks. To say that a function is continuous at a certain point means that the graph of the function has no holes or gaps at that point. For mathematical clarity, a function \( f(x) \) is continuous at a point \( x = c \) if the following three conditions are satisfied:

• The function \( f(x) \) is defined at \( x = c \).
• \( \lim_{x \to c} f(x) \) exists.
• \( \lim_{x \to c} f(x) = f(c) \).

For the piecewise function \( f(x) \) from the original exercise, continuity was tested at \( x = 0 \). It was found non-continuous because the limits from the left \( (x \to 0^-) \) and right \( (x \to 0^+) \) were not the same. This means the graph of \( f(x) \) jumps from one value to another, creating a break. Understanding continuity helps us anticipate how a function behaves across its domain, and whether we can expect smooth progression without interruptions.

Differentiability

Differentiability is closely related to continuity, but a function can be continuous without being differentiable. Differentiability at a point \( x = c \) means that the function has a defined derivative at that point, essentially indicating that the function has a well-behaved tangent that doesn't involve sharp corners or cusps. For a function to be differentiable at a point, it must first be continuous at that point. This is because having an abrupt break or jump disrupts the slope's consistency. For \( f(x) \) in the exercise, differentiability was assessed at various regions, with considerations made at \( x = 0 \). \( f(x) \) was continuous and differentiable in both segments of its piecewise definition individually, but not at \( x = 0 \) due to discontinuity there. Moreover, for \( g(x) \) (also derived by using absolute values), differentiability was found to be problematic around \( x = 0, 1, 2 \) due to sharp transitions in its behavior, a common glitch when dealing with absolute values.

Piecewise Functions

Piecewise functions are functions that have different expressions for different intervals of the domain. These are useful for modeling situations where a rule needs to change depending on the input value. However, this sometimes creates challenges in calculus, particularly at the points where the rules change. In the given exercise, \( f(x) \) is defined piecewise with different rules for \( x < 0 \) and \( x \geq 0 \). Such abrupt shifts can lead to issues in both continuity and differentiability at boundary points, such as \( x = 0 \) in this case. Implementing piecewise functions necessitates checking each boundary for discontinuities or non-differentiability. The function \( g(x) \) compounds this by using absolute values, adding complexity because it influences continuity and differentiability.

• Whenever a function combines segments differently, like with absolute values, verify each point carefully.
• This ensures a complete understanding of how each segment of a piecewise function behaves and interacts.

Piecewise functions train us to approach analysis rigorously, affirming precision as a hallmark of calculus.