Problem 44 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=\left\\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.\) \(g(x)=|f(x)|+f(x))\). Then, in the interval \((-2,2), g\) is: (a) differentiable at all points (b) not continuous (c) not differentiable at two points (d) not differentiable at one point

Step-by-Step Solution

Answer

g is not differentiable at two points, x=0 and x=1.

Step 1: Understand the Piecewise Function

The function \( f(x) \) is defined piecewise with two parts: \(-1\) for \(-2 \leq x < 0\), and \(x^2 - 1\) for \(0 \leq x \leq 2\). We need to analyze these two pieces separately.

Step 2: Analyze the Output of g(x)

Given \( g(x) = |f(x)| + f(x) \), this simplifies to either \( 2f(x) \) when \( f(x) < 0 \), or zero when \( f(x) = 0 \). For \( f(x) \) to be zero, \( x^2 - 1 = 0 \), giving \( x = 1 \) as a zero point, since \( x = -1 \) is outside of the interval considered by that segment.

Step 3: Determine g in the Interval (-2,0)

For \( x \) in \((-2, 0)\), \( f(x) = -1 \). Therefore, \( g(x) = 2(-1) = -2 \). This is constant, so \( g(x) \) is differentiable over this interval.

Step 4: Determine g at x = 0

As \( x \) approaches 0 from the left, \( g(x) = -2 \). Approaching from the right (or at \( x=0 \) because \( f(0) = -1 \)), \( g(x) = 0 \). Therefore, \( g(x) \) has a discontinuity at \( x = 0 \).

Step 5: Determine g at x > 0 (up to x=2)

Within the interval \((0, 2)\), \( f(x) = x^2 - 1 \). Now: If \( x eq 1 \), \( f(x) eq 0 \), hence \( g(x) = 2f(x) = 2x^2 - 2 \). When \( x = 1 \), \( g(1) = 0 \). This forms a sharp corner at \( x = 1 \), thus \( g(x) \) is not differentiable there.

Step 6: Conclusion on Differentiability

Given that \( g(x) \) is not continuous at \( x=0 \) and has a corner at \( x=1 \), it is discontinuous at one point and thus not differentiable at two points: \( x=0 \) and \( x=1 \).

Key Concepts

Piecewise Function

A piecewise function is a mathematical expression defined by different expressions based on different intervals of the domain.
In this context, the function \( f(x) \) is divided into two unique expressions, providing a model that adjusts its output based on the value of \( x \).

• For \( -2 \leq x < 0 \), the function is \( f(x) = -1 \). Here, \( f(x) \) is constant, indicating a horizontal line segment on the graph.


• For \( 0 \leq x \leq 2 \), the function is \( f(x) = x^2 - 1 \). This is a quadratic expression, representing a part of a parabola.

Understanding each segment independently is crucial. By analyzing each piece, we see how they come together to form the overall function. Essential characteristics, like differentiability and continuity of the whole function, can often be determined from examining these segments.

Continuity

Continuity is a fundamental concept in calculus that relates to a function's behavior at a point in its domain. A function is continuous at a point if it can be drawn without lifting the pen at that point.
This concept becomes especially important when dealing with piecewise functions, as they often feature points of transition between the pieces.

• For the function \( g(x) \), we look at how it behaves as \( x \) approaches 0 from both sides. At \( x = 0 \), we see that the function changes significantly. From the left, \( g(x) = -2 \), while from the right, \( g(x) = 0 \). Because these values do not match, \( g(x) \) is not continuous at \( x = 0 \).
• Similarly, continuity might be affected in intervals like around \( x = 1 \), where changes in the slope or direction (sharp corners) occur.

Identifying points of discontinuity helps in analyzing further properties like differentiability, since a function cannot be differentiable at a point where it is not continuous.

Absolute Value Function

The absolute value function, denoted as \( |x| \), returns the non-negative value of \( x \), effectively "flipping" any negative input to positive.
When applied in function manipulations such as \( g(x) = |f(x)| + f(x) \), it emphasizes the behavior of \( f(x) \) where \( f(x) < 0 \).

• For \( f(x) < 0 \), \( |f(x)| = -f(x) \), meaning \( g(x) = 0 \). Thus, the negative component cancels, resulting in a zero value.


• In cases where \( f(x) \) is non-negative, \( |f(x)| = f(x) \), and so \( g(x) = 2f(x) \) becomes active.

This functionality becomes pivotal in examining how \( g(x) \) behaves across its domain, especially near critical points that exhibit changes in direction or slope, leading to non-differentiability. By breaking down the absolute value's role, it's easier to understand where its effects cause alterations in the derivative.