Problem 19 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=-1+|x-2|\), and \(g(x)=1-|x| ;\) then the set of all points where fog is discontinuous is (a) \(\\{0,2\\}\) (b) \(\\{0,1,2\\}\) (c) \(\\{0\\}\) (d) an empty set

Step-by-Step Solution

Answer

The set of discontinuities is \(\{0\}\), hence (c).

Step 1: Understanding Composition of Functions

The composition of two functions \(f\) and \(g\) is denoted as \((f \circ g)(x) = f(g(x))\). Here, we need to find where \(f(g(x))\) is discontinuous.

Step 2: Analyzing Function g(x) Discontinuity

The function \(g(x) = 1 - |x|\) involves an absolute value, which potentially leads to points of non-differentiability. The critical point here is \(x = 0\), where the behavior can change.

Step 3: Analyzing Function f(x) Discontinuity

The other function \(f(x) = -1 + |x - 2|\) also includes an absolute value, possibly causing non-differentiability at \(x = 2\).

Step 4: Evaluating Discontinuity in f(g(x))

To find where \(f(g(x))\) is discontinuous, we substitute \(g(x)\) into \(f(x)\), which gives \(f(g(x)) = -1 + |1 - |x| - 2|\). We need to check around points where \(g(x)\) is not differentiable (\(x=0\)), and where \(f\) is not (\(x=g^{-1}(2)\) if applicable).

Step 5: Determining Discontinuity Points

Substitute \(g(x) = 2\) in the expression \(1 - |x| = 2\) gives no real solution, so \(x=2\) doesn't contribute to \(f(g(x))\) discontinuities. Evaluate at \(x = 0\): \(f(g(0)) = f(1) = -1 + |1 - 2| = 0\), handling possible step changes at \(x = 0\) for function behavior.

Key Concepts

Composition of Functions

Composing functions involves taking two functions and combining them into one. This is done by using the output of one function as the input for another. In notation, the composition of functions \( f \) and \( g \) is written as \( (f \circ g)(x) = f(g(x)) \).

When composing functions, it's crucial to understand how each function behaves individually. **Step-by-step behavior is key.**

• Determine the output of the inner function \( g(x) \).
• Substitute this output into the outer function \( f \).
• Analyze the result to identify any potential discontinuities or irregularities.

Composition is all about understanding how the change in one function affects the other when combined. This exercise uses two functions involving absolute values, which often introduce potential discontinuity points.

Absolute Value Functions

Absolute value functions can be tricky because they fundamentally change how we think about numbers on the number line.

An absolute value function has the form \( |x| \), which is always non-negative. No components of \( |x| \) can be less than zero, as it simply measures a number's distance from zero on the number line.

• For \( g(x) = 1 - |x| \), the function outputs values by subtracting the absolute value of \( x \) from 1.
• For \( f(x) = -1 + |x - 2| \), it shifts the absolute value function horizontally on the number line.
• Both functions contain critical points at the arguments that make the absolute value zero, like \( x=0 \) or \( x=2 \).

Absolute value functions can result in non-differentiability at certain points, where the graph of the function has sharp corners (points where the slope isn't defined smoothly). These sharp corners are potential sources of discontinuity when functions are composed.

Non-differentiability

Non-differentiability refers to points in a function's graph where there is no well-defined tangent line. This occurs at sharp turns or corners, common in absolute value functions.

In the exercise, we examine points where \( f(x) \) and \( g(x) \) lose their smoothness and check if these affect the composition \( f(g(x)) \).

• \( g(x) = 1 - |x| \) is non-differentiable at \( x = 0 \). The output abruptly changes direction, creating a corner.
• \( f(x) = -1 + |x - 2| \) shows non-differentiability at \( x = 2 \), due to the structure of its absolute value component.
• In the composition \( -1 + |1 - |x| - 2| \), checking each critical point \( x = 0 \), reveals if they cause discontinuous behavior.

Observing these specific points allows us to determine if the composed function has any discontinuity, which can contribute to understanding the function's overall behavior more comprehensively.