Problem 37 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(S\) be the set of points where the function, \(f(x)=|2-| x-3 \|, x \in \boldsymbol{R}\), is not differentiable. Then \(\sum_{x \in S} f(f(x))\) is equal to-

Step-by-Step Solution

Answer

The sum is 3.

Step 1: Identify Points of Non-Differentiability

The function, given by \( f(x) = |2 - |x - 3|| \), involves the composition of absolute functions. These functions are typically non-differentiable at points where the expression inside the absolute value equals zero. Thus, we first identify these points: 1. \( |x - 3| = 2 \), which gives \(x - 3 = 2\) or \(x - 3 = -2\). Solving both, we get \(x = 5\) and \(x = 1\).2. At points where \( |x-3| = 0 \), which happens at \(x = 3\).Hence, the function \(f(x)\) is not differentiable at \(x = 1, 3,\) and \(5\).

Step 2: Calculate f(f(x)) for Non-Differentiable Points

Next, we evaluate \(f(f(x))\) for each non-differentiable point \(x = 1, 3, 5\):1. For \(x = 1\), first \(f(1) = |2 - |1 - 3|| = |2 - 2| = 0\), so \(f(f(1)) = f(0) = |2 - |0 - 3|| = |2 - 3| = 1\).2. For \(x = 3\), \(f(3) = |2 - |3 - 3|| = |2 - 0| = 2\), and \(f(f(3)) = f(2) = |2 - |2 - 3|| = |2 - 1| = 1\).3. For \(x = 5\), \(f(5) = |2 - |5 - 3|| = |2 - 2| = 0\), and \(f(f(5)) = f(0) = 1\).

Step 3: Sum the Values of f(f(x))

After computing \(f(f(x))\) for each of the identified points, we sum these values: \[ f(f(1)) + f(f(3)) + f(f(5)) = 1 + 1 + 1 = 3. \]

Step 4: Final Step: State the Result of the Sum

The sum of \(f(f(x))\) at the points where \(f(x)\) is not differentiable is 3.

Key Concepts

Absolute Functions

Absolute functions refer to expressions that involve the absolute value operation, represented by vertical bars, |x|. This operation transforms any real number into its non-negative equivalent, by flipping negative values while leaving positive ones unchanged.
When working with absolute functions, it's crucial to understand that at any point where the expression inside the absolute value becomes zero, the behavior of the function might change. This change can often lead to points where the function is not smooth, potentially resulting in non-differentiability.
For instance, in the function given as part of our exercise, \(f(x) = |2 - |x - 3||\), the inner absolute value \(|x - 3|\) and the outer absolute value \(|2 - \, |x - 3||\) create nested cases where each needs to be carefully examined. These nested absolute functions are crucial to identify underlying critical points where changes in smoothness, and hence differentiability, occur.

Non-Differentiable Points

Non-differentiability in a function refers to points where the function does not possess a derivative. Typically, such points occur where the function graph has sharp turns, vertical tangents, or discontinuities.
For absolute functions like \( f(x) = |2 - |x - 3|| \), non-differentiability arises at the zero points of the inner and outer parts of the absolute value. To find these points, we solve for when these expressions equal zero:

• \( |x - 3| = 2 \): Solving yields \(x = 1\) and \(x = 5\).
• \( |x - 3| = 0 \): Solving gives \(x = 3\).

At \(x = 1\), \(x = 3\), and \(x = 5\), the function \(f(x)\) becomes non-differentiable. Knowing these points helps identify intervals where the function may behave irregularly, essential for understanding the overall analysis of any function problem.

Function Composition

Function composition involves applying one function to the results of another. It's like nesting operations, where the output of one function becomes the input of another.
Unlike simple functions, composed functions can create more complex behaviors. In the exercise example, \( f(f(x)) \) is an application of composition, where \( f(x) \) is applied twice. To analyse this, we first determine \( f(x) \) for each non-differentiable point identified, and then re-evaluate \( f(x) \) again for these outputs:

• Start with \(x = 1\): Evaluate \(f(1)\), then use this result for another \(f\). \( f(1) = 0 \), leading to \(f(f(1)) = f(0) = 1\).
• For \(x = 3\): \(f(3) = 2\), continue with \(f(f(3)) = f(2) = 1\).
• Finally, for \(x = 5\): \(f(5) = 0\) translates to \(f(f(5)) = f(0) = 1\).

This composition helps in understanding how iterating operations on values from initial non-differentiable points delivers outputs crucial for problem solving and analysis of result patterns.