Problem 50 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=|x-2|\) and \(g(x)=f\\{f(x)\\}\), then \(g^{\prime}(x)\) for \(x>2\) is (A) \(-1\) (B) 1 (C) 0 (D) does not exist

Step-by-Step Solution

Answer

(B) 1

Step 1: Understand the Function f(x)

The function given is \(f(x) = |x-2|\). This is the absolute value function, which can be expressed in two cases: \(f(x) = x-2\) for \(x \geq 2\) and \(f(x) = 2-x\) for \(x < 2\). For the domain \(x > 2\), \(f(x)\) simplifies to \(x - 2\).

Step 2: Express g(x) in Terms of f(x)

The function \(g(x) = f(f(x))\). Since we are considering \(x > 2\), \(f(x) = x - 2\). Thus, \(g(x) = f(x-2) = |(x-2) - 2| = |x - 4|\).

Step 3: Differentiate g(x) for x > 2 to Find g'(x)

Since \(x > 2\) and more specifically considering values where \(x > 4\), we use the simplified form \(g(x) = x - 4\). The derivative of \(x - 4\) with respect to \(x\) is \(g'(x) = 1\).

Step 4: Identify the Correct Option

For \(x > 2\), considering that \(g(x) = x - 4\) for \(x > 4\), we have determined that \(g'(x) = 1\). Therefore, the correct option is (B) 1.

Key Concepts

absolute value function

The absolute value function is a fascinating concept in mathematics, known for measuring the distance of a number from zero on the number line. It's often represented as

• If the expression inside the absolute value is positive or zero, the function returns that same expression,
• If the expression is negative, the function returns the opposite (or the positive version) of that expression.

In a more mathematical notation, for a function given by \( f(x) = |x| \), we have:

• \( f(x) = x \) if \( x \geq 0 \)
• \( f(x) = -x \) if \( x < 0 \)

This feature of switching between different forms makes the absolute value function a piecewise function, something vital to understand when exploring further calculus concepts like differentiation.

composite functions

Composite functions occur when one function is applied to the outcome of another. This is a quintessential part of understanding more complex mathematical models. If you have two functions, say \( f(x) \) and \( g(x) \), a composite function \( g(f(x)) \) means you're applying \( g(x) \) to what \( f(x) \) outputs.
These functions are usually symbolized by \( g(f(x)) \), representing that the argument of \( g(x) \) is replaced by the function \( f(x) \), not a simple \( x \). This makes composite functions quite flexible, as they introduce a way to "chain" the processes of multiple functions together.
For example, in the given scenario, the composite function \( g(x) = f(f(x)) \) requires understanding how the inner function \( f(x) = |x - 2| \) transforms the input first. When delineating these layers correctly, differentiating the composite function becomes much more straightforward.

differentiation rules

Differentiation rules form the backbone of calculus. They allow us to find the rate at which one quantity changes with respect to another. The basic rules, such as the power rule, constant rule, and sum rule, provide straightforward methods for handling straightforward functions. However, working with more complex expressions, especially those involving absolute value and composite functions, requires careful attention and sometimes, additional tools like piecewise differentiation.

• Power Rule: Often expressed as if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).
• Constant Rule: If \( f(x) = c \) (where \( c \) is a constant), \( f'(x) = 0 \).
• Chain Rule: For composite functions like \( h(x) = g(f(x)) \), the derivative \( h'(x) = g'(f(x)) \cdot f'(x) \).

In applying these rules to an absolute value function or a composite of such, one typically breaks down the function into segments to handle each piece separately. This entails determining the domains where each piece of the function's definition applies and then differentiating each segment distinctly. By doing so, we derive the correct outputs like \( g'(x) = 1 \) for \( x > 4 \), obtained by differentiating \( g(x) = x - 4 \), resulting from the simplification of \( |x - 4| \) over the correct domain.