Problem 126 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Consider the function \(f(x)=|x-2|+|x-5|, x \in R\). Statement 1: \(f^{\prime}(4)=0\) Statement \(\mathbf{2}: f\) is continuous in \([2,5]\), differentiable in \((2,5)\) and \(f(2)=f(5)\). (A) Statement 1 is false, statement 2 is true (B) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (C) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (D) Statement 1 is true, statement 2 is false

Step-by-Step Solution

Answer

(C) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Step 1: Analyze the Function Definition and Continuity

The function given is \(f(x) = |x-2| + |x-5|\). This function is continuous because absolute value functions are continuous for all real numbers \(x\). Specifically, \(f(x)\) will be continuous on the interval \([2,5]\) as it is composed of continuous functions.

Step 2: Check Differentiability

To understand differentiability, observe that \(f(x)\) involves absolute values, which can cause non-differentiability at certain points. Near points 2 and 5, the derivative can change suddenly due to the corners at these values. In \((2,5)\), \(f(x)\) is defined by linear functions on either side, thus differentiable in \((2,5)\) but not at \(x=2\) and \(x=5\).

Step 3: Evaluate \(f'(x)\) and \(f'(4)\)

Inside the interval \((2,5)\), specifically for \(x=4\), the function can be written as \(f(x) = (x-2) + (5-x) = 3\). Therefore, \(f(x)\) is constant between 2 and 5, and its derivative is 0 in this interval, including at \(x=4\). Therefore, \(f'(4) = 0\).

Step 4: Check the Values of \(f(x)\) at the Endpoints

Calculate \(f(2) = |2-2| + |2-5| = 3\) and \(f(5) = |5-2| + |5-5| = 3\). Since both values are equal, \(f(2) = f(5)\).

Step 5: Evaluate Statements

Statement 1 is true, as \(f'(4) = 0\). Statement 2 is true, as the function is continuous on \([2,5]\), differentiable in \((2,5)\), and \(f(2) = f(5)\). However, statement 2 being true is not the explanation for statement 1 about the derivative at 4.

Key Concepts

Continuity

Continuity in mathematics refers to a function being smooth without any interruptions or jumps across a particular interval. For a function to be continuous at a point, three conditions must be fulfilled:
- The function is defined at the point.
- The limit of the function as it approaches the point exists.
- The value of the function at the point equals the limit.

Absolute value functions are inherently continuous across all real numbers, including the function in our exercise, \( f(x) = |x-2| + |x-5| \). Such functions do not break or jump and smoothly connect all points on their graphs. Specifically, \( f(x) \) is continuous in the interval \([2,5]\). Due to the continuity of the individual absolute value terms, their summation results in the same continuous behavior. This means there are no breaks in \([2,5]\), making statement 2 in our problem true regarding continuity.

Differentiability

Differentiability is a concept in calculus that indicates whether a function has a derivative at each point within an interval. It relates to the idea of a function being "smooth" without any sharp turns or cusps. For a function to be differentiable at a certain point, it must first be continuous at that point. However, unlike continuity, differentiability requires no kinks or abrupt changes in direction at that point.

While absolute value functions are globally continuous, they may not be differentiable everywhere. Specifically, they can have sharp corners or cusps at points where the expression inside the absolute value changes sign. For the function \( f(x) = |x-2| + |x-5| \), the differentiability needs to be reviewed at the points \( x=2 \) and \( x=5 \), since these are the points where the expressions inside the absolute values switch sign. In the interval \((2,5)\), \( f(x) \) is differentiable because it can be expressed as a simple linear combination of \( x-2 \) and \( 5-x \), without any sharp changes. However, it remains non-differentiable exactly at \( x=2 \) and \( x=5 \) due to the abrupt angle changes at these points, making the function non-smooth there.

Absolute Value Function

An absolute value function is a common mathematical function that returns the magnitude of a number, effectively removing any negative sign. It is expressed as \( |x| \), and mathematically defined as:

• \( |x| = x \) when \( x \geq 0 \)
• \( |x| = -x \) when \( x < 0 \)

These functions introduce unique characteristics to continuity and differentiability due to their piecewise nature. The involvement of absolute value functions, like in our exercise, \( f(x) = |x-2| + |x-5| \), often creates piecewise-defined scenarios.

For instance, \( f(x) \) could be broken down depending on whether \( x \) is less than 2, between 2 and 5, or greater than 5. These points of change influence the derivative as they inherently imply shifts from one linear behavior to another, not affecting continuity but potentially affecting differentiability. Understanding how absolute values contribute sharply at their parameter shifts can aid in realizing why differentiability is hindered at those points unlike continuity, which remains unaffected.