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

Question

Consider the function \(f(x)=|x-2|+|x-5|, x \in R\). [2012] Statement 1: \(f^{\prime}(4)=0\) Statement 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

(A) Statement 1 is false, statement 2 is true.

Step 1: Understanding the Function

The function given is \(f(x) = |x-2| + |x-5|\). This function is composed of two absolute value components, which create different linear functions in different intervals of \(x\).

Step 2: Identify Critical Points

The critical points arise from the values inside the absolute values equalling zero. These occur when \(x = 2\) and \(x = 5\). These values will change the expression inside the absolute value, affecting the slope.

Step 3: Function in Different Intervals

Divide the function into intervals relative to the critical points: 1. \(x < 2\): \(f(x) = -(x-2) - (x-5) = -2x + 7\)2. \(2 \leq x < 5\): \(f(x) = (x-2) - (x-5) = x - 3\)3. \(x \geq 5\): \(f(x) = (x-2) + (x-5) = 2x - 7\)

Step 4: Calculate the Derivative

Find \(f'(x)\) in the differentiable interval \((2, 5)\): 1. For \(2 < x < 5\): \(f(x) = x - 3\) so \(f'(x) = 1\).2. There is no interval within \((2, 5)\) where \(f'(x) = 0\).

Step 5: Check Continuity and Differentiability

\(f(x)\) is continuous on \([2, 5]\) because absolute value functions are continuous everywhere. The derivative is \(1\) in \((2, 5)\), so \(f(x)\) is differentiable in this interval.

Step 6: Evaluate Function at Critical Points

Find \(f(2)\) and \(f(5)\):- \(f(2) = |2-2| + |2-5| = 3\)- \(f(5) = |5-2| + |5-5| = 3\)Thus, \(f(2) = f(5)\).

Step 7: Evaluate Statement 1 and 2

Statement 1 is false because \(f'(4) = 1\), not \(0\). Statement 2 is true based on the checks for continuity, differentiability inside \((2,5)\), and the equality \(f(2) = f(5)\). Therefore, the answer is (A).

Key Concepts

Absolute Value Functions

Absolute value functions are a fascinating part of calculus due to their piecewise linear nature. The absolute value of a number is simply its distance from zero on the number line, which makes it always non-negative. In mathematical terms, the absolute value function is defined as:

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

For the function \(f(x) = |x-2| + |x-5|\), this showcases how absolute values can create different expressions based on the ranges of \(x\). By exploring the behavior of the function in these different intervals, you can separate it into linear pieces. The points where \(x=2\) and \(x=5\) are especially important as these are where the expression inside the absolute value changes sign. This leads to changes in the behavior or 'pieces' of the function, making it essential to evaluate them separately for complete understanding.

Continuity in Calculus

Continuity in calculus is a fundamental concept that describes a function without any breaks, jumps, or holes over an interval. A function is continuous at a point if the limit as you approach the point is equal to the function's value at that point. For functions defined over an interval, like \(f(x) = |x-2| + |x-5|\), continuity means that these joining points (or critical points) do not cause a disruption.Using the property of absolute value functions, we can say that \(f(x)\) is continuous everywhere because absolute value functions are inherently continuous. As it turns out, \(f\) is specifically continuous on the interval \([2,5]\). Each linear interval joins smoothly due to the nature of absolute values, ensuring that there is no sudden jump or break in the function.

Critical Points in Calculus

Critical points are specific values of \(x\) in the domain of a function where the behavior of the function changes—either a maximum, a minimum, or a point of inflection. For absolute value functions or even in differentiable functions, critical points are often where the derivative is zero or undefined.In our function \(f(x) = |x-2| + |x-5|\), critical points arise at \(x = 2\) and \(x = 5\). These are where the absolute value expressions change direction, and thus, they affect the slope of the corresponding linear pieces of the function. Evaluating the behavior at these points can reveal transitions in the function's slope, guiding us in examining whether the function contains maxima, minima, or potential cusps.In this specific problem, the derivative is consistently 1 in the interval \((2,5)\), indicating neither a maximum nor a minimum occurs within this domain. That result leads to Statement 1 in the exercise being false, as there is no \(f'(4) = 0\) due to constant slope throughout these points.