Problem 43 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Consider the fucntion, \(f(x)=|x-2|+|x-5|, x \in R\) 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

(B) Statement-1 is true, Statement-2 is true, and Statement-2 correctly explains Statement-1.

Step 1: Understanding f(x)

The function is given as \( f(x) = |x-2| + |x-5| \), which is a piecewise function composed of two absolute value expressions. On the interval \( x \in \mathbb{R} \), this function breaks into different linear components in different ranges.

Step 2: Determine Critical Points

The critical points, where the behavior of the function changes, occur at the points \( x = 2 \) and \( x = 5 \). The function can be split into different expressions according to the intervals \( x < 2 \), \( 2 \leq x < 5 \), and \( x \geq 5 \).

Step 3: Evaluate f(x) in Different Intervals

- For \( x < 2 \): \( f(x) = -(x-2) + -(x-5) = -2x + 7 \).- For \( 2 \leq x < 5 \): \( f(x) = x - 2 + -(x-5) = 5 - 2 = 3 \).- For \( x \geq 5 \): \( f(x) = x - 2 + x - 5 = 2x - 7 \).

Step 4: Check Continuity and Differentiability

The function \( f(x) \) is continuous everywhere, including the interval \([2, 5]\). It is linear in its segments, making it differentiable everywhere except possibly at the points \( x = 2 \) and \( x = 5 \) where the absolute values change expressions.

Step 5: Calculate f'(x) and f'(4)

The derivative \( f'(x) \) in the interval \( 2 < x < 5 \) is 0 because the function is constant: \( f(x) = 3 \). Specifically, at \( x = 4 \), \( f'(4) = 0 \).

Step 6: Verify f(2) = f(5)

- Calculate \( f(2) = |2-2| + |2-5| = 0 + 3 = 3 \).- Calculate \( f(5) = |5-2| + |5-5| = 3 + 0 = 3 \).- Since \( f(2) = f(5) = 3 \).

Step 7: Evaluate Given Statements

- Statement-1: "\( f'(4) = 0 \)" is true.- Statement-2: "\( f \) is continuous in \([2,5]\), differentiable in \((2,5)\), and \( f(2) = f(5) \)" is true.

Key Concepts

Piecewise Functions

Piecewise functions are an important concept in calculus. These functions are defined by different expressions depending on the input value or the range of the input. A piecewise function is essentially multiple sub-functions, each applying to a certain interval of the main function's domain.

In the given exercise, the function \( f(x) = |x-2| + |x-5| \) is a piecewise function composed of two absolute value expressions. Absolute values result in changes of behavior or pieces because they break the real line at their critical points.

To better understand such a function, especially when it involves absolute values, it's helpful to identify the critical points where the function's expression changes. For \( f(x) \), these points are \( x = 2 \) and \( x = 5 \).

The function's behavior changes accordingly in different intervals:

• For \( x < 2 \), the expression is \( f(x) = -2x + 7 \).
• For \( 2 \le x < 5 \), the expression simplifies to \( f(x) = 3 \).
• For \( x \ge 5 \), it changes to \( f(x) = 2x - 7 \).

Piecewise functions can often look complicated, but breaking them into recognizable parts makes them easier to analyze. Understanding this is crucial in calculus, where such functions often appear.

Continuity

Continuity in mathematics means that a function has no sudden jumps, breaks, or holes—it's smooth over its domain. A function is continuous at a point if small changes in the input produce small changes in the output.

For the function \( f(x) = |x-2| + |x-5| \):

• Since it is defined and behaves consistently across all real numbers, it is continuous everywhere.
• In particular, the function is continuous in the interval \([2, 5]\), which is crucial for solving the given problem.

To prove this in our exercise, we check the function values at critical points. We find:

• At \( x = 2 \), \( f(2) = 3 \).
• At \( x = 5 \), \( f(5) = 3 \).

These equal values confirm continuity at the boundaries and throughout the interval. The continuity ensures that \( f(x) \) does not leap abruptly, allowing us to evaluate its differentiability next.

Differentiability

Differentiability ensures that a function has a well-defined tangent and slope at every point within a certain interval. If a function is differentiable at a point, it must also be continuous there. However, the reverse isn't always true—a function can be continuous without being differentiable.

For the piecewise function \( f(x) \) discussed:

• It is differentiable in the interval \((2, 5)\) because in this region, \( f(x) \) remains constant. Hence, its derivative \( f'(x) = 0 \), showing a zero slope.
• At \( x = 4 \), specifically, \( f'(4) = 0 \) since this is within the interval where \( f(x) = 3 \) is constant.

Yet, the points \( x = 2 \) and \( x = 5 \) require special attention. Here, the absolute value operation introduces "sharp corners" where the derivative does not exist, meaning \( f(x) \) isn’t differentiable at these points. Differentiability is often a local property, highlighting the function's smoothness in specific regions rather than across its entire domain.