Problem 55 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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

Statement 1 is false, Statement 2 is true. (a)

Step 1: Understand the Function

The function given is \( f(x) = |x-2| + |x-5| \). This function represents the sum of two absolute value functions, each creating a V-shape with vertexes at \(x = 2\) and \(x = 5\).

Step 2: Analyze Differentiability

Since the absolute value function creates sharp points at its vertices, \(f(x)\) is not differentiable at these points: \(x = 2\) and \(x = 5\). However, \(f(x)\) is differentiable everywhere else in the interval \((2, 5)\).

Step 3: Compute the Derivative in \((2,5)\)

Between \(2 < x < 5\), the function can be split into linear pieces due to the absolute values: 1. For \(x < 2\), \(f(x) = -(x-2) + (x-5) = -3\).2. For \(2 < x < 5\), \(f(x) = (x-2) + (x-5) = 2x - 7\), thus \(f'(x) = 2\).3. For \(x > 5\), \(f(x) = (x-2) - (x-5) = 3\).

Step 4: Evaluate \(f^{\prime}(4)\)

For \(x = 4\), which falls within the interval \((2,5)\), we use the derivative found: \(f'(x) = 2\). Thus, \(f'(4) = 2\). Statement-1 is therefore false because \(f'(4) ≠ 0\).

Step 5: Verify Continuity and Equal Values

The function \(f(x) = |x-2| + |x-5|\) is continuous over the entire real line, including \([2,5]\). Evaluate \(f(2) = |2-2| + |2-5| = 0 + 3 = 3\) and \(f(5) = |5-2| + |5-5| = 3 + 0 = 3\). Hence \(f(2) = f(5)\). Statement-2 is true.

Key Concepts

Absolute Value Function

An absolute value function is a function that describes the distance of a number from zero on the number line. It is characterized by its V-shaped graph. For instance, the function \(f(x) = |x-a|\) has a vertex at \(x = a\) and opens upwards. The graph signifies how values of \(x\) transform through the absolute value.In the given exercise, the function \(f(x) = |x-2| + |x-5|\) combines two separate absolute value expressions. The first term \( |x-2| \) has a vertex at \(x = 2\) and the second term \( |x-5| \) has a vertex at \(x = 5\). This creates sharp corners at \(x = 2\) and \(x = 5\) on the graph of \(f(x)\), making the function appear as a series of two V-shapes connected. These sharp points significantly affect the differentiability of the function, as further explored in the analysis of continuity and derivatives.

Continuity

Continuity, in mathematical terms, refers to a function that has no discontinuities over its domain. This means that the function doesn't have any jumps, breaks, or holes in its graph. Essentially, you can draw the graph without lifting your pencil.In the exercise, the function \(f(x) = |x-2| + |x-5|\) was stated to be continuous. This is indeed true because absolute value functions themselves are continuous over their entire domain and so, a sum of such functions is also continuous. \[f(x)\] is continuous from \[x = 2\] to \[x = 5\] and beyond, without any interruption.It is crucial to understand that even though a function is continuous, it doesn't necessarily mean it is differentiable everywhere. As seen, even though \(f(x)\) is continuous at \(x = 2\) and \(x = 5\), because of the sharp turns, it is not differentiable at these points.

Derivatives

The derivative of a function provides the rate at which the function's value changes with respect to its input value. In simpler terms, it tells us how steep the graph is at any given point.For the given function \(f(x) = |x-2| + |x-5|\), differentiability is affected by the sharp points at \(x = 2\) and \(x = 5\). These points, known as non-differentiable points, occur because the direction of the graph changes abruptly. Thus, the derivative does not exist at these points. However, between \(x = 2\) and \(x = 5\), the function is linear and differentiable, as both absolute values have been resolved to either positive or negative linear equations.According to the step-by-step solution, the derivative on this interval is a constant 2, due to the linear segment represented by \(2x - 7\). Hence, \(f'(4) = 2\) not 0. This analysis reflects that while finding derivatives might seem straightforward, identifying differentiable regions and non-differentiable points is essential for accurate results.