Problem 32 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\int_{0}^{1} \frac{d t}{1+|x-t|}\), then \(f^{\prime}\left(\frac{1}{2}\right)\) is equal two (A) 1 (B) \(-1\) (C) \(\frac{1}{2}\) (D) 0

Step-by-Step Solution

Answer

The value of \( f'\left(\frac{1}{2}\right) \) is 0.

Step 1: Understand the given function

The function given is an integral function: \[ f(x) = \int_{0}^{1} \frac{dt}{1+|x-t|} \] This represents an integral of the function \( \frac{1}{1+|x-t|} \) with respect to \( t \) from 0 to 1.

Step 2: Differentiate under the integral sign

To find \( f'(x) \), use Leibniz's Rule for differentiating under the integral sign: \[ f'(x) = \frac{d}{dx} \int_{0}^{1} \frac{dt}{1+|x-t|} = \int_{0}^{1} \frac{-\partial}{\partial x}\left(\frac{1}{1+|x-t|}\right) dt \] This produces two cases depending on the sign of \( x-t \).

Step 3: Handle absolute value cases

For absolute values, consider: - **Case 1**: \( x - t \geq 0 \Rightarrow |x-t| = x-t \). - **Case 2**: \( x - t < 0 \Rightarrow |x-t| = t-x \). These two cases provide piecewise functions to differentiate as necessary.

Step 4: Compute derivative for Case 1

In Case 1, we have \( |x-t| = x-t \): \[ \frac{1}{1 + x - t} \Rightarrow \frac{-1}{(1 + x - t)^2} \] Here, integrate from \( x \) to 1 since \( x-t \geq 0 \).

Step 5: Compute derivative for Case 2

In Case 2, we have \( |x-t| = t-x \): \[ \frac{1}{1 + t - x} \Rightarrow \frac{1}{(1 + t - x)^2} \] Here, integrate from 0 to \( x \) since \( x-t < 0 \).

Step 6: Evaluate derivative with respect to specific point

Identify \( f'(x) \) as \[ f'(x) = \int_{0}^{x} \frac{1}{(1 + t - x)^2} dt - \int_{x}^{1} \frac{1}{(1 + x - t)^2} dt \] and evaluate this at \( x = \frac{1}{2} \).

Step 7: Simplify the expression

Calculate each integral: - For \( x = \frac{1}{2} \), evaluate \( \int_{0}^{\frac{1}{2}} \frac{1}{(1 + t - \frac{1}{2})^2} dt \) and \( -\int_{\frac{1}{2}}^{1} \frac{1}{(1 + \frac{1}{2} - t)^2} dt \).

Step 8: Calculate the definite integrals

By evaluating both integrals, find that each results in expressions that simplify to zero at the point \( x = \frac{1}{2} \).

Step 9: Conclude based on simplification

Since all components cancel, the derivative \( f'\left(\frac{1}{2}\right) \) evaluates to zero entirely.

Key Concepts

Leibniz's Rule for Differentiation

Leibniz's Rule is a crucial tool in the field of calculus, especially when dealing with integral functions that depend on a variable parameter. It allows us to differentiate an integral with respect to a parameter—usually a variable within the limits of integration. This rule states that if you have a function defined by the integral \[ G(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt \]then its derivative with respect to \( x \) is given by\[G'(x) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt\]This formula effectively handles scenarios where either or both limits of integration might change with \( x \), or the integrand itself depends directly on \( x \). In the given exercise, however, the limits are constants (0 and 1), simplifying the formula to focus solely on differentiating the integrand with respect to \( x \). It's important to note that when dealing with absolute value expressions, special attention is required to correctly handle the derivative based on different intervals, as seen in the piecewise function scenario.

Understanding Piecewise Functions

Piecewise functions are types of functions defined by different expressions, each valid over a particular interval of the domain. They are particularly useful for managing cases where the behavior of the function changes at certain points—often indicated by jumps, breaks, or changes in slope. In the original exercise, the function \( \frac{1}{1+|x-t|} \) can be expressed as a piecewise function based on whether \( x \) is greater than or less than \( t \).

• Case 1: If \( x - t \geq 0 \), then \( |x-t| = x-t \). This affects the integrand, leading to a simplification within a certain range. This is where the form \( \frac{1}{1 + x - t} \) is valid for integration.
• Case 2: If \( x - t < 0 \), then \( |x-t| = t-x \). Here, the expression transforms to \( \frac{1}{1 + t - x} \), which again leads to a distinct function to integrate over a different range.

This approach is instrumental when solving problems involving absolute values in the integrals, as the handling of these expressions often requires implementing piecewise analysis to correctly evaluate the derivative.

Definite Integral Evaluation Techniques

Evaluating definite integrals involves calculating the integral of a function within specified bounds. This is a key concept in calculus, widely used to determine the accumulated area under a curve between two points. In the original exercise, the integral must be evaluated over two intervals determined by the piecewise function:

• Integrate \( \frac{1}{(1 + t - x)^2} \) from 0 to \( x \) for when \( t-x > 0 \).
• Integrate \( -\frac{1}{(1 + x - t)^2} \) from \( x \) to 1 for when \( x-t\geq 0 \).

The evaluation of these integrals at the specific value of \( x = \frac{1}{2} \) involves straightforward antidifferentiation followed by substitution of the bounds. In calculus, when the integral results in functions that cancel each other out or sum to zero, it demonstrates the zero net area between curves intersecting at a midpoint or critical point. This highlights the importance of accurate evaluation and simplification to reach the final result, like finding that \( f'(\frac{1}{2}) = 0 \) in this scenario.