Problem 79 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\int_{-2}^{2} \max \\{(1-x),(1+x), 2\\} d x\) is (A) 8 (B) \(-8\) (C) 9 (D) \(-9\)

Step-by-Step Solution

Answer

The value of the integral is 8, which is option (A).

Step 1: Understanding the Max Function

First, we need to understand the function inside the integral: \( \max\{(1-x), (1+x), 2\} \). This function will take the maximum value out of \((1-x)\), \((1+x)\), and \(2\) for each \(x\) from \(-2\) to \(2\). Hence, we need to find the points where each function is equal to each other or changes dominance.

Step 2: Finding Points of Dominance Change

Let's solve for when two of the functions are equal:1. Set \(1-x = 2\) to find when it equals 2: \\(1-x = 2\Rightarrow x = -1 \).2. Set \(1+x = 2\) to find when it equals 2: \\(1+x = 2\Rightarrow x = 1 \).3. Set \(1-x = 1+x\) for when they are equal:\(1-x = 1+x \Rightarrow x = 0\).The points \(-1\), \(0\), and \(1\) are where changes in dominance happen.

Step 3: Analyzing the Segments

We split the integral at points \(-1\), \(0\), and \(1\) to analyze each segment:1. **For \(x \leq -1\)**: \((1-x) > (1+x)\), so \( \max(1-x, 1+x, 2) = 2\) for \(x = -2\) to \(-1\).2. **For \(-1 < x < 0\)**: \( (1+x) < (1-x) < 2 \), therefore \( \max(1-x, 1+x, 2) = 2\).3. **For \(0 \leq x < 1\)**: \( (1+x) < (1-x) \), so still \( \max(1-x, 1+x, 2) = 2\).4. **For \(x \geq 1\)**: \(1 + x = 2\ and \max(1-x, 1+x, 2) = 2\).Thus, the integrand is 2 for all partitions.

Step 4: Evaluating the Integral

Now that we know the maximum value is 2 over the entire interval, the integral simplifies to: \[ \int_{-2}^{2} 2 \, dx \].We solve this as:\[ 2 \int_{-2}^{2} 1 \, dx = 2 \times [x]_{-2}^{2} = 2 \times (2 + 2) = 8.\]

Step 5: Concluding the Solution

The value of the integral is 8, as derived by evaluating the integral over the interval \([-2, 2]\) where the maximum function remains 2. Thus, the correct answer is option (A).

Key Concepts

Max Function

In the context of definite integrals, the Max Function \((\text{max})\) selects the highest value among multiple functions for each input on an interval. For the integral \(\int_{-2}^{2} \max \{(1-x),(1+x), 2\} \, dx\), we determine at each point \(x\) which of the given expressions \(1-x\), \(1+x\), or \(2\) is dominant. By determining this, we can solve the integral by partitioning the interval and calculating based on this function's dominance.

• For every segment of the interval \([-2, 2]\), identify which expression dominates.
• This involves comparing the expressions across the interval to see when one overtakes others in value.

Recognizing which function dominates at every point helps simplify the integration process, as we only integrate the dominant function over each segment.

Points of Dominance

To find the Points of Dominance, we analyse where two functions are equal or change dominance. This helps in solving the max function effectively.

• Calculate when \(1-x = 2\) which is \(x = -1\).
• Calculate when \(1+x = 2\) which is \(x = 1\).
• Calculate when \(1-x = 1+x\) which is \(x = 0\) for equal value.Thus, the functions shift dominance at \(-1\), \(0\), and \(1\). This division helps in partitioning the interval appropriately for simpler integration.

Partitioning Intervals

In integration, Partitioning Intervals is breaking the main interval into smaller segments based on dominance changes. This method allows for a simplified calculation when dealing with complex functions. Here, since the function shift happens at \(-1\), \(0\), and \(1\), these critical points help in deciding how the function behaves.

• For \(x \, \leq \, -1\): Dominance of \(2\), thus \(\max(1-x, 1+x, 2) = 2\).
• For \(-1 < x < 0\): Dominance remains with \(2\), so \(\max(1-x, 1+x, 2) = 2\).
• For \(0 \leq x < 1\): Dominance stays as \(2\), hence \(\max(1-x, 1+x, 2) = 2\).
• For \(x\, \geq \, 1\): On analyzing, \(\max(1-x, 1+x, 2) = 2\).

These partitions show that \(2\) is the dominant function throughout, allowing a straightforward solution for \(\int_{-2}^{2} \, 2 \, dx = 8\).