Problem 128 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(g(x)=f(x)+f(1-x)\) and \(f^{\prime \prime}(x)<0\) for \(0 \leq x \leq 1\), then (A) \(g(x)\) increases in \(\left(-\infty, \frac{1}{2}\right)\) (B) \(g(x)\) increases in \(\left(0, \frac{1}{2}\right)\) (C) \(g(x)\) decreases in \(\left(\frac{1}{2}, 1\right)\) (D) \(g(x)\) decreases in \(\left(\frac{1}{2}, \infty\right)\)

Step-by-Step Solution

Answer

(B) and (C): \(g(x)\) increases in \((0, 0.5)\) and decreases in \((0.5, 1)\).

Step 1: Understand the Problem

We are given a function \(g(x) = f(x) + f(1-x)\) and told that \(f''(x) < 0\) for the interval \(0 \leq x \leq 1\). We need to determine the behavior of \(g(x)\) in specified intervals.

Step 2: Analyze Second Derivative Condition

Since \(f''(x) < 0\), this means that \(f(x)\) is concave down for \(0 \leq x \leq 1\). This implies its slope (\(f'(x)\)) is decreasing in this interval.

Step 3: Find the Derivative of g(x)

Calculate the derivative of \(g(x)\): \(g'(x) = f'(x) - f'(1-x)\). The subtraction arises because the derivative of \(f(1-x)\) is \(-f'(1-x)\) due to the chain rule.

Step 4: Determine the Sign of g'(x) for x < 0.5

For \(x < 0.5\), \(f'(x)\) is decreasing, and due to concavity, \(f'(x) > f'(1-x)\). Thus, \(g'(x) = f'(x) - f'(1-x) > 0\), suggesting \(g(x)\) is increasing.

Step 5: Determine the Sign of g'(x) for x > 0.5

For \(x > 0.5\), since \(f'(x) < f'(1-x)\), then \(g'(x) = f'(x) - f'(1-x) < 0\). So, \(g(x)\) is decreasing.

Step 6: Conclusion on Intervals for g(x)

Combine the insights: \(g(x)\) increases in \(0 \le x < 0.5\) and decreases for \(0.5 < x \le 1\). Thus, (B) and (C) describe the behavior correctly.

Key Concepts

Understanding Concavity

The concept of concavity is crucial in understanding how a function behaves. When we say a function is concave down over an interval, like in our original exercise where \(f''(x) < 0\) for \(0 \leq x \leq 1\), it means that the function forms a 'frown' shape or a downward arc. This occurs because the second derivative, \(f''(x)\), indicates how the slope of the tangent line, given by the first derivative \(f'(x)\), is changing.

• If the second derivative is negative, the slope \(f'(x)\) is decreasing.
• This implies that as you move along the x-axis, the rate at which \(f(x)\) increases slows down until it begins to decrease, hence the frown shape.

For the given function \(f(x)\), knowing \(f''(x) < 0\) means it's consistently bending downwards between 0 to 1. Such a property helps predict how other functions related to \(f(x)\) will behave, as we see with \(g(x) = f(x) + f(1-x)\) in the exercise.

Applying the First Derivative Test

To determine where our function \(g(x) = f(x) + f(1-x)\) is increasing or decreasing, we use the first derivative test. We'll begin by finding its first derivative: \(g'(x) = f'(x) - f'(1-x)\). This derivative tells us how the function \(g(x)\) is changing at any point in the interval. Next, we analyze the sign of \(g'(x)\) to determine how \(g(x)\) behaves:

• If \(g'(x) > 0\), then \(g(x)\) is increasing at that interval.
• If \(g'(x) < 0\), then \(g(x)\) is decreasing.

In the original exercise, for \(x < 0.5\), due to concavity, we found that \(f'(x) > f'(1-x)\), leading to \(g'(x) > 0\); thus \(g(x)\) increases. Conversely, for \(x > 0.5\), since \(f'(x) < f'(1-x)\), \(g'(x) < 0\); therefore, \(g(x)\) decreases. By knowing these intervals, we can accurately describe where the function rises and falls.

Comprehensive Function Behavior Analysis

Analyzing the overall behavior of a function, like \(g(x)\), involves synthesizing information from concavity and the first derivative test.By understanding these properties:

• The concavity lets us know the general 'shape' of \(f(x)\) and thus how \(f'(x)\) behaves across an interval.
• The first derivative test provides the increasing or decreasing nature of \(g(x)\).

In our exercise, \(g(x) = f(x) + f(1-x)\), we find a balanced approach by combining these insights: - We determined that \(g(x)\) is increasing for \(0 \le x < 0.5\) since the derivative \(g'(x) = f'(x) - f'(1-x) > 0\).- It then decreases for \(0.5 < x \le 1\) as \(g'(x) < 0\).This behavior tells us precisely where the function is rising and falling across its domain, clarifying that responses (B) and (C) are correct.