Problem 119 - A Complete Resource Book in Mathematics for JEE Main

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) \(g(x)\) increases in \((0, 0.5)\); (C) \(g(x)\) decreases in \((0.5, 1)\).

Step 1: Understand the Function g(x)

The given function is defined as \( g(x) = f(x) + f(1-x) \). This means that \( g(x) \) is a combination of \( f(x) \) and \( f(1-x) \).

Step 2: Differentiate g(x)

To understand the behavior of \( g(x) \), we need to differentiate it with respect to \( x \):\[ g'(x) = f'(x) - f'(1-x) \]. This is the derivative of \( g(x) \).

Step 3: Use Given Condition on f(x)

We know that \( f''(x) < 0 \) for \( 0 \leq x \leq 1 \). This implies that \( f'(x) \) is decreasing in this interval.

Step 4: Analyze g'(x) when x < 0.5

For \( x < 0.5 \), \( 1-x > 0.5 \), and since \( f'(x) \) is decreasing, we have \( f'(x) > f'(1-x) \). Therefore, \( g'(x) = f'(x) - f'(1-x) > 0 \), meaning \( g(x) \) is increasing in this interval.

Step 5: Analyze g'(x) when x > 0.5

For \( x > 0.5 \), \( 1-x < 0.5 \), and using the fact that \( f'(x) \) is decreasing, we have \( f'(x) < f'(1-x) \). Therefore, \( g'(x) = f'(x) - f'(1-x) < 0 \), which indicates that \( g(x) \) is decreasing in this interval.

Step 6: Conclusion from Analysis

Based on the analysis, \( g(x) \) increases on the interval \( (0, 0.5) \) and decreases on the interval \( (0.5, 1) \).

Key Concepts

Differentiation

Differentiation is a fundamental tool in calculus that allows us to determine how a function changes at any point along its domain. It involves finding the derivative of a function, which essentially tells us the slope of the function at any given point. To differentiate a function like \( g(x) = f(x) + f(1-x) \), we need to apply the rules of differentiation to each part of the equation. When we differentiate \( g(x) \) with respect to \( x \), we get \( g'(x) = f'(x) - f'(1-x) \). This expression helps us understand how \( g(x) \) behaves in terms of increasing or decreasing on its domain by analyzing the sign of the derivative. If \( g'(x) > 0 \) in some interval, \( g(x) \) is increasing in that interval, whereas if \( g'(x) < 0 \), \( g(x) \) is decreasing.

Decreasing Functions

A function is considered decreasing in an interval if, as you move from left to right across that interval, the function's value declines. This occurs when the derivative of the function is negative over the interval you are observing. In the case of our function \( g(x) \), to determine where it is decreasing, we calculated the derivative \( g'(x) = f'(x) - f'(1-x) \). For \( x > 0.5 \), we found that \( g'(x) < 0 \), because \( f'(x) < f'(1-x) \). As a result, \( g(x) \) decreases in the interval \( (0.5, 1) \). This means that as \( x \) moves from 0.5 to 1, the outputs of \( g(x) \) consistently lower.

Increasing Functions

In contrast to a decreasing function, an increasing function's value goes up as you move from left to right across an interval. This is indicated by a positive derivative in that interval. From our example, we see that for \( x < 0.5 \), the derivative \( g'(x) = f'(x) - f'(1-x) \) is positive because \( f'(x) > f'(1-x) \). Thus, we can conclude that \( g(x) \) is increasing in the interval \( (0, 0.5) \). This tells us that if you pick any two points \( x_1 < x_2 \) within this interval, then \( g(x_1) < g(x_2) \), confirming the increasing nature of \( g(x) \) between 0 and 0.5.

Concavity

Concavity refers to the curvature of the graph of a function. Simply put, it helps us to understand if the graph is "bending" upwards or downwards. A function is concave down if its second derivative is less than zero, indicating that the graph of the function curves downwards, like an upside-down "U". In our exercise, it is given that \( f''(x) < 0 \) for \(0 \leq x \leq 1\), which tells us that \( f(x) \) is concave down in this interval. Concavity impacts the function's behavior in terms of its rate of increase or decrease, particularly when it comes to determining intervals of increasing or decreasing nature. This property of \( f(x) \) contributes significantly to why \( g(x) \) has the increasing and decreasing intervals we calculated earlier.