Problem 91 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\phi(x)=f(x)+f(1-x)\) and \(f^{\prime \prime}(x)<0\) in \((-1,1)\), then \(\phi(x)\) strictly increases in the interval (A) \(\left(0, \frac{1}{2}\right)\) (B) \(\left(\frac{1}{2}, 1\right)\) (C) \((-1,0)\) (D) \((0,1)\)

Step-by-Step Solution

Answer

Option (A): \((0, \frac{1}{2})\) is the interval where \( \phi(x) \) is strictly increasing.

Step 1: Understanding the Given Information

We are given a function \( \phi(x) = f(x) + f(1-x) \) where the second derivative \( f''(x) < 0 \) for \( x \in (-1, 1) \). This implies that the function \( f(x) \) is concave on the interval \((-1, 1)\).

Step 2: Finding the Derivative of \( \phi(x) \)

To determine where \( \phi(x) \) is strictly increasing, first find its derivative. Differentiate \( \phi(x) \):\[\phi'(x) = f'(x) - f'(1-x)\]

Step 3: Analyzing the Derivative's Behavior

Given \( f''(x) < 0 \), \( f'(x) \) is decreasing in \((-1, 1)\). Therefore, if \( x > 1-x \) (i.e., \( x > \frac{1}{2} \)), then \( f'(x) < f'(1-x) \), making \( \phi'(x) < 0 \), which means \( \phi(x) \) is decreasing.

Step 4: Identifying the Interval of Increase

From the previous analysis, when \( x < \frac{1}{2} \), it implies that \( 1-x > x \). Hence, \( f'(x) > f'(1-x) \) within that range, leading to \( \phi'(x) > 0 \). Therefore, \( \phi(x) \) is strictly increasing on \( \left(0, \frac{1}{2}\right) \).

Step 5: Concluding the Correct Interval

The derivative analysis shows \( \phi(x) \) strictly increases on \((0, \frac{1}{2})\). Thus, the answer is option (A).

Key Concepts

Concave Functions

When a function is concave, it has a specific curvature kind, generally bending downwards like an upside-down bowl. A mathematical way to determine the concavity of a function is by analyzing its second derivative. If the second derivative, denoted as \( f''(x) \), is less than zero \( f''(x) < 0 \), then the function is concave over the interval considered.
This characteristic impacts how the function grows and shrinks on the interval. For a concave function \( f(x) \), as in this exercise, its slope decreases as you move along the x-axis, which means its rate of change (first derivative) is lessening. This lets us understand the general shape and behavior of the function, aiding in solving problems related to growth and maxima/minima of functions.
Understanding the role of concavity is crucial for many applications in mathematics, especially in optimization problems where knowing the shape of the function helps in identifying the best solutions. In this problem, realizing that \( f(x) \) is concave tells us about the nature of the change in the function and can guide us in analyzing other properties of related functions such as \( \phi(x) = f(x) + f(1-x) \).

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative itself represents the rate of change of the function concerning its variable. In simpler terms, it tells you how fast or slow a function's value changes as the input changes.
In this exercise, we are interested in finding the first derivative of \( \phi(x) = f(x) + f(1-x) \). By applying the rules of differentiation to each part, we find:

• The derivative of \( f(x) \) is simply \( f'(x) \).
• For \( f(1-x) \), using the chain rule, it becomes \(-f'(1-x)\).

Thus, the derivative of \( \phi(x) \) is \( \phi'(x) = f'(x) - f'(1-x) \). This derivative gives us key insights into where the function is growing or shrinking over its domain.
Differentiation, especially when combined with rules such as the chain rule, product rule, and quotient rule, provides powerful tools for analyzing the behavior of functions. It's a crucial skill for solving and understanding higher-level mathematics problems.

Analysis of Derivatives

Once you have a function's derivative, you can analyze it to determine where the original function increases or decreases. This is done by examining the sign of the derivative.

• When the derivative \( \phi'(x) \) is positive, the function is increasing on that interval.
• When it's negative, the function is decreasing.
• If the derivative equals zero, the function could either be flat (constant) or at a turning point (local maximum or minimum).

In our exercise, the derivative \( \phi'(x) = f'(x) - f'(1-x) \) is used to understand where \( \phi(x) \) is increasing.
By leveraging the information that \( f(x) \) is a concave function, which means \( f''(x) < 0 \) and thus \( f'(x) \) is decreasing, we deduce that for \( x < \frac{1}{2} \), the inequality \( f'(x) > f'(1-x) \) holds, resulting in a positive \( \phi'(x) \).
As a result, \( \phi(x) \) is strictly increasing in the interval \( (0, \frac{1}{2}) \). Analyzing derivatives is a crucial part of understanding the detailed behavior of functions, especially in finding intervals of increase or decrease, which has implications in many areas ranging from physics to economics.