Problem 91 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(A) \(\left(0, \frac{1}{2}\right)\)

Step 1: Understanding the Problem

We are given a function \( \phi(x) = f(x) + f(1-x) \), and we know that the second derivative of \( f(x) \), denoted by \( f''(x) \), is less than zero for \( x \) in \((-1,1)\). This means \( f(x) \) is concave down in this interval. We need to determine in which interval \( \phi(x) \) is strictly increasing.

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

The derivative of \( \phi(x) \) with respect to \( x \) is \( \phi'(x) = f'(x) - f'(1-x) \), using the chain rule for \( f(1-x) \). We need to analyze the sign of \( \phi'(x) \) to determine where \( \phi(x) \) is increasing.

Step 3: Analyzing the Sign of \( \phi'(x) \)

Because \( f''(x) < 0 \), \( f'(x) \) is a decreasing function. Hence, as \( x \) increases, \( f'(x) \) decreases. At the mid-point \( x = \frac{1}{2} \), \( f'(x) \) switches from being larger than \( f'(1-x) \) to smaller. Thus, \( f'(x) - f'(1-x) > 0 \) for \( x < \frac{1}{2} \) and \( f'(x) - f'(1-x) < 0 \) for \( x > \frac{1}{2} \).

Step 4: Determining the Interval Where \( \phi(x) \) is Increasing

From the analysis in Step 3, \( \phi(x) \) is increasing when the derivative \( \phi'(x) > 0 \). This occurs when \( x \) is less than \( \frac{1}{2} \) within the given domain. Therefore, the interval where \( \phi(x) \) is strictly increasing is \( \left(0, \frac{1}{2}\right) \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, helping us understand the rate of change of functions. When you take the derivative of a function, you're essentially looking at how it changes in response to changes in its input. This concept allows us to find the slope of the tangent line at any point on a function.

For a function like \[f(x) = x^2,\]its derivative, denoted as \(f'(x),\)would be \(2x.\)This tells us that at any point on the curve, the slope of the tangent line is double the x-value.

In the original exercise, we found the derivative of a composite function \(\phi(x) = f(x) + f(1-x)\)by applying the chain rule. The resultant expression, \(\phi'(x) = f'(x) - f'(1-x),\)shows the rate of change of \(\phi\) with respect to \(x.\)

Function Analysis

Function analysis involves studying a function's properties to understand its behavior and characteristics. This process typically involves finding derivatives to determine a function's concavity and identifying critical points and intervals of increase or decrease.

In the given context, the second derivative \(f''(x) < 0\)tells us that \(f(x)\)is concave down within \((-1, 1).\)Concavity gives us insight into the function's shape; if a second derivative is negative, the original function curves downward, resembling an upside-down bowl.

By examining \(\phi(x) = f(x) + f(1-x),\)we used its first derivative to identify where the function increases or decreases. Since we know \(f'(x)\)is decreasing, this analysis helps in confirming intervals of strict increase or decrease for \(\phi(x).\)

Increasing Intervals

Increasing intervals of a function are sections within its domain where, as the input increases, the output also increases. In other words, the graph of the function moves upwards as you move from left to right along this interval.

For \(\phi(x)\), we determined where it is strictly increasing by finding where its first derivative \(\phi'(x)\)is positive. During our function analysis, we saw that \(f'(x) - f'(1-x) > 0\) within \((0, \frac{1}{2})\). This tells us that \(\phi\) rises for any \(x\) in this interval.

Recognizing these intervals helps in sketching functions accurately and understanding their true nature. This knowledge aids in numerous applications such as predicting function behavior and solving optimization problems.