Problem 128 - 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 \(\left(0, \frac{1}{2}\right)\)

Step-by-Step Solution

Answer

The function \( g(x) \) increases in the interval \( (0, \frac{1}{2}) \). Choice (B) is correct.

Step 1: Understand the Given Condition

The function \( f(x) \) has a property that its second derivative, \( f''(x) \), is less than zero for \( 0 \leq x \leq 1 \). This implies that \( f(x) \) is a concave function over this interval.

Step 2: Differentiate \( g(x) = f(x) + f(1-x) \)

To understand how \( g(x) \) behaves, differentiate it: \[ g'(x) = f'(x) - f'(1-x) \] This is derived using the chain rule on \( f(1-x) \).

Step 3: Analyze \( g'(x) \)

From \( f''(x) < 0 \), we know \( f'(x) \) is decreasing. Therefore, for \( 0 < x < 1 \), we have \( f'(x) > f'(1-x) \). This means \( g'(x) = f'(x) - f'(1-x) > 0 \), indicating \( g(x) \) is increasing.

Step 4: Determine the Interval of Increase

Since \( g'(x) > 0 \) holds for \( 0 < x < 1 \), this indicates \( g(x) \) is increasing in this interval. Specifically, it is increasing until but not including \( \frac{1}{2} \) since we test in the interval \( 0 < x < \frac{1}{2} \).

Step 5: Conclusion of Intervals

Given \( g(x) \) is shown to increase in \( 0 < x < \frac{1}{2} \), the correct choice reflecting this behavior is choice (B): \( g(x) \) increases in \( (0, \frac{1}{2}) \).

Key Concepts

Concave functions

A concave function is one where the line segment between any two points on its graph will lie below or on the graph itself. In simple terms, the graph of a concave function curves downwards. This can be visualized as the surface of a bowl held upside down.
To mathematically identify a concave function:

• The second derivative of the function, denoted as \( f''(x) \), must be less than zero \( (f''(x) < 0) \). This indicates that the slope of the tangent line to the function is decreasing, contributing to the downward curvature.
• For example, if given \( f(x) \) in a problem, and you know that its second derivative \( f''(x) < 0 \) in a certain interval, you can directly conclude that \( f(x) \) is concave over that interval.

Concavity is crucial in understanding how a function behaves over an interval and is especially helpful in optimization problems and when analyzing the rate of change. In our exercise, knowing \( f(x) \) is concave helps us determine the behavior of another function, \( g(x) \).

Chain rule

The chain rule is an essential technique in calculus used to differentiate compositions of functions. Simply put, it helps when a function is nested within another, allowing us to "unpack" these layers effectively. Use the chain rule when you have a function \( h(x) = f(g(x)) \), where both \( f \) and \( g \) are differentiable.
The formula for the chain rule is:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)

In practice, you:

• First differentiate the outer function \( f \).
• Then multiply it by the derivative of the inner function \( g \).

In our context, when we differentiated \( g(x) = f(x) + f(1-x) \), we applied the chain rule to \( f(1-x) \). Knowing that \( f'(1-x) \) requires differentiating \( f(u) \) with respect to \( u = 1-x \), results in \( -f'(1-x) \) when multiplied by the derivative of \( 1-x \), which is \(-1\).
This careful application of the chain rule helped find \( g'(x) = f'(x) - f'(1-x) \), essential for analyzing how \( g(x) \) increases or decreases.

Second derivative test

The second derivative test is a systematic way to determine local maxima or minima of a function, providing deeper insights into its behavior. This involves investigating the second derivative of a function at critical points, where the first derivative is zero \((f'(x) = 0)\).
Here's how you apply the second derivative test:

• If \( f''(x) > 0 \) at a critical point, the point is a local minimum; the function is concave up there.
• If \( f''(x) < 0 \) at a critical point, the point is a local maximum; the function is concave down.
• If \( f''(x) = 0 \), the test is inconclusive; the point may be a point of inflection and further investigation is needed.

For students working through problems like the given exercise, understanding that \( f''(x) < 0 \) for \( 0 \leq x \leq 1 \) means that \( f(x) \) is decreasingly sloped throughout, aligning with conditions for concavity. While in this scenario, the second derivative test isn't used directly for finding maxima or minima, recognizing \( f''(x) < 0 \) aids significantly in assessing the overall decreasing behavior of the derivative, \( f'(x) \), leading to understanding \( g(x) \)'s increasing nature over a specific interval.