Problem 119 - 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

\(g(x)\) decreases in \(\left(\frac{1}{2}, 1\right)\) (Option C).

Step 1: Understanding Given Functions

We have the function \(g(x) = f(x) + f(1-x)\). The expression involves \(f(x)\) and its transformation \(f(1-x)\). We need to analyze how \(g(x)\) behaves based on the properties of \(f(x)\).

Step 2: Analyzing Concavity of \(f(x)\)

The condition \(f''(x) < 0\) implies that \(f(x)\) is concave down on the interval \(0 \leq x \leq 1\). This means for any interval within \([0, 1]\), \(f(x)\) is decreasing.

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

To find where \(g(x)\) increases or decreases, we compute \(g'(x) = f'(x) - f'(1-x)\). This derivative will tell us the monotonicity of \(g(x)\).

Step 4: Evaluating \(g'(x)\) at \(x = \frac{1}{2}\)

Since \(f(x)\) is concave down, \(f'(x)\) decreases as \(x\) increases from 0 to 1. Therefore, \(f'(x) < f'(1-x)\) for \(x < \frac{1}{2}\), which makes \(g'(x) < 0\), indicating that \(g(x)\) decreases in \(\left(0, \frac{1}{2}\right)\) and \(g(x)\) is undefined for \(x < 0\).

Step 5: Analyzing \(g(x)\) for \(x > \frac{1}{2}\)

For \(x > \frac{1}{2}\), the inequality reverses and \(f'(x) > f'(1-x)\), making \(g'(x) > 0\), which implies \(g(x)\) increases in \(\left(\frac{1}{2}, 1\right)\). However, options related to this need consideration.

Step 6: Conclusion on Behavior of \(g(x)\)

Putting this together, only option \(B\) (\(g(x)\) increases in \(0, \frac{1}{2}\)) is incorrect because \(g(x)\) actually decreases in this range. Thus, \(g(x)\) increases for \(x < \frac{1}{2}\), not strictly for \(0 < x < \frac{1}{2}\), which is not a suitable option, and it decreases for \(x > \frac{1}{2}\).

Key Concepts

Concavity and Derivatives

A concave function is one that curves downward, like the shape of an upside-down bowl. When we have a function where the second derivative, denoted as \( f''(x) \), is less than zero, we identify it as a concave down function over a specific interval. This tells us that as we move along the graph, the slope of the tangent line to the curve is decreasing.

For a function \( f(x) \), if \( f''(x) < 0 \) in a region \( a \leq x \leq b \), it means the function is bending downwards within this region. Each point on this curve has a tangent slope that gets smaller as \( x \) increases. This essentially means the function is flattening out or decreasing in steepness. In the context of our exercise, this property affects the behavior of \( g(x) = f(x) + f(1-x) \), particularly in assessing how the slopes \( f'(x) \) and \( f'(1-x) \) interact.

Monotonicity

Monotonicity refers to the behavior of a function as either consistently increasing or decreasing. To determine the monotonicity of a function like \( g(x) \) from the exercise, we calculate the first derivative, \( g'(x) \). This derivative helps us see how \( g(x) \) changes across its domain.

In our context, \( g(x) = f(x) + f(1-x) \), its derivative is \( g'(x) = f'(x) - f'(1-x) \). We assess this derivative to understand the intervals over which the function increases or decreases:

• When \( g'(x) > 0 \), \( g(x) \) is increasing.
• When \( g'(x) < 0 \), \( g(x) \) is decreasing.

Using the nature of the concave function (where \( f'(x) < f'(1-x) \) for \( x < \frac{1}{2} \)), we realize \( g'(x) \) is negative, making \( g(x) \) decrease in the interval \( 0 < x < \frac{1}{2} \). Conversely, for \( x > \frac{1}{2} \), \( g'(x) \) becomes positive, indicating \( g(x) \) increases here.

Analyzing Transformations

Analyzing transformations helps us understand how modifying a function, through shifting, scaling, or combining functions, changes its properties. Here, the function \( g(x) = f(x) + f(1-x) \) involves a transformation of \( f(x) \) into \( f(1-x) \).

Transformation affects the function in several ways:

• It involves a horizontal flip across the vertical line \( x = \frac{1}{2} \), due to replacing \( x \) with \( 1-x \). This symmetry plays a crucial role in balancing increases and decreases within \( g(x) \).
• Since \( f(x) \) is concave down, both \( f(x) \) and \( f(1-x) \) project this feature on either side of \( x = \frac{1}{2} \), influencing \( g(x) \).

Understanding such transformations is key, as they help us predict how the behavior of more complex functions reveals properties of initial simpler functions.