Problem 107 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f^{\prime \prime}(x)>0 \forall x \in R\) and \(g(x)=f(2-x)+f(4+x)\). Then, \(g(x)\) is increasing in (A) \((-\infty,-1)\) (B) \((-\infty, 0)\) (C) \((-1, \infty)\) (D) None of these

Step-by-Step Solution

Answer

The function \(g(x)\) is increasing in \((-1, \infty)\), which is option (C).

Step 1: Understand Second Derivative Condition

The function \(f^{\prime \prime}(x) > 0\) means that \(f(x)\) is a convex function over the entire real line. The first derivative \(f'(x)\) is increasing for all \(x\in \mathbb{R}\).

Step 2: Analyze Function g(x)

The function is defined as \(g(x) = f(2-x) + f(4+x)\). We need to find \(g'(x)\) and analyze where \(g(x)\) is increasing.

Step 3: Differentiate g(x)

To find \(g'(x)\), apply the chain rule. We have \(g'(x) = -f'(2-x) + f'(4+x)\).

Step 4: Determine Increasing Intervals Based on g'(x)

The function \(g(x)\) is increasing if \(g'(x) > 0\). Therefore, we need \(-f'(2-x) + f'(4+x) > 0\), which simplifies to \(f'(4+x) > f'(2-x)\). Since \(f'(x)\) is increasing, for any \(x\) in the interval \((-1, \infty)\), \(4+x\) is always greater than \(2-x\), hence \(f'(4+x) > f'(2-x)\).

Step 5: Conclusion on Increasing Interval

For \(-f'(2-x) + f'(4+x) > 0\), the function is increasing in the interval \((-1, \infty)\). Therefore, the correct answer is option (C): \((-1, \infty)\).

Key Concepts

Second Derivative

The second derivative of a function, often denoted as \(f''(x)\), offers critical insight into the function's curvature. It measures how the rate of change of the slope of the function itself changes. If \(f''(x) > 0\), this suggests that the function is curving upwards, like a smiley face. Visually, this means the function is concave up, or convex.
When applied to the entire real line, as in this exercise where \(f''(x) > 0\), it indicates that the original function \(f(x)\) behaves consistently across all real numbers. This is crucial for further analysis, as it implies that the first derivative, \(f'(x)\), is strictly increasing everywhere.
Understanding this can help simplify complex relationships, making it easier to predict behavior without calculating endless derivative values for each case.

Convex Function

A function is labeled as convex if its second derivative is greater than zero over its domain. This property is significant because it ensures that the function curves upwards across its entire length.
In simpler terms, think of a bowl shape: the curve goes up as you move along from left to right. This shape guarantees only one local minimum globally if any exist, simplifying analysis particularly in optimization problems where finding global minima is often a goal.
Convex functions have interesting features:

• They have no local maxima.
• Their tangent lines give underestimates to the function values.
• Adding and multiplying convex functions usually results in another convex function.

Understanding convexity allows mathematicians to make generalized conclusions about the behavior of functions over large intervals, much like in this task.

Increasing Function

An increasing function has values that move upwards as you progress along the x-axis. In mathematical terms, for every pair of numbers \(x_1\) and \(x_2\) where \(x_1 < x_2\), it follows that \(f(x_1) < f(x_2)\).
In our given exercise, the focus is on the function \(g(x) = f(2-x) + f(4+x)\). To determine where this function is increasing, we assess the derived function \(g'(x)\).
The condition for \(g(x)\) to be increasing is that its derivative, \(g'(x)\), must be greater than zero. By applying the given conditions and using the nature of convex functions, we confirmed the interval where \(g(x)\) rises, leading us to choose the option \((-1, \infty)\).

Chain Rule

The chain rule is a fundamental rule in calculus used to differentiate compositions of functions. Essentially, if you have a function \(y = f(u)\) and another function \(u = g(x)\), the chain rule helps find the derivative of \(y\) with respect to \(x\).
The rule states that:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

In our problem, where \(g(x) = f(2-x) + f(4+x)\), we employ the chain rule to find \(g'(x)\). Using the differentiation formula, we take:

• For \(f(2-x)\), its derivative becomes \(-f'(2-x)\).
• For \(f(4+x)\), its derivative becomes \(f'(4+x)\).

Hence, \(g'(x) = -f'(2-x) + f'(4+x)\). This derivative helps determine the intervals where \(g(x)\) is increasing, bringing clarity to the function's behavior.