Problem 63 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f\) be a function which is continuous and differentiable for all real \(x\). If \(f(2)=-4\) and \(f^{\prime}(x) \geq 6\) for all \(x \in[2,4]\), then (A) \(f(4)<8\) (B) \(f(4) \geq 8\) (C) \(f(4) \geq 12\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (B) \(f(4) \geq 8\).

Step 1: Understand the Problem

We are given a function \(f\) that is continuous and differentiable for all real \(x\), with specific values and conditions: \(f(2) = -4\) and \(f'(x) \geq 6\) for \(x \in [2, 4]\). Our task is to determine the possible value of \(f(4)\).

Step 2: Integrate the Rate of Change

Since \(f'(x)\) represents the derivative, or the rate of change, of \(f(x)\), we can estimate the change in \(f(x)\) from \(x = 2\) to \(x = 4\). The minimum increase in \(f(x)\) over \([2, 4]\) can be estimated by integrating the minimum value of \(f'(x)\) over the interval. \[\int_2^4 f'(x) \, dx = f(4) - f(2)\]

Step 3: Calculate the Minimum Increase

Given that \(f'(x) \geq 6\), the smallest possible increase from \(f(2)\) to \(f(4)\) is when \(f'(x) = 6\). Calculate this as follows: \[\int_2^4 6 \, dx = 6(4 - 2) = 12\]

Step 4: Find the Value of \(f(4)\)

Using the result from Step 3, substitute back to our equation: \[f(4) - f(2) = 12\]Substitute \(f(2) = -4\): \[f(4) - (-4) = 12 \]\[f(4) + 4 = 12 \]\[f(4) = 12 - 4 = 8 \]

Step 5: Determine the Correct Option

From the calculation, \(f(4) = 8\). Therefore, the correct choice based on the given options is (B) \(f(4) \geq 8\).

Key Concepts

Continuity

Continuity in mathematics, specifically in calculus, refers to a function that does not have any abrupt changes, jumps, or breaks. For a function \( f \) to be continuous at a certain point \( x = a \), it must satisfy the condition that \( \lim_{{x \to a}} f(x) = f(a) \). Therefore, the value of the function at \( a \) should be equal to the limit of the function as it approaches \( a \) from either side.

• Mathematically, if \( \lim_{{x \to a^-}} f(x) = \lim_{{x \to a^+}} f(x) = f(a) \), then \( f \) is continuous at \( x = a \).
• An intuitive way to understand this is imagining drawing the graph of the function \( f \) without lifting your pencil.

Understanding continuity is crucial in calculus as it is the foundation for defining other concepts like differentiability. In our example, since \( f \) is continuous over all real \( x \), it assures us there are no sudden jumps or holes across the interval \([2, 4]\).
This property is essential when calculating changes and differences throughout intervals, paving the way for integration and differentiation.

Differentiability

Differentiability is a concept that indicates a function has a derivative at a certain point. This means the function is smooth enough so that its rate of change, or slope, can be determined at that point. If a function \( f \) is differentiable at \( x = a \), it implies smoothness and is an extension of the concept of continuity.A differentiable function should not have any sharp corners or discontinuities. This is because, at any point where a function is differentiable, it is also continuous. However, a function can be continuous without being differentiable. Consider the function \( f \) is differentiable for all real \( x \) as stated in the problem:

• \( f'(x) \) represents the rate of change of \( f(x) \) per unit change in \( x \).
• This rate of change is crucial for understanding how fast or in which direction the function values are moving.

In the exercise, we know that \( f'(x) \geq 6 \) over \([2,4]\), meaning the slope of the function is never below 6 across this interval. This provides us with the information needed to infer how the function behaves between these points.

Integration

Integration is the process in calculus used to compute the accumulation of quantities, such as areas under a curve, total change, and more. It is often considered as the reverse operation of differentiation.In the exercise, integration helps us find the total change in the function \( f(x) \) across a specific interval. This can be visualized as the area under the curve of \( f'(x) \), which graphically represents the change in \( f(x) \). Here's how integration is applied in our scenario:

• We integrate \( f'(x) \) over the interval from \( x = 2 \) to \( x = 4 \) to find \( f(4) - f(2) \).
• Given that \( f'(x) \) is at least 6 across \([2, 4]\), the calculation \( \int_2^4 6 \ dx = 6 \times (4 - 2) = 12 \).

This means that the smallest possible increase from \( f(2) \) to \( f(4) \) is 12, ensuring that the integration has accurately accounted for the minimum change over the interval. Understanding integration allows students to comprehend how various functions accumulate differences over a range and relate to their rates of change.