Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(a+b=4, a<2\) and \(g(x)\) be a monotonically increasing function of \(x\). Then, \(f(a)=\int_{0}^{a} g(x) d x+\int_{0}^{b} g(x) d x\) (A) increases with increase in \((b-a)\) (B) decreases with increase in \((b-a)\)

Step-by-Step Solution

Answer

(A) and (B) are false; since \(f(a)\) is constant, it doesn't change with \(b-a\).

Step 1: Analyze Given Equations

We are given that \(a + b = 4\) and \(a < 2\). From these conditions, we infer that both \(a\) and \(b\) must be less than 4 since their sum is fixed.

Step 2: Express f(a) in Terms of a

The function \(f(a)\) is defined as: \(f(a) = \int_{0}^{a} g(x) \, dx + \int_{0}^{b} g(x) \, dx\). By using the condition \(a + b = 4\), we can express \(b\) in terms of \(a\) such that \(b = 4 - a\). Thus, \(f(a) = \int_{0}^{a} g(x) \, dx + \int_{0}^{4-a} g(x) \, dx\).

Step 3: Substitute b = 4-a

Substituting \(b = 4 - a\) into the expression for \(f(a)\), we have: \(f(a) = \int_{0}^{a} g(x) \, dx + \int_{0}^{4-a} g(x) \, dx\). This can be written as the sum of integrals \(\int_{0}^{a} g(x)\, dx + \int_{0}^{4-a} g(x)\, dx\).

Step 4: Combine Integrals

By properties of integrals, \(\int_{0}^{a} g(x) \, dx + \int_{0}^{4-a} g(x) \, dx = \int_{0}^{a} g(x) \, dx + \int_{a}^{4} g(x) \, dx = \int_{0}^{4} g(x) \, dx\). Thus, \(f(a)\) simplifies to a constant value, \(\int_{0}^{4} g(x) \, dx\), independent of \(a\).

Step 5: Evaluate Effect of (b-a)

Given \(a + b = 4\) and expressing \(b-a = (4-2a)\), we see that the expression doesn't change \(f(a)\)'s value. Hence, \(f(a)\) remains constant irrespective of the value of \(b - a\). Thus, \(f(a)\) neither increases nor decreases with \(b-a\).

Key Concepts

Monotonic Functions

A function is called monotonically increasing if its value either stays the same or increases as the input increases. In simple terms, as you move to the right on the graph (increasing x values), the function does not ever decrease.
This property is essential because it guarantees that after integrating over an interval, the result doesn't depend on the path. For any range

• If you have two points, say a and b, where a < b, then for a monotonically increasing function g(x), it follows that g(a) ≤ g(b).
• This ensures that contributions from higher values of x are accounted for in the integrals, maintaining a non-decreasing sum.

Understanding monotonicity is crucial for determining the behavior of integrals and, consequently, any functions derived from them. This concept simplifies complex expressions and allows predictions on how changing limits affect integral values.

Integrals

Integrals are a significant concept in calculus. They help measure the area under curves between specified limits. When you integrate a function, essentially, you are adding up an infinite number of infinitesimally small quantities.
For any function g(x), the integral from a to b, represented as \( \int_{a}^{b} g(x) \, dx \), calculates the area under g(x) over

• The range [a, b] on the x-axis.
• This aids in understanding accumulation, like tracking total distance from speed over time or total charge from current.

Key to applications, understanding integrals allows you to address various types of problems, from physics to economics.

Properties of Definite Integrals

Definite integrals, unlike indefinite integrals, yield a specific numerical value. For the function \(f(a)\) in the exercise, the definite integral properties allow you to treat separate integrals as a combination over the shared interval.
Consider these useful properties:

• The integral of a sum is the sum of the integrals, \( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \).
• Integrals can be broken into sub-intervals,\( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \), which simplified finding \( f(a) = \int_{0}^{4} g(x) \, dx \) in the solution.
• Most importantly, even if you adjust a and b, as long as their sum remains constant, the constant integral value \( \int_{0}^{4} g(x) \, dx \) does not change. This property facilitated proving that \( f(a) \) remains unaffected by changes in \( b-a \).

Recognizing these properties simplifies evaluations and helps justify your results in exercises or real-world applications.