Problem 120 - A Complete Resource Book in Mathematics for JEE Main

Question

Instructions: In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(x_{1}=3\) and \(x_{n+1}=\sqrt{2+x_{n}}, n \geq 1\), then \(\lim _{n \rightarrow \infty} x_{n}=2\) Reason: A monotonically decreasing sequence which is bounded below is convergent

Step-by-Step Solution

Answer

Option (A): Both Assertion and Reason are True; Reason is a correct explanation.

Step 1: Identify the Given Elements

We have an assertion concerning a sequence given by the initial term \(x_1 = 3\) and the recursive formula \(x_{n+1} = \sqrt{2 + x_n}\). The sequence is suggested to converge to 2. The reason provided is about a monotonically decreasing sequence that is bounded below being convergent.

Step 2: Analyze the Assertion

Firstly, we need to evaluate the assertion that the sequence converges to 2. Consider \(x_{n+1} = \sqrt{2 + x_n}\). To find the limit, assume \(x_n\) approaches a limit \(L\) such that \(L = \sqrt{2 + L}\). Solving this gives \(L^2 = 2 + L\), which simplifies to the quadratic equation \(L^2 - L - 2 = 0\), giving us solutions for \(L = 2\) and \(L = -1\). However, since \(x_n\) represents a sequence of positive numbers starting at 3, the limit \(L = 2\) is suitable.

Step 3: Verify Sequence Convergence

To prove convergence to \(L = 2\), analyze if the sequence is bounded and monotonic. Starting from \(x_1 = 3\), compute a few terms. Notice that \(x_2 = \sqrt{2 + 3} = \sqrt{5}\), a number between 2 and 3. Now analyze \(\sqrt{2 + x_n} < x_n\) for \(x_n > 2\) to prove it decreases and approaches 2. Verify it is bounded: \(x_n > 2\) as every term generates a value greater than 2 but less than itself if it starts at 3.

Step 4: Evaluate Monotonicity and Boundedness

For \(x_{n+1} = \sqrt{2 + x_n}\), if we assume \(x_n > 2\), \(x_{n+1} < x_n\). Thus, the sequence \(x_n\) is decreasing after the initial term. The sequence is also bounded below by \(x = 2\). A sequence that is bounded and monotonic (in this case, decreasing) is convergent.

Step 5: Assess the Reason

The reason states that a monotonically decreasing sequence which is bounded below is convergent. This is a true statement and a general property of sequences.

Step 6: Combine the Analysis

The assertion that the sequence has the limit 2 is true, and the reason gives a correct explanation that a bounded and monotonically decreasing sequence converges. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concepts

Convergence of Sequences

In mathematics, a sequence is said to converge if it approaches a specific value, known as the limit, as the terms progress towards infinity. Convergence happens when each successive term becomes closer to the limit until the difference between them becomes negligible. The notion of convergence plays a central role because it allows us to understand the long-term behavior of sequences in mathematical analysis.
To determine if a sequence converges, you need to identify the potential limit. For example, assume a sequence is represented by the formula \(x_{n+1} = \sqrt{2 + x_n}\). If we want the sequence to converge, we must solve for the limit \(L\) such that \(L = \sqrt{2 + L}\). Upon solving this equation, you find solutions \(L = 2\) and \(L = -1\). However, since the sequence only takes positive values, the acceptable limit is \(L = 2\).
Thus, to confirm convergence, check if the sequence is steadily approaching this value without oscillating or diverging over time.

Monotonicity in Sequences

Monotonicity refers to the property of a sequence where it consistently moves in one direction, either entirely non-increasing or non-decreasing. Understanding monotonicity is crucial for analyzing a sequence's convergence properties, as a monotonic sequence, by definition, does not change its direction.
In the context of the given problem \(x_{n+1} = \sqrt{2 + x_n}\), the sequence is said to be monotonically decreasing once \(x_n > 2\). This implies that each new term is smaller than the preceding one, ensuring the sequence is "pushing" itself towards its lower bound until it stabilizes at the limit. If \(x_n > 2\), then \(x_{n+1} < x_n\). This pattern suggests that the sequence will continue to decrease until it reaches the limiting value of 2.
Monotonic sequences that are also bounded offer a straightforward path to establishing convergence. Their predictability in movement simplifies proving that they converge to a particular value.

Bounded Sequences

A sequence is considered bounded if it is contained within a specific interval, meaning it doesn't go off to infinity in either direction. Essentially, a bounded sequence has an upper or a lower bound—or both—which constrains how large or small the terms can be.
Regarding the example from our solution, the sequence represented by the formula \(x_{n+1} = \sqrt{2 + x_n}\), starting from 3, is bounded below by 2. This means while the sequence progresses, it can never dip below 2. Such a property is crucial, as bounded sequences are easier to analyze for convergence.
Why is this significant? If a sequence is bounded and monotonic, convergence is usually guaranteed. For instance, if you know that all terms of a sequence are above 2 and are decreasing, they must converge to a number that is not less than the lower bound, which would be 2 in this case. Understanding bounded sequences helps predict the limits and behavior of sequences easily.