Problem 89 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The largest term in the sequence \(a_{n}=\frac{n}{n^{2}+10}, n \in N\) is (A) \(\frac{4}{26}\) (B) \(\frac{3}{19}\) (C) \(\frac{7}{18}\) (D) None of these

Step-by-Step Solution

Answer

The largest term in the sequence is \( \frac{3}{19} \). Option (B) is correct.

Step 1: Identify the Sequence Function

The given sequence is defined by \( a_n = \frac{n}{n^2 + 10} \). We need to find the value of \( n \) that maximizes this expression.

Step 2: Find the First Derivative

To find the maximum, use calculus to find the derivative of \( a_n \) with respect to \( n \). Start by finding the first derivative using the quotient rule: \( f(n) = n \), \( g(n) = n^2 + 10 \). With the quotient rule, \( a_n' = \frac{g(n)f'(n) - f(n)g'(n)}{[g(n)]^2} = \frac{(n^2 + 10) \cdot 1 - n \cdot 2n}{(n^2 + 10)^2} \).

Step 3: Simplify the Derivative

Simplify the derivative: \( a_n' = \frac{n^2 + 10 - 2n^2}{(n^2 + 10)^2} = \frac{-n^2 + 10}{(n^2 + 10)^2} \). We need to find where this derivative equals zero to find the critical points.

Step 4: Solve for Critical Points

Set the derivative to zero to find the critical points: \( -n^2 + 10 = 0 \). Solving for \( n \), we find \( n^2 = 10 \), so \( n = \sqrt{10} \). However, since \( n \) is a natural number, we cannot use \( n = \sqrt{10} \).

Step 5: Evaluate Sequence at Values Around \( \sqrt{10} \)

Since \( \sqrt{10} \approx 3.162 \), check the integer values of \( n \) around \( 3 \) and \( 4 \): Evaluate \( a_3 \) and \( a_4 \).

Step 6: Calculate \( a_3 \) and \( a_4 \)

Calculate \( a_3 = \frac{3}{3^2 + 10} = \frac{3}{19} \) and \( a_4 = \frac{4}{4^2 + 10} = \frac{4}{26} \). Compare these values to determine the largest term.

Step 7: Conclusion

Between \( a_3 = \frac{3}{19} \) and \( a_4 = \frac{4}{26} \), note that \( \frac{3}{19} \) is greater than \( \frac{4}{26} \). Hence, \( \frac{3}{19} \) is the largest term in the sequence among the given options.

Key Concepts

Sequence Maximization

In calculus, sequence maximization involves finding the value within a sequence that attains the highest possible value. The sequence in this problem is defined by the formula \(a_n = \frac{n}{n^2 + 10}\). This sequence represents the relationship between \(n\) and the value of each term. To maximize a sequence, identify the point where the function reaches its peak. This often involves finding the critical points where the derivative of the sequence function equals zero or is undefined, as these points indicate where the function changes behavior.
Maximization typically involves either using calculus or evaluating numerical approaches. Here, calculus gives a systematic method to pinpoint such points accurately, but numerical evaluation allows for checking integer values around critical points. Both methods together ensure a comprehensive approach to finding maximum values within sequences.
By exploring these methods, students can grasp how sequence maximization reveals important insights about how sequences behave and evolve.

Derivative Calculus

Derivative calculus is a fundamental tool in finding the maximum or minimum values of functions. In this exercise, we begin by finding the derivative of the sequence function \(a_n = \frac{n}{n^2 + 10}\) using the quotient rule. The quotient rule is applicable here because the function is a ratio of two expressions. It states: if \( f(n) = \frac{u(n)}{v(n)}\), then \(f'(n) = \frac{u'(n)v(n) - u(n)v'(n)}{[v(n)]^2}\).
Applying this rule, we derive \(a_n' = \frac{(n^2 + 10) \cdot 1 - n \cdot 2n}{(n^2 + 10)^2}\). Simplifying leads to \(a_n' = \frac{-n^2 + 10}{(n^2 + 10)^2}\). The significance of a derivative is that it provides the slope of the tangent to the graph of the function at any point \(n\). When this derivative equals zero, it suggests potential maxima or minima, essential for optimization tasks.
Understanding derivative calculus equips students with the ability to systematically analyze and interpret the behavior of complex functions beyond simple static evaluations.

Critical Points Analysis

The critical points of a function occur where its derivative equals zero or is undefined. These points are crucial since they can indicate where a function changes from increasing to decreasing, or vice versa.
In our exercise, solving the derivative \( -n^2 + 10 = 0 \) gives \( n^2 = 10 \), which leads to \( n = \sqrt{10} \). Since \( n \) must be a natural number, \(n = \sqrt{10}\) is not usable directly; however, it informs us about nearby integer candidates to evaluate. Thus, checking around \( \sqrt{10} \approx 3.162\) leads to evaluating \(n = 3\) and \(n = 4\).
Analyzing function values at these critical-point data points, like \( a_3 = \frac{3}{19} \) and \( a_4 = \frac{4}{26} \), helps determine where the function is maximized. By checking these specific points, we establish which value is the largest, ensuring accurate conclusions about the sequence's behavior around these important points.