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

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 is \(\frac{3}{19}\) (Option B).

Step 1: Identify the Function

The function we are analyzing is given by \(a_n=\frac{n}{n^2+10}\). To find the largest term in this sequence, we need to examine the behavior of this function as \(n\) changes.

Step 2: Calculate the Derivative

First, calculate the derivative of the function \(a_n = \frac{n}{n^2+10}\). Use the quotient rule: if \(f(x) = \frac{g(x)}{h(x)}\), then \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \].For our function:- \(g(x) = n\) and \(g'(x) = 1\).- \(h(x) = n^2 + 10\) and \(h'(x) = 2n\).Applying the quotient rule gives us:\[a_n' = \frac{(1)(n^2 + 10) - (n)(2n)}{(n^2 + 10)^2} = \frac{n^2 + 10 - 2n^2}{(n^2 + 10)^2} = \frac{-n^2 + 10}{(n^2 + 10)^2}.\]

Step 3: Set the Derivative to Zero

To find the critical points where the function might have a maximum, set the derivative equal to zero:\[\frac{-n^2 + 10}{(n^2 + 10)^2} = 0.\]This fraction equals zero when the numerator is zero:\(-n^2 + 10 = 0\).Solving for \(n\) gives:\[n^2 = 10,\Rightarrow n = \sqrt{10}.\] Since \(n\) must be a natural number, evaluate around this point.

Step 4: Evaluate Nearby Natural Numbers

Since \(n = \sqrt{10}\) is not an integer, check natural numbers near \(n\). Evaluate \(a_n\) for \(n = 3\) and \(n = 4\):- \(a_3 = \frac{3}{3^2 + 10} = \frac{3}{19}\).- \(a_4 = \frac{4}{4^2 + 10} = \frac{4}{26} = \frac{2}{13}\).Compare both values, where \(\frac{3}{19} \approx 0.1579\) and \(\frac{4}{26} \approx 0.1538\). Therefore, \(a_3\) is the larger value.

Step 5: Conclusion

Since \(a_3 = \frac{3}{19}\) produces the largest value compared to \(a_4\), the largest term in the sequence is \(\boxed{\frac{3}{19}}\).

Key Concepts

Derivative Calculation

Derivatives help us understand how a function changes as its input changes. In calculus, we calculate derivatives to study the rate of change or the slope of a function's graph. Here, we focus on how to compute the derivative of a specific function to find important characteristics such as local maxima or minima.
When given a function, the derivative provides a mathematical way to analyze its behavior. In our exercise, we have the sequence described by the function \(a_n = \frac{n}{n^2+10}\). Calculating the derivative is a step toward finding where this function reaches its largest value.
Talking about derivatives, think of the graph of a function. The derivative at a point is essentially the slope of the tangent line at that point. If you were to draw a curve for the function \(a_n\), moments where the derivative equals zero could signify a peak or trough. That's because a zero-slope line means no change at that instant.

Quotient Rule

The Quotient Rule is crucial when working with derivatives of functions represented as one expression divided by another. It's one of the basic techniques in calculus that help take derivatives of quotients easily. The formula used fits the scenario where a function \(f(x)\) is expressed as \(\frac{g(x)}{h(x)}\).
The rule states: if \(f(x) = \frac{g(x)}{h(x)}\), then the derivative \(f'(x)\) is given by:\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.\]Let's break it down:

• \(g(x)\) is the numerator, and we take its derivative \(g'(x)\)
• \(h(x)\) is the denominator, and its derivative is \(h'(x)\)

The formula helps capture how both functions' rates of change affect the overall quotient's rate of change.
In the solution, \(g(x) = n\) leading to \(g'(x) = 1\), and \(h(x) = n^2 + 10\) making \(h'(x) = 2n\). Plug these into the Quotient Rule to find \(a_n'\). This computation is a key step to identifying critical points for locating maximum values.

Critical Points

Critical points are the values of \(x\) where the derivative of a function is zero or undefined. They are potential spots for local maxima, minima, or points of inflection. Finding these critical points involves setting the derivative to zero and solving for \(x\).
In our exercise, we set the derivative \(-n^2 + 10 = 0\) from our function to zero. Solving this, we get \(n^2 = 10\). The solutions \(n = \sqrt{10}\) are critical points, although not integers in this context. However, since \(n\) is meant to represent natural numbers, we check values near these points to determine maxima.
Evaluating the function at these critical points helps decide which specific integer values yield the largest term. By comparing the calculated results of \(a_n\) for \(n = 3\) and \(n = 4\), we determine that the largest value of the sequence close to \(\sqrt{10}\) is at \(n = 3\), making \(\frac{3}{19}\) the largest value.