Question

The largest term of the sequence:
${{a}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}+200}$ is:
(a) $\dfrac{{{400}^{\dfrac{2}{3}}}}{600}$
(b) $\dfrac{8}{700}$
(c) $\dfrac{49}{543}$
(d) $\dfrac{{{89}^{\dfrac{1}{3}}}}{800}$

Answer 1

Hint: Find the differentiation of the function and equate it to 0. Find the value of n where the maximum or minimum lies and then find the maximum value.

Complete step-by-step answer:
In mathematics, if derivative of a function is 0. Then those points are special to us.
The points may be maximum or minimum.
It is found by number line or second degree differential equation.
Here we have a sequence represented by:
${{a}_{n}}=\dfrac{{{n}^{2}}}{{{n}^{3}}+200}$
So every term of this sequence follows this equation.
Now substitute x in place of n:
Let n = x,
By this the given sequence can be treated as a function.
Let the function be named as f(x):
f(x) = $\dfrac{{{x}^{2}}}{{{x}^{3}}+200}$
Now we need the maximum value of f(x).
For that we need to differentiate the function f(x).
By differentiating, we get:
For differentiating, we use the formula:
$d\left( \dfrac{u}{v} \right)=\dfrac{vdu-udv}{{{v}^{2}}}$
$d\left( f\left( x \right) \right)={f}'\left( x \right)$
By substituting the above 2 equations, we get:
$\text{{f}'}\left( x \right)=x.\dfrac{\left( 400-{{x}^{3}} \right)}{{{\left( {{x}^{3}}+200 \right)}^{2}}}$
Now for finding maximum or minimum, we must equate differentiation to 0.
By equating to 0, we get:
$x.\dfrac{\left( 400-{{x}^{3}} \right)}{{{\left( {{x}^{3}}+200 \right)}^{2}}}=0$
The possible solutions are:
x = 0…..(1), and

& 400-{{x}^{3}}=0 \\\ & \Rightarrow x={{400}^{\dfrac{1}{3}}}.....\left( 2 \right) \\\ \end{aligned}$$ So now we have 2 points (1) and (2). We need to find which of them, is the point where maximum occurs. So, we will use the number line of $$\text{{f}'}\left( x \right)$$ here. Case – 1:- If x < 0: $$\text{{f}'}\left( x \right)$$ < 0 Case – 2:- If 0 < x < $${{400}^{\dfrac{1}{3}}}$$: $$\text{{f}'}\left( x \right)$$ > 0 Case – 3:- If x > $${{400}^{\dfrac{1}{3}}}$$ : $$\text{{f}'}\left( x \right)$$ < 0 So by above we can say that at x = $${{400}^{\dfrac{1}{3}}}$$ the value of $$\text{{f}'}\left( x \right)$$turns from positive to negative. So the point, x = $${{400}^{\dfrac{1}{3}}}$$ is maximum. But in a sequence we need integers. We know that 7 < $${{400}^{\dfrac{1}{3}}}$$ < 8. So the value of n for the largest term will be 7 or 8. So we will find the 7th and 8th terms of the given sequence. $$\begin{aligned} & {{a}_{7}}=\dfrac{{{7}^{2}}}{{{7}^{3}}+200}=\dfrac{49}{543} \\\ & {{a}_{8}}=\dfrac{{{8}^{2}}}{{{8}^{3}}+200}=\dfrac{64}{712} \\\ & {{a}_{7}}>{{a}_{8}} \\\ \end{aligned}$$ So, the 7th term is the longest term. Option (c) is correct. Note: The main point in this question is that n should only be integer. As part of sequence the terms will be 1st, 2nd, 3rd,….. all integers. If you don’t take care you may end up getting option (a) which is wrong.