Q101 / 200 · Manipal (MET) 2014 · Mathematics
The largest term in the sequence $a_n=\frac{n^2}{n^3+200}$ is given by
- A$\frac{529}{49}$
- B$\frac{8}{89}$
- Correct answerC$\frac{49}{543}$
- DNone of these
Solution
Consider the function $f(x)=\frac{x^2}{x^3+200}$
$\begin{aligned}
& & f^{\prime}(x) & =x \frac{\left(400-x^3\right)}{\left(x^3+200\right)^2}=0 \\
\Rightarrow & & x^3 & =400 \\
\Rightarrow & & x & =(400)^{1 / 3}
\end{aligned}$
When $x \lt 0, f^{\prime}(x)\gt0$ $x\gt0, f^{\prime}(x) \lt 0$ $\therefore f(x)$ has maximum at $x=(400)^{1 / 3}$ Since, $7 \lt (400)^{1 / 3} \lt 8$, either $a_7$ or $a_8$ is the greatest term of the sequence. $\because \quad a_7=\frac{49}{543}, a_8=\frac{8}{89}$ and $\quad \frac{49}{543}\gt\frac{8}{89}$ $a_7=\frac{49}{543}$ is the greatest.
When $x \lt 0, f^{\prime}(x)\gt0$ $x\gt0, f^{\prime}(x) \lt 0$ $\therefore f(x)$ has maximum at $x=(400)^{1 / 3}$ Since, $7 \lt (400)^{1 / 3} \lt 8$, either $a_7$ or $a_8$ is the greatest term of the sequence. $\because \quad a_7=\frac{49}{543}, a_8=\frac{8}{89}$ and $\quad \frac{49}{543}\gt\frac{8}{89}$ $a_7=\frac{49}{543}$ is the greatest.
Hence, the correct answer is option: 'C'.