QJEEM23Ma51cd: JEE Main 2023 Mathematics - Class 12 JEEM23Ma51cd Solution: If is the greatest term in the sequence , then is equal to…

$\begin{aligned}& \text { Let } y=\frac{x^3}{x^4+147} \\\\& \Rightarrow \frac{d y}{d x}=\frac{\left(x^4+147\right) \times 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\\\& =\frac{3 x^6+441 x^2-4 x^6}{\left(x^4+147\right)^2}=\frac{441 x^2-x^6}{\left(x^4+147\right)^2}\end{aligned}$

For maxima/minima, put $\frac{d y}{d x}=0$

$\begin{aligned}& \Rightarrow 441 x^2-x^6=0 \Rightarrow x^4=441 \\\\& \Rightarrow x= \pm \sqrt{21}, \pm \sqrt{21} i\end{aligned}$

Now, by descrates rule on number line we have



Since sign changes from negative to positive at 0 .

$\therefore$ Maximum value of is at $x=\sqrt{21}=4.58$

Now, 4<4.5<5

$\begin{aligned}& \therefore \text { yat } x=4=\frac{64}{403}=0.159 \\\\& y \text { at } x=5=\frac{125}{772}=0.162\end{aligned}$

So, $y$ is maximum at $x=5$

$\therefore \alpha=5$