JEE 2023 — Mathematics PYQ

Let $y=\frac{x^3}{x^4+147}$
$$\Rightarrow \frac{dy}{dx}=\frac{(x^4+147)\times 3x^2-x^3(4x^3)}{(x^4+147{)}^2}$$
$$=\frac{3x^6+441x^2-4x^6}{(x^4+147{)}^2}=\frac{441x^2-x^6}{(x^4+147{)}^2}$$

For maxima/minima, put $\frac{dy}{dx}=0$
$\Rightarrow 441x^2-x^6=0\Rightarrow x^4=441\Rightarrow x=\pm \sqrt{21}$

Maximum value is at $x=\sqrt{21}\approx 4.58$
Now, 4 < 4.58 < 5
$y\text{ at }x=4\Rightarrow \frac{64}{403}\approx 0.159$
$y\text{ at }x=5\Rightarrow \frac{125}{772}\approx 0.162$

So, $y$ is maximum at $x=5\Rightarrow \alpha =5$