JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Sets Relations and Functions

The log needs $\frac{2x+3}{4x^2+x-3}>0$, i.e. $\frac{2x+3}{(4x-3)(x+1)}>0$. Testing the sign across the critical points $-\frac{3}{2},-1,\frac{3}{4}$ gives $x\in \left(-\frac{3}{2},-1\right)\cup \left(\frac{3}{4},\infty \right)$.

The inverse cosine needs $-1\le \frac{2x-1}{x+2}\le 1$.

$\frac{2x-1}{x+2}\ge -1\Rightarrow \frac{3x+1}{x+2}\ge 0\Rightarrow x\in (-\infty ,-2)\cup \left[-\frac{1}{3},\infty \right)$.

$\frac{2x-1}{x+2}\le 1\Rightarrow \frac{x-3}{x+2}\le 0\Rightarrow x\in (-2,3]$.

Intersecting these two gives $x\in \left[-\frac{1}{3},3\right]$.

Intersecting with the log condition: $\left(-\frac{3}{2},-1\right)\cup \left(\frac{3}{4},\infty \right)$ meets $\left[-\frac{1}{3},3\right]$ only in the second piece, since $\left(-\frac{3}{2},-1\right)$ lies entirely below $-\frac{1}{3}$. This leaves $\left(\frac{3}{4},3\right]$.

So the domain is $\left(\frac{3}{4},3\right]$, giving $\alpha =\frac{3}{4}$, $\beta =3$, and $5\beta -4\alpha =15-3=12$.