JEE Main 2024 — Mathematics Algebra

Given,

$$f(x)={\mathrm{log}}_{e}(\frac{2x+3}{4{x}^{2}+x-3})+{\mathrm{cos}}^{-1}(\frac{2x-1}{x+2})$$f(x)=loge​(4x2+x−32x+3​)+cos−1(x+22x−1​)

Now, using the definition of $\mathrm{log}$log function and ${\mathrm{cos}}^{-1}$cos−1 function we get,

$\frac{2x+3}{4{x}^{2}+x-3}>0$4x2+x−32x+3​>0 and $-1\leq \frac{2x-1}{x+2}\leq 1$−1≤x+22x−1​≤1

\Rightarrow \frac{2x+3}{(4x-3)(x+1)}>0,\frac{3x+1}{x+2}\geq 0&\frac{x-3}{x+2}\leq 0

Now, solving $\frac{2x+3}{(4x-3)(x+1)}>0$(4x−3)(x+1)2x+3​>0 we get,

$$\Rightarrow x\in (\frac{-3}{2},-1)\cup (\frac{3}{4},\infty )....(i)$$⇒x∈(2−3​,−1)∪(43​,∞)....(i)

Similarly for $\frac{3x+1}{x+2}\geq 0$x+23x+1​≥0 we get,

$$x\in (-\infty ,-2)\cup [\frac{-1}{3},\infty )...(ii)$$x∈(−∞,−2)∪[3−1​,∞)...(ii)

And for $\frac{x-3}{x+2}\leq 0$x+2x−3​≤0 we get,

$$x\in (-2,3]...(iii)$$x∈(−2,3]...(iii)

Now, taking intersection of $(i)\cap (ii)\cap (iii)$(i)∩(ii)∩(iii) we get,

$$(\frac{3}{4},3]$$(43​,3]

So, on comparing with $(\alpha ,\beta ]$(α,β] we get, \alpha =\frac{3}{4}&\beta =3

Hence, $5\beta -4\alpha =15-3=12$5β−4α=15−3=12