JEE Main 2024 — Mathematics Algebra

Given: $f(x)=\frac{\sqrt{{x}^{2}-25}}{4-{x}^{2}}+\mathrm{log}({x}^{2}+2x-15)$f(x)=4−x2x2−25​​+log(x2+2x−15)

$$\Rightarrow {x}^{2}-25\geq 0$$⇒x2−25≥0

$$\Rightarrow (x-5)(x+5)\geq 0$$⇒(x−5)(x+5)≥0

$$\Rightarrow x\in (-\infty ,-5]\cup [5,\infty )...(i)$$⇒x∈(−∞,−5]∪[5,∞)...(i)

$$4-{x}^{2}\neq 0$$4−x2=0

$$\Rightarrow x\neq 2,-2...(ii)$$⇒x=2,−2...(ii)

$$\Rightarrow {x}^{2}+2x-15>0$$⇒x2+2x−15>0

$$\Rightarrow (x+5)(x-3)>0$$⇒(x+5)(x−3)>0

$$\Rightarrow x\in (-\infty ,-5)\cup (3,\infty )...(iii)$$⇒x∈(−∞,−5)∪(3,∞)...(iii)

Using $(i),(ii)\text{and}(iii)$(i),(ii)and(iii)

$$\Rightarrow x\in (-\infty ,-5)\cup [5,\infty )$$⇒x∈(−∞,−5)∪[5,∞)

So, on comparing with $x\in (-\infty ,\alpha )\cup [\beta ,\infty )$x∈(−∞,α)∪[β,∞) we get,

$$\Rightarrow {\alpha }^{2}+{\beta }^{3}={(-5)}^{2}+{(5)}^{3}$$⇒α2+β3=(−5)2+(5)3

$$\Rightarrow {\alpha }^{2}+{\beta }^{3}=150$$⇒α2+β3=150