If the domain of the function f(x) = √x^2-25 / (4-x^2) + log10(x^2 + 2x - 15) is

Question

If the domain of the function \(f(x)=\frac{\sqrt{x^{2}-25}}{\left(4-x^{2}\right)}+\log _{10}\left(x^{2}+2 x-15\right)\) is \((-\infty, \alpha) \cup[\beta, \infty)\), then \(\alpha^{2}+\beta^{3}\) is equal to :

(1) 140

(2) 175

(3) 150

(4) 125

Answer 1

Correct option is (3) 150

\(f(\mathrm{x})=\frac{\sqrt{\mathrm{x}^{2}-25}}{4-\mathrm{x}^{2}}+\log _{10}\left(\mathrm{x}^{2}+2 \mathrm{x}-15\right)\)

Domain: \(\mathrm{x}^{2}-25 \geq 0 \Rightarrow \mathrm{x} \in(-\infty,-5] \cup[5, \infty)\)

\(4-\mathrm{x}^{2} \neq 0 \Rightarrow \mathrm{x} \neq\{-2,2\}\)

\(\mathrm{x}^{2}+2 \mathrm{x}-15>0 \Rightarrow(\mathrm{x}+5)(\mathrm{x}-3)>0\)

\(\Rightarrow \mathrm{x} \in(-\infty,-5) \cup(3, \infty)\)

\(\therefore \mathrm{x} \in(-\infty,-5) \cup[5, \infty)\)

\(\alpha=-5 ; \beta=5\)

\(\therefore \alpha^{2}+\beta^{3}=150\)