If the domain of the function\ $f(x)=\frac{\sqrt{x^{2}-25}}{\left(4-x^{2} ight)}+\log _{10}\left(x

Question

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

Professor PC Thomas Classes · Answer

(d): For domain, $4-x^{2} 
eq 0 \Rightarrow x 
eq \pm 2$ $$ x^{2}-25 \geq 0 ; x^{2} \geq 25 $$ $\Rightarrow \quad x \in(-\infty,-5] \cup[5, \infty)$\ Also, $x^{2}+2 x-15>0 \Rightarrow(x+5)(x-3)>0$\ $\Rightarrow \quad x \in(-\infty,-5) \cup(3, \infty)$\ $	herefore \quad D_{f}=(-\infty,-5) \cup[5, \infty)$\ So, $\alpha=-5$ and $\beta=5$\ $	herefore \quad \alpha^{2}+\beta^{3}=25+125=150$\