Jee Main 2024 01 02 2024 Shift 2 - Question15
Question 15
If the domain of the function $f(x)=\frac{\sqrt{x^{2}-25}}{4-x^{2}}+$ $\log \left(x^{2}+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^{2}+\beta^{2}$ is equal to
Show Answer
Answer: (50)
Solution:
$x^{2}-25 \geq 0 \Rightarrow x \in(-\infty,-5) \cup[5, \infty)$
$4-x^{2} \neq 0$
$\Rightarrow x \neq-2,2$
$x^{2}+2 x-15>0$
$(x+5)(x-3)>0$
$x \in(-\infty,-5) \cup(3, \infty)$
So,
$x \in(-\infty,-5) \cup[5, \infty)$
$\alpha=-5$
$\beta=5$
$\alpha^{2}+\beta^{2}=50$
BETA
