If the domain of the function
$f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is $(a, b)$, then $(1+a)^2+b^2$ is equal to :

Question

If the domain of the function $f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is $(a, b)$, then $(1+a)^2+b^2$ is equal to :

Aicharya · Accepted Answer

$\mathrm{x}+|\mathrm{x}| \gt 0 \quad \Rightarrow \mathrm{x} \in(0, \infty)$ ...(1)
$\begin{aligned} & \& 10+3 x-x^2 \gt 0 \\ & \Rightarrow x^2-3 x-10 \lt 0\end{aligned}$
$\Rightarrow x \in(-2,5)$ ...(2)
$\begin{aligned} & \text { from }(1) \&(2) \quad \mathrm{x} \in(0,5) \\ & \therefore \mathrm{a}=0 \& \mathrm{~b}=5 \\ & \therefore\left(1+\mathrm{a}^2\right)+\mathrm{b}^2=1+25=26\end{aligned}$

If the domain of the function $f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}$ is $(a, b)$, then…

Solution