If the domain of the function f(x)=1√10+3x-x^2+1x+|x| is (a,b), then (1+a)^2+b^2 is equal …

Question

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

Accepted Answer

$x+|x|= $$2x, \& x\ge 0 \ 0, \& x$\Rightarrow \frac{1}{\sqrt{x+|x|}}$, domain is $x>0$, as $2x\ne 0$$$ Similarly, $$$\begin{gathered} \frac{1}{\sqrt{3x+10-{x}^{2}}}3x+10-{x}^{2}>0 \ \Rightarrow {x}^{2}-3x-10<0 \ (x-5)(x+2)<0 \ \Rightarrow x\in (-2,5) \ \Rightarrow (0,\infty)\cap (-2,5)=(0,5) \ \Rightarrow (1+a{)}^{2}+{b}^{2}=1+25=26 \end{gathered}$$

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