If the domain of the function \(f\left(x\right)=\frac{1}{\sqrt{10+3\mathrm{x}-{\mathrm{x}}^{2}}}+\frac{1}{\sqrt{\mathrm{…

Consider \(\frac{1}{\sqrt{10+3 x-x^2}}\)

For this term to be defined,

\(10+3 x-x^2>0\)

\((x-5)(x+2)<0\)

\(-2<x<5\)

Now, consider \(\frac{1}{\sqrt{x+|x|}}\)

we need \(x+|x|>0\)

If \(x \geq 0\), then \(|x|=x\), so

\(x+|x|=2 x>0 \Rightarrow x>0\)

If \(x<0\), then \(|x|=-x\), so

\(x+|x|=0\)

which is not allowed in denominator.

Hence, \(x>0\)

Therefore, the domain is:

\((-2,5) \cap(0, \infty)=(0,5)\)

So, \(a=0, b=5\)

\((1+a)^2+b^2=(1+0)^2+5^2=1+25=26\)