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 :

The domain of a sum is the intersection of the individual domains.
Hence we analyse each term of $$f(x)=\frac{1}{\sqrt{10+3x-x^{2}}}+\frac{1}{\sqrt{x+\lvert x\rvert}}$$ separately.

Term 1: $$\dfrac{1}{\sqrt{\,10+3x-x^{2}\,}}$$ is defined only when the radicand is positive:

$$10+3x-x^{2}\gt 0$$

Rewrite the quadratic in standard form:

$$-(x^{2}-3x-10)\gt 0\quad\Longrightarrow\quad x^{2}-3x-10\lt 0$$

Factor the quadratic:

$$x^{2}-3x-10=(x-5)(x+2)$$

Because the coefficient of $$x^{2}$$ is positive, the quadratic is negative between its roots. Thus

$$-2\lt x\lt 5\qquad -(1)$$

Term 2: $$\dfrac{1}{\sqrt{\,x+\lvert x\rvert\,}}$$ requires $$x+\lvert x\rvert\gt 0$$ and the denominator non-zero.

Consider both cases for $$x$$:

• If $$x\ge 0$$ then $$\lvert x\rvert=x$$, so $$x+\lvert x\rvert=2x\gt 0\;\Longrightarrow\;x\gt 0$$.

• If $$x\lt 0$$ then $$\lvert x\rvert=-x$$, giving $$x+\lvert x\rvert=0$$, which is not >0.

Hence the second term is defined only for

$$x\gt 0\qquad -(2)$$

Overall domain: Intersect $$(1)$$ and $$(2)$$:

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

Therefore the domain of $$f(x)$$ is $$(a,b)=(0,5)$$, so $$a=0$$ and $$b=5$$.

Compute the required expression:

$$(1+a)^{2}+b^{2}=(1+0)^{2}+5^{2}=1+25=26$$

Hence $$(1+a)^{2}+b^{2}=26$$, which corresponds to Option A.