1. Continuity at $x=0$ means $\lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x)=b.$ So we compute the left-hand and right-hand limits.

2. Left-hand limit $(x<0)$: $f(x)=\frac{\sin((a+1)x)+\sin 2x}{2x}.$ Using $\displaystyle \lim_{x\to 0}\frac{\sin kx}{x}=k$, $\lim_{x\to 0^-} f(x)=\frac{1}{2}\left((a+1)+2\right)=\frac{a+3}{2}.$ For continuity, $b=\frac{a+3}{2}. \qquad (1)$

3. Right-hand limit $(x>0)$: $f(x)=\frac{\sqrt{x+bx^3}-\sqrt{x}}{bx^{5/2}}.$ Factor $\sqrt{x}$ from the numerator: $\sqrt{x+bx^3}=\sqrt{x(1+bx^2)}=\sqrt{x}\sqrt{1+bx^2}.$ Hence f(x)=\frac{\sqrt{x}(\sqrt{1+bx^2}-1)}{bx^{5/2}}=rac{\sqrt{1+bx^2}-1}{bx^2}. Now rationalize:

   =\frac{1+bx^2-1}{bx^2(\sqrt{1+bx^2}+1)} =\frac{1}{\sqrt{1+bx^2}+1}.$$

Therefore, $\lim_{x\to 0^+} f(x)=\frac{1}{1+1}=\frac{1}{2}.$ For continuity, $b=\frac{1}{2}. \qquad (2)$

4. Substitute $b=\frac12$ into (1): $\frac12=\frac{a+3}{2}$ $a+3=1$ $a=-2.$ Thus, $a+b=-2+\frac12=-\frac32.$

5. Option check:

• A: $-3$ ❌
• B: $-2$ ❌
• C: $-\frac52$ ❌
• D: $-\frac32$ ✅

Therefore, the correct answer is D.