If the function f: R rightarrow R defined by | vedclass

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(A) Given the function $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x  0

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-$1$

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(A) Given the function $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x  0 \end{cases}$ is continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = b$. First,calculate the left-hand limit: $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) = (a+1) + 1 = a+2$. Next,calculate the right-hand limit: $\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \sqrt{x}} = \lim_{x \rightarrow 0^+} \frac{\sqrt{1+x} - 1}{x}$. Multiplying by the conjugate: $\lim_{x \rightarrow 0^+} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)} = \lim_{x \rightarrow 0^+} \frac{1+x-1}{x(\sqrt{1+x} + 1)} = \lim_{x \rightarrow 0^+} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{1+1} = \frac{1}{2}$. Since the function is continuous,$b = \frac{1}{2}$ and $a+2 = b$. Substituting $b = \frac{1}{2}$ into $a+2 = b$: $a + 2 = \frac{1}{2} \implies a = \frac{1}{2} - 2 = -\frac{3}{2}$. Therefore,$a+b = -\frac{3}{2} + \frac{1}{2} = -1$.

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $a+b =$

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