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

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

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$-1$

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(A) Given: $f(x) = \begin{cases} \frac{\sin(a + 1)x + \sin x}{x}, & x  0 \end{cases}$ Since $f(x)$ is continuous on $R$,it must be continuous at $x = 0$. Therefore,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = b$. For the left-hand limit: $\lim_{x 	o 0^-} \frac{\sin(a + 1)x + \sin x}{x} = \lim_{x 	o 0^-} \left( \frac{\sin(a + 1)x}{x} + \frac{\sin x}{x} ight) = (a + 1) + 1 = a + 2$. So,$a + 2 = b$ . . . $(i)$. For the right-hand limit: $\lim_{x 	o 0^+} \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{1 + x} - 1)}{x \cdot \sqrt{x}} = \lim_{x 	o 0^+} \frac{\sqrt{1 + x} - 1}{x}$. Multiplying by the conjugate: $\lim_{x 	o 0^+} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \lim_{x 	o 0^+} \frac{1 + x - 1}{x(\sqrt{1 + x} + 1)} = \lim_{x 	o 0^+} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}$. Thus,$b = \frac{1}{2}$. Substituting $b$ into $(i)$: $a + 2 = \frac{1}{2} \Rightarrow a = \frac{1}{2} - 2 = -\frac{3}{2}$. Finally,$a + b = -\frac{3}{2} + \frac{1}{2} = -\frac{2}{2} = -1$.

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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