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(D) Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$.First,find $f(0) = b$.

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(D) Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$.First,find $f(0) = b$.Next,calculate the left-hand limit:$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sin(a+1)x + \sin 2x}{2x} = \lim_{x \rightarrow 0^{-}} \left( \frac{\sin(a+1)x}{2x} + \frac{\sin 2x}{2x} \right) = \frac{a+1}{2} + \frac{2}{2} = \frac{a+1}{2} + 1$.Now,calculate the right-hand limit:$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}} = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{1+bx^2} - 1)}{bx^{5/2}} = \lim_{x \rightarrow 0^{+}} \frac{\sqrt{1+bx^2} - 1}{bx^2}$.Using the expansion $\sqrt{1+u} \approx 1 + \frac{u}{2}$,we get $\lim_{x \rightarrow 0^{+}} \frac{1 + \frac{bx^2}{2} - 1}{bx^2} = \frac{1}{2}$.Equating the limits:$b = \frac{1}{2}$ and $\frac{a+1}{2} + 1 = \frac{1}{2}$.From $\frac{a+1}{2} = -\frac{1}{2}$,we get $a+1 = -1$,so $a = -2$.Therefore,$a+b = -2 + \frac{1}{2} = -\frac{3}{2}$.

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