Let f(x) = begin{cases} a sin(x + b) & x ge 0 | vedclass

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(B) For $f(x)$ to be differentiable at $x = 0$,it must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = f(0) \implies 6(0)^7 - 0 + 1 = a \sin(b

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$2$

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(B) For $f(x)$ to be differentiable at $x = 0$,it must be continuous at $x = 0$. Thus,$\lim_{x 	o 0^-} f(x) = f(0) \implies 6(0)^7 - 0 + 1 = a \sin(b) \implies a \sin b = 1$. Also,the left-hand derivative must equal the right-hand derivative at $x = 0$. $f'(x) = \begin{cases} a \cos(x + b) & x > 0 \ 42x^6 - 1 & x Equating the derivatives at $x = 0$: $a \cos b = -1$. Squaring and adding the two equations: $(a \sin b)^2 + (a \cos b)^2 = 1^2 + (-1)^2 \implies a^2(\sin^2 b + \cos^2 b) = 2 \implies a^2 = 2 \implies a = \pm \sqrt{2}$. Case $I$: If $a = \sqrt{2}$,then $\sin b = \frac{1}{\sqrt{2}}$ and $\cos b = -\frac{1}{\sqrt{2}}$. This corresponds to $b = \frac{3\pi}{4}$. Case $II$: If $a = -\sqrt{2}$,then $\sin b = -\frac{1}{\sqrt{2}}$ and $\cos b = \frac{1}{\sqrt{2}}$. This corresponds to $b = \frac{7\pi}{4}$. Thus,the ordered pairs are $(\sqrt{2}, \frac{3\pi}{4})$ and $(-\sqrt{2}, \frac{7\pi}{4})$. There are $2$ such ordered pairs.

Let $f(x) = \begin{cases} a \sin(x + b) & x \ge 0 \\ 6x^7 - x + 1 & x < 0 \end{cases}$ be differentiable for all real $x$. If $a \in \mathbb{R}$ and $b \in [0, 2\pi]$,then the number of ordered pairs $(a, b)$ is:

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