Let a, b in R, b neq 0. Define a function f(x | vedclass

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(B) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First,find $f(0)$ and the left-hand lim

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$14$

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(B) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First,find $f(0)$ and the left-hand limit:$f(0) = a \sin \frac{\pi}{2}(0-1) = a \sin(-\frac{\pi}{2}) = -a$.Next,find the right-hand limit:$\lim_{x 	o 0^+} \frac{	an 2x - \sin 2x}{bx^3} = \lim_{x 	o 0^+} \frac{	an 2x(1 - \cos 2x)}{bx^3} = \lim_{x 	o 0^+} \frac{	an 2x \cdot 2 \sin^2 x}{bx^3}$.Using standard limits $	an 	heta \approx 	heta$ and $\sin 	heta \approx 	heta$ as $	heta 	o 0$:$\lim_{x 	o 0^+} \frac{(2x) \cdot 2(x^2)}{bx^3} = \frac{4x^3}{bx^3} = \frac{4}{b}$.Equating the limits:$-a = \frac{4}{b} \implies -ab = 4$.Finally,calculate $10 - ab$:$10 - ab = 10 + 4 = 14$.

Let $a, b \in R, b \neq 0$. Define a function $f(x) = \begin{cases} a \sin \frac{\pi}{2}(x-1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & \text{for } x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $10 - ab$ is equal to ...... .

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