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(B) For $f$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = \alpha$.First,calculate t

Vedclass · Accepted Answer

$12$

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(B) For $f$ to be continuous at $x = 0$,we must have $\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0) = \alpha$.First,calculate the left-hand limit:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^-} \frac{1 - \cos 2x}{x^2} = \lim_{x ightarrow 0^-} \frac{2 \sin^2 x}{x^2} = 2 	imes 1^2 = 2$.Thus,$\alpha = 2$.Next,calculate the right-hand limit:$\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0^+} \frac{\beta \sqrt{1 - \cos x}}{x} = \lim_{x ightarrow 0^+} \frac{\beta \sqrt{2 \sin^2 (x/2)}}{x} = \lim_{x ightarrow 0^+} \frac{\beta \sqrt{2} |\sin(x/2)|}{x}$.Since $x > 0$,$\sin(x/2) > 0$,so $|\sin(x/2)| = \sin(x/2)$.$\lim_{x ightarrow 0^+} \frac{\beta \sqrt{2} \sin(x/2)}{x} = \lim_{x ightarrow 0^+} \frac{\beta \sqrt{2} \sin(x/2)}{2(x/2)} = \frac{\beta \sqrt{2}}{2} = \frac{\beta}{\sqrt{2}}$.Setting this equal to $\alpha = 2$,we get $\frac{\beta}{\sqrt{2}} = 2$,which implies $\beta = 2\sqrt{2}$.Finally,$\alpha^2 + \beta^2 = (2)^2 + (2\sqrt{2})^2 = 4 + 8 = 12$.

Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to:

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