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

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$(-2, -1)$

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(B) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0)$.First,consider the $LHL$: $\lim_{x ightarrow 0^-} \frac{a+2 \cos x}{x^2}$.For this limit to exist and be finite,the numerator must approach $0$ as $x ightarrow 0$. Thus,$a + 2 \cos(0) = 0 \implies a + 2 = 0 \implies a = -2$.Using the Taylor series expansion for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$,we get:$LHL = \lim_{x ightarrow 0} \frac{-2 + 2(1 - \frac{x^2}{2} + \dots)}{x^2} = \lim_{x ightarrow 0} \frac{-x^2}{x^2} = -1$.Now,consider the $RHL$: $\lim_{x ightarrow 0^+} b 	an \frac{\pi}{[x+4]}$.As $x ightarrow 0^+$,$[x+4] = 4$. So,$RHL = b 	an \frac{\pi}{4} = b(1) = b$.Since the function is continuous,$LHL = RHL$,which implies $b = -1$.Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f(x)= \begin{cases} \frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0 \end{cases}$ is continuous at $x=0$,then the ordered pair $(a, b)$ is equal to

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