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

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$(-2, -1)$

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(B) For $f(x)$ 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, $a + 2 \cos(0) = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.Substituting $a = -2$ into the limit: $\lim_{x ightarrow 0^-} \frac{-2 + 2 \cos x}{x^2} = \lim_{x ightarrow 0^-} \frac{-2(1 - \cos x)}{x^2} = \lim_{x ightarrow 0^-} \frac{-2(2 \sin^2(x/2))}{x^2} = -2 \lim_{x ightarrow 0^-} \frac{\sin^2(x/2)}{(x/2)^2 	imes 2} = -2 	imes \frac{1}{2} = -1$.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 $LHL = RHL$, we have $-1 = b$.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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