If f(x) = begin{cases} frac{a + 3cos x}{x^2}, | vedclass

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(B) For $f(x)$ to be continuous at $x = 0$,$	ext{LHL} = 	ext{RHL} = f(0)$.$	ext{LHL} = \lim_{x 	o 0^-} \frac{a + 3\cos x}{x^2}$.For the limit to e

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$a = -3, b = \frac{-\sqrt{3}}{2}$

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(B) For $f(x)$ to be continuous at $x = 0$,$	ext{LHL} = 	ext{RHL} = f(0)$.$	ext{LHL} = \lim_{x 	o 0^-} \frac{a + 3\cos x}{x^2}$.For the limit to exist,the numerator must approach $0$ as $x 	o 0$,so $a + 3(1) = 0 \Rightarrow a = -3$.Using the Taylor expansion $\cos x \approx 1 - \frac{x^2}{2}$,we get $	ext{LHL} = \lim_{x 	o 0^-} \frac{-3 + 3(1 - \frac{x^2}{2})}{x^2} = \lim_{x 	o 0^-} \frac{-3/2 x^2}{x^2} = -\frac{3}{2}$.$	ext{RHL} = \lim_{x 	o 0^+} b 	an \left( \frac{\pi}{[x + 3]} ight) = b 	an \left( \frac{\pi}{3} ight) = b\sqrt{3}$.Since $	ext{LHL} = 	ext{RHL}$,we have $b\sqrt{3} = -\frac{3}{2} \Rightarrow b = -\frac{\sqrt{3}}{2}$.Thus,$a = -3$ and $b = -\frac{\sqrt{3}}{2}$.

If $f(x) = \begin{cases} \frac{a + 3\cos x}{x^2}, & x < 0 \\ b\tan \left( \frac{\pi}{[x + 3]} \right), & x \geqslant 0 \end{cases}$ is continuous at $x = 0$,then:

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