Let f(x)=lim _{theta rightarrow 0}left(frac{c | vedclass

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(A) Given $f(x) = \lim_{	heta ightarrow 0} \frac{\cos \pi x - x^{(2/	heta)} \sin(x-1)}{1 + x^{(2/	heta)}(x-1)}$.Case $1$: $|x| < 1$. As $	heta \

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Neither $(I)$ nor $(II)$ is True

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(A) Given $f(x) = \lim_{	heta ightarrow 0} \frac{\cos \pi x - x^{(2/	heta)} \sin(x-1)}{1 + x^{(2/	heta)}(x-1)}$.Case $1$: $|x| Case $2$: $|x| > 1$. As $	heta ightarrow 0$, $x^{(2/	heta)} ightarrow \infty$. Dividing numerator and denominator by $x^{(2/	heta)}$, we get $f(x) = \frac{-\sin(x-1)}{x-1}$.At $x=1$: $LHL = \lim_{x ightarrow 1^-} \cos \pi x = -1$. $RHL = \lim_{x ightarrow 1^+} \frac{-\sin(x-1)}{x-1} = -1$. Since $LHL = RHL = f(1) = -1$, $f(x)$ is continuous at $x=1$. Statement $(I)$ is False.At $x=-1$: $LHL = \lim_{x ightarrow -1^-} \frac{-\sin(x-1)}{x-1} = \frac{-\sin(-2)}{-2} = \frac{-\sin 2}{2}$. $RHL = \lim_{x ightarrow -1^+} \cos \pi x = \cos(-\pi) = -1$. Since $LHL 
eq RHL$, $f(x)$ is discontinuous at $x=-1$. Statement $(II)$ is False.Therefore, neither $(I)$ nor $(II)$ is true.

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$. Consider the following two statements: $(I)$ $f(x)$ is discontinuous at $x=1$. $(II)$ $f(x)$ is continuous at $x=-1$. Then,

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