Let f(x)= _ θ arrow 0 ( π x-x^ ((2)/(θ)) (x-1) 1+x^ ((2)/(θ)) (x-1)...

Taking the limit as $\theta \to 0$, we note $x^{(2/\theta)} \to 0$ for $|x| < 1$ and $\to \infty$ for $|x| > 1$. This gives:

$f(x) =$\begin{cases} \cos \pi x & x \to 1^- \\ \dfrac{-\sin(x-1)}{(x-1)} & x \to 1^+ \end{cases}

Continuity at $x = 1$:
$\text{RHL} = \lim_{x \to 1^+} \frac{-\sin(x-1)}{(x-1)} = -1$
$\text{LHL} = \lim_{x \to 1^-} \cos \pi x = -1$, $f(1) = -1$
$\Rightarrow f(x)$ is continuous at $x = 1$. So Statement (I) is false.

Continuity at $x = -1$:
$f(x) =$\begin{cases} \dfrac{-\sin(x-1)}{-(x-1)} & x \to -1^- \\ \cos \pi x & x \to -1^+ \end{cases}

$\text{RHL} = \lim_{x \to -1^+} \cos \pi x = -1$
$\text{LHL} = \lim_{x \to -1^-} \frac{-\sin(x-1)}{-(x-1)} = \frac{\sin 2}{-2}$

Since LHL $\neq$ RHL, $f(x)$ is discontinuous at $x = -1$. So Statement (II) is also false.

Hence, neither (I) nor (II) is true.