Q. Let

$ f(x)=\lim_{\theta\to 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\mathbb{R}. $

Consider the following two statements :

(I) $f(x)$ is discontinuous at $x=1$.

(II) $f(x)$ is continuous at $x=-1$.

Then,

Correct Answer: Neither (I) nor (II) is True

Explanation

We analyze the behavior of the term

$x^{\left(\frac{2}{\theta}\right)}$ as $\theta \to 0$

As $\theta \to 0$ :

If $|x|<1$, then $x^{\left(\frac{2}{\theta}\right)} \to 0$
If $|x|>1$, then $x^{\left(\frac{2}{\theta}\right)} \to \infty$
If $x=\pm 1$, the value is $1$

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At $x=1$ :

$x^{\left(\frac{2}{\theta}\right)} = 1$

Substitute $x=1$ in $f(x)$ :

$f(1)=\dfrac{\cos(\pi)-\sin(0)}{1-0} = \dfrac{-1-0}{1}=-1$

The limit exists and equals the function value, hence

$f(x)$ is continuous at $x=1$.

So, Statement (I) is false.

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At $x=-1$ :

$(-1)^{\left(\frac{2}{\theta}\right)} = 1$

Substitute $x=-1$ :

$f(-1)=\dfrac{\cos(-\pi)-\sin(-2)}{1-(-2)} =\dfrac{-1+\sin(2)}{3}$

The limit exists but the function is not continuous at $x=-1$ due to mismatch with nearby values.

So, Statement (II) is false.

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Therefore, the correct answer is

Neither (I) nor (II) is True

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