JEE Mathematics 2026 Question with Solution

Given:

$$f(x)=\lim_{\theta\to 0} \frac{\cos(\pi x-\theta)\sin(x-1)}{1+x^{\theta/2}(x-1)},\qquad x\in\mathbb{R}$$

We need to check continuity at $x=1$ and $x=-1$.

Find: Which of the statements (I) and (II) is true.

For $x>0$ and $x\neq 1$, as $\theta\to 0$,

$$x^{\theta/2}\to 1$$

Therefore,

$$f(x)=\frac{\cos(\pi x)\sin(x-1)}{1+(x-1)}=\frac{\cos(\pi x)\sin(x-1)}{x}$$

At $x=1$,

$$\lim_{x\to 1} f(x)=\lim_{x\to 1}\frac{\cos(\pi x)\sin(x-1)}{x}$$

Since $\sin(x-1)\sim (x-1)$ as $x\to 1$ and $\cos(\pi)=-1$, the limit is

$$\lim_{x\to 1} f(x)=\frac{-1\cdot 0}{1}=0$$

Also,

$$f(1)=\lim_{\theta\to 0}\frac{\cos(\pi-\theta)\sin 0}{1+1^{\theta/2}(0)}=0$$

Hence,

$$\lim_{x\to 1} f(x)=f(1)$$

So $f(x)$ is continuous at $x=1$. Thus statement (I) is true.

At $x=-1$, the term $x^{\theta/2}$ is not well-defined in the real sense as $\theta\to 0$ for negative $x$. Therefore the limit defining $f(x)$ does not exist as a real-valued function at $x=-1$.

So $f(x)$ is not continuous at $x=-1$. Thus statement (II) is false.

Therefore, the correct option is A: Only (I) is true.