JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Continuity and Differentiability

We have $f(x)=[x-1]\cos \left(\frac{(2x-1)\pi }{2}\right)=[x-1]\cos \left(x\pi -\frac{\pi }{2}\right)=[x-1]\sin (x\pi )$.

At integer values $x=n$: $[n-1]=n-1$ and $\sin (n\pi )=0$, so $f(n)=0$.

For $x$ slightly less than $n$: $[x-1]=n-2$ and $\sin (x\pi )\approx \sin (n\pi )=0$, so ${\lim }_{x\to n^{-}}f(x)=0$.

For $x$ slightly greater than $n$: $[x-1]=n-1$ and $\sin (x\pi )\approx 0$, so ${\lim }_{x\to n^{+}}f(x)=0$.

At $x=1$: $f(1)=0\cdot \sin (\pi )=0$. Both one-sided limits equal 0, so $f$ is continuous at $x=1$.

At other integers $n\ne 1$: The greatest integer function $[x-1]$ has a jump discontinuity, but since it's multiplied by $\sin (x\pi )$ which is 0 at integers, the product is continuous at all integers.

However, at non-integer points near integers (except 1), the function has discontinuities due to the greatest integer function jumps. The correct answer is that $f$ is discontinuous at all integral values except at $x=1$.