Question

Let the function $f:R\to R$ be defined as
$f\left(x\right)=\left[\sin x\right]$
where for any real number $t,\left[t\right]$ denotes the greatest integer not exceeding $t$. Then $f$ is continuous at

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{3\pi }{2}$
• D. $2\pi$

Answer 1

The correct option is C $\frac{3\pi }{2}$

The given function is $f\left(x\right)=\left[sinx\right]$

We know that for any real number t, $\left[t\right]$ denotes the greatest integer not exceeding $t$.

From the given graph of the function $f\left(x\right)=\left[sinx\right]$, we can see that:

When $0<x<\frac{\pi }{2}$ $\to 0<sinx<1$ hence $\left[sinx\right]=0$

At $x=\frac{\pi }{2}$, $\left[sinx\right]=1$

Again When $\frac{\pi }{2}<x<\pi$ $\to 1<sinx<0$ hence $\left[sinx\right]=0$

At $x=\pi$, $\left[sinx\right]=0$

Now When $\pi <x<\frac{3\pi }{2}$ $\to 0<sinx<-1$ hence $\left[sinx\right]=-1$

Also $x=\frac{3\pi }{2}$, $\left[sinx\right]=-1$

Again When $\frac{3\pi }{2}<x<2\pi$ $\to -1<sinx<0$ hence $\left[sinx\right]=-1$

At $x=2\pi$, $\left[sinx\right]=0$

Now When $2\pi <x<\frac{5\pi }{2}$ $\to 0<sinx<1$ hence $\left[sinx\right]=0$

Here we can see, the left hamd value from $\frac{3\pi }{2}$ is $-1$ and right hand value at $\frac{3\pi }{2}$ is also $-1$, also we can see that value of function $\left[sinx\right]$ at $\frac{3\pi }{2}$ is $-1$ too.

This behaviour doesn't follow at $\frac{\pi }{2}$, $\pi$ or $2\pi$. Hence the function is only continuous at $\frac{3\pi }{2}$.

So the correct answer is $C$.