Question

Let $[x]$ be the greatest integer less than or equals to $x$. Then, at which of the following point(s) the function $f(x)=x\cos (\pi (x+[x]))$ is discontinuous?

• A. $x=1$
• B. $x=-1$
• C. $x=0$
• D. $x=2$

Answer 1

Answer: (A), (B), (D)

$f(x)=x\cos (\pi (x+[x]))$
at $x=1\underset{x\to 1^{+}}{\lim }f(x)=\underset{h\to 0^{+}}{\lim }(1+h)\cos (\pi (1+h+[1+h]))$
$=\underset{h\to 0}{\lim }(1+h)\cos (\pi (2+h))$
$=\underset{h\to 0}{\lim }(1+h)\cos 2h=1$
Similarly
$\underset{x\to 1^{-}}{\lim }f(x)=-1$
$\Rightarrow$ discontinuous at $X=1$
at $x=-1\underset{x\to -1}{\lim }f(x)=\underset{h\to 0}{\lim }f(-1+h)$
$=\underset{h\to 0}{\lim }(-1+h)\cos (\pi (-1+h+[-1+h])$
$=\underset{h\to 0}{\lim }(-1+h)\cos (\pi (-2+h))$
$=\underset{h\to 0}{\lim }(-1+h)\cos (2\pi -\pi h)=-1$
$\underset{x\to -1^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(-1-h)=1$
$\Rightarrow$ discontinuous at $x=-1$
at $x=0$
$\underset{t\to 0^{-}}{\lim }f(x)=0$ and $f(0)=0$
$\Rightarrow$ continuous at $x=0$
at $x=2$
$\underset{x\to 2^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(2+h)=\underset{h\to 0}{\lim }(2+h)\cos (\pi (2+h+[2+h]))$
$=\underset{h\to 0}{\lim }(2+h)\cos (4\pi +\pi h)=2$
$\underset{x\to 2^{-}}{\lim }f(x)=-2$
discontinuous at $x=2$