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