Limit Continuity and Differentiability 4 Question 7
7. 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? $\quad$ (2017 Adv.)
(a) $x=-1$
(b) $x=1$
(c) $x=0$
(d) $x=2$
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Answer:
Correct Answer: 7. $(a, b, d)$
Solution:
- Given, $f(x)=\frac{1}{2} x-1$ for $0 \leq x \leq \pi$
$$ \begin{aligned} & \therefore \quad[f(x)]=\begin{array}{cc} -1, & 0 \leq x<2 \\ 0, & 2 \leq x \leq \pi \end{array} \\ & \Rightarrow \quad \tan [f(x)]=\begin{array}{cc} \tan (-1), & 0 \leq x<2 \\ \tan 0, & 2 \leq x \leq \pi \end{array} \\ & \therefore \quad \lim _{x \rightarrow 2^{-}} \tan [f(x)]=-\tan 1 \end{aligned} $$
and $\quad \lim _{x \rightarrow 2^{+}} \tan [f(x)]=0$
So, $\tan f(x)$ is not continuous at $x=2$.
Now, $f(x)=\frac{1}{2} x-1 \Rightarrow f(x)=\frac{x-2}{2} \Rightarrow \frac{1}{f(x)}=\frac{2}{x-2}$
Clearly, $1 / f(x)$ is not continuous at $x=2$.
So, $\tan [f(x)]$ and $\frac{1}{f(x)}$ are both discontinuous at $x=2$.
