Question

Which of the following functions is/are continuous?
I. Every rational function in its domain.
II. Sine function.
III. Cosine function.
IV. Tangent function is continuous in their domain.

• A. Only I is continuous
• B. Only II is continuous
• C. I and II are continuous
• D. All are continuous

Answer 1

Answer: (D)

(i) Recall that every rational function $f$ is given by
$f(x)=\frac{p(x)}{q(x)}, q(x) \neq 0$
where, $p$ and $q$ are polynomial functions. The domain of $f$ is all real numbers except those points at which $q$ is zero. Since, polynomial functions are continuous, $f$ is continuous.
(ii) To seeth is, we use the following facts
$\displaystyle\lim _{x \rightarrow 0} \sin x=0$
We have not proved it, but is intuitively clear from the graph of $\sin x$ near 0 .
Now, observe that $f(x)=\sin x$ is defined for every real number. Let $c$ be a real number. Put $x=c+h$. If $x \rightarrow c$ we know that, $h \rightarrow 0$. Therefore,
$\displaystyle\lim _{x \rightarrow c} f(x)=\displaystyle\lim _{x \rightarrow c} \sin x$
$=\displaystyle\lim _{h \rightarrow 0} \sin (c+h)$
$=\displaystyle\lim _{h \rightarrow 0}(\sin c \cos h+\cos c \sin h)$
$=\displaystyle\lim _{h \rightarrow 0}[\sin c \cos h]+\displaystyle\lim _{h \rightarrow 0}[\cos c \sin h]$
$=\sin c+0=\sin c=f(c)$
Thus, $\displaystyle\lim _{x \rightarrow c} f(x)=f(c)$ and hence $f$ is a continuous function.
(iii) Let $f(x)=\cos x$ and let $c$ be any real number.
Then, $ \displaystyle\lim _{x \rightarrow c^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(c+h)$
$\Rightarrow \displaystyle\lim _{x \rightarrow c^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} \cos (c+h) $
$ \Rightarrow \displaystyle\lim _{x \rightarrow c^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0}(\cos c \cos h-\sin c \sin h)$
$ \Rightarrow \displaystyle\lim _{x \rightarrow c^{+}} f(x)=\cos c \displaystyle\lim _{h \rightarrow 0} \cos h-\sin c \displaystyle\lim _{h \rightarrow 0} \sin h$
$ =\cos c \times 1-\sin c \times 0 $
$ \displaystyle\lim _{x \rightarrow c^{+}} f(x)=\cos c$
$ \left(\because \displaystyle\lim _{h \rightarrow 0} \cos h=1 \text { and }\displaystyle \lim _{h \rightarrow 0} \sin h=0\right) $
Similarly, we have
$\displaystyle\lim _{x \rightarrow c^{-}} f(x)=f(c) $
$\therefore \displaystyle\lim _{x \rightarrow c^{-}} f(x)=\displaystyle\lim _{x \rightarrow c^{+}} f(x)=f(c)$
$f(x)$ is continuous at $x=c$.
$\because c$ is arbitrary real number, so $f(x)$ is everywhere continuous.
(iv) Let $f(x)=\tan x$
Clearly, domain $f=R-\left\{(2 n+1) \frac{\pi}{2} n \in Z\right\}$
We have, $ f(x)=\tan x=\frac{\sin x}{\cos x}$
$\because \sin x$ and $\cos x$ are everywhere continuous.
Therefore, $f(x)=\tan x$ is continuous for all $x \in R$ such that
$ \cos x \neq 0 $
But $ \cos x \neq 0$
$\Rightarrow x \neq(2 n+1) \frac{\pi}{2}, n \in Z$
Hence, $f(x)=\tan x$ is continuous for all
$x \in R-\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$