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
$f(x)=\frac{p(x)}{q(x)},q(x)\ne 0$
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
$\underset{x\to 0}{\lim }\sin x=0$
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 $x\to c$ we know that, $h\to 0$. Therefore,
$\underset{x\to c}{\lim }f(x)=\underset{x\to c}{\lim }\sin x$
$=\underset{h\to 0}{\lim }\sin (c+h)$
$=\underset{h\to 0}{\lim }(\sin c\cos h+\cos c\sin h)$
$=\underset{h\to 0}{\lim }[\sin c\cos h]+\underset{h\to 0}{\lim }[\cos c\sin h]$
$=\sin c+0=\sin c=f(c)$
Thus, $\underset{x\to c}{\lim }f(x)=f(c)$ and hence $f$ is a continuous function.
(iii) Let $f(x)=\cos x$ and let $c$ be any real number.
Then, $\underset{x\to c^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(c+h)$
$\Rightarrow \underset{x\to c^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }\cos (c+h)$
$\Rightarrow \underset{x\to c^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }(\cos c\cos h-\sin c\sin h)$
$\Rightarrow \underset{x\to c^{+}}{\lim }f(x)=\cos c\underset{h\to 0}{\lim }\cos h-\sin c\underset{h\to 0}{\lim }\sin h$
$=\cos c\times 1-\sin c\times 0$
$\underset{x\to c^{+}}{\lim }f(x)=\cos c$
$\left(\because \underset{h\to 0}{\lim }\cos h=1\text{ and }\underset{h\to 0}{\lim }\sin h=0\right)$
Similarly, we have
$\underset{x\to c^{-}}{\lim }f(x)=f(c)$
$\therefore \underset{x\to c^{-}}{\lim }f(x)=\underset{x\to c^{+}}{\lim }f(x)=f(c)$
$f(x)$ 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 $f=R-\left\{(2n+1)\frac{\pi }{2}n\in Z\right\}$
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
$\cos x\ne 0$
But $\cos x\ne 0$
$\Rightarrow x\ne (2n+1)\frac{\pi }{2},n\in Z$
Hence, $f(x)=\tan x$ is continuous for all
$x\in R-\left\{(2n+1)\frac{\pi }{2}:n\in Z\right\}$