Question

If $f\left(x\right)=\left[\tan x\right]+\sqrt{\tan x-\left[\tan x\right]},0\le x<\frac{\pi }{2}$, where $[.]$ denotes thegreatest integer function, then

• A. $f\left(x\right)$ is continuous in $[0,\frac{\pi }{2})$
• B. $f\left(x\right)$ is not continuous at $x=0$
• C. $f\left(x\right)$ is continuous at $x=0,\frac{\pi }{4}$
• D. $f\left(x\right)$ has infinite points of discontinuity

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is continuous in $[0,\frac{\pi }{2})$
$f\left(x\right)$ is continuous at $x=0,\frac{\pi }{4}$

$f\left(x\right)=\left[\tan x\right]+\sqrt{\tan x-\left[\tan x\right]}=\left[t\right]+\sqrt{t-\left[t\right]},,$

where $t=\tan x.$

Clearly, $0\le t<\infty$ at $0\le x<\frac{\pi }{2}.$

Possible points of discontinuity may be, at which $t\in N.$

Let $t=k\in N$

$L.H.L.$ at $t=k=\underset{t\to k^{-}}{lim}\left[t\right]+\sqrt{t-\left[t\right]}$

$=\underset{h\to 0}{lim}[k-h]+\sqrt{(k-h)-[k-h]}$

$=\underset{h\to 0}{lim}\{k-1+\sqrt{[k-h-k+1]}\}=k$

$R.H.L.$ at $t=k=\underset{t\to k^{+}}{lim}\left[t\right]+\sqrt{t-\left[t\right]}$

$=\underset{h\to 0}{lim}[k+h]+\sqrt{k+h-[k+h]}$

$=\underset{h\to 0}{lim}k+\sqrt{k+h-k}=k$

$\therefore$ The function is continuous at $t=k\in N$

Thus, function $f\left(x\right)$ is continuous for all $[0,\frac{\pi }{2})$