Question

If $f\left(x\right)=[\sin x+[\cos x+[\tan x+[\sec x\left]\right]\left]\right]$, $x\in (0,\frac{\pi }{4}),$ where $[\cdot ]$ denotes the greatest integer less than or equal to $x$, then

• A. $f$ is one-one function
• B. $f$ is many-one function
• C. Range of $f$ is $\{0,1\}$
• D. Range of $f$ is$\{2,3\}$

Answer 1

The correct option is B $f$ is many-one function

Given $f\left(x\right)=[\sin x+[\cos x+[\tan x+[\sec x\left]\right]\left]\right]$
$=[\sin x+p],$ where $p=[\cos x+[\tan x+\left[\sec x\right]\left]\right]$
$=\left[\sin x\right]+p,$ (as $p$ is an integer)
$=\left[\sin x\right]+[\cos x+[\tan x+\left[\sec x\right]\left]\right]$
$=\left[\sin x\right]+\left[\cos x\right]+\left[\tan x\right]+\left[\sec x\right]$
Now, for $x\in (0,\frac{\pi }{4}),\sin x\in (0,\frac{1}{\sqrt{2}}),\cos x\in (\frac{1}{\sqrt{2}},1),\tan x\in (0,1),\sec x\in (1,\sqrt{2})$
$\Rightarrow \left[\sin x\right]=0,\left[\cos x\right]=0,\left[\tan x\right]=0$ and $\left[\sec x\right]=1$
$\Rightarrow$ The range of $f\left(x\right)$ is $\left\{1\right\}.$
So, for the given interval $f\left(x\right)$ has only one value.
Hence, it will be a many-one function.