Name: Number of Onto Functions
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: In this video, I'll show you how to prove the number of onto functions formula. This will prove handy for a better understanding of this concept.

Question

Find the range of

f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]},

where

[\cdot]

denotes the greatest integer function.

Accepted Answer

Given, $f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]},$

where $[\cdot]$ denotes greatest integer function.

Period of $\sin x$ will be $2 π$ .

Period of $\sin 2 x$ and $\cos 2 x$ will be $\frac{2 π}{2} = π$ .

Case I: $x \in (0, \frac{π}{4})$

$[\sin 2 x] = 0$

$[\cos 2 x] = 0$

$y = f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]}$

$\Rightarrow y = \sqrt{0 - 0}$

$\Rightarrow y = 0$

Case II: $x \in (\frac{π}{4}, \frac{π}{2})$

$[\sin 2 x] = 0$

$[\cos 2 x] = - 1$

$y = f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]}$

$\Rightarrow y = \sqrt{0 - (- 1)}$

$\Rightarrow y = 1$

Case III: $x \in (\frac{π}{2}, \frac{3 π}{4})$

$[\sin 2 x] = - 1$

$[\cos 2 x] = - 1$

$y = f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]}$

$\Rightarrow y = \sqrt{(- 1) - (- 1)}$

$\Rightarrow y = 0$

Case IV: $x \in (\frac{3 π}{4}, π)$

$[\sin 2 x] = - 1$

$[\cos 2 x] = 0$

$[\sin 2 x] - [\cos 2 x] < 0$

So, $x \in (\frac{3 π}{4}, π)$ is not in domain of $f (x)$ .

Case V: $x = 0, π$

$[\sin 2 x] = 0$

$[\cos 2 x] = 1$

$[\sin 2 x] - [\cos 2 x] < 0$

So, $x = 0, 2 π$ are not in domain of $f (x)$ .

Case VI: $x = \frac{π}{4}, \frac{3 π}{4}$

At $x = \frac{π}{4}, [\sin 2 x] = 1$

and $[\cos 2 x] = 0$

So, $y = f (x) = \sqrt{1 - 0}$

$\Rightarrow y = 1$

At $x = \frac{3 π}{4}, [\sin 2 x] = - 1$

and $[\cos 2 x] = 0$

So, $x = \frac{3 π}{4}$ is not in the domain.

Combining all cases,

$y \in \{0, 1\}$

Find the range of $f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]},$ where $[\cdot]$ denotes the greatest integer function.

Important Questions on Functions

Learn and Prepare for any exam you want !

Find the range of fx=sin2x-cos2x, where · denotes the greatest integer function.

Important Questions on Functions

Learn and Prepare for any exam you want !

Find the range of $f (x) = \sqrt{[\sin 2 x] - [\cos 2 x]},$ where $[\cdot]$ denotes the greatest integer function.