Question

find the domain and the range of 1/(4-cos3x)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x)=\dfrac{1}{4-cos(3x)} }$

The given function will be defined, iff $\tt{4-cos(3x)\ne0}$

$\tt{\implies\,cos(3x)\ne4}$

This is always true, because $\rm{cos(\theta)\in[-1,1]}$

So,

$\sf{\large{\bold{\green{dom(f)\in\,\mathbb{R}}}}}$

Now, let $\sf{y=\dfrac{1}{4-cos(3x)}}$

$\sf{\implies4-cos(3x)=\dfrac{1}{y}}$

$\sf{\implies\,cos(3x)=4-\dfrac{1}{y}}$

$\sf{\implies\,3x=cos^{-1}\bigg(4-\dfrac{1}{y}\bigg)}$

According to the definition of an inverse trigonometric function,

$\sf{-1\le4-\dfrac{1}{y}\le1}$

$\sf{\implies-1-4\le-\dfrac{1}{y}\le1-4}$

$\sf{\implies-5\le-\dfrac{1}{y}\le-3}$

$\sf{\implies5\ge\dfrac{1}{y}\ge3}$

$\sf{\implies\dfrac{1}{5} \le\,y\le\dfrac{1}{3} }$

Hence,

$\sf{\large{\bold{\green{range(f)\in\,\bigg[\dfrac{1}{5},\dfrac{1}{3} \bigg]}}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{4 - cos3x}$

Domain of function is defined as the set of those values of x for which function is well defined.

We know,

$\rm :\longmapsto\: - 1 \leqslant cos3x \leqslant 1$

$\bf\implies \:4 - cos3x \ne \: 0$

$\bf\implies \:x \: \in \: R$

Now, To find Range

We know,

$\rm :\longmapsto\: - 1 \leqslant cos3x \leqslant 1$

$\rm :\implies\: - 1 \leqslant - cos3x \leqslant 1$

Adding 4 in each term

$\rm :\implies\: 4- 1 \leqslant 4 - cos3x \leqslant 1 + 4$

$\rm :\implies\: 3\leqslant 4 - cos3x \leqslant 5$

$\bf\implies \:\dfrac{1}{5} \leqslant \dfrac{1}{4 - cos3x} \leqslant \dfrac{1}{3}$

$\bf\implies \:\dfrac{1}{5} \leqslant f(x) \leqslant \dfrac{1}{3}$

$\bf\implies \:f(x) = \bigg[\dfrac{1}{5}, \: \dfrac{1}{3} \bigg]$

Additional Information :-

$\boxed{ \tt{ \: |x| < y \: \: \tt\implies\: - y < x < y \: \: }}$

$\boxed{ \tt{ \: |x| \leqslant y \: \: \tt\implies\: - y \leqslant x \leqslant y \: \: }}$

$\boxed{ \tt{ \: |x - z| \leqslant y \: \: \tt\implies\: z- y \leqslant x \leqslant y + z\: \: }}$

$\boxed{ \tt{ \: |x| \geqslant y \: \: \tt\implies\: x \leqslant - y \: \: or \: x \geqslant y\: \: }}$

$\boxed{ \tt{ \: |x - z| \geqslant y \: \: \tt\implies\: x \leqslant z- y \: \: or \: x \geqslant \: z + y\: \: }}$

$\boxed{ \tt{ \: a > b \: \tt \: \implies \: \frac{1}{a} < \frac{1}{b} \: \: }}$

$\boxed{ \tt{ \: a \geqslant b \: \tt \: \implies \: \frac{1}{a} \leqslant \frac{1}{b} \: \: }}$