Question

1/1-sin3x ??Domain and range??

Answer 1

Answer:

Hey there!!

__________

◆Given function is :

\begin{gathered}f(x) = \frac{1}{2 - sin3 x} \\\end{gathered}

f(x)=

2−sin3x

1

●For DOMAIN :

-----------------------

Function f(x) is not defined when

=> 2 - sin3x = 0

so , sin3x = 2 ---------(1)

but we know that range of sinx € [ -1, 1 ]

so maximum value of sin3x = 1

therefore sin3x ≠ 2 ( not possible)

=> 2 - sin3x ≠ 0 ------(2)

so from equation (2) we can se tha function is defined for all values of x

=> Domain € R

#FOR RANGE :

----------------------

\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered}

weknowthat

−1⩽sin(3x)⩽1

now multiplying by -1 we get,

=> 1 ≥ - sin3x ≥ -1

adding 2 we get,

=> 2+1 ≥ 2 - sin3x ≥ 2-1

=> 3 ≥ 2-sin3x ≥ 1

now taking inverse we get,

\frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1

3

1

⩽

2−sin(3x)

1

⩽1

so Range € [ ⅓ , 1 ]