Question

The range of the function f (x) = 2 - sin 3x

Answer 1

●For DOMAIN :

●For DOMAIN :Function f(x) is not defined when

●For DOMAIN :Function f(x) is not defined when=> 2 - sin3x = 0

●For DOMAIN :Function f(x) is not defined when=> 2 - sin3x = 0so , sin3x = 2 ---------(1)

●For DOMAIN :Function f(x) is not defined when=> 2 - sin3x = 0so , sin3x = 2 ---------(1)but we know that range of sinx € [ -1, 1 ]

t we know that range of sinx € [ -1, 1 ]so maximum value of sin3x = 1

t we know that range of sinx € [ -1, 1 ]so maximum value of sin3x = 1threfore sin3x ≠ 2 ( not possible)

t we know that range of sinx € [ -1, 1 ]so maximum value of sin3x = 1threfore sin3x ≠ 2 ( not possible)=> 2 - sin3x ≠ 0 ------(2)

t we know that range of sinx € [ -1, 1 ]so maximum value of sin3x = 1threfore sin3x ≠ 2 ( not possible)=> 2 - sin3x ≠ 0 ------(2)so from equation (2) we can se tha function is defined for all values of x

t we know that range of sinx € [ -1, 1 ]so maximum value of sin3x = 1threfore 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 :

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

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

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1=> 3 ≥ 2-sin3x ≥ 1

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1=> 3 ≥ 2-sin3x ≥ 1now taking inverse we get,

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1=> 3 ≥ 2-sin3x ≥ 1now taking inverse we get,\frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1=> 3 ≥ 2-sin3x ≥ 1now taking inverse we get,\frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1 3

#FOR RANGE :\begin{gathered}we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\\end{gathered} weknowthat−1⩽sin(3x)⩽1now multiplying by -1 we get,=> 1 ≥ - sin3x ≥ -1adding 2 we get,=> 2+1 ≥ 2 - sin3x ≥ 2-1=> 3 ≥ 2-sin3x ≥ 1now taking inverse we get,\frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1 31

⩽

⩽ 2−sin(3x)

⩽ 2−sin(3x)1

⩽ 2−sin(3x)1 ⩽1

⩽ 2−sin(3x)1 ⩽1so Range € [ ⅓ , 1 ]