Question

find the range of following functions :
$\left(i\right)f\left(x\right)=\frac{1}{8-3\sin x}$

Answer 1

We have,

$f\left(x\right)=\frac{1}{8-3\sin x}$ …… (1)

Therefore,

$8-3\sin x\ne 0$

$f\left(x\right)$ is defined for all real value $x$.

Which means,

Domain of $f\left(x\right)$ is real number $R$.(Set of all real numbers.)

Now, from equation (1),

We have,

$8-3\sin x=\frac{1}{f\left(x\right)}$

$\sin x=\frac{8}{3}-\frac{1}{3f\left(x\right)}$

Now, we know that,

$\sin x\in (1,-1)$

So,

$\frac{8}{3}-\frac{1}{3f\left(x\right)}\in (-1,1)$

$\Rightarrow -\frac{1}{3f\left(x\right)}\in (-\frac{11}{3},\frac{-5}{3})$

$\Rightarrow \frac{1}{3f\left(x\right)}\in (\frac{11}{3},\frac{5}{3})$

$\Rightarrow \frac{1}{f\left(x\right)}\in (11,5)$

$\Rightarrow f\left(x\right)\in (\frac{1}{11},\frac{1}{5})$

Hence, the range of this function is $(\frac{1}{11},\frac{1}{5})$

Hence, this is the answer.