Mathematics - Functions Question with Solution | TestHub

Given,

$f\left(x\right)={\sin }^{-1}\left(\frac{x^2-3x+2}{x^2+2x+7}\right)$

We know that domain of ${\sin }^{-1}\left(f\left(x\right)\right)$ is $-1\le f\left(x\right)\le 1$

So, $\frac{x^2-3x+2}{x^2+2x+7}\ge -1$ and $\frac{x^2-3x+2}{x^2+2x+7}\le 1$

Now solving $\frac{x^2-3x+2}{x^2+2x+7}\ge -1$

Since denominator $x^2+2x+7$ is always positive,

So, $x^2-3x+2\ge -\left(x^2+2x+7\right)$

$\Rightarrow 2x^2-x+9\ge 0$

$\Rightarrow x\in R .....\left(1\right)$

Now solving $\frac{x^2-3x+2}{x^2+2x+7}\le 1$

$\Rightarrow x^2-3x+2\le x^2+2x+7$

$\Rightarrow 5x\ge -5$

$\Rightarrow x\ge -1 .....\left(2\right)$

So from equation  $\left(1\right) \& \left(2\right)$ we get domain $x\in [-1,\infty )$