JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 14)
The domain of the function $$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$$ is :
$$[1, \infty)$$
$$[-1,2]$$
$$[-1, \infty)$$
$$(-\infty, 2]$$
Explanation
$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$
$$ -1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 $$
$$ \begin{aligned} & \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\ & x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\ & 5 x \geq-5 \\\\ & x \geq-1 \end{aligned} $$
And $$ \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1 $$
$$ x^{2}-3 x+2 \geq-x^{2}-2 x-7 $$
$2 x^{2}-x+9 \geq 0$
$$ x \in R $$
(i) $\cap$ (ii)
Domain $\in[-1, \infty)$
$$ -1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 $$
$$ \begin{aligned} & \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\ & x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\ & 5 x \geq-5 \\\\ & x \geq-1 \end{aligned} $$
And $$ \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1 $$
$$ x^{2}-3 x+2 \geq-x^{2}-2 x-7 $$
$2 x^{2}-x+9 \geq 0$
$$ x \in R $$
(i) $\cap$ (ii)
Domain $\in[-1, \infty)$
Comments (0)
