Question

Domain of the function $f\left(x\right)={\sin }^{-1}(1+3x+2x^2)$ is

• A. $(-\infty ,\infty )$
• B. $(-1,1)$
• C. $[-\frac{3}{2},0]$
• D. $(-\infty ,\frac{-1}{2})\cup (2,\infty )$

Answer 1

Answer: (C)

$[-\frac{3}{2},0]$

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

as ${\sin }^{-1}x$ lies between $(-1,11)$

$-1\le 2x^2+3x+1\le 1$

solving the inequalities

$-1\le 2x^2+3x+1$ and

$2x^2+3x+1\le 1$

common solution of the inequalities

$-1\le 2x^2+3x+1$

$2x^2+3x+2\ge 0$ $⟶\left(1\right)$

Now, $2x^2+3x+1\le 1$ $⟶\left(2\right)$

$2x^2+3x\le 0$

$x(2x+3)\le 0$

$\Rightarrow x=0,$ $x=\frac{-3}{2}$

on solving $\left(1\right)$

$\Rightarrow 2x^2+3x+2=0$

$\Rightarrow a=2$ $b=3$ $c=2$

$\Rightarrow D=b^2-4ac=9-16=-7$

since D is negative so it is discarded

$\therefore x=0,\frac{-3}{2}$

so the internal is $[\frac{-3}{2},0]$

Hence, the answer is $[\frac{-3}{2},0].$