logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The domain of $y=sin^{-1}(1+3x+2x^2)$ is

$\begin{array}{1 1}(-\infty,\infty) \\ (-\infty,-\frac{3}{2}]\cup [0,\infty) \\ [-\frac{3}{2},0] \\(-\infty,-\frac{1}{2}]\cup [2,\infty) \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Domain of $y=sin^{-1}x\:\:is\:\:[-1,1]$
Ans- (C)
$-1\leq (1+3x+2x^2)\leq 1$
$\Rightarrow\:0\leq (2x^2+3x+2)\leq 2$
$\Rightarrow 2x^2+3x+2\geq 0\:\:and\:\:2x^2+3x\leq 0$
$\Rightarrow\:x\in R\:\:and\:\:x\in[-\frac{3}{2},0]$
$\Rightarrow\:$ domain $=[-\frac{3}{2},0]$
answered May 7, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...