Functions Question 570
Question: Domain of the function $ f(x)={{\sin }^{-1}}(1+3x+2x^{2}) $ is
[Roorkee 2000]
Options:
A) $ (-\infty ,\ \infty ) $
B) $ (-1,\ 1) $
C) $ [ -\frac{3}{2},\ 0 ] $
D) $ ( -\infty ,\ \frac{-1}{2} )\cup (2,\ \infty ) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ -1\le 1+3x+2x^{2}\le 1 $ Case I : $ 2x^{2}+3x+1\ge -1 $ ; $ 2x^{2}+3x+2\ge 0 $ $ x=\frac{-3\pm \sqrt{9-16}}{6} $ $ =\frac{-3\pm i\sqrt{7}}{6} $ (imaginary). Case II : $ 2x^{2}+3x+1\le 1 $
Þ $ 2x^{2}+3x\le 0 $
Þ $ 2x,( x+\frac{3}{2} )\le 0 $
Þ $ \frac{-3}{2}\le x\le 0 $
Þ $ x\in [ -\frac{3}{2},0 ] $ In case I, we get imaginary value hence, rejected \ Domain of function = $ [ \frac{-3}{2},,0 ] $ .
BETA
