A) \[(-\infty ,\ \infty )\]
B) \[(-1,\ 1)\]
C) \[\left[ -\frac{3}{2},\ 0 \right]\]
D) \[\left( -\infty ,\ \frac{-1}{2} \right)\cup (2,\ \infty )\]
Correct Answer: C
Solution :
\[-1\le 1+3x+2{{x}^{2}}\le 1\] Case I : \[2{{x}^{2}}+3x+1\ge -1\]; \[2{{x}^{2}}+3x+2\ge 0\] \[x=\frac{-3\pm \sqrt{9-16}}{6}\] \[=\frac{-3\pm i\sqrt{7}}{6}\] (imaginary). Case II : \[2{{x}^{2}}+3x+1\le 1\] Þ \[2{{x}^{2}}+3x\le 0\] Þ \[2x\,\left( x+\frac{3}{2} \right)\le 0\] Þ \[\frac{-3}{2}\le x\le 0\] Þ \[x\in \left[ -\frac{3}{2},\,\,0 \right]\] In case I, we get imaginary value hence, rejected \ Domain of function = \[\left[ \frac{-3}{2},\,0 \right]\].
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