A) \[(-1, 1)\]
B) \[(-\infty ,\infty )\]
C) \[\left[ -\frac{3}{2},0 \right]\]
D) \[\left( -\infty ,-\frac{1}{2} \right)\cup (2,\infty )\]
Correct Answer: C
Solution :
[c] : The given function is \[f(x)=si{{n}^{-1}}\left( 2{{x}^{2}}+3x+1 \right)\] f(x) will be defined when \[-1\le 2{{x}^{2}}+3x+1\le 1\] \[\Rightarrow -\frac{1}{2}\le {{x}^{2}}+\frac{3}{2}x+\frac{1}{2}\le \frac{1}{2}\] \[\Rightarrow -\frac{1}{2}\le {{(x)}^{2}}+2x\frac{3}{4}+{{\left( \frac{3}{4} \right)}^{2}}+\frac{1}{2}-\frac{9}{16}\le \frac{1}{2}\] \[\Rightarrow -\frac{1}{2}\le {{\left( x+\frac{3}{4} \right)}^{2}}-\frac{1}{16}\le \frac{1}{2}\] \[\Rightarrow -\frac{1}{2}+\frac{1}{16}\le {{\left( x+\frac{3}{4} \right)}^{2}}\le \frac{1}{2}+\frac{1}{16}\] \[\Rightarrow -\frac{7}{16}\le {{\left( x+\frac{3}{4} \right)}^{2}}\le \frac{9}{16}\Rightarrow 0\le {{\left( x+\frac{3}{4} \right)}^{2}}\le \frac{9}{16}\] \[\Rightarrow -\frac{3}{4}\le \left( x+\frac{3}{4} \right)\le \frac{3}{4}\Rightarrow -\frac{3}{2}\le x\le 0\]
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