A) \[[1,2]\]
B) \[\left[ \frac{3}{2},\infty \right)\]
C) \[\left[ \frac{3}{2},2 \right]\]
D) None of these
Correct Answer: C
Solution :
\[-1\le {{x}^{2}}-3x+3\le 1\] |
\[\Rightarrow {{x}^{2}}-3x+4\ge 0\] and \[{{x}^{2}}-3x+2\ge 0\] |
\[\Rightarrow x\in [1,2]\] |
\[f'(x)=\frac{2x-3}{\sqrt{1-{{({{x}^{2}}-3x+3)}^{2}}}}\ge 0\Rightarrow x\ge \frac{3}{2}\] |
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