A) \[1\le b\le 2\]
B) \[b=\{1,2\}\]
C) \[b\in \left( -\infty ,-1 \right)\]
D) None of these
Correct Answer: D
Solution :
[d] for \[x\le 1\] \[f'\left( x \right)=3{{x}^{2}}-2x+10=3\left[ {{\left( x-\frac{1}{3} \right)}^{2}}+\frac{24}{3} \right]>0\] \[\therefore \text{ }f\left( x \right)\text{ }is\uparrow f{{x}^{n}}\] For \[x>1,\text{ }f'\left( x \right)<0\] \[\therefore \text{ }f\left( x \right)\] has greatest value at x = 1 \[\underset{x\to {{1}^{+}}}{\mathop{lim}}\,f\left( x \right)\le f\left( 1 \right)\] \[\Rightarrow -2+lo{{g}_{2}}\left( {{b}^{2}}-2 \right)\le 5\] \[lo{{g}_{2}}\left( {{b}^{2}}-2 \right)\le 7\] \[{{b}^{2}}\le 130\text{ }but\text{ }{{b}^{2}}>2\] \[2\le {{b}^{2}}\le 130\] \[b\in \left[ \left( -\sqrt{\text{1}30},-\sqrt{2} \right)\cup \left( \sqrt{2},\sqrt{\text{1}30} \right) \right]\]
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