A) \[K=3\]
B) \[0\le K<3\]
C) \[K\le 4\]
D) \[K=0\]
Correct Answer: A
Solution :
[a] Let \[\int\limits_{1}^{2}{\left\{ {{K}^{2}}+\left( 4-4K \right)x+4{{x}^{3}} \right\}dx\le 12}\] \[\Rightarrow {{K}^{2}}x+\left. \frac{(4-4K){{x}^{2}}}{2}+\frac{4{{x}^{4}}}{4} \right|_{1}^{2}\le 12\] \[\Rightarrow [2{{K}^{2}}+(2-2K)(4)+16]-[{{K}^{2}}+(2-2K)\] \[+1]\le 12\] \[\Rightarrow (2{{K}^{2}}+8-8K+16)-({{K}^{2}}-2K+3)\le 12\] \[\Rightarrow {{K}^{2}}-6K+21\le 12\] \[\Rightarrow {{K}^{2}}-6K+9\le 0\Rightarrow {{(K-3)}^{2}}\le 0\] \[\Rightarrow K=3\]
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