A) 1
B) 5
C) 10
D) both [A] and [C]
Correct Answer: D
Solution :
\[x\in [-1,0]\] \[x+\frac{1+{{x}^{2}}}{2}=-2x\] \[{{x}^{2}}+6x+1=0\] \[x=2\sqrt{2}-3\Rightarrow |10a|=[|20\sqrt{2}-30|]\]\[=30-20\sqrt{2}\] \[x\in [0,1]\] \[x+\frac{1+{{x}^{2}}}{2}=2x\] \[1+{{x}^{2}}=2x\Rightarrow x=1\Rightarrow |10a|=10\] \[|10a|=10,|20\sqrt{2}-30|\] \[\Rightarrow [|10a|]=1,10\]You need to login to perform this action.
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