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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