A) \[-1\]
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
(b): \[2K\left( 2K-2 \right)+K\left( 2-3K \right)+1\left( 3K-2K \right)=0\] \[\Rightarrow \]\[4{{K}^{2}}-4K+2k-3{{K}^{2}}+3K-2K=0\] \[\Rightarrow \]\[{{K}^{2}}-K=0\] \[\Rightarrow \]\[K\left( K-1=0 \right)\] \[\Rightarrow \]\[K=0,1\]You need to login to perform this action.
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