A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{3}\]
D) None of these
Correct Answer: C
Solution :
Since \[\mathbf{c}=(x-2)\mathbf{a}+\mathbf{b}\] and \[\mathbf{d}=(2x+1)\mathbf{a}-\mathbf{b}\] are collinear, therefore \[\mathbf{c}=\lambda \mathbf{d}\] \[\Rightarrow (x-2)\mathbf{a}+\mathbf{b}=\lambda (2x+1)\mathbf{a}-\lambda \mathbf{b}\] or \[[(x-2)-\lambda (2x+1)]\mathbf{a}+(\lambda +1)\mathbf{b}=0\] \[(x-2)-\lambda (2x+1)=0,\lambda +1=0\] \[(\because \,\,\,\mathbf{a},\,\mathbf{b}\] are linearly independent) \[\Rightarrow x-2+2x+1=0\Rightarrow x=\frac{1}{3}.\]
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