A) \[a\in (-\infty ,\,\,2)\]
B) \[a\in (2,\,\,\infty )\]
C) \[a\in (-\infty ,\,\,1)\]
D) \[a\in (1,\,\,\infty )\]
Correct Answer: D
Solution :
\[\mathbf{a}\cdot \mathbf{b}=[x\mathbf{i}+(x-1)\mathbf{j}+\mathbf{k}]\cdot [(cx+1)\mathbf{i}+\mathbf{j}+a\mathbf{k}]\] \[=x(x+1)+x-1+a={{x}^{2}}+2x+a-1\] We must have\[\mathbf{a}\cdot \mathbf{b}>0,\,\,\forall x\in R\] \[\Rightarrow \] \[{{x}^{2}}+2x+a-1>0,\,\,\forall x\in R\] \[\Rightarrow \] \[4-4(a-1)<0\] \[\Rightarrow \] \[a>2\]You need to login to perform this action.
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