A) no value of a
B) more than two values of a
C) exactly one value of a
D) exactly two values of a
Correct Answer: A
Solution :
Given equation of lines are \[({{a}^{3}}+3)x+ay+a-3=0\]and \[({{a}^{5}}+2)x+(a+2)y+2a+3=0\](a real) Since point of intersection of lines lies ony-axis. \[\therefore \]Put x = 0 in each equation, we get\[ay+a-3=0\] and\[(a+2)y+2a+3=0\] On solving these we get \[(a+2)(a-3)-a(2a+3)=0\] \[\Rightarrow \]\[{{a}^{2}}-a-6-2{{a}^{2}}-3a=0\] \[\Rightarrow \]\[-{{a}^{2}}-4a-6=0\Rightarrow {{a}^{2}}+4a+6=0\] \[\Rightarrow \]\[a=\frac{-4\pm \sqrt{16-24}}{2}=\frac{=4\pm \sqrt{-8}}{2}\](not real) This shows that the point of intersection of the lines lies on the .y-axis for no value of 'a?
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