A) FFT
B) TFT
C) TFF
D) TTT
Correct Answer: B
Solution :
\[{{\text{S}}_{\text{1}}}\]: if \[{{x}^{2}}+ax+7=0\]has imaginary roots with positive real parts then D < 0 and sum of roots > 0 \[\Rightarrow \]\[{{a}^{2}}-28<0\]and\[-a>0\] \[\Rightarrow \]\[-\sqrt{28}<a<\sqrt{28}\]and a < 0 \[\Rightarrow \]\[a=-1,-2,3,-4-5\] \[{{\text{S}}_{\text{2}}}\]:\[{{x}^{2}}-(a+3)x+5=0\]has roots\[\alpha ,a,\beta \]If \[\alpha ,a,\beta \]are in AP. Then \[2a=\alpha +\beta \Rightarrow 2a=a+3\Rightarrow a=3\] The equation becomes\[{{x}^{2}}-6x+5=0\]which has roots 1 and 5. \[{{\text{S}}_{\text{3}}}\]: Case-I: If \[0<x<1,\]then \[2+x\ge 6-x>0\Rightarrow 2x\ge 4\]and\[x<6\Rightarrow x\ge 2\]and\[x<6\Rightarrow x\in [2,6)\] \[\therefore x\in (0,1)\cap [2,6)=\phi ,\therefore x\in \phi \] Case II: If\[x>1,\]then\[0<2+x\le 6-x\] \[\Rightarrow x>-2\]and\[x\le 2,\] \[\therefore x\in (1.2]\]You need to login to perform this action.
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