| \[{{S}_{1}}:\]Number of integrals values of \['a'\] for which the roots of the equation \[{{x}^{2}}+ax+7=0\] are imaginary with positive real parts is \[5\]. |
| \[{{S}_{2}}:\]Let \[\alpha ,\,\,\beta \] are roots \[{{x}^{2}}-(a+3)x+5=0\] and \[\alpha ,\,\,a,\,\,\beta \] are in \[A.P.\] then roots are \[2\] and \[5/2\] |
| \[{{S}_{3}}:\] Solution set of \[{{\log }_{x}}(2+x)\le {{\log }_{x}}(6-x)\]is \[\text{is}\]\[(1,\,\,2]\] |
A) \[FFT\]
B) \[TFT\]
C) \[TFF\]
D) \[TTT\]
Correct Answer: B
Solution :
\[{{S}_{1}}:\]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\[a<0\] \[\Rightarrow \] \[a=-1,\,\,-2,\,\,-3,\,\,-4,\,\,-5\] \[{{S}_{2}}:\]\[{{x}^{2}}-(a+3)x+5=0\]has roots\[\alpha ,\,\,a,\,\,\beta \] If\[\alpha ,\,\,a,\,\,\beta \]are in\[A.P.\]then \[2a=\alpha +\beta \Rightarrow 2a=a+3\Rightarrow a=3\] The equation becomes \[{{x}^{2}}-6x+5=0\] which has roots \[1\] and \[5\]. \[{{S}_{3}}:\]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\[x\le 2\]\[\therefore \,\,x\in (1,\,\,2]\]
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