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JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

  • question_answer
    Consider the following statements:                       \[{{S}_{1}}\]: 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 (1, 2] State, in order, whether \[{{S}_{1}},{{S}_{2}},{{S}_{3}}\]are true or false.

    A)  FFT                       

    B)  TFT

    C)  TFF                                       

    D)  TTT

    Correct Answer: B

    Solution :

     \[{{S}_{1}}\]:if x2+ 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 \[a<0\] \[\Rightarrow \]\[a=-1,-2,-3,-4,-5\] \[{{S}_{2}}\]: x2 - (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. \[{{S}_{3}}\]: Case-1: 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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