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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • question_answer
    Consider the following statements : \[{{\text{S}}_{\text{1}}}\]: Number of integrals values of 'a' for which the roots of the equation x2 + ax + 7 = 0 are imaginary with positive real parts is 5. \[{{\text{S}}_{2}}\]: Let \[\alpha ,\beta \] are roots x2 - (a + 3) x + 5 = 0 and \[\alpha ,a,\beta \] pare in P. then roots are 2 and 5/2 \[{{\text{S}}_{\text{3}}}\]: Solution set of \[(2+x)\le {{\log }_{x}}(6-x)\]is (1, 2] State, in order, whether \[{{\text{S}}_{\text{1}}}\text{,}{{\text{S}}_{\text{2}}}\text{,}{{\text{S}}_{\text{3}}}\]are true or false.

    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\[x\le 2,\]                           \[\therefore x\in (1.2]\]


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