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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    The number of integral values of m for which the quadratic expression, \[(1+2m){{x}^{2}}-2\]\[(1+3m)x+4(1+m),x\in R,\]is always positive is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 8                                             

    B) 7    

    C) 3                                 

    D)   6

    Correct Answer: B

    Solution :

    The given quadratic expression\[(1+2m){{x}^{2}}-2(1+3m)x+4(1+m),\]\[x\in R\]is always positive if \[1+2m>0\Rightarrow m>\frac{-1}{2}\]                           ...(i) And, D < 0 \[\Rightarrow \]\[4{{(1+3m)}^{2}}-4(1+2m)\times 4(1+m)<0\] \[\Rightarrow \]\[(1+9{{m}^{2}}+6m)-4(1+3m+2{{m}^{2}})<0\] \[\Rightarrow \]\[{{m}^{2}}-6m-3<0\] \[\Rightarrow \]\[3-\sqrt{12}<m<3+\sqrt{12}\]                  ?(ii) From (i) and (ii), common interval is\[3-\sqrt{12}<m<3A+\sqrt{12}\] So, integral values of m are 0, 1, 2, 3,4, 5, 6.


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