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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    If \[{{e}^{\lambda }}+1,{{e}^{-\lambda }}+1\] are the roots of the equation \[{{x}^{2}}-2(p+1)x+5p-{{p}^{2}}=0\] where \[X,p\in R\] then number of integral values of 'p' is

    A)  2      

    B)  3

    C)  4                                

    D)  infinite

    Correct Answer: A

    Solution :

    \[D\ge 0\] \[4{{(p+1)}^{4}}-4(5p-{{p}^{2}})\ge 0\] \[2{{p}^{2}}-3p+1\ge 0\Rightarrow (2p-1)(p-1)\ge 0\] \[\therefore p\in (-\,\infty ,\,1/2)\cup [1,\infty )\]            ?(1) \[\alpha +\beta \ge \,\Rightarrow \,2(p+1)\ge 4\Rightarrow p\ge 1\]                ?(2) \[\alpha \beta \ge 4\Rightarrow 5p-{{p}^{2}}\ge 4\Rightarrow {{p}^{2}}-5p+4\le 0\] \[\therefore P\in [1,4]\]                                     ?(3) Since, sum of the roots \[({{e}^{\lambda }}+1)+({{e}^{-\lambda }}+1)\] Also product of the roots \[({{e}^{\lambda }}+1)\times ({{e}^{-\lambda }}+1)\] Hence, sum is equal to product. \[\Rightarrow 2p+2=5-{{p}^{2}}\Rightarrow {{p}^{2}}-3p+2=0\] \[\therefore p=1\,\,\text{or}\,\,2\] hence, number of integral values of p is 2.                              


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