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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 17

  • question_answer
    Let a and b be two arbitrary real numbers. The smallest natural number b for which the equation \[{{x}^{2}}+2(a+b)x+(a-b+8)=0\] has unequal real roots \[\forall \,\,a\in R\] is

    A) 4                     

    B)        5         

    C) 6                     

    D)        7

    Correct Answer: B

    Solution :

       [b] Equation \[{{x}^{2}}+2(a+b)x+(a-b+8)=0\] has unequal real roots \[\forall \,\,x\in R\] \[\Rightarrow \,\,D>0\,\,\forall a\in R\] \[\Rightarrow \,\,4{{(a+b)}^{2}}-4(a-b+8)>0,\,\,\forall \,a\in R\] \[\Rightarrow \,\,{{a}^{2}}+a(2b-1)+({{b}^{2}}+b-8)>0,\,\,\forall a\in R\] \[\Rightarrow \,\,D<0\] \[\Rightarrow \,\,{{(2b-1)}^{2}}-4({{b}^{2}}+b-8)<0\] \[\Rightarrow \,\,\,\,b>\frac{33}{8}\]        


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