The value of all values of a belong to R such that x^2+2ax+a=square root (a^2+x-1/16) -1/16 , x is >= -a has two distinct roots contains the set .

Answers (1)

Solution:   We have ,  

                   x^2+2ax+a=sqrta^2+x-frac116-frac116

           Rightarrow        x^2+2ax+frac116=sqrta^2+x-frac116-a

          Let             y=f(x)=x^2+2ax+frac116

          Rightarrow               y=x^2+2ax+frac116

          Rightarrow                x^2+2ax+frac116-y=0

          Rightarrow                   x=frac-2apm sqrt4a^2-4(frac116-y)2

          Rightarrow                   x=-apm sqrta^2-frac116+y

          Rightarrow                f ^-1(y)=x=-a+sqrta^2-frac116+y

         Rightarrow                 f^-1(x)=-a+sqrta^2-frac116+x

             ecause             f(x)=f^-1(x)   ,  f(x) is symmetrical about  y=x

             Hence ,      f(x)=x   Rightarrow      x^2+2ax+frac116=x

        Rightarrow                   x^2+x(2a-1)+frac116=0  

        Rightarrow              D> 0    Rightarrow   (2a-1)^2-frac14 > 0

        Rightarrow               (2a-1)^2> frac14        Rightarrow       (2a-1)> frac12   or (2a-1)< -frac12

        Rightarrow               a> frac34        or      a< frac14        Hence ,    ain (frac34,infty )                  

                    @hspace0.1cmjayant

 

 

 

 

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