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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 .

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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 , $ain (frac34,infty )$

$@hspace0.1cmjayant$

Posted by

Deependra Verma

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