Consider the two functions f(x)=x^2+2bx+1 and g(x) =2a(x+b), where the varible x and the constants a and b are real numbers. Each such pair of the constants a and b may be considered as a point (a,b) in an ab-plane. Let S be the set ......?

Consider the two functions $f \left(x\right) = {x}^{2} + 2 b x + 1$ and $g \left(x\right) = 2 a \left(x + b\right)$, where the varible x and the constants a and b are real numbers. Each such pair of the constants a and b may be considered as a point (a,b) in an ab-plane. Let S be the set of such points (a,b) for which the graaphs of $y = f \left(x\right) \mathmr{and}$y=g(x)# do not intersect (in the xy - plane). The area of S is?

Mar 6, 2017

$\pi$

Explanation:

Solving for $x$

${x}^{2} + 2 b x + 1 = 2 a \left(x + b\right)$ we get

$x = a - b \pm \sqrt{{a}^{2} + {b}^{2} - 1}$

There is not intersection for the values $a , b$ such that

${a}^{2} + {b}^{2} - 1 < 0$ defining $S$ as the interior of circle ${a}^{2} + {b}^{2} = 1$

Thus area $S$ is $\pi$