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 ......?

Question

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 of such points (a,b) for which the graaphs of #y=f(x) and #y=g(x)# do not intersect (in the xy - plane). The area of S is?

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Answer 1 · 2017-03-06T18:34:41

#pi#

Solving for #x#

#x^2 + 2 b x + 1 = 2 a (x + b)# 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#