SAT1000 - P878

Algebra Level pending

Given the function y = f ( x ) ( x R ) y=f(x)\ (x \in \mathbb R) , for function y = g ( x ) ( x I ) y=g(x)\ (x \in I) , let's define symmetric function of g ( x ) g(x) respect to f ( x ) f(x) as y = h ( x ) ( x I ) y=h(x)\ (x \in I) , y = h ( x ) y=h(x) is such that x I \forall x \in I , point ( x , h ( x ) ) , ( x , g ( x ) ) (x,h(x)), (x,g(x)) are symmetric about point ( x , f ( x ) ) (x,f(x)) .

GIven that h ( x ) h(x) is the symmetric function of g ( x ) = 4 x 2 g(x)=\sqrt{4-x^2} respect to f ( x ) = 3 x + b ( b R ) f(x)=3x+b\ (b \in \mathbb R) , h ( x ) > g ( x ) h(x)>g(x) is always true for all x x on the domain of g ( x ) g(x) , then find the range of b b .

The range can be expressed as ( L , + ) (L,+\infty) , submit L 2 L^2 .


Have a look at my problem set: SAT 1000 problems


The answer is 40.

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1 solution

We have 2 b 10 2 > 2 \dfrac {2b}{\sqrt {10}}-2>2

b > 2 10 \implies b>2\sqrt {10} . So, L = 2 10 L 2 = 40 L=2\sqrt {10}\implies L^2=\boxed {40} .

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