Let f(x) = ₂ (a ^2 x + b x x + c ^2 x) be a function whose range is [1, 4] . Then the value of (a-c)^2 + b^2 is

Step-by-step solution

Let the inner function be g(x) = a ^2 x + b x x + c ^2 x . Using the double angle identities ^2 x = 1- 2x 2 , ^2 x = 1+ 2x 2 , and x x = 2x 2 , we can rewrite g(x) as: g(x) = a ( 1- 2x 2 ) + b ( 2x 2 ) + c ( 1+ 2x 2 ) g(x) = a+c 2 + b 2 2x + c-a 2 2x The expression A + B has a maximum value of A^2+B^2 and a minimum value of - A^2+B^2 . Thus, the maximum and minimum values of g(x) are: g_ max = a+c 2 + ( b 2 )^2 + ( c-a 2 )^2 = a+c 2 + 1 2 b^2 + (a-c)^2 g_ min = a+c 2 - 1 2 b^2 + (a-c)^2 The function f(x) = ₂(g(x))