Let the range of the function f(x) = 1 ^2 x + 4 x x + 5 ^2 x be [a, b] . The quadratic equation whose roots are a and b is

Step-by-step solution

The given function is f(x) = 1 ^2 x + 4 x x + 5 ^2 x . First, we simplify the denominator using double-angle identities: ^2 x + 5 ^2 x + 4 x x = 1 - 2x 2 + 5 ( 1 + 2x 2 ) + 2(2 x x) = 1 - 2x + 5 + 5 2x 2 + 2 2x = 6 + 4 2x 2 + 2 2x = 3 + 2 2x + 2 2x The expression 2 2x + 2 2x is of the form A + B , which has a range of [- A^2+B^2 , A^2+B^2 ] . Here, the range is [- 2^2+2^2 , 2^2+2^2 ] = [-2 2 , 2 2 ] . Thus, the range of the denominator is [3 - 2 2 , 3 + 2 2 ] . Since 3 - 2 2 > 0 , the range of f(x) is the reciproca