range of (x-1)^½ + (5-x)^{1/2 ….................................................................................................}

Let y= √(x-1) + √(5-x).Now we will square both sides.y²= x-1 + 5-x + 2√(x-1)√(5-x)Y²= 4 + 2 √-(x²-6x+5)Here,z= x²-5x+6 is a quadratic expression whose range can be calculated by the formula {-D/4a} . We will get range of z ≥ -4And hence -z ≤ 4 ... But since -z cannot be negative.Then, -z € [0,4]Now y²= 4 + 2 √-z Putting maximum and minimum values of -z we will get,Y²= 4 + 2×0 =4Y=2Y²= 4 + 2×2 = 8Y= 2√2Hence y belongs to [2,2√2]