What is the maximum value of (x + 1) + (10-2x) under y = 5 radical Use inequality to solve

What is the maximum value of (x + 1) + (10-2x) under y = 5 radical Use inequality to solve

From 1 ≤ x ≤ 5
Let x = 3 + 2cos α, 0 ≤ α ≤ π, 0 ≤ α / 2 ≤ π / 2
y=5√(3+2cosα-1)+√(10-2(3+2cosα))
=5√2(1+cosα)+√(4(1-cosα))
=5√4(cos²α/2)+√(8(sin²α/2))
=10cos(α/2)+2√2sin(α/2)
=2√27sin(α/2+β)≤2√27=6√3
So the function y = 5 × root (x-1) + root (10-2x)
The maximum is 6 √ 3

If x is a real number, then=_____________ The maximum value of X radical (1-2x ^ 2) is
The answer is 1 / 2 and 2 / 4, all three of them are wrong
Ask for detailed explanation

X is squared with (1-2x ^ 2) to get x ^ 2 (1-2x ^ 2), so when x = 1 / 2, the maximum value is root 2 / 4

Let a = {x | X & # 178; + BX + C = x} B = {x | (x-1) & # 178; + B (x-1) + C = x + 5}, if a = {2}, find B

A = {2} then x & # 178; + (B-1) x + C = (X-2) & # 178;, B = - 3, C = 4 (x-1) & # 178; + B (x-1) + C = x + 5 (x-1) & # 178; - 3 (x-1) + 4 = X-1 + 6 (x-1) & # 178; - 4 (x-1) + 4 = 6 (x-3) & # 178; = 6x = 3 + √ 6 or x = 3 - √ 6 set B = {3 + √ 6, 3 - √ 6}