About the function, just answer the fourth question

About the function, just answer the fourth question

① If ∠ CPQ = 90 °, then PQ ‖ ob, then AP / AQ = AO / AB, but AP / AQ = 1 / 2, AO / AB = 3 / 5, ≠ CPQ ≠ 90 °, ② if ∠ PCQ = 90 °, then point q is on the extension line of AB, which does not conform to the meaning of the problem. ③ if ∠ PQC = 90 ° after T seconds, make QH ⊥ CP on H, ∫ AQ = 2T, ∧ QH = 8t / 5, ah = 6t / 5, ∧ P (6-T, 0) Q (6-6t / 5

It is known that a and B are two real roots of the equation (X-2) (x-m +) = (P-2) (P-M) about X
(1) Finding the value of a and B
(2) If a and B are the lengths of the two right sides of a right triangle, ask what conditions the real numbers m and P satisfy, then the area of the right triangle is the largest, and find out its maximum
Dear brothers and sisters, I'm stuck in this question
Sorry, that equation is wrong
Is (X-2) (x-m) = (P-2) (P-M)

(1) From (X-2) (x-m) = (P-2) (P-M), we can get: (X-P) * (x-m-2 + P) = 0, so x = P, or x = m + 2-P, that is, a = P, B = m + 2-P or a = m + 2-P, B = P (2) s = P * (M + 2-P) / 2 = - P & sup2 / / 2 + (M + 2) * P / 2 = - [P - (M + 2) / 2] & sup2 / / 2 + (M + 2) & sup2 / / 8

Given that the function f (x) defined on positive integer set satisfies the following conditions: F (1) = 2, f (2) = - 2, f (n + 2) = f (n + 1) - f (n), then the value of F (2011) is?

From the known easy to find f (3) = - 4, f (4) = - 2, f (5) = 2, f (6) = 4, f (7) = 2, f (8) = - 2
So f (n) is a function with period 6
So f (2011) = f (6 * 335 + 1) = f (1) = 2

The function f (x) defined on a positive integer set satisfies f (1) = 2011 and f (1) + F (2) + +F (n) = n square f (n) (n is greater than or equal to 1) find f (2011)=

From F (1) + F (2) + +f(n-1)+f(n)=n^2 × f(n)
So f (1) + F (2) + +f(n-1) =(n-1)^2 × f(n-1)
The two formulas can be obtained by subtracting
f(n)=n^2 × f(n) - (n-1)^2 × f(n-1)
That is, (n ^ 2-1) × f (n) = (n-1) ^ 2 × f (n-1)
So f (n) = (n-1) ^ 2 × f (n-1) / (n ^ 2-1)
= f(n-1) (n-1) / (n+1)
Therefore, it is concluded that
f(2011)=f(2010)× 2010/2012
=f(2009)× (2010×2009)/(2012×2011)
……
=f(1)×(2010×2009×2008…… ×3×2×1)/(2012×2011×2010×…… ×5×4×3)
=f(1)×2/(2012×2011)
=2011×2/(2012×2011)
=1/1006