Question

Salut, ma puteti ajuta la problema numarul 8...?

Answer 1

$f = X^{100}+aX^{99}+bX+1 \\ \\ f(X) = (X^3-X^2-X+1)\cdot Q(X)+X^2+X+1\\\\X^3-X^2-X+1 = X^2(X-1)-(X-1) = (X-1)(X^2-1) = \\ = (X-1)^2(X+1)\\ \\ f(X) = (X-1)^2(X+1)\cdot Q(X)+X^2+X+1\\ \\ f(1) = 0+1^2+1+1 \Rightarrow 1+a+b+1 = 3 \Rightarrow a+b = 1 \\ f(-1) = (-1)^2-1+1 \Rightarrow 1-a-b+1 = 1 \Rightarrow a+b = 1\\ \\ \text{Derivam relatia teoremei impartirii cu rest:}\\ \\ f'(X) = [(X-1)^2]'(X+1)Q(X)+(X-1)^2(X+1)'Q(X)+\\+(X-1)^2(X+1)Q'(X)+2X+1$

$f'(X) = \\ =2(X-1)(X+1)Q(X)+(X-1)^2Q(X)+(X-1)^2(X+1)Q'(X)+\\+2X+1 \\ \\ f'(1) = 0+0+0+2+1 \Rightarrow 100+99a+b = 3\Rightarrow \\ \\ 99a+b = -97\\a+b = 1\\ \\ 98a = -98 \Rightarrow \boxed{a = -1}\text{ si } \boxed{b = 2}$