Let f (x) and G (x) be the even and odd functions of R respectively in the domain of definition, f (x) = 2F (x) + G (x) = MX ^ 2 + NX + 1, f (1) = 1, f (- 1) = 5, M, n Finding f (2) and G (2)

Let f (x) and G (x) be the even and odd functions of R respectively in the domain of definition, f (x) = 2F (x) + G (x) = MX ^ 2 + NX + 1, f (1) = 1, f (- 1) = 5, M, n Finding f (2) and G (2)

From the meaning of the title: M + N + 1 = 1
m-n+1=5
The solution is m = 2, n = - 2
Even function does not contain odd term, odd function does not contain even term and constant term;
So: 2F (x) = MX & # 178; + 1 = 2x & # 178; + 1, G (x) = NX = - 2x
So: F (x) = x & # 178; + 1 / 2, G (x) = - 2x
So: F (2) = 9 / 2, G (2) = - 4
If you don't understand, please hi me,

If the function f (x) = (m-1) x ^ 5 + (m ^ 2-3m + 2) x ^ 3 + MX ^ 2 + 1 is even, then f (2) =?

∵ f (x) = (m-1) x ^ 5 + (m ^ 2-3m + 2) x ^ 3 + MX ^ 2 + 1 is even function
Ψ M-1 = 0 and m ^ 2-3m + 2 = 0
∴m=1
∴f(x)=x^2+1
∴f(2)=5

If the even function f (x) defined on R satisfies f (x + 2) = f (x), and if x ∈ [0,1], f (x) = x & # 178;,
Then the number of zeros of the function y = f (x) - log with base 5 and true number | X-1 | is

-11 has five zeros
Similarly, X