MATHS

1)sin² x + cosx + a = 0(1 - cos² x) + cosx + a = 0cos² x - cosx - (a+1) = 0For it has real root,
△ = 1 + 4(a+1) = 4a + 5 ≥ 0 , i.e. a ≥ - 5/4 ......(1)
and
0 ≤ cosx ≤ 1 for 0° ≤ x ≤ 90°So 0 ≤ product of roots ≤ 10 ≤ - (a+1) ≤ 1 - 1 ≤ (a+1) ≤ 0- 2 ≤ a ≤ - 1 ....(2)Combining (1) , (2) :- 5/4 ≤ a ≤ - 1
2)ax² + bx + c = 0Sum of roots = - b/a = odd/odd = odd , so one root is odd and the other root is even.Product of roots = c/a = odd/odd = odd ,
but odd root * even root is even.(Contradiction)So if a,b,c are odd numbers, ax^2+bx+c=0 has no integral roots.

2 If there is an integral root, then the root should be either odd or even.

Case 1 Odd

(odd)(odd)(odd) + (odd)(odd) + (odd) = odd

which is a contradiction since 0 is an even number

Case 2 Even

(odd)(even)(even) + (odd)(even) + (odd) = even + even + odd = odd

Again a contradiction

3 f(x) = x

For f(x) = x + n, where n is real
It satisfies the following expression:
f(x+k) = f(x) + k
L.H.S. = x+k+n
R.H.S. = x+n+k
L.H.S. = R.H.S.

2011-02-07 13:37:26 補充：
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