Page 30 - RMGO 4

30 Rezolvarea problemelor din num˘arul anterior

1 3 1 (3 − 2a) + 8a 3
3
Z
2
2
Dar (3x−2a) dx = (3x − 2a) = = 4a −6a+3 s , i, integrˆand
0 9 0 9
Z 1 1 Z 1
00
0
0
0
prin p˘art , i, (3x−2a)f (x) dx = (3x−2a)f (x) − 3f (x) dx = (3−2a)f (1)+

0 0 0
0
3
2
2
2
2af (0) − 3 (f(1) − f(0)) = (3 − 2a)(a − 4a + 3) + 2a − 3(a − 1) = 8a − 12a + 6,
1 2
Z
2
2
2
00
deci ˆınlocuind ˆın (1) obt , inem c˘a (4a − 6a + 3) f (x) dx ≥ 4(4a − 6a + 3) .
0
ˆ Imp˘art , ind prin 4a − 6a + 3 > 0 rezult˘a inegalitatea din enunt , .
2
2
Remarc˘am c˘a egalitatea are loc pentru f(x) = x(x − a) .