If x 0 e(a - 1)x + 2 bx + (c - 2)e-xx x - e(1 + x) = 2, then a2 + b2 + c2 is equal to :
- 3
- 5
- 9
- 7
Answer
(4) 7
Solution
Denominator: x (1 - x22 ) - (x - x22 + x33 ) = x22 +
So the numerator must be x2 + higher powers.
Numerator = [1 + (a - 1)x + (a - 1)2x22 ] + [2 - b2x2 ] + (c - 2) [1 - x + x22 ] +
Constant term = 0: 1 + 2 + c - 2 = 0 c = -1
Coefficient of x = 0: (a - 1) - (c - 2) = 0 a + 2 = 0 a = -2
Coefficient of x2: 92 - b2 - 32 = 3 - b2
Limit = 3 - b21/2 = 2 b2 = 2
a2 + b2 + c2 = 4 + 2 + 1 = 7
Answer: (4)