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共計2.5小時考試時間
此套試卷由兩部分題目組成
Part A共8題,每題5分
Part B共4題,每題10分
共計12題,滿分80分
不可使用任何計算器
完整版下載鏈接見文末
Part A Solutions:
A4)?Solution:Factoring equation 3, (x + 2y - z)(x + 2y + z) = 15.
Substituting x + 2y - z = 5, 5(x + 2y + z) = 15 or, x + 2y + z = 3.
Subtracting this from (2): 2x = 8
Therefore, x = 4.
The average on this question was 3.4.
Part B Solutions:
B2)Assume that we can expand and compare coefficients. Expanding, (x + r)(x2 + px + q) = x3 + (p + r)x2 + (pr + q)x + qr
Comparing coefficeints,
p + r = b (1)
pr + q = c (2)
qr = d (3)
If bd + cd is odd, so is d(b + c). From this, d and b + c are both odd. From (3), if d is odd then q and r are both odd.(4)
Adding (1) and (2), b+c = p+r+pr+q = (q+r)+p(1+r). Since b+c is odd then (q+r)+p(1+r) is also odd. From (4), if q and r are both odd then q+r is even. This implies that p(1 + r) must be odd but this is not possible because r is odd and r + 1 is then even making p(1 + r) both odd and even at the same time. This contradiction implies that our original assumption was incorrect and thus x3 +bx2 +cx+d cannot be expressed in the form (x + r)(x2 + px + q).
The average on this question was 2.1.
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