✔ 最佳答案
1.
Let x cm be the length of AP.
Then, the length of PB = (8 - x) cm
AP2 + PB2
= [x2 + (8 - x)2] cm2
= [x2 + x2 - 16x + 64] cm2
= 2(x2 - 8x + 32) cm2
= [2(x2 - 8x + 16) + 2 x 16] cm2
= [2(x - 4)2 + 32] cm2
(x - 4)2 ≥ 0
Therefore, AP2 + PB2 = 2(x - 4)2 + 32 ≥ 32
AP2 + PB2 is the minimum when x - 4 = 0, i.e. x = 4
Length of AP = 4 cm
2.
Let Q(x) and (ax + b) be the quotient and remainder when P(x) is divided by x2 + x - 6.
P(x) = (x2 + x - 6)Q(x) + (ax + b)
When P(x) is divided by (x - 2), the remainder is 10:
P(2) = 10
(22 + 2 - 6)Q(2) + (2a + b) = 10
2a + b = 10 ...... (1)
P(x) is divisible by (x + 3):
P(-3) = 0
[(-3)2 + (-3) - 6]Q(-3) + (-3a + b) = 0
-3a + b = 0 ...... (2)
(1) - (2):
5a = 10
a = 2
(1):
2(2) + b = 10
b = 6
The remainder when P(x) is divided by (x2 + x - 6) = 2x + 6
3.
z = kx3/√y
When x = xo, y = yo, z = zo
zo = kxo3/√yo
When x = 2xo, y = 3yo, z = z1
z1 = k(2xo)3/√(3yo)
z1 = (8/√3)(kxo3/√yo)
z1 = (8/√3)zo
Percentage change of z
= {[(8/√3)zo - zo]/zo} x 100%
= 100[(8/√3) - 1] x 100%
= +361.88%
4.
(a)
a and b have opposite signs, one positive and one negative.
(b)
One of a, b and c is negative, and the other two are of the same sign (both positive or both negative).
(c)
Case 1: When x < 1,
(x - 1) < 0, (x - 2) < 0 and (x - 3) < 0
Hence, (x - 1)(x -2)(x - 3) < 0
Case 2: When 1 < x < 2,
(x - 1) > 0, (x - 2) < 0 and (x - 3) < 0
Hence, (x - 1)(x -2)(x - 3) > 0
Case 3: When 2 < x < 3,
(x - 1) > 0, (x - 2) > 0 and (x - 3) < 0
Hence, (x - 1)(x -2)(x - 3) < 0
Case 4: When x > 3,
(x - 1) > 0, (x - 2) > 0 and (x - 3) > 0
Hence, (x - 1)(x -2)(x - 3) > 0
Ans:. x < 1 ..or.. 2 < x < 3
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