G12_MATH_P2_13 好難啊! 救命啊!

Let BC = AB = x
AC = √(AB2 + BC2) = √(x2 + x2) = √(2x2) = √2x
so, GC = EC = AC = √2x
BG = √(BC2 + GC2) = √(x2 + 2x2) = √(3x2) = √3x
cos∠BGC = GC / BG = (√2x) / (√3x) = √2/√3


the locus is a straight line which
the slope = (0 - (-2)) / (4 - 0) = 1/2
and the y-int = -3
so the equation is y = (1/2)x - 3
i.e. x - 2y - 6 = 0

(x3 - 3x2- x + 3) / (x3 - 2x2- 3x)
= [x2(x - 3) - (x - 3)] / [x(x2 - 2x - 3)]
= [(x2 - 1)(x - 3)] / [x(x - 3)(x + 1)]
= [(x - 1)(x + 1)(x - 3)] / [x(x - 3)(x + 1)]
= (x - 1) / x

from the graph y = abx, we have b > 1 and a = 4
log2 y = log2 (abx) = log2 a + xlog2 b = log2 4 + xlog2 b = 2 + xlog2 b
the slope of the required straight line = log2 b > 0
and the y-int = 2
so, the answer is D

210 + 28 + 26 + 5
= 210 + 28 + 26 + 22 + 20
= 10101000101