會考a.maths mock paper 2題[超難]

[h should be replaced by a]
[point P instead of Point A]
14)(a)(i)
centre = (2a/2, 0) = (a, 0)
radius = 0.5 * root(4a^2 - 8a + 4) = 0.5 * 2 * root(a^2 - 2a + 1)
= root((a - 1)^2) = a - 1

14(a)(ii)(1) As L2 is a tangent to C1, we have
(3a - 4*0 + 1)^2/25 = (a - 1)^2
9a^2 + 6a + 1 = 25a^2 - 50a + 25
16a^2 - 56a + 24 = 0
a = 3 or 0.5
As a > 1, thus a = 3
sub (a, 0) into L1
2a - 0 - 6 = 2(3) - 6 = 0
Thus, centre of C1 lies on L1

14(a)(ii)(2)radius of C1 = 3-1 = 2units

14)(b)(i)slope L1 = 2, slop L2 = 3/4
tan θ = (2 - 3/4)/(1 + 2 * 3/4) = 1.25/2.5 = 1/2

14)(b)(ii)Point P = Intersection of L1, L2 = (5, 4)
Thus, equation of L is x = 5.

14)(c) As C2 touches L, L2 and C1, which is similar to C1, its centre must lies on L1 like C1
Let radius of C2 be R, centre = (h,k)
Intersection of L1, L2 = (5, 4)
Distance of (5, 4) and (3, 0) = root(2^2+4^2) = 2 * root 5
By simple ratio,
R/2 = (2 * root 5 - R - 2)/2 * root 5
R = 0.764, corr 3 sig fig.

Also, 2/4 = R/(4 - k)
k = 4 - 2R = 2.472
As (h,k) lies on L1,
2h - k - 6 = 0
2h = 6 + k
h = 4.236
Thus, centre = (4.236, 2.472)


15)(a)(Int from -6 to a)pi * x^2 dy = (Int from -6 to a)pi * (36 - y^2) dy
= pi[(36a - 1/3 * a^3) - (36(-6) - 1/3 * (-6)^3)]
= pi(36a-1/3 * a^3 +144)

15)(b)(Int from 0 to b)pi * x^2 dy = (Int from 0 to b)pi * 2y dy
= pi * b^2

15)(c)(i) As water depth in vessel 1 is 5, thus, a = -1
Water in vessel 1 = pi(-36-1/3 * -1 +144) = 323pi/3
As water depth in vessel 2 is 5, thus, b = 4
Water vessel 2 = pi * 4^2 = 16pi

16pi + 323pi/3 = pi * h^2
h^2 = 371/3, h = 11.12

15)(c)(ii)V_2 = pi * b^2
dV_2/dt = 2b * pi * db/dt = 2 * 4 * pi * -3 = -12pi
V_1 = pi(36a-1/3 * a^3 +144)
dV_1 = (-a^2 + 36)pi * da/dt
12pi = (-25 + 36)pi * da/dt
da/dt = 12/11