Maths question(*o*)

22)a) q = 2p(6 - p) + p(6 - 2p) = 12p - 2p² + 6p - 2p² = 18p - 4p²b) 18p - 4p² = 6²/2 ⇒ 2p² - 9p + 9 = (p - 3)(2p - 3) = 0 ⇒ p = 3/2 or 3 (rej.)c) q = 18p - 4p² = 81/4 - (2p - 9/2)²; Max. of q = 81/4 when2p - 9/2 = 0, p = 9/4 25)a) f(x)= 10(x - 1)(x - 5)/[(- 1)(- 5)] = 2(x - 1)(x - 5)b) f(x)= 2x² - 12x+ 10 = (√2 x -3√2)² - 8; √2 p - 3√2 = 0, p = 3; q = - 8; b) r = 2(6- 1)(6 - 5) = 10c) i) g(x)= [(- 8 - 10)/(3 - 6)²](x -6)² + 10 = - 2(x - 6)² + 10c) ii) Let Q = (a, b);c) ii) (a -6)² + (b - 10)² = (6 - 3)² + (10 + 8)² = 333 --- (i)c) ii) b = - 2(a - 6)² + 10 ---(ii)c) ii) Sub.(ii) into (i), (a - 6)² + [- 2(a- 6)² + 10- 10]² = 333c) ii) ⇒ 4(a - 6)⁴ + (a - 6)² - 333 = 0 ⇒ [(a - 6)² - 9][4(a - 6)² + 37] = 0c) ii) ⇒ (a - 6)² = 9 or - 37/4 (rej.) ⇒ a - 6 = ± 3 ⇒ a = 3 (rej.)or 9c) ii) Sub.a = 9 into (ii), b = - 2(9 - 6)² + 10= - 8c) ii)∴ Q = (9, - 8)c) ii) ∵ Mǫʀ = (- 8 - 10)/(9 - 6) = - 6 = (- 8 - 10)/(0 - 3) = Mᴘᴄc) ii)∴ QR // PC 26)a) Let A= (a, 0); (2 - 0)/(6 - a) = 1 ⇒ a = 4; ∴ A = (4, 0)b) Let B = (b, 0); (6- 4)² + 2² = (b - 6)² + 2² ⇒ b = 4(rej.) or 8; ∴ B = (8, 0)c) Equation of BP isy = - 2/(8 - 6) x + 8 ⇒ y = 8 - xd) i) OA = 4 =8 - 4 = ABd) ii) Let C = (0, c), c = 8 - 0 = 8d) ii) A∆ᴏᴀᴄ : A∆ᴀᴘᴄ : A∆ᴀʙᴘ = 4 × 8 : 4 × 8 - 4 × 2 : 4× 2 = 4 : 3 : 1 I HOPE THIS CAN HELPYOU!PLEASE FEEL FREE TO ASKAGAIN IF YOU STILL HAVE ANYPROBLEM!~ ^_^

2012-07-20 23:02:53 補充：

Wong's Q22 c) answer is wrong (and no need calculus!).

Question 22 part c:

dq/dp=18-8p （呢個系涉及到微積分的問題，唔知你學左未...不過應該學左）

let dq/dp=0

18-8p=0 (其實中間仲要證明距系minimum point or maximum point，在d多次，距系less than0，就系maximum point)

p=2.15

25.

(a)

let y=f(x)=ax^2+bx+c (a,b,c are real number)

because f(x) passes through A, B ,C

so

10=0+0+c (1)

0=a+b+c (2)

0=25a+5b+c (3)

solving, a=2 ,b= -12 , c=10

so f(x)=2x^2-12x+10

(b)

f(x)=2x^2-12x+10

=2(x^2-6x+3^2-3^2)+10

=2(x-3)^2-8

so p=3, q=-8

put R(6,r) into y=2x^2-12x+10

r=72-72+10

=10

(c)(1)

because g(x) passes through P and has R as its vertex

let y=a(x-6)^2+10

put P(3,-8) into above

-8= a(9)+10

a= -2

so g(x)=-2(x-6)^2+10

(2)

because QR=PR

obviously, the coordinates of Q is (9,-8) (因為P到R得x-axis系+3，所以R到Q的x-axis都系+3，另外個y-axis同P一樣，原因系R系vertex and QR=PR)

so the slope of QR=(-8-10)/(9-6)=-6

the slope of PC=(-8-10)/(3-0)=-6

because the slope of QR=the slope of PC=-6

so QR is parallel to PC

26.part dii

by(a)(b)(c)

a=4,b=c=8

A(1)=0.5 x OA x OC

=0.5 x 4 x 8

=16

A(3)=0.5 x AB x height(the y-axis of P)

=0.5 x 4 x2

=4

A(2)=the area of triangle OBC - A(1) - A(3)

=0.5 x 8 x 8 -16- 4

=12

A(1):A(2):A(3)=16:12:4=4:3:1

2012-07-20 23:52:19 補充：

correction:

Question 22 part c:

p=2.25