close
標題:
Maths question(*o*)
發問:
Question 22 part c: 圖片參考：http://imgcld.yimg.com/8/n/HA01132001/o/701207200056113873412830.jpg Question 25 and 26(part dii): 圖片參考：http://imgcld.yimg.com/8/n/HA01132001/o/701207200056113873412831.jpg
最佳解答:
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)? + (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) ∵ M?? = (- 8 - 10)/(9 - 6) = - 6 = (- 8 - 10)/(0 - 3) = M??c) 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???? : A???? : A???? = 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
Maths question(*o*)
發問:
Question 22 part c: 圖片參考：http://imgcld.yimg.com/8/n/HA01132001/o/701207200056113873412830.jpg Question 25 and 26(part dii): 圖片參考：http://imgcld.yimg.com/8/n/HA01132001/o/701207200056113873412831.jpg
最佳解答:
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)? + (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) ∵ M?? = (- 8 - 10)/(9 - 6) = - 6 = (- 8 - 10)/(0 - 3) = M??c) 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???? : A???? : A???? = 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!).
此文章來自奇摩知識+如有不便請留言告知
其他解答: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
文章標籤
全站熱搜
留言列表
