O-Level Additional Math

[tutor4kids]

Hi,


Could you help me solve part ii of the following question?

A polynomial p(x) when divided by (x^2 - 1) leaves a quadratic remainder with no remainder. The constant term in p(x) is 10.
Given that p(x) leaves a remainder of 630 and 112 when divided by (x - 4) and (x + 3) respectively, find
i) an expression for p(x) in descending powers of x.
ii) solutions when p(x) = x^2 - x + 10

Ans: i) 3x^4 + x^3 - 13x^2 - x + 10
ii) x = 0, -7/3, 2

[leeven]

1. A number is a called palindrome if it reads the same from left and from right. example 130031 is a palindrome. writing all palindroms we get 1,2,3,4,5,…9,11,22,…what is the 2007th palindrome. do you have formula to find the nth palindrome.


please help

[FrekiWang]

tutor4kids:

Hi,


Could you help me solve part ii of the following question?

A polynomial p(x) when divided by (x^2 - 1) leaves a quadratic remainder with no remainder. The constant term in p(x) is 10.
Given that p(x) leaves a remainder of 630 and 112 when divided by (x - 4) and (x + 3) respectively, find
i) an expression for p(x) in descending powers of x.
ii) solutions when p(x) = x^2 - x + 10

Ans: i) 3x^4 + x^3 - 13x^2 - x + 10
ii) x = 0, -7/3, 2

I assume 'quadratic remainder' is an error and it should be 'quadratic quotient'.
(i)
p(x) = Divisor x Quatient + Remainder = (x^2 -1)(ax^2+bx+c)+0 = ax^4 +bx^3 + cx^2 -ax^2 - bx - c = ax^4 + bx^3 + (c-a)x^2 - bx - c
Since the constant term is 10, we have -c = 10, c = -10
Thus, p(x)=ax^4 + bx^3 + (-10-a)x^2 - bx + 10
According to the remainder thereom,
p(4)=630 and p(-3)=112
256a+64b+16(-10-a)-4b+10=630 and 81a-27b+9(-10-a)+3b+10=112
240a+60b=780 and 72a-24b=192
solve, we get a=3 and b=1
therefore p(x)= 3x^4 + x^3 -13x^2 - x + 10
(ii) p(x)=x^2 - x + 10
3x^4 + x^3 - 13x^2 - x +10 = x^2 - x + 10
3x^4 + x^3 - 14x^2 = 0
x^2(3x^2 + x - 14) = 0
x^2(3x + 7)(x - 2) = 0
x^2 = 0 or 3x + 7 = 0 or x - 2 = 0
x = 0 or x = -7/3 or x = 2

PS: You may need more systematic help for this chapter, not just Q&A

[FrekiWang]

leeven:

1. A number is a called palindrome if it reads the same from left and from right. example 130031 is a palindrome. writing all palindroms we get 1,2,3,4,5,....9,11,22,,....what is the 2007th palindrome. do you have formula to find the nth palindrome.


please help

First of all, I do not believe this is a secondary school question. The standard of this problem is for competition at upper secondary school level, e.g. SMO (Senior)

So, here is my solution, assuming you have some basic knowledge about competition mathematics.

There are 9 1-digit palindroms.
There are 9 2-digit palindroms.
For 3-digit palindroms. You can choose a number from 0 to 9 (10numbers here) and insert into a 2-digit palindrom to form a 3-digit palindrom. Therefore 9 2-digit palindroms can produce 9 x 10 = 90 3-digit palindroms.
For 4-digit palindroms. You can choose a number from 00, 11 to 99(10numbers here) and insert into a 2-digit palindrom to form a 4-digit palindrom.
Therefore 9 2-digit palindroms can produce 9 x 10 = 90 4-digt palindroms.

Similarly, insert a number from 0 ... 9 into a 4-digit palindrom to form a 5-digit palindroms, there are 90 x 10 = 900 of them.
Insert a number from 00 ... 99 into a 4-digit palindrom to form a 6-digit palindroms, there are 90 x 10 = 900 of them.

Now we have 9 + 9 + 90 + 90 + 900 + 900 = 1998 palindroms in totoal, we need 9 more.
Let's start counting the 7-digit palindroms from the least: (inserting 0..9 into 100001)
1000001, 1001001, 1002001, 1003001, 1004001, 1005001, 1006001, 1007001, 1008001 - this is the 9th, so the answer is 1008001.

[Herbie]

hi how to solve this qqn


a team of 8 men can build 3 swimming pools in 5 weeks. Find the no. Of men required to build 6 swimming pools in 4 weeks? Tq

[tianzhu]

Herbie:

hi how to solve this qqn


a team of 8 men can build 3 swimming pools in 5 weeks. Find the no. Of men required to build 6 swimming pools in 4 weeks? Tq

Hi

One way is to use compound units which in this question is \"man-weeks\"

3 swimming pool ------ 8*5 ------ 40 man-weeks
6 swimming pools ------ 2*40 ------- 80 man-weeks

80/4 ------ 20 men

Best wishes

[Herbie]

hi tianzhu, many thanks!


I hv another qn.

15 men working 8 hours a day can build a bridge in 30 days.
How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

B. How many hours a day must 25 men work to build the same bridhe in 16 days!

Tq

[tianzhu]

Herbie:

15 men working 8 hours a day can build a bridge in 30 days.
How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

B. How many hours a day must 25 men work to build the same bridhe in 16 days!

Hi

Your answer of 50 days for part A is right.

For part B,
1 bridge ------ 15*30*8 ------- 3600 man-hours

Number of men*number of days*number of hours ------- 3600 man-hours

25*16*h ------- 3600 where h refers to the number of hours the men must work.

h ------- 9 hours.

Best wishes

[tutor4kids]

Hi, thank you for the solution.

Could you have me with the following question?

In the diagram below, ACB is a tangent to the circle at point C. This tangent is parallel to the line FE. Points F, G and H lie on the circle.
i) Prove that (EC)(CG) = (CF)^2.
ii) The line FE is extended to meet the circle at H. Prove (FE)(EH) = CF^2 - EC^2.

http://i52.tinypic.com/33zd7hv.jpg\">

Thank you.

[FrekiWang]

(i) (this is a very standard A Maths Question, Similarity is one of the standard approaches)

In FEC and GFC,
AngleEFC=AngleFCA=AngleFGC (hopefully you know why)
AngleFCE=AngeGCF
Thus FEC is similar to GFC
FE/GF=EC/FC=FC/GC
Cross-multiply 2nd and 3rd, we have EC x GC = FC^2(proven)

(ii) (usually (ii) is related to (i))
RHS=CF^2-EC^2
=EC x GC - EC^2 (from part (i))
=EC x (GC - EC)
=EC x GE
=FE x EH (intersecting chords thereom)
=LHS (proven)
Seriously, part(ii) is much easier. In the exam/test, if you cant prove part(i), you can always assume part(i) result and use it to prove part(ii)