JC 1 math question:equations

Jia Xun SG

A cubic polynomial has equation given by y=ax^3 + bx^2 + cx + d.where a b c d are integers(the sign Z)

Find one such cubic polynomial whose graph passes through the points (1,1),(2,2) and (3,3).How many such cubic polynomials intersect the x-axis at exactly one point?

And also one more...

A factory manufactures three different designs of bracelets,A,B and C.Bracelet A is made up of 4 white beads and 4 pink beads,bracelet B is made up of 6 white beads and 3 pink beads while bracelet C is made up of 3 white beads and 4 pink beads.A total of 70 white beads and 52 pink beads are used per day.How many bracelets of each design can be made per day?

Really need help on these questions as I have a test next week.I'll appreciate any relavant help given.Thanks!:)

frekiwang SG

lol, are you sure these are JC1 H1/H2 Math questions? To me, more like competition type, or Uni maths content. Anyway, Q2 first:

4A+6B+3C=70 ..(1) and 4A+3B+4C=52..(2)

(1)-(2), we have 3B-C=18, B>=6

B=6, C=0, A=38/4(reject)

B=7, C=3, A=19/4(reject)

B=8, C=6, A=4/4=1

B>=9, C>=9, 4A<=52-3(9)-4(9)=-9, no more solution.

So A=1, B=8, C=6 is the only solution

frekiwang SG

Q1:

Part (i)

subst the points into the equation, we have

a+b+c+d=1 ...(1)

8a+4b+2c+d=2 ...(2)

27a+9b+3c+d=3 ...(3)

(2)-(1), we have 7a+3b+c=1 ...(4)

(3)-(2), we have 19a+5b+c=1...(5)

(5)-(4), we have 12a+2b=0 or b=-6a

subst into (4), we have c=1-3(-6a)-7a=1+11a

subst into (1), we have d=1-(1+11a)-(-6a)-a=-6a

the general solution is b=-6a, c=1+11a, d=-6a

One possible set of solution a=1,b=-6,c=12,d=-6, i.e x^3-6x^2+12x-6=0

Part (ii), (direct, but requires higher level knowledge)

discriminant

=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2

=18a(-6a)(11a+1)(-6a)-4(-6a)^3(-6a)+(-6a)^2(11a+1)^2-4a(11a+1)^3-27a^2(-6a)^2

=7128a^4+648a^3-5184a^4+4356a^4+792a^3+36a^2-5324a^4-1452a^3-132a^2-4a-972a^4

=4a^4-12a^3-96a^2-4a<0

Solve we have

-3.6<a<-0.04 or 0<a<6.6

a=-3,-2,-1,1,2,3,4,5,6. no. of such cubic polynomial = 9.

frekiwang SG

ok, short-cut:

ax^3-6ax^2+(11a+1)x-6a=0

re-arrange,

x^3-6x^2+11x-6=-x/a

investigate the no. of intersections between the line and the curve using GC.

For one point of intersection and a being an integer,

-3<=a<0 or 0<a<=6

a=-3,-2,-1,1,2...6

Jia Xun SG

Haha nope,these are RI's tutorial questions.:)

I'm revising though on every question given on this chapter.

By the way,is your method related to gaussian's elimination?It looks similar.

Thanks for the help!^_^

frekiwang SG

lol, RI is really teaching something crazy...

wee_ws SG

Hi Jia Xun,

Cambridge sets questions on SLE with unique solutions, so you need not worry. Your school is probably stretching all of you at the beginning of the year or to scare students into studying hard.

Past-year problems that have appeared are:

- formulate a system of 3 linear equations involving weights and total amounts provided in table form (in 2007);

- formulate a system of 3 linear equations involving terms of a sequence that has a quadratic form (in 2009);

- formulate a system of 3 linear equations involving a parabola passing through points (in 2011).

Schools typically set harder problems that students face difficulty forming the equations because they cannot interpret English into mathematical representation in a competent way.

Hope these tips help, good luck!

Cheers,
Wen Shih

Jia Xun SG

No worries,I'm doing tutorial questions set by RI only,not studying there.:)

Thanks.:)

FireIce SG

arbo call RI?

wee_ws SG

Hi Jia Xun,

Choice of questions for practice matters. Consider those that are reflective of Cambridge exams and your school in particular, and those that assess, reasonably and objectively, relevant mathematical skills and concepts. Avoid tedious and irrelevant questions.

Thanks.

Cheers,
Wen Shih

Jia Xun SG

Alright,shall take notice.:)Thanks!:D

eagle SG

Part (i)

subst the points into the equation, we have

a+b+c+d=1 ...(1)

8a+4b+2c+d=2 ...(2)

27a+9b+3c+d=3 ...(3)

(2)-(1), we have 7a+3b+c=1 ...(4)

(3)-(2), we have 19a+5b+c=1...(5)

(5)-(4), we have 12a+2b=0 or b=-6a

subst into (4), we have c=1-3(-6a)-7a=1+11a

subst into (1), we have d=1-(1+11a)-(-6a)-a=-6a

the general solution is b=-6a, c=1+11a, d=-6a

One possible set of solution a=1,b=-6,c=12,d=-6, i.e x^3-6x^2+12x-6=0

Part (ii), (direct, but requires higher level knowledge)

discriminant

=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2

=18a(-6a)(11a+1)(-6a)-4(-6a)^3(-6a)+(-6a)^2(11a+1)^2-4a(11a+1)^3-27a^2(-6a)^2

=7128a^4+648a^3-5184a^4+4356a^4+792a^3+36a^2-5324a^4-1452a^3-132a^2-4a-972a^4

=4a^4-12a^3-96a^2-4a<0

Solve we have

-3.6<a<-0.04 or 0<a<6.6

a=-3,-2,-1,1,2,3,4,5,6. no. of such cubic polynomial = 9.

We can use GC in Part 1, similar to how we solve for line of intersection between two vector planes.

Given

a+b+c+d=1 ...(1)

8a+4b+2c+d=2 ...(2)

27a+9b+3c+d=3 ...(3)

Use PolySimul in GC for 3 equations 4 unknowns

Everything will be given in terms of d.

Find a value for d that makes a, b and c. If there are fractions, one way would be to find the LCM of the denominators of these fractions.

Solving,

a = -d/6, b = d, c = 1 - 11d/6

We can let d be 6, and a = -1, b = 1, c = -10

eagle SG

lol, are you sure these are JC1 H1/H2 Math questions? To me, more like competition type, or Uni maths content. Anyway, Q2 first:

4A+6B+3C=70 ..(1) and 4A+3B+4C=52..(2)

(1)-(2), we have 3B-C=18, B>=6

B=6, C=0, A=38/4(reject)

B=7, C=3, A=19/4(reject)

B=8, C=6, A=4/4=1

B>=9, C>=9, 4A<=52-3(9)-4(9)=-9, no more solution.

So A=1, B=8, C=6 is the only solution

Same as previous

We can use GC to solve this, by using 2 equations and 3 unknowns

4A+6B+3C=70 .....(1)

4A+3B+4C=52 .....(2)

Using GC,

A = 17/2 - 5C/4
B = 6 + C/3

If we try C = 12 (LCM), we get A = -1, which is rejected.
Since there's a 17/2 in A, we can try C = 6 to get 17/2 - 15/2, of which we get A = 1
Then B = 8

so A = 1, B = 8, C = 6