Mathematics Extension 1 Predictions/Thoughts (1 Viewer)

[WeiWeiMan]

IT WAS

“It’d be funny”

 

[leehuan]

14(b)(ii)
The graphs suggest that as c increases from c=0.8 to c=1, there should be some point where the graph has exactly one x-intercept. Furthermore, that x-intercept will be a double root, as indicated by there is also a stationary point.

Therefore, let f(x) = x^4 - 2cx^3 + 1, and f'(x) = 4x^3 - 6cx^2. Setting f'(x) = 0 gives the condition 2x^2(2x-3c) = 0, and note that x=0 cannot be a valid solution. Setting f(x) = 0 gives the condition x^4 - 2cx^3 + 1 = 0.

The simultaneous equations turns out to yield the solution c = 2 / 3^(3/4).

14(c)(i)
Use area = 1/2 * b * h, where the base is OA, and the height is the projection of (b1,b2) onto (a2, -a1)

14(c)(ii) this may be wrong with the details, but it gives an idea on how to proceed.
Given the hint that OP = OI + IP, we can first note that OI is the vector (r, 0) whilst IP is the vector (r cos(t), r sin(t)). Therefore
p1 = r + r cos(t)
p2 = r sin(t)

Whereas OJ is the vector (-R, 0) and JQ is the vector (R cos(2t), R sin(2t)). Therefore
q1 = -R + R cos(2t)
q2 = R sin(2t)

Then area = 1/2 |p2q1 - q2p1| = 1/2 r R |sin(t) * (cos(2t)-1) - sin(2t) * (cos(t) - 1)|. Apparently this is maximised when t = 2pi/3 or -2pi/3.

 

[carrotsss]

I took derivative and got x in terms of c then subbed it in and solved for c but idk

ohh that makes sense

 

[cantdefeatben]

only knew how to do 14 ci) cause of 2021 bos trials lol

 

[WeiWeiMan]

I got C = 2

WHAT I GOT 2/(27^1/4)

 

[Aminza21]

@pdawg only 67? bro im thinking i got like 68 or 69

 

[gunnerfan99]

c= its a root of multiplicity 2 so u differentiate and let it =0 and sub it back in

Yep

 

[WeiWeiMan]

@pdawg only 67? bro im thinking i got like 68 or 69

/70 or %

 

[carrotsss]

14(b)(ii)
The graphs suggest that as c increases from c=0.8 to c=1, there should be some point where the graph has exactly one x-intercept. Furthermore, that x-intercept will be a double root, as indicated by there is also a stationary point.

Therefore, let f(x) = x^4 - 2cx^3 + 1, and f'(x) = 4x^3 - 6cx^2. Setting f'(x) = 0 gives the condition 2x^2(2x-3c) = 0, and note that x=0 cannot be a valid solution. Setting f(x) = 0 gives the condition x^4 - 2cx^3 + 1 = 0.

The simultaneous equations turns out to yield the solution c = 2 / 3^(3/4).

14(c)(i)
Use area = 1/2 * b * h, where the base is OA, and the height is the projection of (b1,b2) onto (a2, -a1)

14(c)(ii) this may be wrong with the details, but it gives an idea on how to proceed.
Given the hint that OP = OI + IP, we can first note that OI is the vector (r, 0) whilst IP is the vector (r cos(t), r sin(t)). Therefore
p1 = r + r cos(t)
p2 = r sin(t)

Whereas OJ is the vector (-R, 0) and JQ is the vector (R cos(2t), R sin(2t)). Therefore
q1 = -R + R cos(2t)
q2 = R sin(2t)

Then area = 1/2 |p2q1 - q2p1| = 1/2 r R |sin(t) * (cos(2t)-1) - sin(2t) * (cos(t) - 1)|. Apparently this is maximised when t = 2pi/3 or -2pi/3.

ok maybe things aren’t as bad as I thought, turns out I actually somehow got 14c)ii), just messed up 14b)ii) and 13a)iii)

 

[thomas mcnamee]

14(b)(ii)
The graphs suggest that as c increases from c=0.8 to c=1, there should be some point where the graph has exactly one x-intercept. Furthermore, that x-intercept will be a double root, as indicated by there is also a stationary point.

Therefore, let f(x) = x^4 - 2cx^3 + 1, and f'(x) = 4x^3 - 6cx^2. Setting f'(x) = 0 gives the condition 2x^2(2x-3c) = 0, and note that x=0 cannot be a valid solution. Setting f(x) = 0 gives the condition x^4 - 2cx^3 + 1 = 0.

The simultaneous equations turns out to yield the solution c = 2 / 3^(3/4).

14(c)(i)
Use area = 1/2 * b * h, where the base is OA, and the height is the projection of (b1,b2) onto (a2, -a1)

14(c)(ii) this may be wrong with the details, but it gives an idea on how to proceed.
Given the hint that OP = OI + IP, we can first note that OI is the vector (r, 0) whilst IP is the vector (r cos(t), r sin(t)). Therefore
p1 = r + r cos(t)
p2 = r sin(t)

Whereas OJ is the vector (-R, 0) and JQ is the vector (R cos(2t), R sin(2t)). Therefore
q1 = -R + R cos(2t)
q2 = R sin(2t)

Then area = 1/2 |p2q1 - q2p1| = 1/2 r R |sin(t) * (cos(2t)-1) - sin(2t) * (cos(t) - 1)|. Apparently this is maximised when t = 2pi/3 or -2pi/3.

pretty sure its +pi/3 and -pi/3 but may be wrong

 

[WeiWeiMan]

c= its a root of multiplicity 2 so u differentiate and let it =0 and sub it back in

LES GOOOOOO

 

[WeiWeiMan]

ok maybe things aren’t as bad as I thought, turns out I actually somehow got 14c)ii), just messed up 14b)ii) and 13c)iii)

What’s 13ciii
If it’s the projectile then that question was fricked for the third part

 

[scaryshark09]

what are these 2?

 

[breadcrumbs]

i feel like ive been reincarnated aftee rhat paper that was so hard im moving on with my life after that

 

[rainbowspoon1]

Yeah surely harder than 2022 I reckon look at how accessible 13 & 14 there are compared to 23.

i feel like 2022 wasnt that rough outside of the proj motion and the flights question while this like consistently sucked from the last part of q12

 

[Aminza21]

yeah i agree, it was kinda fricked

 

[rainbowspoon1]

View attachment 40827
what are these 2?

i got c bc if u draw it and then draw the inv function it looks like it crosses x = 1 3 times. for the last one i guessed c or d i dont remember

 

[carrotsss]

What’s 13ciii
If it’s the projectile then that question was fricked for the third part

wait I meant a)iii) whoops

 

[gunnerfan99]

View attachment 40827
what are these 2?

 

[WeiWeiMan]

wait I meant a)iii) whoops

What was that