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Mathematics Extension 1 Predictions/Thoughts (2 Viewers)

leehuan

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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

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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

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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

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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
 

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