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March 29, 2024, 07:28:50 am

Author Topic: 3U Maths Question Thread  (Read 1230375 times)  Share 

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MysteryMarker

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Re: 3U Maths Question Thread
« Reply #585 on: August 16, 2016, 05:56:42 pm »
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Just came across a geometry question and was wondering if, just as with similar triangles their sides are in corresponding ratios, can this be applied to a parallelogram?

Just curious. Cheers.

RuiAce

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Re: 3U Maths Question Thread
« Reply #586 on: August 16, 2016, 06:00:34 pm »
+1
Just came across a geometry question and was wondering if, just as with similar triangles their sides are in corresponding ratios, can this be applied to a parallelogram?

Just curious. Cheers.
Similar polygons with number of sides greater than 3 are not in the course. Hence, no.

Also, keep in mind that you need more than all 4 sides to be in corresponding ratios. You also need a few ANGLES to be equal for a pair of quadrilaterals. (Of course, if you know it's a parallelogram you would only need two adjacent sides and one angle)

(After all, if you take two parallelograms with sides in matching ratios, but then just "squash" one, well...)
« Last Edit: August 16, 2016, 06:07:09 pm by RuiAce »

MysteryMarker

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Re: 3U Maths Question Thread
« Reply #587 on: August 16, 2016, 06:43:16 pm »
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When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1)

Always been confused with these types of questions. Cheers.

EDIT: Another question,

find the exact value

sin(cos-12/3 + tan-1-3/4)



« Last Edit: August 16, 2016, 06:49:53 pm by MysteryMarker »

Jakeybaby

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Re: 3U Maths Question Thread
« Reply #588 on: August 16, 2016, 07:01:47 pm »
+1
When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1)

Always been confused with these types of questions. Cheers.
From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5.
Also, division with (x+1) leaves 8, therefore P(-1) = 8.

Using our general formula:

P(x) = (x-1)(x+1)q(x)+Ax+B
Substituing in the particular values for x, in this case, 2 and -1.
P(-1) = 8, therefore:
8 = -A+B

P(2) = 5
5 = 2A+B
 
Solving the simultaneous equations, we get:
A = -1 & B = 7

Therefore the remainder present upon the division by (x-2)(x+1) is -x + 7
« Last Edit: August 16, 2016, 07:15:37 pm by RuiAce »
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RuiAce

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Re: 3U Maths Question Thread
« Reply #589 on: August 16, 2016, 07:14:57 pm »
+1
When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1)

Always been confused with these types of questions. Cheers.

EDIT: Another question,

find the exact value

sin(cos-12/3 + tan-1-3/4)






Using our general formula:

P(x) = (x-1)(x+1)q(x)+Ax+B
Substituing in the particular values for x, in this case, 2 and -1.
P(-1) = 8, therefore:
8 = -A+B
Comment: Would like to mention that the "formula" is known as the "division transformation". :)

MysteryMarker

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Re: 3U Maths Question Thread
« Reply #590 on: August 16, 2016, 07:41:44 pm »
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From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5.
Also, division with (x+1) leaves 8, therefore P(-1) = 8.

Using our general formula:

P(x) = (x-1)(x+1)q(x)+Ax+B
Substituing in the particular values for x, in this case, 2 and -1.
P(-1) = 8, therefore:
8 = -A+B

P(2) = 5
5 = 2A+B
 
Solving the simultaneous equations, we get:
A = -1 & B = 7

Therefore the remainder present upon the division by (x-2)(x+1) is -x + 7

Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B

Cheers.


jakesilove

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Re: 3U Maths Question Thread
« Reply #591 on: August 16, 2016, 07:45:31 pm »
+1
Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B

Cheers.

The remainder will always be in the form that has a power ONE LESS than the divisor

So, if you divide by ax, the remainder will be a constant. If you divide by ax^2+bx+c, the remainder will be Ax+C (ie. x is one power lower than the divisor!). If you divide by x^n, the remainder will have a factor including x^(n-1)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #592 on: August 16, 2016, 07:53:00 pm »
+1
Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B

Cheers.
Furthering onto what Jake said now.






« Last Edit: August 16, 2016, 07:55:44 pm by RuiAce »

Loki98

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Re: 3U Maths Question Thread
« Reply #593 on: August 16, 2016, 09:22:35 pm »
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How would you do the following questions attached?

RuiAce

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Re: 3U Maths Question Thread
« Reply #594 on: August 16, 2016, 10:01:24 pm »
+1
How would you do the following questions attached?
See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii).

I don't have the time to do lengthy questions right now.
« Last Edit: August 16, 2016, 10:10:15 pm by RuiAce »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #595 on: August 16, 2016, 11:14:17 pm »
+2
See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii).

I don't have the time to do lengthy questions right now.

I do ;D

For the first part of the question, we can equate the x and y displacements:



Skipped a few steps of algebra at the end there, but that should give you the idea ;D

Second bit, we set y=0:



That last line was dividing by sine alpha ;D

The last one is a little tricky to spot, but consider a rearranged form of part (b):



Now remember that T2 happens after T1! This means that the ratio of T2/T1 must be greater than 1, therefore:



Solving that quadratic on the open domain of 0 to pi/2, will give you the answer you need ;D

RuiAce

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Re: 3U Maths Question Thread
« Reply #596 on: August 16, 2016, 11:21:06 pm »
+1
Aha thanks Jamon :)




jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #597 on: August 16, 2016, 11:25:16 pm »
+3
Aha thanks Jamon :)





Teamwork! Lucky I checked back in, I was about to write the quadratic formula bit ;)

MysteryMarker

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Re: 3U Maths Question Thread
« Reply #598 on: August 17, 2016, 07:40:28 am »
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If cosx = -3/5 for 0<x<pi then tan(x/2) = ?

RuiAce

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Re: 3U Maths Question Thread
« Reply #599 on: August 17, 2016, 08:12:27 am »
+1
If cosx = -3/5 for 0<x<pi then tan(x/2) = ?


« Last Edit: August 19, 2016, 09:31:18 pm by RuiAce »