December 12, 2019, 02:06:26 pm

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#### Sine

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« Reply #4290 on: July 19, 2019, 10:00:44 pm »
+1
Hey there, i am unsure on what i'm doing wrong in the second part of this question, and the working out is confusing me. I keep getting the final answer as -1 but it's not. Hope you can help
What particular aspect of this question is troubling you?
The "hence" in the second part pretty much directly tells you that you need to use something from the first part to solve it. Hopefully, you can recognise it is integration by recognition.

#### Kombmail

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« Reply #4291 on: July 19, 2019, 11:00:32 pm »
0
Hey Guys!
I just did a question from the 2006 paper 7a and didnt understand how the answer was three?
The question was to let 'a' and' b' be the solutions of x squared minus 3x +1= 0 and find a + 1/a
this was a stepwise question with part a being find the product of the roots which equals to 1 when i solved it.

I tried solving it a rearranged it to a plus b /ab divided by 1/b.
This further equals 3/1 divided by 1/b and b becomes equal to three which was the correct answer.
Is this the right process?
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#### fun_jirachi

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« Reply #4292 on: July 20, 2019, 12:24:00 am »
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Hey Guys!
I just did a question from the 2006 paper 7a and didnt understand how the answer was three?
The question was to let 'a' and' b' be the solutions of x squared minus 3x +1= 0 and find a + 1/a
this was a stepwise question with part a being find the product of the roots which equals to 1 when i solved it.

I tried solving it a rearranged it to a plus b /ab divided by 1/b.
This further equals 3/1 divided by 1/b and b becomes equal to three which was the correct answer.
Is this the right process?

I'm not quite sure what you've done there, seems a tad dodgy on inspection (and tbh got me a bit confused).

An easier method (by easier I mean less convoluted) would be to note that the roots a and b satisfy the following set of equations (by sum and product of roots):
a+b=3 (1)
ab=1 (2)
From (2), note that b=1/a. Substituting this into (1), we get that a+1/a=3.

The fact that the first part forced you to find the product of roots indicates you should use that result in some way, which should be the case for all 'step-wise questions'.

Hope this helps
« Last Edit: July 20, 2019, 12:26:28 am by fun_jirachi »
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#### boulos

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« Reply #4293 on: July 20, 2019, 10:50:04 am »
0
What particular aspect of this question is troubling you?
The "hence" in the second part pretty much directly tells you that you need to use something from the first part to solve it. Hopefully, you can recognise it is integration by recognition.

Yea i recognised it was integration, I rearranged to get the integration of xe^-x, but i am some how not getting the answer. And the working out is pretty confusing.

#### fun_jirachi

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« Reply #4294 on: July 20, 2019, 11:11:20 am »
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As with many previous posts, the key here is that 'every derivative gives us another integral' and vice versa. If a question asks us to differentiate something, then asks us to integrate a similar function to the derivative, it's pretty much asking us to manipulate the integral and use the result in some way.
$ie. \ \text{If} \ \frac{d}{dx} \ F(x) = f(x), \ \int f(x) \ dx = F(x)+C$
From part a),
$\frac{d}{dx} \ 2xe^{-x} = 2e^{-x}-2xe^{-x} \\ \text{Then,} \ \int_0^1 xe^{-x} \ dx \\ = -\frac{1}{2} \int_0^1 -2xe^{-x} \ dx \\ = -\frac{1}{2} \int_0^1 2e^{-x}-2xe^{-x}-2e^{-x} \ dx \\ = -\frac{1}{2}[2xe^{-x}+2e^{-x}]_0^1 \\ = -\frac{1}{2}\left(\frac{2}{e}+\frac{2}{e}-2\right) \\ = 1-\frac{2}{e}$
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#### boulos

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« Reply #4295 on: July 20, 2019, 11:52:00 am »
0
As with many previous posts, the key here is that 'every derivative gives us another integral' and vice versa. If a question asks us to differentiate something, then asks us to integrate a similar function to the derivative, it's pretty much asking us to manipulate the integral and use the result in some way.
$ie. \ \text{If} \ \frac{d}{dx} \ F(x) = f(x), \ \int f(x) \ dx = F(x)+C$
From part a),
$\frac{d}{dx} \ 2xe^{-x} = 2e^{-x}-2xe^{-x} \\ \text{Then,} \ \int_0^1 xe^{-x} \ dx \\ = -\frac{1}{2} \int_0^1 -2xe^{-x} \ dx \\ = -\frac{1}{2} \int_0^1 2e^{-x}-2xe^{-x}-2e^{-x} \ dx \\ = -\frac{1}{2}[2xe^{-x}+2e^{-x}]_0^1 \\ = -\frac{1}{2}\left(\frac{2}{e}+\frac{2}{e}-2\right) \\ = 1-\frac{2}{e}$

Got it, saw where i went wrong. Thanks mate

#### LoneWolf

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« Reply #4296 on: July 22, 2019, 07:32:45 am »
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Hi PPl
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*

*For a mathematics journal i am keeping as part of an assessment!

#### RuiAce

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« Reply #4297 on: July 22, 2019, 09:58:19 am »
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Hi PPl
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*

*For a mathematics journal i am keeping as part of an assessment!
Without spoiling your assignment too much, in my opinion the main benefit of the chain rule lies in a certain concept that happens to be taught in MX1. It is the concept of analysing rates of change, among variables that are somehow related.

The chain rule says that $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$. So if you want to analyse how $y$ changes with respect to $x$, but you know stuff about $u$....

#### LoneWolf

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« Reply #4298 on: July 22, 2019, 10:17:17 am »
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Thanks very much.
I would be glad of any other contributions from anyone!!!!
It all helps!

#### blyatman

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« Reply #4299 on: July 22, 2019, 10:25:08 am »
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Hi PPl
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*

*For a mathematics journal i am keeping as part of an assessment!

Hi there, engineer here.

In the real world, everything is measured in rates of change, and we can make various predictions based on these rates. For example, suppose you drop a ball from a cliff, and you want to know where the ball will be after a certain time. To calculate this, you need to know the forces are acting on the ball (which will tell you how the speed of the ball is changing). This is, of course, it's weight, which is the force due to gravity. Now, the force acting on an objective is equal to it's RATE OF CHANGE of momentum with respect to time. The fact that it's a rate of change allows us to use elementary calculus to determine the position of the ball as a function of time.

Basic differentiation provides us with the tools to calculate these rates of change. However, unlike the idealised problems you deal with in high school calculus, problems in the real world are multivariable, meaning that they depend on a multitude of variables. In the above example, other factors that come into play are air resistance, which depends on the density of the air as well as the geometry of the ball. If you were to drop the ball from a significant altitude, the density of the air would change as the ball falls through the earth's atmosphere, which would in turn change the air resistance. Thus, you need to take into account the rate of change of numerous variables. The chain rule allows us to calculate the rate of change of a function that depends on numerous variables.
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#### KingTings

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« Reply #4300 on: July 22, 2019, 01:01:24 pm »
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I have a question from the 2018 CSSA Trials!
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#### annabeljxde

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« Reply #4301 on: July 22, 2019, 01:23:11 pm »
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I have a question from the 2018 CSSA Trials!

I'm not too sure if this is the proper way to go about the question, but this is how I would end up with the answer C:

Since f(-x) = -f(x), this means that f(x) has odd symmetry. On the other hand, if f(x)=f(-x), the graph would have even symmetry.

I would recommend doing a rough sketch if you're still unsure, but basically, since (3,7) is a maximum turning point, there will be a minimum turning point at (-3,-7), which is symmetrical to the maximum but negative. (If the condition was even symmetry, there would be a maximum turning point again at (-3,7) )

Hopefully this makes some sort of sense... If not, let me know and I'll demonstrate with a diagram
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#### KingTings

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« Reply #4302 on: July 22, 2019, 04:41:43 pm »
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Ahhhh I totally get it.
Thanks so much!
And if it was f(x) = f(-x) there would be a max turning point at (-3,7)??
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#### RuiAce

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« Reply #4303 on: July 22, 2019, 04:43:39 pm »
+1
Ahhhh I totally get it.
Thanks so much!
And if it was f(x) = f(-x) there would be a max turning point at (-3,7)??
Correct, because even function symmetry is about the $y$-axis, so it's just a reflection.

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