May 23, 2019, 11:30:07 am

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#### therese07

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« Reply #4185 on: April 20, 2019, 06:58:47 pm »
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Hi!

So I was doing some revision on curve sketching, and I came across this question (attached below).

I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)

u = 2x +1           v = (x-2)^4
u' = 2                 v' = 4(x-2)^3

dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4

I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.

HSC 2019: Studies of Religion 1 | English Advanced | Mathematics | Chemistry | Investigating Science | PDHPE |

#### jamonwindeyer

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« Reply #4186 on: April 20, 2019, 07:11:20 pm »
+1
Hi!

So I was doing some revision on curve sketching, and I came across this question (attached below).

I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)

u = 2x +1           v = (x-2)^4
u' = 2                 v' = 4(x-2)^3

dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4

I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.

Smashed the first bit! If you pull out a factor of $\left(x-2\right)^3$, you'll get:

$\frac{dy}{dx}=\left(x-2\right)^3\left(4(2x+1)+2(x-2)\right)=\left(x-2\right)^3\left(10x\right)$

From there you can solve for the $x=2$ and $x=0$ stationary points, and determine their nature using a table (the second derivative test is probably a bit cumbersome here)

#### Thankunext

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« Reply #4187 on: April 22, 2019, 10:14:21 pm »
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Help!
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?

#### RuiAce

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« Reply #4188 on: April 22, 2019, 10:42:47 pm »
+3
Help!
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?

$\text{You can just use the sum and product of roots.}\\ \text{In the first case, suppose that the roots are negatives of each other, i.e. }\boxed{\beta = -\alpha}.$
$\text{But since }\alpha+\beta = -(k+15),\text{ we therefore have}\\ \alpha-\alpha = -k-15 \implies \boxed{0 = -k-15}.$
You can do something similar for the next question.

#### spnmox

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« Reply #4189 on: April 23, 2019, 09:29:44 pm »
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2014 HSC Q7 Multiple Choice

How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?

#### jamonwindeyer

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« Reply #4190 on: April 23, 2019, 09:35:02 pm »
+1
2014 HSC Q7 Multiple Choice

How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?

Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!

$\sin{x}-1=0\\\sin{x}=1\\x=\frac{\pi}{2}\\\tan{x}+2=0\\\tan{x}=-2$

And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, $x=\frac{\pi}{2}$ isn't a solution! Why? If you substitute $x=\frac{\pi}{2}$ into $\tan{x}$, it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees

Really tricky, but you have to ignore that solution. Which means only the two solutions from the $\tan$ part of the equation work - The answer is two

#### spnmox

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« Reply #4191 on: April 23, 2019, 10:09:00 pm »
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Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!

$\sin{x}-1=0\\\sin{x}=1\\x=\frac{\pi}{2}\\\tan{x}+2=0\\\tan{x}=-2$

And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, $x=\frac{\pi}{2}$ isn't a solution! Why? If you substitute $x=\frac{\pi}{2}$ into $\tan{x}$, it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees

Really tricky, but you have to ignore that solution. Which means only the two solutions from the $\tan$ part of the equation work - The answer is two

oh, thank you so much! haha now it seems obvious

#### Karlamineeeee

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« Reply #4192 on: April 24, 2019, 01:01:59 pm »
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I hope this question is allowed here, I'm trying out the first series and sequence test in the ATAR notes' HSC mathematics topic tests.

For the first question, why do the answers calculate for the 15th term instead of the 30th?

#### RuiAce

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« Reply #4193 on: April 24, 2019, 01:40:48 pm »
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I hope this question is allowed here, I'm trying out the first series and sequence test in the ATAR notes' HSC mathematics topic tests.

For the first question, why do the answers calculate for the 15th term instead of the 30th?
I had a look. That's definitely some accidental error. Should certainly sub in $n=30$ in $T_n = 6(2)^{n-1}$

#### Karlamineeeee

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« Reply #4194 on: April 24, 2019, 07:30:00 pm »
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I had a look. That's definitely some accidental error. Should certainly sub in $n=30$ in $T_n = 6(2)^{n-1}$

Thank you so much!!

Also, for the final question of that test, where did the 500 value come from? Wasn't $1000 deposited? Sorry for asking so late!! DX #### RuiAce • HSC Lecturer • Honorary Moderator • Great Wonder of ATAR Notes • Posts: 8323 • "All models are wrong, but some are useful." • Respect: +2229 ##### Re: Mathematics Question Thread « Reply #4195 on: April 24, 2019, 10:31:58 pm » 0 Thank you so much!! Also, for the final question of that test, where did the 500 value come from? Wasn't$1000 deposited? Sorry for asking so late!! DX
Only just arrived back at home and yea, that's another accidental error I think. Should be 1000.

#### jamonwindeyer

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« Reply #4196 on: April 25, 2019, 12:45:36 am »
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Confirming both of those are errors, and both have corrections at this link: https://atarnotes.com/product-updates/. There's a full fixed solution to that second one too!

Apologies for the confusion caused - It's been fixed for future prints, but we must still be working through existing stock!

#### Karlamineeeee

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« Reply #4197 on: April 25, 2019, 01:41:17 am »
+1
Confirming both of those are errors, and both have corrections at this link: https://atarnotes.com/product-updates/. There's a full fixed solution to that second one too!

Apologies for the confusion caused - It's been fixed for future prints, but we must still be working through existing stock!

Thank you so much!!

#### therese07

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« Reply #4198 on: May 07, 2019, 08:01:07 pm »
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Hi there!

I was doing some homework tasks for math, and I got stuck on 10ii, proving that the stationary point is a maximum. I was wondering how would I be able to do this question. Would I need to second derivative at all?

Thank you!!!

HSC 2019: Studies of Religion 1 | English Advanced | Mathematics | Chemistry | Investigating Science | PDHPE |

#### RuiAce

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« Reply #4199 on: May 07, 2019, 08:28:17 pm »
+1
Hi there!

I was doing some homework tasks for math, and I got stuck on 10ii, proving that the stationary point is a maximum. I was wondering how would I be able to do this question. Would I need to second derivative at all?

Thank you!!!
The question wants you to avoid using the second derivative but rather "consider the gradient on each side of $x=e$".
$\text{So we just sub some numeric values in.}\\ \text{For a point on the left, we can try }x=2.\\ \text{When }x=2\text{, the calculator says that }\frac{dy}{dx} = 0.0767\dots > 0$
Therefore on the left of $x=e$, the curve is increasing.
$\text{For a point on the right, we can try }x=3.\\ \text{When }x=3\text{, the calculator says that }\frac{dy}{dx} = -0.0109\dots < 0$
Therefore on the right of $x=e$, the curve is decreasing.
$\text{So the curve increases, until it reaches }x=e.\\ \text{Then after }x=e\text{, it decreases.}\\ \text{That's all that is required to conclude that }x=e\\ \text{must therefore be a local maximum.}$
This method (which is really just a table of values) should be noted, because sometimes computing the second derivative is just nastily exhausting effort. The whole point of this method is to get you out of that trouble.
Remark: Second derivative
$\text{If you insist on using the second derivative}\\ \text{you more or less need the quotient rule }\textit{again}.$
\begin{align*}\frac{dy}{dx} &= \frac{x\cdot \frac1x - 1 \cdot \ln x}{x^2}\\ &= \frac{1-\ln x}{x^2}\\ \therefore \frac{d^2y}{dx^2} &= \frac{x^2\cdot -\frac1x - 2x(1-\ln x)}{x^4}\end{align*}
$\text{At this point, you can now sub }x=e\text{ in straight away.}$
This can optionally be simplified to $\frac{d^2y}{dx^2} = \frac{-3x+2x\ln x}{x^4} = \frac{-3+2\ln x}{x^3}$ but you don't really need to. You can just sub $x=e$ into the unsimplified expression.

For this question though, in theory you should not do this. This is because they told you to use the other method.