November 13, 2019, 07:49:50 am

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#### fun_jirachi

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« Reply #4380 on: October 20, 2019, 01:01:36 pm »
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Hey there!

I think it might be an issue with interpreting the question.

I agree with you, when the question is worded like that, the answer for picking a heart would be 13/51 given Gerry picks the queen of spades then picks any heart (thus implying no replacement). However, if we try to justify the question's answer, there would be two possibilities I can think of (which are basically one possibility)

a) There's replacement
b) They're two separate events ie What is the probability of Gerry picking a queen of spades as opposed to a heart

Just make sure you check the wording of the question very carefully, because one or two words alter what the question implies entirely.

Hope this helps
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#### Hawraa

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« Reply #4381 on: October 20, 2019, 01:10:15 pm »
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Hey there!

I think it might be an issue with interpreting the question.

I agree with you, when the question is worded like that, the answer for picking a heart would be 13/51 given Gerry picks the queen of spades then picks any heart (thus implying no replacement). However, if we try to justify the question's answer, there would be two possibilities I can think of (which are basically one possibility)

a) There's replacement
b) They're two separate events ie What is the probability of Gerry picking a queen of spades as opposed to a heart

Just make sure you check the wording of the question very carefully, because one or two words alter what the question implies entirely.

Hope this helps

But which one do you think U should go with if I get a question like this in the exam?

#### spnmox

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« Reply #4382 on: October 20, 2019, 02:30:46 pm »
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Hey guys, need help with attached MC. I integrated as normal and just put some abs value signs around it, I got A but answer is C

Also, for v=2- 4/(t+1). Find the exact distance travelled by the particle in the first 7 seconds.

Thanks so much 5 more days !!
« Last Edit: October 20, 2019, 03:00:49 pm by spnmox »

#### fun_jirachi

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« Reply #4383 on: October 20, 2019, 03:48:54 pm »
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But which one do you think U should go with if I get a question like this in the exam?

Always go with what the question implies But there's no way the HSC will be this ambiguous ever, so you shouldn't have to worry about it at all.

Hey guys, need help with attached MC. I integrated as normal and just put some abs value signs around it, I got A but answer is C

Also, for v=2- 4/(t+1). Find the exact distance travelled by the particle in the first 7 seconds.

Thanks so much 5 more days !!

Hey there!

You can't just 'integrate as normal' here and then stick some absolute value signs; that's always very risky and nearly always wrong. A rule of thumb is to never integrate absolute values conventionally

There's a few better ways to consider this:
a) Draw up the graph, then calculate the area using triangles
b) Split up the integral using the definition of the absolute value like so
$\int_{-3}^2 |x+1| dx = \int_{-3}^{-1} (-x-1) dx + \int_{-1}^2 x+1 dx$

For your second question, log always denotes ln ie. they are the same thing except on your calculator, where log is log base 10, while ln is log base e. They are the same answer

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#### violet123

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« Reply #4384 on: October 20, 2019, 07:20:17 pm »
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Hi,
why does sin (x+) = -sin(x)?

#### DrDusk

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« Reply #4385 on: October 20, 2019, 07:26:33 pm »
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Hi,
why does sin (x+) = -sin(x)?

$\sin(x + \pi) = \sin x \cos\pi + \cos x \sin\pi$

$= -\sin x$
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#### Kombmail

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« Reply #4386 on: October 21, 2019, 10:36:54 am »
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Guys for this question,
Consider the function f(x)= xe^-x it says show that f’’(x) = e^-x(x-2) would you use product rule on the xe^-x?

Ps. Can anyone help me with understanding how to sketch this? Indicating stationary points and nature..
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#### fun_jirachi

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« Reply #4387 on: October 21, 2019, 12:02:37 pm »
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Hey there!

Yes, you would have to use the product rule.

In sketching this:
a) You would've found f'(x) by finding f''(x) --> Note that there is a singular stationary point at x=1, where f'(x)=0, and since f''(1) is negative, this indicates that the stationary point is a local maximum
b) Similarly, there is a point of inflexion at x=2
c) Have a look at the values of f(x) as x approaches positive and negative infinity to find any asymptotes
d) Clearly also, f(0)=0, so put that point on your graph, as well as the inflexion point (2, 2/e2) and the local max (1, 1/e)

Hope this helps

EDIT: Typo as kindly pointed out by Hawraa and Rui
« Last Edit: October 21, 2019, 02:14:41 pm by fun_jirachi »
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#### Hawraa

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« Reply #4388 on: October 21, 2019, 12:44:30 pm »
+1
Hey there!

Yes, you would have to use the product rule.

In sketching this:
a) You would've found f'(x) by finding f''(x) --> Note that there is a singular stationary point at x=1, where f'(x)=0, and since f''(1) is positive, this indicates that the stationary point is a local maximum
b) Similarly, there is a point of inflexion at x=2
c) Have a look at the values of f(x) as x approaches positive and negative infinity to find any asymptotes
d) Clearly also, f(0)=0, so put that point on your graph, as well as the inflexion point (2, 2/e2) and the local max (1, 1/e)

Hope this helps

Isn't like when the second derivative is positive then it is a minimum, not maximum?

#### RuiAce

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« Reply #4389 on: October 21, 2019, 12:48:41 pm »
+1
Isn't like when the second derivative is positive then it is a minimum, not maximum?
Yeah. Think he just made a typo. The second derivative should be negative when $x=1$, and hence we have a local maximum.

$\sin(x + \pi) = \sin x \cos\pi + \cos x \sin\pi$

$= -\sin x$
This formula is not taught in the 2U course. It's a valid cheat from a 3U context, but we cannot assume 2U students know it.
Hi,
why does sin (x+) = -sin(x)?
Your LaTeX isn't showing up, but if like DrDusk said you meant $\sin(x+\pi)$ on the left, you basically need to think back to ASTC.

Throughout the course, you've done things like $\sin \left( \frac{5\pi}{4} \right) = -\sin \frac\pi4$, and $\sin \frac{3\pi}{4} = \sin \frac\pi4$. These were done identifying things like $\frac{5\pi}4$ is in the 3rd quadrant and $\frac{3\pi}{4}$ is in the 2nd quadrant. But you may have learnt it as nothing more but a trick, without thinking about why they work.

In reality, the identity $\boxed{\sin(\pi + x) = -\sin x}$ for the 3rd quadrant, along with other identities for the 2nd and 4th quadrants, are the formulas that let you do this in the first place. In general, depending on which quadrant you're in, you can use the `ASTC' trick to figure out whether the trig ratio is positive or negative. But the actual formulae being used are:
\begin{align*}
\sin(\pi - x) &= \sin x\tag{+'ve in 2nd quad}\\
\sin(\pi + x) &= -\sin x\\
\sin (2\pi - x) &= -\sin x
\end{align*}
\begin{align*}
\cos(\pi - x) &= -\cos x\\
\cos(\pi + x) &= -\cos x\\
\cos(2\pi - x) &= \cos x\tag{+'ve in 4th quad}
\end{align*}
\begin{align*}
\tan(\pi - x) &= -\tan x\\
\tan (\pi + x) &= \tan x\tag{+'ve in 3rd quad}\\
\tan(2\pi - x) &= -\tan x
\end{align*}
If you're genuinely interested in derivations of such results, you have to go back to the unit circle definition of the trigonometric ratios. This video can help get you started.
« Last Edit: October 21, 2019, 01:00:07 pm by RuiAce »

#### Youssefh_

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« Reply #4390 on: October 21, 2019, 03:20:19 pm »
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Hi please just in need of quick assistance, this question is so easy and I know it is easy, but I don't know why I dont know how to find the radius from the equation of the tangent, can you please quickly explain

#### fun_jirachi

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« Reply #4391 on: October 21, 2019, 03:25:48 pm »
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If a line is tangent to a circle, we know that the distance from the centre to the point of contact is the radius, which meets the tangent at a right angle. Basically, we substitute (3, -2) into the perpendicular distance formula, then use the equation of a circle (x-h)2+(y-k)2=r2 to find the answer (given the centre (h, k)).
Hope this helps

Hey there!

Hope this helps
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#### annabeljxde

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« Reply #4392 on: October 21, 2019, 08:39:34 pm »
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Hi!
I'm stuck on this 2010 HSC Question (8d)

I got the required k value, which was 3, but the answer said the inequality was k≥3, while I got k≤3.

The marker's notes states that many students were unable to recognise the necessary sign change. Can someone please explain why the signs need to change?

Thank you.
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#### DrDusk

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« Reply #4393 on: October 21, 2019, 08:58:41 pm »
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Hi!
I'm stuck on this 2010 HSC Question (8d)

I got the required k value, which was 3, but the answer said the inequality was k≥3, while I got k≤3.

The marker's notes states that many students were unable to recognise the necessary sign change. Can someone please explain why the signs need to change?

Thank you.

$f(x) = x^3 - 3x^2 +kx + 8 \\ \therefore f'(x) = 3x^2 - 6x + k \\ \text{Now this is a quadratic so we can solve for it's roots}\hspace{2mm}x = \dfrac{6 \pm \sqrt{36 - 12k}}{6} \\ \text{For a quadratic the way we can always guarantee that it is never negative is if it has no real roots} \\ \text{Because then it will always be above the x axis} \\ \text{We must therefore set the discriminant to be less than or equal to 0} \\ \therefore 36 - 12k \leq 0 \Rightarrow k \geq 3$

Something my tutor always told me was always consider what everything means physically. Like what does discriminant mean and how does it affect the roots of the polynomial? What is it's physical significance?
That way of thinking will help with more conceptual questions or this one for example where you have to link the fact that the discriminant implies the nature of the roots, i.e. are they real or not which will allow you to visualize the polynomial.
« Last Edit: October 21, 2019, 09:03:09 pm by DrDusk »
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#### Kombmail

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