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January 29, 2020, 01:34:24 pm

### AuthorTopic: 3U Maths Question Thread  (Read 605137 times) Tweet Share

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#### JkbC

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« Reply #4260 on: January 03, 2020, 11:40:32 am »
0
Are we supposed to assume that $g = 9.8$ or something here?

Yes, I probably should have given you this info as well. Cambridge states this:

19 NEWTONS SECOND LAW AND THE UNITS OF FORCE
 One newton, written in symbols as 1 N, is the force required to accelerate a body at a rate of 1 m/s^2 .
 Newtons second law of motion says that F = ma.
If a body of mass m kg is accelerating at a m/s^2 , then the sum of all the forces acting on the body has magnitude F = ma newtons, and acts in the same direction as the acceleration.
 One kilogram weight is the downward force due to gravity on a mass of 1 kg at the Earths surface.
 Acceleration due to gravity at the Earths surface has the symbol g, whose approximate value is 9.8 m/s^2
 One newton is therefore about kg weight  about the downward force due to gravity of a 100-gram apple on your open hand.

(Thanks a lot for your help with the previous question as well!)

#### JkbC

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« Reply #4261 on: January 03, 2020, 11:45:08 am »
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 One newton is therefore about kg weight  about the downward force due to gravity of a 100-gram apple on your open hand.

(I can't edit my previous message for some reason) This is meant to say: One newton is therefore about 1/10 kg weight  about the downward force due to gravity of a 100-gram apple on your open hand.

#### RuiAce

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• Respect: +2490 ##### Re: 3U Maths Question Thread
« Reply #4262 on: January 03, 2020, 12:20:54 pm »
+4
For instance, here's one way of doing Q14. Notes:
- There's many different ways of doing b). This is just my preferred method ala-new MX1 syllabus style. (But all methods require somehow splitting the vector up into its $\mathbf{i}$ and $\mathbf{j}$ components. This is almost always done through right-angled trigonometry.)
- c) could've been done more quickly for this question by using right-angled trigonometry. The dot product is just the more generic way of going about angles involving vectors.

You should still have an attempt at Q6 nonetheless, and come back if you're stuck somewhere. Or of course, feel free to ask for some elaboration with Q14.
« Last Edit: January 03, 2020, 12:26:13 pm by RuiAce »  #### shekhar.patel

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« Reply #4263 on: January 08, 2020, 03:37:15 pm »
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Hi, I need some assistance in this question. I am able to draw the diagram and see that its in the ratio 1:2 but I am not sure how to prove it.
Prove that the area of the curve y=x^2 divides the rectangle enclosed by the x-axis, the y-axis, the lines x=a and y=a^2 in the ratio 1:2

#### DrDusk ##### Re: 3U Maths Question Thread
« Reply #4264 on: January 08, 2020, 04:29:41 pm »
0
Hi, I need some assistance in this question. I am able to draw the diagram and see that its in the ratio 1:2 but I am not sure how to prove it.
Prove that the area of the curve y=x^2 divides the rectangle enclosed by the x-axis, the y-axis, the lines x=a and y=a^2 in the ratio 1:2
Just integrate both of the areas.

HSC/Prelim Physics tutor

#### JkbC

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« Reply #4265 on: January 13, 2020, 07:08:35 pm »
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Hi, I'm stuck on this MI from Cambridge (Y12 3U: Ex 2A Q10).

"Prove by mathematical induction that for all positive integers n,
(n + 1) (n + 2) (n + 3) × · · · × 2n = 2^n × ( 1 × 3 × 5 × · · · × (2n − 1) ) ."

I'm used to seeing the · · ·'s only on one side, so I'm not sure what to do when they're on both sides. Thanks in advance!

#### RuiAce

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« Reply #4266 on: January 13, 2020, 10:06:23 pm »
+2
Hi, I'm stuck on this MI from Cambridge (Y12 3U: Ex 2A Q10).

"Prove by mathematical induction that for all positive integers n,
(n + 1) (n + 2) (n + 3) × · · · × 2n = 2^n × ( 1 × 3 × 5 × · · · × (2n − 1) ) ."

I'm used to seeing the · · ·'s only on one side, so I'm not sure what to do when they're on both sides. Thanks in advance!
(Also from the 1999 exam if I recall. It appears both there and Cambridge.)
$\text{When }n=1,\\ LHS = 1\times 2 = 2,\\ RHS = 2^1(1) = 2.$
$\text{Assume that}\\ (k+1)(k+2)(k+3)\cdots 2k = 2^k (1\times 3\times 5\times \cdots \times (2k-1))$
Everyone always gets confused at this question because it's a bit tricky writing out what you need to prove for the inductive step. On the $RHS$, this should be a bit more obvious - simply increment the power by 1, and add in the next term.

Now on the left, both the first term and the last term are affected here, since they both have $k$'s in them. When $n=k$, the first term was $k+1$. When $n=k+1$, the first term is $k+2$.

When $n=k$ the last term was $2k$. When $n=k+1$, the last term thus must be $2(k+1)$, i.e. $2k+2$.

So in the inductive step, we wish to prove that
$(k+2)(k+3)\cdots (2k)(2k+1)(2k+2) = 2^{k+1} (1\times 3\times 5\times \cdots \times (2k-1)(2k+1))$
\begin{align*}
LHS &= (k+2)(k+3)\cdots (2k)(2k+1)(2k+2)\\
&= 2(k+1)(k+2)(k+3)\cdots (2k)(2k+1)\\
&= 2\left[2^{k} (1\times3\times5\times\cdots\times(2k-1) \right](2k+1)\\
&= 2^{k+1} (1\times3\times5\times\cdots\times(2k-1)(2k+1))
\end{align*}  