# 3U Maths HSC 2019 Solutions (1 Viewer)

##### New Member
So I was bored and I decided to do the HSC 3U Maths for 2019. Please tell me if any of the answers are wrong. I am not a HSC student lol. In my opinion the exam was harder than last year so hopefully it scales better. Other than that have a look a through it

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#### Arrowshaft

##### Active Member
So I was bored and I decided to do the HSC 3U Maths for 2019. Please tell me if any of the answers are wrong. I am not a HSC student lol. In my opinion the exam was harder than last year so hopefully it scales better. Other than that have a look a through it
lol I made the exact same silly mistake, the integral isn’t tan inverse, it’s a natural log, as u=cos^2x, otherwise everything looks correct to me

#### integral95

##### Well-Known Member
So I was bored and I decided to do the HSC 3U Maths for 2019. Please tell me if any of the answers are wrong. I am not a HSC student lol. In my opinion the exam was harder than last year so hopefully it scales better. Other than that have a look a through it
You can't justify your proof in 14bii) with just that

$\bg_white f(a)

But that's only true if the function is strictly increasing (which is not true ripp)

#### integral95

##### Well-Known Member
So I was bored and I decided to do the HSC 3U Maths for 2019. Please tell me if any of the answers are wrong. I am not a HSC student lol. In my opinion the exam was harder than last year so hopefully it scales better. Other than that have a look a through it
So you actually would also have to include the fact that
$\bg_white f'(x) >0, \forall x > \frac{2k}{3}$
Which then you know the function is increasing for between the 3 values

##### New Member
You can't justify your proof in 14bii) with just that

$\bg_white f(a)

But that's only true if the function is strictly increasing (which is not true ripp)
but what if you proved that for a continuous function,

$\bg_white f(a) <0 < f(c)$ and $\bg_white a < c$
Therefore b , which is the only root, such that f(b)=0 has to lie between a and c for the curve to be both smooth and continuous

therefore $\bg_white a < b< c$

##### New Member
lol I made the exact same silly mistake, the integral isn’t tan inverse, it’s a natural log, as u=cos^2x, otherwise everything looks correct to me
Yeah I realised that mistake, thanks for pointing it out

#### Drongoski

##### Well-Known Member
$\bg_white \int \frac{sin 2x}{4+cos^2 x} dx =\int \frac {sin 2x}{4 + \frac{1}{2} (1 + cos 2x)}dx = \int \frac {2sin 2x}{9 + cos 2x}dx \\ \\ = \int \frac {-d(9 + cos 2x)}{9 + cos 2x} = -ln|9+cos 2x| + C\\ \\ \therefore \int _0 ^{\frac {\pi}{4}} = -[ln(9 + 0) - ln(9 + 1)] = ln \frac {10}{9}$

Is this right??

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#### jathu123

##### Active Member
Does anyone have the pdf of the exam? Can't seem to find it on any thread (or im just blind haha)

#### integral95

##### Well-Known Member
but what if you proved that for a continuous function,

$\bg_white f(a) <0 < f(c)$ and $\bg_white a < c$
Therefore b , which is the only root, such that f(b)=0 has to lie between a and c for the curve to be both smooth and continuous

therefore $\bg_white a < b< c$
This is why you shouldn't look at math at 1am lel..
Yeah you're right cause I was thinking of a counter example where you could have you roots

$\bg_white \alpha < a< c$

But your idea would negate that.

#### Arrowshaft

##### Active Member
$\bg_white \int \frac{sin 2x}{4+cos^2 x} dx =\int \frac {sin 2x}{4 + \frac{1}{2} (1 + cos 2x)}dx = \int \frac {2sin 2x}{9 + cos 2x}dx \\ \\ = \int \frac {-d(9 + cos 2x)}{9 + cos 2x} = -ln|9+cos 2x| + C\\ \\ \therefore \int _0 ^{\frac {\pi}{4}} = -[ln(9 + 0) - ln(9 + 1)] = ln \frac {10}{9}$

Is this right??
Oooh I love that merhod! But /sin and /cos lel

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#### Arrowshaft

##### Active Member
[/QUOTE]
Yeah I realised that mistake, thanks for pointing it out
How many marks do you reckon I’d lose if I made the exact same mistake as you?