2025 HSC Extension 2 Exam Paper (1 Viewer)

ivanradoszyce · Saturday at 8:31 AM

Hi all,

Is there anywhere I can get a PDF of this year's exam paper?

Cheers

preptile6 · Saturday at 8:55 AM

See attached file for pdf copy.

ivanradoszyce · Saturday at 11:50 AM

Cheers.

coolcat6778 · Saturday at 3:07 PM

Copied from VCE

I literally saw this question before but didn't bother with it as I didn't think it could be asked in HSC extension 2.
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Copied from a 2025 WME paper

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They fucking plagiarized the entire test

tywebb · 2025-10-26T04:58:03+1100

and this one

in terry lee book

tywebb · 2025-10-26T05:20:32+1100

coolcat6778 said:
They fucking plagiarized the entire test

well maybe not the entire test

so what wasn't plagiarised?

how about a WRONG question? I bet that wasn't plagiarised.

like this

so why wrong?

consider this solution

$A=\int_2^4\frac{e^x}{x-1}\ \!dx\text{ is a non-zero constant}$

$\int_{m-4}^{m-2}\frac{e^{-x}}{x-m+1}\ \!dx\text{ is a non-zero constant for a constant }m$

$\therefore\int_{m-4}^{m-2}\frac{e^{-x}}{x-m+1}\ \!dx=kA,\text{ where }k\text{ and }m\text{ are constants}$

or perhaps reimagining this as a primary school question

$\text{Show that }676767\text{ is a constant multiplied by }67$

$\text{Answer: }676767=10101\times67\text{ and }10101\text{ is a constant}$

What they might have meant was something like

$\text{Show that for a constant }m, \int_{m-4}^{m-2}\frac{e^{-x}}{x-m+1}\ \!dx=kA\text{ where }k=-e^{-m}$

so bets are on that they will change the question when they put it on their website

coolcat6778 · 2025-10-27T01:23:48+1100

Notice the different font especially the brackets, likely indicative of a different group of exam authors.

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2025 HSC Extension 2 Exam Paper (1 Viewer)

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