Spec Exam 1

[Carlamaker]

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[AAGlue]

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Why is every single maths exam this year harder than last years’ ? 

[99.95_goal]

It wasn't easy but it was expected as compared to methods(looool)
except for that modulus question tho, definitely threw me a bit

[hairs9]

I found it manageable but it took me longer than it usually takes to complete the exam. There was a lot of algebra on it and it's super easy to make mistakes in working or by using mathematical conventions of writing. But I thought it was fair overall

[AAGlue]

Same lol

Was anyone able to finish the last question?

[Xu ZhongYu]

easier than Methods 

[fish12]

Same lol

Was anyone able to finish the last question?

yes but only just. I managed to figure out it was DOPs which saved time

[AAGlue]

easier than Methods 

True tho

[AAGlue]

yes but only just. I managed to figure out it was DOPs which saved time

Omg it took me five whole minutes to find the simplified expression
And idk if my final answer is even accurate lol

[ZelezReich]

Will the A+ cutoff be higher or lower than previous years?

[99.95_goal]

Will the A+ cutoff be higher or lower than previous years?

It was harder than last year's - the cut off was 37, so this year it might be 33? the modulus question definitely threw many people(including me lol) and there was so much ugly algebra so

[S_R_K]

Suggested solutions - please let me know of any errors:

UPDATED: worked solutions attached to this post.

Question 1a

\(\frac{4g-10\sqrt{3}-5}{2}\)

Question 1b

\(\frac{10-5\sqrt{3}}{4}\)


Question 1c

\(20-10\sqrt{3}\)


Question 2

\(-\frac{10}{3}+\frac{8\sqrt{2}}{3}\)

Question 3

cis\(\left(\frac{7\pi}{12}\right)\), cis\(\left(-\frac{\pi}{12}\right)\), cis\(\left(-\frac{3\pi}{4}\right)\)

Question 4

\(x \in \left(-\infty, \frac{7-\sqrt{5}}{2}\right)\)

Question 5a

\(m=4\)

Question 5b

\(\frac{1}{18}\left(47\vec{i}-10\vec{j}+7\vec{k}\right)\)

Question 6a

By chain rule, \(f'(x)=3\times \frac{1}{(3x-6)^2+1}=\frac{3}{9x^2-36x+37}\) as required.


Question 6b

\(f''(x)=\frac{-3(18x-36)}{\left(9x^2-36x+37\right)^2}\)
\(f''(x)=0 \rightarrow 18x-36=0 \rightarrow x=2\).
Since \(\left(9x^2-36x+37\right)^2 > 0\) for all \(x\), \(f''(x) > 0\) when \(x<2\)and \(f''(x) < 0\) when \(x>2\), so \(f''(x)\) changes sign at \(x=2\).

Question 6c

Graph of arctan function with horizontal asymptotes \(y=\frac{\pi}{2},\frac{3\pi}{2}\), and point of inflection at \((2,\pi)\)

Question 7a

\(f(x)\) is continuous everywhere if \(m+n=\frac{4}{1+1^2} \rightarrow m+n=2\).
\(f'(x)\) is continuous everywhere if \(m=\frac{-8\times 1}{(1+1^2)^2} \rightarrow m=-2\).
Hence, \(m=-2,n=4\)

Question 7b

\(3+\frac{\pi}{3}\)

Question 8

\(2\pi\left(\log_e\left(2\sqrt{3}+2\right)+\frac{\pi}{3}\right)\)

Question 9a

\(\begin{aligned}[t]\left(\frac{dy}{dt}\right)^2 &= \left(\frac{1}{1+t}-\frac{1}{4(1-t)}\right)^2 \\
&= \frac{1}{(1+t)^2}-\frac{1}{2(1-t)(1+t)}+\frac{1}{16(1-t)^2} \\
&= \frac{1}{(1+t)^2}-\frac{1}{2(1-t^2)}+\frac{1}{16(1-t)^2}\end{aligned}\)
So \(a=1, b=-2, c=16\) as required.

Question 9b

\(\log_e\left(\frac{3}{2}\right)-\frac{1}{4}\log_e\left(\frac{1}{2}\right)\)

[AAGlue]

Suggested solutions - please let me know of any errors:

Question 1a

\(\frac{10\sqrt{3}+5}{2}\)

Question 1b

\(\frac{10-5\sqrt{3}}{4}\)


Question 1c

\(20-10\sqrt{3}\)


Question 2

\(-\frac{10}{3}+\frac{8\sqrt{2}}{3}\)

Question 3

cis\(\left(-\frac{7\pi}{12}\right)\), cis\(\left(\frac{\pi}{12}\right)\), cis\(\left(\frac{3\pi}{4}\right)\)

Question 4

\(x \in \left(-\infty, \frac{7-\sqrt{5}}{2}\right)\)

Question 5a

\(m=4\)

Question 5b

\(\frac{1}{18}\left(47\vec{i}-10\vec{j}+7\vec{k}\right)\)

Question 6a

By chain rule, \(f'(x)=3\times \frac{1}{(3x-6)^2+1}=\frac{3}{9x^2-36x+37}\) as required.


Question 6b

\(f''(x)=\frac{-3(18x-36)}{\left(9x^2-36x+37\right)^2}\)
\(f''(x)=0 \rightarrow 18x-36=0 \rightarrow x=2\).
Since \(\left(9x^2-36x+37\right)^2 > 0\) for all \(x\), \(f''(x) > 0\) when \(x<2\)and \(f''(x) < 0\) when \(x>2\), so \(f''(x)\) changes sign at \(x=2\).

Question 6c

Graph of arctan function with horizontal asymptotes \(y=\frac{\pi}{2},\frac{3\pi}{2}\), and point of inflection at \((2,\pi)\)

Question 7a

\(f(x)\) is continuous everywhere if \(m+n=\frac{4}{1+1^2} \rightarrow m+n=2\).
\(f'(x)\) is continuous everywhere if \(m=\frac{-8\times 1}{(1+1^2)^2} \rightarrow m=-2\).
Hence, \(m=-2,n=4\)

Question 7b

\(3+\frac{\pi}{3}\)

Question 8

\(2\pi\left(\log_e\left(2\sqrt{3}+2\right)+\frac{\pi}{3}\right)\)

Question 9a

\(\begin{aligned}[t]\left(\frac{dy}{dt}\right)^2 &= \left(\frac{1}{1+t}-\frac{1}{4(1-t)}\right)^2 \\
&= \frac{1}{(1+t)^2}-\frac{1}{2(1-t)(1+t)}+\frac{1}{16(1-t)^2} \\
&= \frac{1}{(1+t)^2}-\frac{1}{2(1-t^2)}+\frac{1}{16(1-t)^2}\end{aligned}\)
So \(a=1, b=-2, c=16\) as required.

Question 9b

\(\log_e\left(\frac{3}{2}\right)-\frac{1}{4}\log_e\left(\frac{1}{2}\right)\)

Shouldn’t Q3 be -pi/12?
Because the cartesian form of the cube was like 1/sqrt(2) - 1/sqrt(2)i or smth

[S_R_K]

Shouldn’t Q3 be -pi/12?
Because the cartesian form of the cube was like 1/sqrt(2) - 1/sqrt(2)i or smth

Yes, I copied the question down incorrectly. Ouch.

[AAGlue]

Yes, I copied the question down incorrectly. Ouch.

Tysm for the answers tho