Evaluation of HSC 2015 Maths Extension 1 Exam - Prelim versus HSC Content

I took a look at the HSC 2015 Math Extension 1 paper and concluded that about 1/3 is Prelim, and 2/3 is HSC.

The Prelim questions are easier in general than those with HSC content.
Strategy wise, I think it would be good to get the Prelim questions 100% correct! so it's a good idea to ensure proficiency in all the preliminary topics.
Holidays are a good time to do that sort of thing.
Here are the BOSTES 3U syllabus topics. 
We have essentially finished the preliminary content.

Here is my break down of Prelim/HSC content for those interested :-)

Merry Christmas, Happy New Year, and Fantastic Festive Season to everyone. 

Regards, :-) JH

EXTRA-CURRICULAR EXTENSION - Combining Binomial Theorem, Differentiation and Induction !! Differentiation of the product of two functions $n$ times - A Formula

This is outside of syllabus but many of you should extend yourself beyond the syllabus a bit. Is there a formula for differentiating the product of two functions $$y=uv$$  many times over (say, $n$ times)

To get an idea of what the answer to this might be, actually perform the differentiation several times over to see what you get. So find the first derivative, the second derivative, the third derivative, and the fourth derivative. Is there a pattern? Here are the answers.

First derivative,
${d \over dx } (uv) = u'v+uv'$

Second derivative,
$\begin{array}{rl}{d^2 \over dx^2 } (uv) & = u''v+u'v'+u'v'+uv''\\ &= u''v+2u'v'+uv''\end{array}$

Third derivative,
$\begin{array}{rl}{d^3 \over dx^3 } (uv)
& = u'''v+u''v'+2u''v'+2u'v''+u'v''+uv'''\\
& = u'''v+3u''v'+3u'v''+uv'''\end{array}$

Fourth derivative,
$\begin{array}{rl}{d^4 \over dx^4 } (uv)
& = u''''v+u'''v'+3u'''v'+3u''v''+3u''v''+3u'v'''+u'v'''+v''''\\
& = u''''v+4u'''v'+6u''v''+4u'v'''+v''''\end{array}$ 

Can you see that the coefficients (numbers in front) follow the pattern of Pascal's Triangle below? 

1

1   1

1  2  1

1  3   3  1

1  4  6   4   1

Here is the general case. It uses the binomial theorem and sigma notation, so hopefully you have seen these before.

THEOREM:
If $f(x) = uv$ where $u$ and $v$ are functions of $x$ then the
nth derivative of this product is given by
$$f^{(n)}(x)={d^n \over dx^n } (uv)=\sum^n_{r=0} {n\choose r} u^{(r)}v^{(n-r)}$$
where we use the notation for the $r^{th}$ derivative

$\displaystyle u^{(r)}={ d^r\over dx^r} u$

(When it's a derivative there are brackets around the 'index')

Using this notation differentiating one more time looks like this
$$\displaystyle{ d\over dx}\left( u^{(r)}\right) = { d\over dx} { d^r\over dx^r} u={ d^{r+1}\over dx^{r+1}}=u^{(r+1)}$$
PROOF: 
By induction on $n$.
For $n=0$, $f^{(0)}(x)=f(x)=uv$ true. and

$ {0\choose 0} u^{(0)}v^{(0)}=uv$ and so it is true for $n=0$.

[Remark, $n=0$ corresponds to not differentiating at all, so it should be the original product $uv$.]

For $n=1$,

$ {1\choose 0} u^{(0)}v^{(1)}+{1\choose 1} u^{(1)}v^{(0)}=uv'+u'v$ and so it is true for $n=1$.


[Remark, $n=1$ corresponds to differentiating once, the result is nothing more than the usual product rule for two functions.]

Assume the statement is true for $n=k$, prove it is true for $n=k+1$.


$\begin{array}{rcl}f^{(k+1)}(x)&=&{ d\over dx} f^{(k)}(x)\\
{ \;}&=&\displaystyle{ d\over dx} \sum^k_{r=0} {k\choose r} u^{(r)}v^{(k-r)}\\
&=&\displaystyle
\sum^k_{r=0} {k\choose r} u^{(r+1)}v^{(k-r)}+
\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\
&=&\displaystyle
\sum^{k+1}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}+

\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\\end{array}$

where we changed variables to $s=r+1$ in the first summation.

$\begin{array}{lcl}
f^{(k+1)}(x) &= &\displaystyle{k\choose k} u^{(k+1)}v^{(0)}+{k\choose 0} u^{(0)}v^{(k+1)}\\&&\displaystyle +\sum^{k}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}\\&&\displaystyle +
\sum^k_{r=1} {k\choose r} u^{(r)}v^{(k+1-r)}\\&= &\displaystyle u^{(k+1)} + v^{(k+1)}\\&&\displaystyle +\sum^{k}_{r=1} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)} \\
&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} {k+1\choose r}u^{(r)}v^{(k+1-r)}\end{array}$
and so it is proven true for $n=k+1$. Therefore by induction the statement is true for all integers $n\ge 0$.

SUGGESTED EXTRA PROBLEM.

Can you apply this formula (or otherwise) and find an expression for the nth derivative of the function $\displaystyle y={u\over v}$?

BOSTES Formula Reference Sheet 2U/3U/4U

http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths-ref-sheet.pdf

GROVE HSC 3U EX2.8 - REGARDING Q15

JUst regarding finding the second derivative of this (good luck to you!)

WOLFRAM ALPHA SAYS

AND THIS CAN NEVER BE ZERO, SO THERE ARE NO INFLEXIONS.

(yes x here should be h)

GROVE HSC 3U - EX10.3 - Q16

GROVE HSC 3U - EX9.2 - Q5

CAMB Y12 - 3U - EX6E - Q4ab

GROVE HSC 3U - EX2.8 - SIMILAR QUESTION

Question: Find any non horizontal inflexions.

(by finding second derivative)

[this is not Q15, but looks similar.

It's good practice for derivatives]

GROVE HSC 3U TEST YOURSELF 10 - Q2

GROVE HSC 3U TEST YOURSELF 10 - Q13

GROVE HSC 3U. TEST YOURSELF 10 - Q19

GROVE HSC 3U - TEST YOURSELF 10 - Q13

Q13

GROVE HSC 3U - Q15 test yourself 10

GROVE HSC 3U - TEST YOURSELF 10 - Q5

(a)  Put $x=0$  on LHS and RHS!

(b)  Put $x=-1$  on LHS and RHS!

Keep in mind $(-1)^2=1$, $(-1)^3=-1$, $(-1)^4=1$, ...

GROVE HSC 3U - TEST YOURSELF 9 (APPROX OF ROOTS) - Q6

It's up to you to find to values of $x$ such that the function value changes sign. 
It will be easy trial and error. (unlikely in an actual test)

Let's randomly (with a bit of intuition) try $x=0$ and $x=2$! 

$f(x)=x^3+3x-7$

$f(0)=-7<0$
$f(2)=2^3+3(2)-7=8+6-7=7>0$

Since $f(0)<0$ and $f(2)>0$, then there is at least one root  between $x=0$ and $x=2$.  

Now proceed to half the interval 3 times.....

GROVE HSC 3U (APPROX OF ROOTS) EX 9.1 - Q10 -

CAMBRIDGE 3U YEAR 12 - EX 6E Q8a

CAMBRIDGE 3U YEAR 12 - EX 5E Q7c Binomial Theorem - coefficient of greatest absolute value

Grove HSC 3U EX 10.4 Q20c Binomial Theorem - Greatest Coefficient (a second version by Ellie & Tat :-))

This is a second version of this solution.

Thanks Tat & Ellie

Regarding which variables get $k$ and which get $9-k$ in the expansion of $(x+3)^9$

Regarding



QUESTION

Hi sir in this question why can you swap the position of the $3$ and the $x$? because when Bec and I were trying to do it now, we put the power of $(9-k)$ on the $x$ term....​

ANSWER

Good question. The coefficients are totally symmetric just like Pascals Triangle, and can be seen in that rule we have proved before, namely


$\Large{n\choose k}={n\choose n-k}$


Now I am going to write the expansion of $\displaystyle (x+3)^9$ in two ways. They are equivalent and it doesn't matter which one you use, the answer is the same. I would suggest you put the $k$ on the more complicated expression as this makes algebra easier.

$$\normalsize\displaystyle (x+3)^9=\sum_{k=0}^n {n\choose k}x^{9-k}\, 3^k = \sum_{k=0}^n {n\choose k}x^{k}\,3^{9-k}$$

If you expand either version out fully the two expansions will be identical. The $k$ is just a dummy variable.

FACT SHEET BINOMIAL THEOREM

REMARK: IF YOU CHOOSE TO MEMORISE FORMULAS LIKE F.6 PLEASE DO MEMORISE THE WHOLE LINE!

WORKSHEET - REVISION OR NOT - SOLUTIONS

WORKSHEET - REVISION OR NOT - SOLUTIONS

Topics: Approximation of roots, Newtons method, halving the interval method, Binomial Theorem, Further graphs - involving stationary points, inflexions, horizontal, vertical asymptotes.

Binomial theorem GROVE HSC 3u EX 10.4 Q3

Fanks Tatiana :-)

Further Graphs Differentiating fun.

GROVE HSC 3U EX10.5 Q6 Finding the coefficient of $x^2$

BINOMIAL THEOREM WORSHEET MATHS PLUS 2001 + SOLUTIONS

BINOMIAL THEOREM WORSHEET MATHS PLUS 2001 + SOLUTIONS

GROVE HSC 3U EX2.8 Q1

GROVE HSC 3U EX 10.5 Q3

tHANKS hOLSTA :-)

GROVE HSC 3U Q21a Greatest coefficient

GROVE HSC 3U EX10.4 Q20a

GROVE HSC 3U EX10.4 Q18a

Thanks Juliette :-)

GROVE HSC 3U EX10.4 - Binomial Theorem

GROVE HSC 3U EX 10.4 Q11c Binomial Theorem - constant term

GROVE HSC 3U EX 10.4 Q11c

CF GROVE HSC 3U EX10.5 Binomial Theorem - Differentiating to get results

GROVE HSC 3U EX 10.5 Q2 Binomial Theorem, Equating coefficients

Grove HSC 3U EX10.3 Q9

Grove HSC 3U EX10.3 Q5g

GROVE HSC 3U EX10.3 Q5j

Grove HSC 3U EX 10.4 Q21e Coefficient of greatest absolute value

Grove HSC 3U EX 10.4 Q20c Binomial Theorem - Greatest Coefficient

Grove HSC 3U EX 10.4 Q20c Binomial Theorem - Greatest Coefficient

Grove HSC 3U EX 10.4 Q21b Coefficient of greatest absolute value

Grove HSC 3U EX 10.4 Q21b