A Level H2 Maths 2012 Paper 2 Q3 Solution; H2 Maths Tuition

A Level H2 Maths 2012 Paper 2 Q3 Solution

(i)

(The graph above is drawn using the Geogebra software 🙂 )

(ii)

$x^3+x^2-2x-4=4$

$x^3+x^2-2x-8=0$

By GC, $x=2$

By long division, $x^3+x^2-2x-8=(x-2)(x^2+3x+4)$

The discriminant of $x^2+3x+4$ is

$D=b^2-4ac=3^2-4(1)(4)=-7<0$

Hence, there are no other real solutions (proven).

(iii) $x+3=2$

$x=-1$

(iv)

(v)

$|x^3+x^2-2x-4|=4$

$x^3+x^2-2x-4=4$ or $x^3+x^2-2x-4=-4$

$x^3+x^2-2x-8=0$ or $x^3+x^2-2x=0$

$x^3+x^2-2x-8=0 \implies x=2$ (from part ii)

$x^3+x^2-2x=x(x^2+x-2)=x(x-1)(x+2)=0$

$x=0,1,-2$

In summary, the roots are $-2,0,1,2$

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H2 JC Maths Tuition Foot of Perpendicular 2007 Paper 1 Q8

One of my students asked me how to solve 2007 Paper 1 Q8 (iii) using Foot of Perpendicular method.

The answer given in the TYS uses a sine method, which is actually shorter in this case, since we have found the angle in part (ii).

Nevertheless, here is how we solve the question using Foot of Perpendicular method.

(Due to copyright issues, I cannot post the whole question here, so please refer to your Ten Year Series.)

Firstly, let F be the foot of the perpendicular.

Then, $\vec{AF}=k\begin{pmatrix}3\\-1\\2\end{pmatrix}$ ——– Eqn (1)

$\vec{OF}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$ ——– Eqn (2)

From Eqn (1), $\vec{OF}-\vec{OA}=\begin{pmatrix}3k\\-k\\2k\end{pmatrix}$

$\begin{array}{rcl}\vec{OF}&=&\vec{OA}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1\\2\\4\end{pmatrix}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\end{array}$

Substituting into Eqn (2),

$\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$

$14k+9=17$

$k=4/7$

Substituting back into Eqn (1),

$\displaystyle\vec{AF}=\frac{4}{7}\begin{pmatrix}3\\-1\\2\end{pmatrix}$

$\displaystyle|\vec{AF}|=\frac{4}{7}\sqrt{14}$

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H2 Maths Tuition: Complex Numbers Notes

H2 Maths: Complex Numbers 1 Page Notes

	Modulus	Argument
Cartesian Form		Draw diagram first, then find the appropriate quadrant and use (can use GC to double check)
Polar Form
Exponential Form

☺ When question involvespowers, multiplication or division, it may be helpful toconvert to exponential form.

☺ Please write Ƶ and 2 differently.

☺ De Moivre’s Theorem

Equivalent to

☺

Memory tip: Notice that arg behaves similarly to log.

☺ Locusof z is aset of pointssatisfying certain given conditions.

☺ in English means:The distance between (the point representing)and (the point representing)

☺ Means the distance offromis a constant,.

So this is acircular loci.

Centre:, radius =

☺ means that the distance offromis equal to its distance from

In other words, the locus is theperpendicular bisectorof the line segment joiningand.

☺ represents ahalf-linestarting frommaking an anglewith the positive Re-axis.

(Exclude the point (a,b) )

Common Errors

– Some candidates thought thatis the same asand thatis the same as.

– The “formula”for argumentsdoes not workfor points in the 2^ndand 3^rdquadrant.

– Very many candidates seem unaware that their calculators will work in radians mode and there were many unnecessary “manual” conversions from degrees to radians.

Ten Year Series: How many questions or papers to practice for Maths O Levels / A Levels?

This is a question to ponder about, how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series?

If you searched Google, you will find that there is no definitive answer of how many questions to practice for Maths O Levels/ A Levels anywhere on the web.

For O Level / A Level, practicing the Ten Year Series is really helpful, as it helps students to gain confidence in solving exam-type questions.

Here are some tips about how to practice the Ten Year Series (TYS):

1) Do a variety of questions from each topic. This will help you to gain familiarity with all the topics tested, and also revise the older topics.

2) Fully understand each question. If necessary, practice the same question again until you get it right. There is a sense of satisfaction when you finally master a tough question.

3) Quality is more important than quantity. It is better to do and understand 1 question completely than do many questions but not understanding them.

Back to the original query of how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series, I will attempt to give a rough estimate here, based on personal experience.

5 Questions done (full questions worth more than 5 marks) will result in an improvement of roughly 1 mark in the final exam.

(The 5 Questions must be fully understood. )

So, if a student wants to improve from 40 marks to 70 marks, he/she should try to do 30×5=150 questions (around 7 years worth of past year papers). Repeated questions are counted too, so doing 75 questions (around 3 years worth of past year papers) twice will also count as doing 150 questions. In fact, that is better for students with weak foundation, as the repetition reinforces their understanding of the techniques used to solve the question.

If the student starts revision early, this may work out to just 1 question per day for 5 months. Of course, the 150 questions must be varied, and from different subject topics.

Marks improved by	Long Questions to be done	Approx. Number of years of TYS	OR (even better)
10	50	2	1 year TYS practice twice
20	100	4	2 year TYS practice twice
30	150	6	3 year TYS practice twice
40	200	8	4 year TYS practice twice
50	250	10	5 year TYS practice twice

This estimate only works up to a certain limit (obviously we can’t exceed 100 marks). To get the highest grade (A1 or A), mastery of the subject is needed, and the ability to solve creative questions and think out of the box.

When a student practices TYS questions, it is essential that he/she fully understands the question. This is where a tutor is helpful, to go through the doubts that the student has. Doing a question without understanding it is essentially of little use, as it does not help the student to solve similar questions should they come out in the exam.

Hope this information will help your revision.

H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I)

Foot of Perpendicular is a hot topic for H2 Prelims and A Levels. It comes out almost every year.

There are two versions of Foot of Perpendicular, from point to line, and from point to plane. However, the two are highly similar, and the following article will teach how to understand and remember them.

H2: Vectors (Foot of perpendicular)

From point (B) to Line ( $l$ )

(Picture)

Equation (I):

Where does F lie? F lies on the line $l$ .

$\overrightarrow{\mathit{OF}}=\mathbf{a}+\lambda \mathbf{m}$

Equation (II):

Perpendicular:

$\overrightarrow{\mathit{BF}}\cdot \mathbf{m}=0$

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}})\cdot \mathbf{m}=0$

Final Step

Substitute Equation (I) into Equation (II) and solve for $\lambda$ .

Example:

[CJC 2010 P1Q7iii]

Relative to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors $\left(\begin{matrix}1\\2\\1\end{matrix}\right)$ , $\left(\begin{matrix}2\\1\\3\end{matrix}\right)$ and $\left(\begin{matrix}-1\\2\\3\end{matrix}\right)$ Find the shortest distance from $C$ to $\mathit{AB}$ . Hence or otherwise, find the area of triangle $\mathit{ABC}$ .

[Note: There is a 2nd method to this question. (cross product method)]

Solution:

Let the foot of perpendicular from C to AB be F.

Equation (I):

$\overrightarrow{\mathit{OF}}=\overrightarrow{\mathit{OA}}+\lambda \overrightarrow{\mathit{AB}}=\left(\begin{matrix}1+\lambda \\2-\lambda \\1+2\lambda \end{matrix}\right)$

Equation (II):

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}})\cdot \overrightarrow{\mathit{AB}}=0$

$\left(\begin{matrix}2+\lambda \\-\lambda \\-2+2\lambda \end{matrix}\right)\cdot \left(\begin{matrix}1\\-1\\2\end{matrix}\right)=0$

$\lambda =\frac{1}{3}$

$\overrightarrow{\mathit{CF}}=\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}}=\left(\begin{matrix}2\frac{1}{3}\\-{\frac{1}{3}}\\-1\frac{1}{3}\end{matrix}\right)$

$\left|{\overrightarrow{{\mathit{CF}}}}\right|=\sqrt{\frac{22}{3}}$

Area of $\Delta \mathit{ABC}=\frac{1}{2}\left|{\overrightarrow{\mathit{AB}}}\right|\left|{\overrightarrow{\mathit{CF}}}\right|=\sqrt{11}$

For the next part, please read our article on Foot of Perpendicular (from point to plane).

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Is d/dx (a^x)=x a^{x-1}? (a conceptual error in O/A Level Math)

In O Level, students are taught that $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$

So naturally, students may think that $\displaystyle\frac{d}{dx}(a^{x})=xa^{x-1}??$ (a is a constant)

Well, actually that is good pattern spotting, but unfortunately it is incorrect. Do not be too disheartened if you make this mistake, it is a common mistake.

The above is a conceptual error as $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$ only holds when n is a constant.

Fortunately, this question is rarely tested, though it is quite possible that it can come up in A Levels.

To fully understand the following steps, it would help read my other post (Why is e^(ln x)=x?) first.

First, we write $\displaystyle a^x=e^{\ln a^x}=e^{x\ln a}$ .

Hence
$\displaystyle \begin{array}{rcl} \frac{d}{dx} (a^x)&=&\frac{d}{dx} (e^{x\ln a})\\ &=&e^{x\ln a}(\ln a)\\ &=&e^{\ln a^x}(\ln a)\\ &=&a^x(\ln a) \end{array}$

After fully understanding the above steps, you may memorize the formula if you wish:

$\boxed{\frac{d}{dx} (a^x)=a^x(\ln a)}$

Memory Tip: If you let a=e, you should get $\boxed{\frac{d}{dx} (e^x)=e^x(\ln e)=e^x}$

The above steps involve the chain rule, which I will cover in a subsequent post.