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A-Maths Tuition

Linear Law,Just 1 Strategy Makes A Grade Difference


What is Linear Law?

It is a tool which will allow you to transform non-straight line equation to straight line equation so that you can plot a straight line. Most of the time, the axis will consist of x and/or y

One of the techniques involving taking lg (log base 10) on both sides of the given equation.

linearlawquestion1.PNG

Common Mistake! STOP
y=Ax^b=>lgy=blgAx=>blgA+blgx

This is one of the common mistakes I have highlighted in Top 7 Commonly Made Mistakes in Logarithm

How To Counter This Mistake

Very simple. Just add in a bracket on the right hand side of the equation.
y=(Ax^b)<br>  =>lgy=lg(Ax^b)<br>  => lgy=lgA+lgx^b<br>  => lgy=lgA+blgx<br>

Now, your y-axis will be lg y and x-axis will be lg x. Gradient = b and y-intercept = lg A

Filed Under: A-Maths Tuition, scorers mindset, Weekly Question Tagged With: Exam Questions, logarithm equations, straight line graphs

Quadratic Equation Discriminant Proving Question


I realize many students have a challenge with presenting the solution of proving question. I am going to illustrate the correct way using the following question.

Show that is always positive for all real values of x.

=> what is always +ve? the values of y not x ( x can be of any values)

=> The graph will not touch x-axis; it is "hanging" above the x-axis

=> No roots, hence discriminant is less than 0

TIPS FOR SHOWING/PROVING QUESTION

  • Have an end in your mind. Be clear of the underlying concepts that the question is asking you to prove. Work towards that. For example, in the question, we want to prove that the discriminant is less than 0. <== this is the end.
  • Sketch to have a clearer idea.

quadraticprovingqn.jpgquadraticprovingqn.jpg

Filed Under: A-Maths Tuition Tagged With: discriminant, proving, quadratic equations, showing

Quadratic Equation And Discriminant


quadratic-eqns-roots.jpg
Click on picture for larger view

Siginifance of Discriminant b^2-4ac : Type of Discriminant Determines The Types of Roots/Solutions ( No root, 1 real root or 2 real and distinct roots)

Filed Under: A-Maths Tuition Tagged With: discriminant, quadratic equations, quadratic graphs, roots

What Has My Left Palm Got To Do With Math?


Palm Reading for my Destiny? Nope.

I was recently doing graph topics with some students. And some were complaining how much they have to memorize until I revealed to them some of the Underground Secrets on Mastery Of Graphs. They were amazed and amused.

expographhand.jpg

expographhand.PNG

Click on pictures for clearer view

Our palms have the exponential graph y=e^x embedded on them! Now you have 1 less graph to take care of. Just ensure you keep your hands nice and clean so that the palm lines stay on for you forever.

When the need to draw y=e^x arises, flip that palm up :)

Warning: You ought to do it tactfully in case the invilgator suspects you of cheating. Haha

I will be writing on more secrets on Mastery of Graphs. If you like these secrets, subscribe to singaporeolevelmaths for free to be posted of more updates.

Filed Under: A-Maths Tuition, E-Maths Tagged With: expo, graphs, sketch

Logarithm Equation Question 3


This is an interesting question which I came across under Additional Mathematics (A-Math):

Find the value of x.

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

Who is courageous to work on this question?

I will respond to this question when someone discusses about this question first :)

Update:

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(Apply Power Law)

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(multiply by 4 on both sides of equal sign)

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(Apply Power Law)
(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(Apply Quotient Law)
(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}
(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(Factorise)

(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}

(log_x8)=0 (NA)or log_x3+1=0

log_x3=-1,3=x^-^1,3=\frac {1}{x},3x=1,x= \frac {1}{3}

Filed Under: A-Maths Tuition, Weekly Question Tagged With: Exam Questions, logarithm equations

Logarithm Equation Question 2 [Checking validity of answers in logarithms equations]


Solve the equation

\lg(3a+2)+ lg(5a-3)=1+ 2lg (a)

Ans : a = 1 or -6/5

Question : Do we need to include -6/5

Contributed by YC

It is important for you to check if your answers are valid i.e are there any answers you would need to reject?
This is what I called checking for validity of answers.

Filed Under: A-Maths Tuition, Weekly Question Tagged With: logarithm equations, validity of answers

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