Student’s Question #3

JC Mathematics

IMG_6450This questions is on vectors, in particular, finding the shortest distance of a point (0, 0, 3) to the plane. My student was able to do it but had a terrible careless mistake. But nevertheless, I want to share this question as there are a few ways to go about solving it.

First method will be to consider the vector
And then find the shortest distance using |a \times \hat{b}|.

Second method is slightly longer which involves finding the foot of perpendicular of (0, 0, 3) to the line. This will require finding the projection vector first.

I personally recommend the first method. As I always discuss in class, students need to appreciate the โ€œgoldenโ€ triangle for vectors as it can solve everything involving projection and perpendicular distance.

Normal Distribution Question #1

JC Mathematics

This is an interesting question from ACJC’2014 Prelim Exam. I feel that it test a lot on students’ understanding of the normal distribution curve.

The height X , in centimetres, of a randomly chosen girl in HTAM Junior College is normally distributed with mean \mu cm and standard deviation \sigmacm.

Given that $latex P(X > \mu – \sigma) + P(x > \mu + \sigma) + P(\mu < X < \mu +2 \sigma)=1.38$, find the $latex P(X > \mu + 2\sigma)$. Hint: Draw the curve and shade it out according. ๐Ÿ™‚

Completing the square

JC Mathematics

This is a very old-school and technical method we learnt in secondary school. However, many students are unable to do it accurately still. The trick of completing the square is realising that it depends on our (even older) (a\pm b)^2=a^2 \pm 2ab + b^2 formula. We are always introducing the -b^2 after we complete the square to ensure the expression is kept the same.


Students should know the completing the square is a technique, not confined to just quadratic expressions.

For example,
5-4\sqrt{x}+x=(\sqrt{x}-2)^2 +1

APG Question #2 (Interest Rate)

JC Mathematics

The following is a good APGP practice question from AJC. Personally, I feel that its well set and tests students on their concepts and understanding.

Mary has a monthly income of $4,000. She is considering applying for a car loan of $40,000 for 6 years which charges an interest rate of 3.00% per annum, compounded monthly. Interest is chargeable immediately when the loan sum is drawn out. The monthly repayment, $m, is fixed throught out the loan tenure.

  1. Show that the calculated loan balance at the end of the n^(th) loan month, after the monthly repayment is made, is given by 40000(\frac {(401)}{400})^n-400m((\frac {401}{400})^n-1).
  2. By legislation, banks can approve a car loan only if the monthly repayment does not exceed 15% of an applicant’s monthly income. Prove that Mary will not be able to apply for the car loan.

Do give it an attempt and see if you can show as required. Share with me if there are any doubts.

Let’s try some calculus questions

JC Mathematics

Many students often overlook that the coefficient of x in the integration or differentiation formulas in MF15 is 1!!! When it is not 1, many things changes. I’ll let the examples do the talking. ๐Ÿ™‚

Differentiation (recall chain rule)

\frac {d}{dx}(sin^{-1}(3x^2)) = \frac {1}{\sqrt{1-(3x^2)^2}}(6x)


\int \frac {1}{4+9x^2} dx = \frac {1}{2} {tan^{-1}(\frac {3x}{2})}\times\frac{1}{3}

For my careless students, I usually recommend they make the case of the coefficient of x be ONE instead. So \int \frac {1}{4+9x^2} dx = \frac{1}{9}\int \frac {1}{\frac{4}{9}+x^2} dx and after applying formula gives, (\frac{1}{9})(\frac{1}{\frac{2}{3}}){tan^{-1}(\frac {x}{\frac{2}{3}})} which will give the same answers after simplifications.


You may practice a few of the following questions!

\int\frac {1}{4-9x^2}dx

\int\frac {1}{9x^{2}-4}dx

\int\frac {1}{\sqrt{4-9x^2}}dx

\int\frac {1}{2x^{2}-2x-10}dx

Let me know if you have problems!