Add math (2010/2A)

Question One

The first question asks the student to solve the following simultaneous equations:

$$ \left\{ \begin{array}{l} x - 2y = 7\\xy - x = 9y \end{array} \right. $$

The first thing we notice is that the second equation can be factorized and rewritten as

$$x(y-1) = 9y$$

Next, we can introduce this expression to the first equation. But before introducing this term, we need to multiply each of the term in the first equation by \(y-1\)

$$\color{brown}{x(y-1)}- 2y(y-1) = 7(y-1) $$ $$\color{brown}{9y}- 2y(y-1) = 7(y-1) $$ $$11y- 2y^2 = 7y - 7 $$ $$2y^2-4y = 7 $$ $$2y^2-4y + 2 = 9 $$ $$2(y-1)^2 = 9 $$ $$ y-1 = \pm \tfrac{3}{\sqrt{2}} $$ $$ y = 1\pm \tfrac{3}{\sqrt{2}} $$ $$ x = \left(2\pm \tfrac{6}{\sqrt{2}}\right) + 7 = 3\left(3 \pm \tfrac{2}{\sqrt{2}}\right)$$
Question Two.
The first part of this question asks the student to sketch a trigonometric function y = 1 + 3 cos x for values of x between 0 and 2π. This problem can be resolved by visualizing the following steps:

Ability to sketch such simple trigonometric function by hand is important although such plot may be generated pretty easily by a computer.

The second part of the question is an extension of the first part, it asks the student to obtain the number of solutions to the equation (but not to solve the given equation):

Successful solution of this problem requires the student to manipulate the given equation so that the previous sketch of 1 + 3 cos x can be utilized. One may proceed as follows, in which the number of solution to the given equation is the number of intersections between the trigonometric function and the straight line y = 3 - 3x/2π.

Question Three.
In this question, the student is given a series of cylinders of same radius but increasing height. It is a simple exercise to reformulate the given information systematically in the following table:

From the table it is extremely simple to compute the volume of the 17th cylinder, which is 324p. The second part of the question supposes that the sum of volume of the first n cylinders is 1620p, and asks the student the compute n. And it can be obtained by solving the following equation for positive value of n, which is 12 in this case.

Question Four.
This question requires the student to translate the graphical information into analytical expression. In this case, the shaded area bordered by y = g(x), x = 3, and y = 0 can be expressed analytically as:

Therefore, the next expression can be evaluated as follows:

which is basically testing the student's skill of manipulating integrands.
In the second part of the question, we are given the fact that the derivative of g(x) is 2x - 8 and the analytical expression of g(x) is required. Thus one may proceed as follow to obtain

Since from the given diagram, we have g(1) = 0, thus the value of c must be 7.
[However, this question is not carefully formulated by the examiner as there is a contradiction between the information given in part (a) and part (b), for if you were to integrate the function of g(x) obtained in part (b) between 1 and 3, you get -28/3 instead of -10. The value of c has to be 20/3 to be consistent with numerical value of the area given in part (a). This value, however, is inconsistent with g(1) = 0, as shown graphically in diagram accompanying the question.]

Question Five.
In this question, the student is given an equation of a straight line AC, 3y = 2x - 15, and that the coordinate of point A is (-3, -7). The first part of the question requires the student to formulate the equation of the straight line which passes through point A and is perpendicular to AC. In this case, the slope of the equation must be -3/2 since the product of the slopes of AC and its perpendicular must be negative unity.

Now the equation of the perpendicular must be of the form y = -3x/2 + c. We were told that this perpendicular passes through point A, which means that we can put the abscissa and ordinate of A into this equation and obtain the value of c, which is -23/2 in this case.

The second part of the question asks for the coordinates of C given that the ratio of AB to BC is 2 to 7. From the equation of AC, it can be easily determined that the coordinates of B is (0,-5). To obtain the abscissa of C, we write

From which we obtain the abscissa of C as 21/2. Writing the same equation for the ordinates of A, B, and C, we can compute the ordinate of C as 2.

Question Six.
The first part of this question requires the student to draw a histogram to represent the given set of data. A feat which I believe is best left to a computer. This skill is no longer important and should be de-emphasized.

1. Several basic mathematical skills are required to solve this problem successfully, all of which are basic algebraic skills, as outlined below:
   
   
   
   The skills outlined above can be easily trivialized by a computer.