At first glance, you maybe tempted to solve ODE using Laplace Transforms. But it is not easy to find the Laplace Transform of 1/(1+t²). Since the ODE is a nonhomogeneous equation y'' + p(t)*y' + q(t)*y = g(t), your Lecturer probably has taught you about the Variation of Parameters formula.



This post has been edited by Critical_Fallacy: Nov 15 2013, 01:58 PM

Because the gas undergoes expansion from L to J

W(L,J) = (0.005 − 0.003)m³ × (100)kPa = 200 J

and then the gas undergoes compression from J to L

W(J,L) = ½ × {(0.005 − 0.003)m³ × (600 + 100)kPa} = 700 J

Therefore, the net work of the system is:

W(L,J) − W(J,L) = 200 − 700 = -500 J

Ans (A)...

This post has been edited by Critical_Fallacy: Nov 14 2013, 12:39 AM

Technically NO because you missed out the Matrix B.

Given a system of linear equations in the form of Ax = B:



Transform the linear system into augmented matrix:



Then, perform Gaussian Elimination to put the augmented matrix into the upper triangular form:

This belongs to the topic called Absolute-Value Inequalities.

To solve this problem, let's understand what is absolute value.

The absolute value of a real number R (denoted “|R|”) may be defined as the distance of R from zero.

For example, both –1 and 1 are one unit from zero, and therefore we have |–1| = |1| = 1.

The absolute value of R can be written as a function of the positive square root: |R| = √(R²).

Come back to the problem: |–2 / (x + 1)| < 1.

Let's define f(x) = –2 / (x + 1), and it follows that |f(x)| < 1.

By the above definition, we can intuitively imagine that f(x) = –1 and f(x) = 1.

To satisfy the condition of the inequality |f(x)| < 1, let's do four possible logic tests:

Test 1: f(x) < –1 :: Let f(x) = –1.1, and |f(x)| = |–1.1| = 1.1 < 1 ~ FALSE

Test 2: f(x) > –1 :: Let f(x) = –0.9, and |f(x)| = |–0.9| = 0.9 < 1 ~ TRUE

Test 3: f(x) < 1 :: Let f(x) = 0.9, and |f(x)| = |0.9| = 0.9 < 1 ~ TRUE

Test 4: f(x) > 1 :: Let f(x) = 1.1, and |f(x)| = |1.1| = 1.1 < 1 ~ FALSE

From the results, because Test 2 and 3 are TRUE, we can conclude that given the inequality |x| < a, the solution is always of the form –a < x < a.

Therefore, we have two inequalities: –1 < |f(x)| < 1.

Solve inequality #1: –1 < –2 / (x + 1) ... → ... we have x > 1.

Solve inequality #2: –2 / (x + 1) < 1 ... → ... we have x < –3.

Once you understand the concept, you can use the same simple approach as ystiang did.

Question: What if you were given the inequality |x| > a?

Even though it is a complex number, you can solve this problem using basic algebraic manipulation skill that you have already learned in Form 2 Factorization & Form 3 Simultaneous equations. Let me show you:

Find √(16 − 30i).

Since (16 − 30i) is a complex number, your developed mathematical intuition should tell you that the expected answer must be also a complex number (a + bi). Therefore it can be written as

√(16 − 30i) = √[(a + bi)²]

Your task now is to find a and b. As simple as that, regardless of the square root sign.

16 − 30i = (a + bi)² = a² + 2abi + b²i².

Since i² = −1, you can rearrange the equation to

16 − 30i = (a² − b²) + (2ab)i.

Eq.(1): a² − b² = 16

Eq.(2): 2ab = −30 ... → ... b = −15/a

You can solve the simultaneous equations by expressing b in term of a and substitute it into Eq.(1):

a² − (−15/a)² = 16

Multiply both sides with a² and turn it into a standard quadratic equation:

(a²)² − 16a² − 225 = 0

(a² − 25) (a² + 9) = 0

By the definition of a complex number (a + bi) where a and b must be Real Numbers, it renders a² = −9 invalid and therefore

a² − 25 = 0 ... → ... a = 5

Back-substituting a = 5 into Eq.(2) to find b:

b = −15/5 = −3

With the values found for a & b, and for all complex z with Re(z) > 0, √(z²) = z, the answer for

√(16 − 30i) = 5 − 3i

This post has been edited by Critical_Fallacy: Nov 15 2013, 01:59 PM

Given a system of linear equations with a set of variables x, y, z. Find k.

R1: 1x − 2y + 2z = 6
R2: 2x + 1y + 3z = 5
R3: 4x + 7y + 5z = k

Let's do some elementary row operations: You're lucky because element a11 is 1. See R1.

To eliminate coefficient of x (2) in R2, therefore 2 × R1: 2x − 4y + 4z = 12
R2 − 2R1:
R2': 5y − 1z = −7 ... eliminated x

To eliminate coefficient of x (4) in R3, therefore 4 × R1: 4x − 8y + 8z = 24
R3 − 4R1:
R3': 15y − 3z = k − 24 ... eliminated x

If you compare carefully R2' with R3', you'll notice that R2' & R3' are the same equation when scaled by a factor of 3, therefore 3 × R2': 15y − 3z = −21

Because it is stated that the equations are consistent, only then you can perform the next operation:
R3' − 3R2': 0 = k − 3

Solve for k and you have k = 3.

P.S. If you analyze the coefficient matrix, its determinant is 0, and thus the matrix is said to be singular, or it doesn't have a matrix inverse. If you plot the three equations in 3D, you'll discover that the linear system has infinitely many solutions because the equations are linearly dependent (see above R2' and R3'). If the system has a single unique solution, the three planes will intersect at a single point.

Like crazywing26 said, you hardly know which conic section is unless you have done many practices. However, it's not the end of maths. With a little quick-witted algebraic manipulation skills plus some knowledge in trigonometric identities, you can derive the General Parametric Equations for parabola, ellipse, and hyperbola. The parametric equations of a circle with radius r can be derived from the parametric equations of an ellipse by setting a = b = r.

There might be a more sinister motive behind the smiley. Well, straight to the point, it's a mathematical fallacy, where it presents an invalid proof for 3 reasons as explained in the following:

Reason #1: s has no sum.
s = 1 + 2 + 4 + 8 + … + 2^n is the infinite series whose terms are the successive powers of two. It is logically obvious that the sum of ever-growing positive-definite terms does not converge at a negative value or s = −1. In other words, it diverges to infinity, and so in the usual sense it has no sum.

Reason #2: Division-by-zero fallacy.
Let's start from here: 2s = s − 1. We know s is said to tend to infinity (∞). Using the notion of infinite limits, we can generalize that s − 1 becomes extremely close to s as s approaches indefinitely to infinity (∞).

And so, it follows that
2s = s
Divide by the non-zero s
2 = 1
The fallacious result is the implicit assumption that dividing by 0 is a legitimate operation
2 × 0 = 1 × 0
Dividing by zero gives:
2 × (0/0) = 1 × (0/0)

Reason #3: Invalid argument of power series.
Using Maclaurin series expansion, one can show that 1/(1 − x) = 1 + x + x² + x³ + …
If we plug in x = 2, the series becomes s and it sums the terms on the RHS to the finite value of −1.
Unfortunately, the expansion is only VALID for arguments |x| < 1. In plain English, invalid argument yields invalid result.

This post has been edited by Critical_Fallacy: Nov 18 2013, 03:43 AM

The information given on your physics question are incomplete. No assumptions are made. However, this is my probable explanation of the elastic collision. Do you get the concept?

Perhaps you can ask your classmates, Just Visiting By or crazywing26 to explain to you. By the way, are you familiar with the fundamentals of Linear Momentum and Elastic Collision in One Dimension?

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, yellowpika, ystiang

Here is my solution for Question 1 ~ [8 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, yellowpika, ystiang

Here is my solution for Question 2 ~ [7 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, yellowpika, ystiang

Here is my solution for Question 3 ~ [5 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, yellowpika, ystiang

Here is my solution for Question 4 ~ [9 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, WoanPing, yellowpika

Here is my solution for Question 5 ~ [9 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, WoanPing, yellowpika

Here is my solution for Question 6 ~ [7 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, WoanPing, yellowpika

Repost of solution for Question 7 ~ [15 marks]:

2013 STPM Maths T paper: Part 1 & Part 2

Hi ailing tan, crazywing26, iChronicles, WoanPing, yellowpika

Repost of solution for Question 8 ~ [15 marks]:

Hi maximR, this is Remainder Theorem. Most likely you will master this in less than 3 minutes.

Hi maximR, this is the Solution of Cubic equations & Quartic equations. Most likely you will master this in about 10~15 minutes.



You have already learned the basic factorization skill in Form 2 and quadratic solution by completing the square & formula in Form 4 Add Math. So, tell me now, are your Math skills sufficient to solve cubic equations and quartic equations with the systematic procedure?

This post has been edited by Critical_Fallacy: Nov 23 2013, 02:20 AM