Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. page 86 Practice 16 Question 1(d)

The laws of logarithms state that,
log_axy = log_ax + log_ay
log_ax/y = log_ax – log_ay
log_axⁿ = n log_ax

First, group all terms with log₁₀.
log₁₀5 – 1 = log₁₀(2x + 7) – log₁₀(x + 15)
log₁₀5 – log₁₀(2x + 7) + log₁₀(x + 15) = 1

We simplify log₁₀5 – log₁₀(2x + 7) + log₁₀(x + 15).
Use these laws of logarithms.
log_axy = log_ax + log_ay
log_ax/y = log_ax – log_ay
We get log₁₀5(x + 15) / (2x + 7) = 1.

Rearranging the equation, we get
5(x + 15) / (2x + 7) = 10¹
5x + 75 = 10 (2x + 7)
5x + 75 = 20x + 70
15x = 5.
x = 5/15 = 1/3

We can then conclude that x = 1/3.

Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. pg 215 Formative Exercise 9.5 Question 8

Recall that y_new = y_initial + δy = y_initial + (dy/dx)δx.
Note that δy/δx = dy/dx.
Hence, δy = (dy/dx)δx.

First, we find dy/dx.
y = 4x² – 5x + 3
dy/dx = (2)4x^{2 – 1} – 5x^{1 – 1} = 8x – 5

δy = (dy/dx)δx.
Find the value of δx .
We know that x increased by 1.5% at the instant when x = 3.
Therefore,
δx = 3 x 0.015 = 0.045.

Find the value of dy/dx at the instant when x = 3
Therefore,
dy/dx = 8(3) – 5 = 19.

Replace dy/dx and δx with the acquired values.
δy = (19)(0.045) = 0.855

At the instant when x = 3,
y = 4(3)² – 5(3) + 3 = 24

Therefore, the percentage of change in y = 0.855 / 24 X 100% = 3.56%

Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. pg 214 Practice 21 Question 3

Recall that y_new = y_initial + δy = y_initial + (dy/dx)δx.
Note that δy/δx = dy/dx.
Hence, δy = (dy/dx)δx.

To solve part the problem, we have to recall the formula to determine the volume of a cylinder.
The volume of a cylinder is given by πr²h, where r is the radius and h is the height.

(a) the approximate change in the volume of the cylinder

We know that V (Volume of cylinder) is defined by πr²h.
Find dV/dr.
Recall that dy/dx = anx^n-1.

dV/dr = 2πr^{2 – 1}h, whereby a = πh and n = 2 (Note that h is a constant!)
Simplify and we get dV/dr = 2πrh.

Given r = 5cm and h= 12cm,
dV/dr = 2π(5)(12)
= 120π cm² (Note the units!)

To find the approximate change, we use the formula δy/δx = dy/dx.
We can rewrite the equation to represent the problem.
δV/δr = dV/dr.

The problem requires us to find the approximate change in volume of the cylinder.
Therefore, we shift δr to the right hand side.

δV = (dV/dr) δr

We are given that δr is -0.05 cm (because the radius shrinks by 0.05 cm).
We have calculated earlier that dV/dr = 120π cm².

Replacing the values into δV = (dV/dr) δr,
δV = (120π cm²) (-0.05 cm)
δV = -6π cm³

Therefore, the approximate change in volume of the cylinder is -6π cm³.

(b) the final volume of the cylinder

Recall that y_new = y_initial + (dy/dx)δx.
We can rewrite it to represent the problem.
V_new = V_initial + (dV/dr)δr.

Given r = 5cm and h = 12cm,

V_initial = πr²h
= π(5 cm)²(12 cm)
= 300 π cm³

(dV/dr)δr = -6π cm³ (Previously calculated in part (a) )

Therefore, the final volume of the cylinder is
300 π cm³ + (-6π cm³)
= 294 π cm³

Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. page 76 Formative Exercise 5.1 Question 4

This question is quite simple as long as you remember the law of indices
The law of indices states that :
a^m X aⁿ = a^{m + n}
a^m ÷ aⁿ = a^{m – n}
(a^m)ⁿ = a^mn
(ab)ⁿ = aⁿbⁿ
(a/b)ⁿ = aⁿ/bⁿ

First, we create similarities between the terms.
For instance, we know that 8^x can be rewritten as 2^3x.
8^x = (2³)^x = 2^3x

We can then rewrite 2^{2x + 1} X 3^{x – 1} = 8^x X 3^2x as 2^{2x + 1} X 3^{x – 1} = 2^3x X 3^2x.

From 2^{2x + 1} X 3^{x – 1} = 2^3x X 3^2x, we shift the terms with twos (2) to the left, and the threes (3) to the right. By doing this, we are grouping them together to simplify or rather enable us to work out the problem.

We will end up with 2^{2x + 1} / 2^3x = 3^2x / 3^{x – 1}.

Recalling the law of indices,
2^{2x + 1} / 2^3x can be rewritten as 2^{2x + 1 – 3x} and simplified to 2^{1 – x}.
3^2x / 3^{x – 1} can be rewritten as 3^{2x – x + 1} and simplified to 3^{x + 1}.

Recalling the index law again,
We write 2^{1 – x} as 2 / 2^x.
We write 3^{x + 1} as 3^x X 3.

We then end up with 2 / 2^x = 3^x X 3.

We move 2^x to the right hand side and 3 to the left hand side.

We get 2 / 3 = 3^x X 2^x.
Simplifying the equation will give us 2/3 = 6^x.
We have then concluded and proved that 6^x = 2/3.

Given px² – p(3x – 1) + 3x – 1, and whereby the two roots of the equation are 2 and m. Both p and m are constants. Find the values of m and p.

First of all, you would have to figure out what special attribute of the quadratic equation to exploit. For this case, we would use the fact that x² – (A + B)x + AB = 0, whereby A and B are roots of the equation.

Working from px² – p(3x – 1) + 3x – 1, we divide the equation with p. We can divide the equation with p because p is a constant. Be careful about this.

px² – p(3x – 1) + 3x -1 …….. divide by p
= x² – 3x + 1 + 3/p x – 1/p …….. rearrange
= x² – 3x + 3/p x + 1 – 1/p …….. group them accordingly
= x² – x(3 – 3/p) + 1 – 1/p

Recalling the equation x² – (A + B)x + AB = 0, we can conclude that A+B = 3 – 3/p (1) and AB = 1 – 1/p (2) by comparison.

Let A = 2, B = m
Recall (1)
A+B = 3 – 3/p
2 + m = 3 – 3/p
m = 1 – 3/p …….. (3)

Recall (2)
AB = 1 – 1/p
2m = 1 – 1/p
Replace m with equation (3)
2(1 – 3/p) = 1 – 1/p
2 – 6/p = 1 – 1/p
1 = 5/p
p = 5

Therefore, we conclude that m = 2/5.

Therefore, the two roots of the equation are 2/5 and 2.

To check the validity of the calculation, we replace the value of p inside px² – p(3x – 1) + 3x – 1. We get :

5x² – 5(3x – 1) + 3x – 1
= 5x² – 15x + 5 + 3x -1
= 5x² – 12x + 4

Now, we know that the roots are 2/5 and 2. We can then conclude :

(x – 2/5)(x – 2) = 0

Expand the equation. We get :

x² – 2x – 2/5 x + 4/5 …….. multiply by 5
= 5x² – 10x – 2x + 4
= 5x² – 12x + 4

We obtain the exact equation.
Therefore, the calculations are valid.

Note that this question was taken from the National Physics Quiz 2007. I got a ratio of 4:9 while the answer states that it should be 9:13. Now, do it for yourself. What do you get? Click the spoiler below to view how I got a 4 : 9.

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Now, if anyone could show me by the means of mathematical working why x:y is 9:13 and not 4:9, please post your working below. I would like to learn too. Heh.

Here’s another brainteaser. If ds/dt is velocity and dv/dt is acceleration, what is da/dt?

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